Chapter 1 Linear Models page Linear Models Part 1 2 Linear Models Activities 1 17 Linear Models Part 2 21 Linear Models Activities 2 28

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1 Table of Contents Chapter 1 Linear Models page Linear Models Part 1 Linear Models Activities 1 17 Linear Models Part 1 Linear Models Activities 8 Chapter Linear Programming Linear Programming Part 1 34 Linear Programming Activities 1 43 Linear Programming Part 49 Linear Programming Activities 61 More LP 64 Chapter 3 Time Value of Money Time value of Money Part 1 67 Time value of Money Activities 1 8 Time value of Money Part 86 Time value of Money Activities 93 Time value of Money Part 3 97 Time value of Money Activities Time value of Money Part 4 11 Time value of Money Activities 4 1 Time value of Money Part 5 15 Time value of Money Activities Chapter 4 Functions&Trendlines Functions and Trendlines Part Functions and Trendlines Activities Functions and Trendlines Part 167 Functions and Trendlines Activities 179 Chapter 5 Measuring Change Measuring Change Part Measuring Change Activities 1 10 Measuring Change Part 17 Measuring Change Activities 39 Measuring Change Part 3 43 Measuring Change Activities 3 57 Measuring Change Part 4 64 Measuring Change Activities 4 74 Excel Tutorial 1 CreditCardPart1 Appendix A Excel Tutorials 79 Excel Tutorial CreditCardPart 86 Excel Tutorial 3 BreakEvenTable 93 Excel Tutorial 4 BreakEvenGraph 98 Linear Models Appendix B Answers to Exercises 301 Linear Programming 306 Time Value of Money 311 Functions and Trendlines 318 Measuring Change 35 Joseph F. Aieta Page 1 Printed 8/14/003 All Rights Reserved

2 Linear Models Part 1 Section 1.1 Page Linear Models Part 1 Linear modeling involves using linear functions to describe the relationships among variables representing data from the real world. Common business applications of models include linear depreciation, salary computations involving commissions, cost functions, revenue functions, profit functions, utility bills, tax computations, volume discounts and other pricing methods. We will examine several types of linear models starting with linear depreciation. In order to investigate and solve business problems that can be modeled by linear functions, you need to be familiar with the algebra and analytic geometry of linear equations and their graphs. To determine if you need a review of function concepts or skills related to various forms of the equation of a line, skim through the Linear Equation Review, starting on page 11. This review contains references to several useful Excel spreadsheets. Linear Depreciation Section 1.1 Each year at tax time, company accountants report company assets that have declined in value since the previous tax year. These assets include motor vehicles, computers, machinery, and other equipment. The factors that determine the value of the asset at a particular point in time (residual value) are the original purchase price, the length of time the company plans to keep the asset (useful life) and its scrap value when it is no longer useful to the company. The depreciation is taken as a tax deduction and accountants can choose among several formulas in computing the amount of this deduction. The simplest method is called straight line depreciation or linear depreciation. This method assumes that the asset depreciates by a constant amount every year. When the residual value of the asset is plotted on the vertical axis with years on the horizontal axis, then the graph displays a series of points that lie on a line which slopes downward from left to right. Example We will examine linear depreciation as it applies to a rental company that owns a collection of trucks. The purchase price of the trucks ranges from $40,000 to $100,000. The useful life of the trucks for rental purposes ranges from 3 years to 15 years. One truck was purchased for $64,000 and was sold by the rental company for $6,000 after 8 years. The spreadsheet in Figure shows a table and graph. Figure Joseph F. Aieta Page 8/14/003

3 Linear Models Part 1 Section 1.1 Page 3 The annual change in residual value, commonly referred to as depreciation, is found by taking the difference between cost and scrap value and then dividing by the useful life. In this case $64,000 $6,000 = $7, 50 so the rate of change is a reduction 8 of 7,50 dollars. As we see in Figure the truck had a residual value (sometimes referred to as book value) at the end of year the first year of $ $7,50 = $56,750. The coordinates of all points on the graph are of the form (year, residual value). $64,000 $6,000 = $7,50 If we calculate the slope of the line containing the points (0, $64000) and (8, $6000), we obtain 0 8 which is equivalent to $ 64,000 $6,000. It is actually more common for accountants to say that the value depreciates by 8 $7,50 per year instead of saying that the slope of the residual value line -$7,50. The formula for residual value can be succinctly stated in function notation as V(t) = 64,000 7,50 t, where t is the number of years since the truck was purchased. This is the formula that was used to create the table of values in Figure This function is entered into Excel as = *t where t refers to the named range containing years and multiplication is explicitly indicated by the asterisk. cost scrap In general, if the annual rate of change is then cost scrap V(t) cost t useful life useful life = Table shows the book value of other trucks at different points in time after they were purchased. Verify the numerical values in the last column, using the above formula with either a calculator or a spreadsheet. Years of Depreciation rate Years since Book value Original cost Scrap value Useful life per year purchase (t) V(t) $80,000 $4, $7, $19,00 $50,000 $9, $8,00.00 $33,600 $100,000 $0, $13, $0,000 Table Using a spreadsheet such as LinearDepreciation.xls, an accountant can enter values for the original cost, the scrap value, and the useful life and have Excel automatically calculate the annual depreciation. Appendix A on Excel Basics contains tutorials for creating tables and graphs. Joseph F. Aieta Page 3 8/14/003

4 Linear Models Part 1 Section 1. Page 4 Example 1.1. For linear depreciation, the slope of the residual value line can be calculated using any two points on the graph. Suppose the book value of a truck is $56,000 in year 4 and $38,000 in year 8. a) If the useful life of the truck is 10 years then what is its scrap value? b) If the truck was sold as scrap for $11,000 then what was its useful life? Solution: Find the slope from two points using the formula (8, $38,000) y = x y x y x 1 1 where one point is (4, $56,000) and a second point is $38,000 $56,000 $18,000 slope = = = 4, dollars per year Using the point-slope form y - y 0 = m(x -x 0 ), we obtain y - 56,000 = -4,500 (x - 4). Solving for y we obtain y = -4,500 x + 74,000. Using t for years instead of x and the function notation V(t) instead of y, we obtain V(t) = -4,500 t + 74,000. a) If the useful life is 10 years then V(10) = -4,500 (10) +74,000 = 9,000 so the scrap value at the end of year 10 is $9,000. b) If the scrap value is $11,000 then 11,000 = -4,500 t + 74,000. When we solve for t we get 11,000 74,000 = -4,500 t or -63,000 = t or t = 14. The useful life of this truck is 14 years. Sales Applications of Linear Models Section 1. Example 1..1 For several months a company tracks its sales and its expenses (in millions of dollars) and observes a linear trend in which expenses, y, are a function of sales, x. Sales of $40 million in one month correspond to expenses of $ million. One month earlier, expenses of $10 million were accompanied by sales of $15 million. Since expenses are a function of sales the coordinates of points on the line are ordered as (sales, expenses). The point (40, ) corresponds to sales of 40 million dollars and expenses of million dollars. Solution: a) How much did sales increase from one month to another? b) How much did selling expenses increase from one month to another? c) Assuming a linear relationship, write the formula for expenses as a function of sales. a) The increase in sales, in millions of dollars, is = 5. b) The increase in expenses, in millions of dollars, is 10 = 1. c) First we find the slope of the line from the two points (40, ) and (15, 10) 10 1 m = = = 0.48 so expenses, y, are related to sales, x, by the linear equation y 10 = 0.48(x -15) This is equivalent to y = 0.48x +.8 where x and y are in millions of dollars. Joseph F. Aieta Page 4 8/14/003

5 Linear Models Part 1 Section 1.3 Page 5 Example 1.. A salesperson named Harry receives a base pay of $,000 per month plus a 5% commission on sales. a) If sales in one month were $5,000 then what was Harry s total pay for that month? b) If total pay was $4,750 then what was the sales volume for that month? c) Answer questions a) and b) if the base salary is unchanged but the commission structure is changed to 5% commission on sales over $30,000. Solution: a) The formula for total pay is y =, x where x is the sales volume in dollars. For sales of 5,000 dollars, Harry s total pay in dollars is, (5,000) = 3,50 b) If 4,750 =, x then 0.05x = 4,750,000 =,750 and x =,750/0.05 = 55,000. Sales of $55,000 correspond to total pay of $4,750. c) Since $5,000 is less than $30,000 total pay for the month will be just base the salary of $,000. For sales above $30,000 commission based upon x dollars in sales becomes 0.05 (x 30,000) so total pay is, (x 30,000). If Harry s total pay is $4,750 then 4,750 =, (x 30,000),750 = 0.05 (x 30,000),750/0.05 = x 30,000 x = 55, ,000 = 85,000 dollars Harry needed sales of $85,000 in order for him to make $4,750 Piecewise Linear Functions Section 1.3 Many types of applications involve functions that can be described by linear segments as shown in Figure This figure is made up of three line segments, and so is called a piecewise linear function. There are even situations when it is useful to approximate a continuous curve as shown in Figure 1.3. by a collection of many linear segments. Figure Figure 1.3. For the piecewise linear function in Figure 1.3.1, we will focus on the left-hand endpoints of each segment. Joseph F. Aieta Page 5 8/14/003

6 Linear Models Part 1 Section 1.3 Page 6 The first segment is defined for a < x < b, the second segment is defined for b < x < c the third segment is defined for c < x (which is equivalent to x > c ) _a b c Example Figure x If we knew (1) the y coordinates that correspond to the breakpoints at x = a, x = b, and x = c and () the slope of each segment, then we could write an equation with three parts { equation equation for segment 1 equation for segment for a < x < b for b < x < c for segment 3 for c < x Let s illustrate how to develop the equations for each segment with the following hypothetical values: The left- hand endpoint for segment 1 is (0, 0) and the slope of the first segment is The left- hand endpoint for segment is (30,000, 750) and the slope of this second segment is The left- hand endpoint for segment 3 is (50,000, 1,300) and the slope of this third linear segment is In the point-slope form, the equation of a line with slope m and containing the point (x 0, y 0 ) is y - y 0 = m(x -x 0 ) Using a slight variation of this form we can define each segment as y = y 0 + m(x -x 0 ) where m is the appropriate slope and (x 0, y 0 ) are the coordinates of the point at the left of each segment. { 0.05 x for 0 < x < 30,000 f(x) = (x - 30,000) for 30,000 < x < 50,000 1, (x -50,000) for 50,000 < x Now study the following verbal interpretation of this function and then try to imagine how such a function may appear in a practical situation. The above function f(x) can be interpreted as follows: First segment: Calculate.5% of any value of x starting at 0 and going up to x = 30,000, inclusive. Middle segment: For values of x above 30,000, but not larger than 50,000, calculate.75% of the amount above 30,000 and add that to 750. Last segment: For values of x greater than 50,000 take 3% of the amount above 50,000 and add that to 1,300. It may not be immediately obvious how we found the constant, 750, for the segment over the interval 30,000 < x < 50,000 or the constant, 1,300 corresponding to the segment in which x is greater than 50,000. As we see in Figure 1.3.3, the function f(x) is continuous with no gaps or jumps. Consequently, the right-hand endpoint of each linear segment must match the left-hand endpoint of the next segment. At the upper limit of the first segment, when x =30,000 we calculate f(30,000) = 0.05(30,000) = 750. Similarly, the value of the function at x = 50,000 can be calculated as: f(50,000)= (50,000 30,000) = (0,000) = = 1,300. Joseph F. Aieta Page 6 8/14/003

7 The calculation of graduated income taxes leads to the following mathematical model: Example Linear Models Part 1 Section 1.3 Page 7 Suppose that the graduated state income tax for a single person in a state on the east coast is a piecewise linear function of Adjusted Gross Income (AGI). The total tax due is computed as follows:.50% of the first $30,000 of AGI plus.75% of the AGI between $30,000 and $50,000 plus 3.00% of any amount of AGI in excess of $50,000 Determine state income tax due as a function of Adjusted Gross Income and state all results to the nearest cent. Solution: a) Find the total state tax for Anna whose AGI is $3, b) Find the total state tax for Betsy, whose AGI is $40, c) Find the total state tax for Cecilia whose AGI is $58, d) If Darlene paid $1,16.50 in taxes then what was her AGI? Let x represent the Adjusted Gross Income. The graph of this three part income tax function is displayed in Figure The tax rate is increasing as AGI changes from one bracket to the next. Consequently the slopes of the segments are increasing from left to right. The break points that define the brackets are 0, 30,000, and 50,000. The rule for calculating state income tax is given by the function: { 0.05 x for 0 < x < 30,000 f(x) = (x - 30,000) for 30,000 < x < 50,000 1, (x -50,000) for 50,000 < x a) For an income level of $3, the first rule applies since 3,000 < 30,000. The tax due is 0.05(3,000) = 575 so Anna owes $575. b) An income level of $40, is between 30,000 and 50,000 so the middle rule applies. Betsy s state tax due is (40,000-30,000) = (10,000) = = 1,05 so Betsy owes $ 1,05. A common mistake in determining the second segment is to overlook the fact that the taxpayer has already paid the tax on the first $30,000 of income, which was $750. The calculation (40,000) is incorrect since the taxpayer would then be paying tax again on the first $30,000 of income. c) For an income level of $58,000, the rule for the third segment applies. Cecilia has already paid a tax of $1,300 on the first $50,000 of her income so her total state tax due is 1, (58,000-50,000) = 1, (8,000) = = 1,540 dollars. d) Darlene s tax of $1,16.50 is less than $1300 but greater than $750 so the middle rule applies. Solve 1,16.50 = (x - 30,000) for x 1, = 0.075(x - 30,000) 41.50= 0.075(x - 30,000) 41.50/0.075 = x 30,000 15,000= x 30,000 15, ,000 = x so Darlene s AGI is $45,000 Joseph F. Aieta Page 7 8/14/003

8 Linear Models Part 1 Section 1.3 Page 8 The utility bill calculation in Example 1.3. shows that a mathematical model may be made up of linear segments whose slopes decrease instead of increase. Example 1.3. An electric utility bills its monthly charges according to the customer s monthly usage of electricity recorded in kilowatt hours. This utility charges a particular class of commercial users according to the following rate structure: A basic monthly charge of $0.00 plus $0.80 per unit for the first 50 units plus $0.70 per unit for all units in excess of 50 a) Write the rule for this cost function, C(x), as a piece-wise linear function of x where x is the number of units recorded during the month. b) What are the total charges for a commercial customer with 30 units recorded? c) What are the total charges for commercial customer with 00 units recorded? d) What is the average charge per unit if 30 units were used in a particular month? Solution: Start by drawing a rough sketch. Figure The rough sketch shows two segments with the slope of the second segment slightly less than the slope of the first segment. The left-hand endpoint of the first segment is (0, 0.00) because of the $0 basic monthly charge. The slope of the first piece is 0.80 and the slope of the second piece is a) The second coordinate of the right-hand endpoint is (50) = = The breakpoint (50, 60.00) is also the left-hand endpoint for the second piece. Since the slope for this second piece is 0.70, the rule for charges that correspond to units over 50 is simply (x 50) C(x) = { x (x - 50) for 0 < x < 50 for 50 < x b) The total cost of 30 units is C(30)= (30) = = dollars. c) The total charges for using 00 units is C(70) = (00-50) = (150) = $ d) In a month when the commercial customer used 30 units, the average charge per unit is the total cost of 30 units C(30) $44.00 divided by 30. To the nearest cent this is = = $1. 47 per unit Joseph F. Aieta Page 8 8/14/003

9 Linear Models Part 1 Section 1.3 Page 9 Another situation in which piecewise linear models may apply is in the volume discounts that wholesale suppliers offer to retailers. As a retailer s order quantity increases beyond certain break points, the wholesale supplier drops the unit price. See VolumeDiscount.xls Example Fashion Designs Inc. sells suits to retail stores. FDI charges a fixed fee for shipping and handling and a volume discount that depends upon the number of suits ordered by the retailer. The shipping and handling charge is $305. The minimum order is 0 suits. The volume discount structure is as follows: $75 per suit for the first 50 suits, $5 per suit for the next 50 suits, $15 per suit for each suit in excess of 100. a) Determine the cost function. b) What are the total and average costs of ordering 90 suits (to the nearest cent)? c) How many suits were ordered if the total charge was $31,555? d) What is the minimum number of suits that must be purchased so that the average charge per suit is $00 per suit or below? Solution: A rough graph should consist of three segments with decreasing slopes as shown in Figure Since the minimum number of suits that can be ordered is 0, the lowest total cost is (0) = 5,805 dollars. $40,000 $35,000 $30,000 $5,000 $0,000 $15,000 A simple list of declining unit cost (price per suit) and the corresponding break points will help us in deriving the three-part rule for total cost. $10,000 $5,000 units $ Figure shipping and handling rate1 breakpoint 1 rate breakpoint rate3 $ $ $ $15.00 a) At the first breakpoint, when x = 50, the total cost for 50 suits is $305 + $75(50) = $14,055. At the second breakpoint, when x = 100, the total cost for 100 suits is $14,055 + $5(100-50) = $5, x For 0 < x < 50 { f(x) = 14, ( x -50) for 50 < x < 100 5, ( x 100) for 100 < x Joseph F. Aieta Page 9 8/14/003

10 Linear Models Part 1 Section 1.3 Page 10 b) The total cost of ordering 90 suits is 14, (90-50) =14, (40) = $3, so the average cost of ordering 90 suits is 3, / 90 = dollars per suit. c) A total cost is $31,555 is above $5,305 so the third equation applies and we must solve the equation 5, (x 100) = 31,555 for x 5, (x 100) = 31,555 15(x 100) =31,555-5,305 15(x 100) = 6,50 (x 100) = 6,50/15 x 100 = 50 x = 150 suits. d) If the average charge per suit is $00, we first need to determine which segment of the function applies. Since the average price for 100 suits is $5,305/100 = $53.05, the retailer must order more than 100 to get the average price down to $00. Therefore, we need to find x such that 5, (x 100) divided by x is $00. Start by multipying both sides of the resulting equation 5, (x -100) = 00 x 5, (x 100) = 00 x 5, x 1,500 = 00 x 5,305 1,500 = 00 x - 15 x 1,805 = 00 x - 15 x (00 15) x = 1, 805 x = 1,805 / 75 = by x to get Since the solution is between 170 and 171, Fashion Designs, Inc. must sell 171 suits in order to bring its average cost per suit to a value at or below $00. Note: If the solution to the equation had been x = then we would not round down to 170 in this context. The average cost of 170 suits is f(170) / 170 5, ( ) 5, (70 ) = = which is not at or below 00 dollars Joseph F. Aieta Page 10 8/14/003

11 Linear Equations Review Page 11 Review The function concept and the different forms for the equation of a line \ input / A numerical function f is a rule with certain inputs and outputs. You could think of a function as behaving like a machine. It produces a unique output f(x), pronounced f of x, for each input x taken from a specific collection of numbers called its domain. The input variable x is often called the independent variable. The output variable (or dependent variable) is often represented by the variable y, as in y = f(x). A function of a single input variable can be described in a variety of ways: / output \ x f(x) numerically: as a table with two rows or two columns In the table above f(4) = ¼ or 0.5, f(8) = 1/8 or 0.15, and so on. The rule for this reciprocal function is f(x) = 1/x. algebraically: as a formula or rule If the function f is given by the rule f ( x) = 3x x 1 then f () = 3 f ( 1) = 3( 1) 1 = = 7 ( 1) 1 = = 4 graphically: as a graph with the input variable on the horizontal axis and the output variable on the vertical axis Suppose the graph of the function f is the line indicated below with double headed arrows that goes through the point (0,). If we start at (0, ) and go over 3 and up we reach the point (3, 4) so f(3) =4, If we start at (3,4) and go over 3 and up we reach the point (6, 6) so f(6) =6 slope of a line change in y divided by change in x Example: The slope of the line through (1, 3) and (3, 4) is given by m = y x y1 x 1 y = x 4 ( 3) 7 m = = = Joseph F. Aieta Page 11 8/14/003

12 Linear Equations Review Page 1 linear functions and linear equations This category of functions turns out to be both relatively easy to work with and useful in a wide variety of business situations. Major forms for the equations of lines slope-intercept form, general form, and point-slope form slope-intercept form y = mx + b where m is the slope and b is the y-intercept. Example: y = 10x + 00 All linear functions can be described algebraically in this form where the parameters m and b are fixed values. The labels 'm' and 'b' are traditional in algebra but these letters are really arbitrary. For example, in statistics a line is often written in the form y = b 0 +b 1 x. The roles of m and b are illustrated in the worksheet slope-intercept contained in the file LinearForms.xls general form ax + by = c Example, 4x + 5y = 10 The x-intercept is c/a provided a 0. If 4x + 5y = 10, x-intercept is (.5, 0) The y-intercept is c/b provided b 0. If 4x + 5y = 10, y-intercept is (0, ) Example The roles of a, b and c are illustrated in the worksheet general form contained in the file LinearForms.xls point-slope form y - y 0 = m(x -x 0 ) where ( x 0, y 0 ) is any point on the line. A slight variation of this form y = y 0 + m(x - x 0 ) is particularly useful for determining formulas for piece-wise linear functions in section 1.3. conversions Any linear function written in general form or point-slope form can be transformed into slope-intercept form. Note: an equation such as x = 5 represents a vertical line and is not a function The equations on the left can be solved for y and transformed into slope intercept form as a linear function of x as shown on the right. 3 3x y +4 = 0 y = x + 6y = 0 y = 0 y 5 = (x 3) y = x -1 3x + 4y = y = x Joseph F. Aieta Page 1 8/14/003

13 Linear Equations Review Page 13 Open the workbook LinearForms.xls and use the Excel worksheet slope intercept in answering questions 1, and In the figure below, the markers for line 1 are diamonds and the markers for line are squares. In the Excel worksheet the parameters of a line can be changed by moving the appropriate slider bars. Slope-intercept y = mx+b slope y_intercept line line line 1 line Y1 = 3-4 *x + 10 x Y1 = -4*x + 10 Y =1*x + 3 Y = 1 *x Y1 = -4*x Y =1*x Joseph F. Aieta Page 13 8/14/003

14 1. a) Manipulate the parameters to illustrate two parallel lines. What are their equations? b) Manipulate the parameters to illustrate identical lines. What are their equations? Linear Equations Review Page 14 c) Perpendicular lines meet at right angles. Manipulate the parameters to illustrate two lines that are perpendicular. What are their slopes? What is the relationship between the slopes of two perpendicular lines if neither one is a vertical line? d) Manipulate the parameters to illustrate two horizontal lines. What are their slopes? e) Explain why a vertical line cannot be displayed in slope-intercept form. f) Display two different lines that intersect at (1,). What are their equations? g) If a line is given in the form y - 3 = (x - 1) then it contains the point (1, 3) and has a slope of. Display this line and state its equation in slope intercept form.. a) Verify that the point (, 1) is on the line y = 3x - 5. To locate another point on this line go over one unit over in the horizontal direction and in the vertical direction. b) Verify that the point (, 4) on the line y = -x + 8. To locate another point on this line go one unit over in the horizontal direction and in the vertical direction. c) Suppose that P (x 0, y 0 ) is any point on the line y = mx + b. To locate another point on this line start at point P and go one unit to the right in the horizontal direction and in the vertical direction. What determines whether you should go up or down? Joseph F. Aieta Page 14 8/14/003

15 Linear Equations Review Page 15 Use the worksheet general form in the Excel workbook LinearForms.xls to help you answer questions 3 and 4. General Form ax + by = c a 1 b 1 c *x + 6*y = line1 a b c *x + 0*y = vertical line at x = a) Display the line x - 5y = 0. What is the slope of this line? b) Display the line x - y = 5. What is the slope of this line? What are the coordinates of its intercepts? c) Display the line x + 5y = 0. What is the slope of this line? What are the coordinates of its intercepts? d) Display a distinct line parallel to x + 5y = 0 by changing only one parameter. What is the equation of this line? e) Display the line 4x - 8y = 40 and then display a line crossing the y axis at 5 that is parallel to the first line. What is the equation of this second line? f) Display the line 5x - y = 0 and state the slope and both intercepts for this line. g) Display the line -5x + y =10 and state the slope and both intercepts for this line. h) Display the line 5x +y = 40 and state the slope and both intercepts for this line. i) Display a line with y-intercept 6 and x-intercept 5. What is its equation? Can you do this more efficiently than a trial and error approach? Joseph F. Aieta Page 15 8/14/003

16 Linear Equations Review Page 16 k) Display the line 0x + 50y = 100. What is the slope of this line? What are the coordinates of its intercepts? l) Write an equation, in general form, for a line with x-intercept 1 and y-intercept 15 j) State the slope, x-intercept, and y -intercept for ax + by = c where a 0 and b a) Under what condition(s) on the parameters a, b, c will the graph of a linear equation of the form ax + by = c be a horizontal line and what is the equation of the line? b) Give an example of the equation of a horizontal line. c) What is the slope of every horizontal line? d) Under what condition(s) on the parameters a, b. c will the graph of a linear equation of the form ax + by = c be a vertical line? e) Give an example of the equation of a vertical line. f) Does a vertical line have a slope? Explain. 5. True or False. If the statement is false then explain why. a) A line that rises to the right and is almost vertical does not have a slope. b) The slope of the x-axis is zero. c) The segment connecting P(x 1, y 1 ) and Q(x 1, y ) where y 1 y is on a vertical line. d) A line segment of negative slope rises to the left. e) The slope of the y-axis is not a number. f) A line segment that is very, very close to vertical has a slope that has a large absolute value. g) No matter how large a number we may write down, there is a line segment whose slope exceeds this number. h) A line segment contained entirely in the second quadrant necessarily has a negative slope. i) The quadrant in which a line segment lies has no necessary relation to the sign of the slope of that segment. For additional practice, see Slope Intercept.xls Joseph F. Aieta Page 16 8/14/003

17 Linear Models Activities 1 Linear Models Activities 1 Page 17 Open the Excel workbook LinearDepreciation.xls 1. Given that the original cost of a piece of equipment = $60,000, scrap value = $8,000, and useful life = 8 years. a) What is the slope of the line and what does it represent? b) What does the vertical intercept of this line represent? c) What is the residual value of the equipment at the end of year two? d) What is the residual value of the equipment at the end of year six?. Given that the original cost of a piece of equipment = $6,000, scrap value = $4,000 and useful life = 8 years. a) What is the slope of the line and what does it represent in this context? b) In this context, give an interpretation of the point where the line crosses the vertical axis? c) Let t represent the time variable in years. Write a simplified formula, for the residual value, V(t), of this equipment as a function of t. 3. A machine, which is purchased for $9,000, will have a scrap value of $1,000 in 5 years. a) What will the machine be worth in years? b) Find a formula for the machine s residual value, V(t), after t years. 4. Suppose the original cost of a piece of equipment is $50,000 and the equipment depreciates $6000 per year until it is sold for $8000. a) What is its useful life? b) Is this problem easier to solve with a spreadsheet or with paper and pencil? Joseph F. Aieta Page 17 Printed 8/14/003

18 5. The chart below plots the book value of an asset assuming straight line depreciation Linear Models Activities 1 Page 18 Straight Line Depreciation $5,000 residual value $0,000 $15,000 $10,000 $5,000 $ years a) What was the purchase price of the asset? b) Estimate the book value of the asset after 6 years. c) What is the yearly rate of depreciation? d) After how many years will the value of the asset be at or below $1000? 6. A copy card costs $ When it is inserted into a campus copy machine, the machine automatically deducts the cost of the duplicating job from the current value of the card. The scatter plot shows the value remaining on the card corresponding to the total number of copies made. remaining value $15.00 $10.00 $5.00 Copy Machine Card copies $ a) Construct a linear model for the relationship between the remaining value of the card, and the total number of copies printed. b) What value remains on the card when 00 copies have been made? c) If the value remaining on the card is $.00, how many copies have been charged to the card? 7. The relationship between total cost and number of units made is linear. Suppose cost increases by $3 for each additional unit made and the total cost of 10 units is $40, find the equation of the relationship between total cost (y) and number of units made (x). 8. Assume that the relationship between the number of units manufactured and the manufacturing cost is linear. As the number of units increases from 100 to 00, the manufacturing cost increases from $350 to $650. Find the equation of the relationship between cost (y) and number of units made (x). Joseph F. Aieta Page 18 Printed 8/14/003

19 Linear Models Activities 1 Page As sales (x) change from $100 to $400, selling expense (y) changes from $75 to $150. Assume that the given data establish a linear relationship between sales and selling expense, state this relationship in the form of a linear equation. 10. If x represents sales and y represents selling expense, then (30, ) would mean $30 in sales were accompanied by $ of selling expense. Suppose that last month s figures were (14, 10) and this month s are (30, ). a) How much did sales increase? b) How much did selling expense increase? c) Make a graph showing the points and label the change in x and the change in y. 11. a) If a taxi fare (y) is 50 cents plus 0 cents per quarter mile, write the equation relating fare to number of miles traveled, x. b) The gross weekly earnings of a salesman are $50 plus 10 percent of the retail value of the goods she sells. Write the equation for earnings, E, in terms of sales volume, V. What is the common name for the slope of this line? 1. A salesperson receives a base pay of $1,000 per month plus a.5 % commission on all sales. a) If sales in one month were $60,000 then what was the total pay for that month? b) If total pay was $3,50 then what was the sales volume for that month? c) and d) Answer questions a) and b) if the base salary was unchanged at $1000 but the commission structure was.5 % commission only on sales over $40, Colonial Gas Company charges its customers according to their usage of gas as follows: An initial $6.00 customer charge $1.180 per unit (100 cubic feet) for the first 0 units and $0.806 pr unit for each unit over 0. a) Determine the cost function and draw its graph. b) What are the total and average charges for using 15 units? c) What are the total and average charges for using 50 units? d) How many units were used if the total charge is $73.93? Joseph F. Aieta Page 19 Printed 8/14/003

20 Linear Models Activities 1 Page The tax rates shown below are from the IRS 003 tax table 003 Married Filing Jointly If Taxable Income Of the is Over But Not > The Tax is: Amount > 0 1, % 0 1,000 47,450 1, % 1,000 47, ,650 6, % 47, , ,700 4, % 114, , ,950 4, % 174, , , % 311,950 a) How much tax would a couple pay on a Taxable Income of $70,000? b) How much tax would a couple pay on a Taxable Income of $150,000? c) What was the couple s Taxable Income if the 003 tax is $3,900? 15. Two fee schedules apply for each item listed and sold on a popular online auction Website. The first fee is called an Insertion Fee and is charged to individuals who place a regular listing. The calculation of the second fee, called the Final Value Fee is based upon the final sale price. The Insertion Fee formula for a regular listing is shown in the following table. Opening Value or Minimum Price Insertion Fee $ $9.99 $0.5 $ $4.99 $0.50 $ $49.99 $1.00 $50.00 and up $.00 a) Sketch the graph of this function with initial price on the horizontal axis and the Insertion Fee on the vertical axis. b) Explain why this graph is described as a step function. c) Refer back to the previous problem #14. Would taxpayers have a legitimate objection if a graduated tax formula was a step function of adjusted gross income (AGI)? Explain. 16. Start with a blank worksheet and create an interactive linear depreciation worksheet, like the table and graph, pictured below. It should be designed in such a way that the rate of depreciation, the table, and the graph change automatically when the entries in the shaded cells for cost, and scrap value are changed. Assume that useful life is fixed at 10 years. annual cost scrap value useful_life depreciation $80,000 $4, $7, residual year value $100,000 0 $80,000 $90,000 1 $7,400 $80,000 $64,800 $70,000 3 $57,00 4 $49,600 $60,000 5 $4,000 $50,000 6 $34,400 $40,000 7 $6,800 $30,000 8 $19,00 9 $11,600 $0, $4,000 $10,000 residual value $ years Joseph F. Aieta Page 0 Printed 8/14/003

21 Linear Models Part Section 1.4 Page 1 Linear Models Part Cost, Revenue, Profit, and Break-even Section 1.4 In this course we examine many situations that involve cost, revenue and profit. When the models for cost, revenue, or profit can be described by linear functions then the strategies for solving problems are the basic algebraic and geometric techniques for manipulating linear equations and drawing the graphs of lines. Example CapeCodTees is a summer business run by college students. They sell t-shirts imprinted with slogans and images associated with the resort area. The owners of the business separate their costs into fixed cost and variable cost. Fixed cost includes such things as the rental of space, equipment and other costs that are incurred before any t-shirts are bought from wholesalers to be imprinted and sold. They estimate that their fixed costs will be about $,000. Variable cost is related to the cost of the plain t- shirts that they buy and the cost of the materials for imprinting each image. They estimate that variable costs will be about $6 per t-shirt. They do not include their labor as a cost since they plan to split the profits when the season is over. They will price the t-shirts at $1.50 per shirt. They believe that the maximum number of t-shirts that they can sell in a summer with good weather is around 800. If the weather is bad then sales may be significantly less. They want to know the minimum number of t-shirts that they must sell to break even. Total cost, revenue, and profit are each functions of quantity, which we will denote by q. Profit equals revenue minus cost. This can be written as P(q) = R(q) C(q) using function notation. A rough picture of the cost function looks like Figure and the revenue function looks like Figure Cost Revenue C(q) R(q) F q q Figure Figure 1.4. Several important concepts are apparent from these rough, unlabeled sketches. The cost function cuts the vertical axis at point F units above the origin since the business has fixed costs of F dollars even if no t-shirts are bought and sold. When quantity q is zero, total cost is the fixed cost, C(0) = F. There is no revenue at q = 0 so R(0) = 0 and consequently the profit at q = 0 will be negative, P(0) = R(0) C(0) = -F and the point (0, -F) on the profit function is located F units directly below the origin (0, 0). The owners of the business want to be confident that they can eventually sell a sufficient number of t-shirts in order for revenue to exceed cost. Profit must move into positive territory, at least far enough to cover the fixed cost. It should also be apparent that the slope of the revenue line in Figure 1.4. must be greater than the slope of the cost line of Figure otherwise revenue could never overtake cost no matter how many t-shirts are sold and the business would always lose money. Joseph F. Aieta Page 1 Printed 8/14/003

22 Linear Models Part Section 1.4 Page Cost, Revenue, and Profit As long as the slope of the revenue line (price per shirt), is greater than the slope of the cost line (variable cost per shirt) the two lines will eventually cross. If the point of intersection where R(q) = C(q) is beyond any feasible sales volume then the owners will not have any profit to share. For example, if breakeven occurs at q = 1000, and there is very low probability that the business can sell that many t-shirts in one season, then the owners will have to find a way to raise prices, lower costs, or both. If this can t be accomplished then they will not have a profitable business this season. F R(q) C(q) P(q) q Figure Figure shows a rough sketch of cost, revenue, and profit functions, all on the same pair of axes, where quantity is the measure on the horizontal axis and dollars is the measure on the vertical axis. We define break-even as the point where revenue = cost or where profit = 0. In function notation, break-even quantity is that value of q such that R(q) = C(q). Finding the break-even quantity is equivalent to solving P (q) = 0. The difference between the slopes of the revenue and cost functions, relative to the size of fixed cost is what ultimately determines the break-even quantity. In Words Revenue is $1.50 times the number of t-shirts sold. Total cost is fixed cost plus variable cost $000 plus $6.00 per shirt. Profit is revenue minus cost. Function notation R(q) = 1.50 q C(q) = q P(q) = R(q) C(q) P(q) = 1.50 q ( q) P(q) = ( ) q 000 P(q) = 6.50 q 000 In order for the company owners to get accurate graphs of the cost, revenue, and profit functions as they relate to the quantity of t-shirts imprinted and sold, they construct tables in Excel. They also use Excel s Chart Wizard to produce the graphs of the cost, revenue, and profit functions on the same coordinate axes as shown in Figure Joseph F. Aieta Page Printed 8/14/003

23 Linear Models Part Section 1.4 Page 3 quantity Cost Revenue Profit q C R P ($,000) ($1,838) ($1,675) ($1,513) ($1,350) ($1,188) ($1,05) ($863) ($700) ($538) ($375) ($13) ($50) $ $ $ $ $ $95 Break-Even Analysis C R P $7,000 $6,000 $5,000 $4,000 $3,000 $,000 $1,000 $0 -$1, $,000 -$3,000 quantity Figure The table in Figure reveals that the break-even quantity is near 300. Determining when profit is equal to zero is equivalent to finding the value of q such that P(q) = 0. We use basic algebra to find the number of t-shirts where revenue is equal to cost. We will use basic algebra to solve this equation. Since P(q) = 6.50 q 000 we must solve 6.50 q 000 = q = 000 q = 000/ [In later examples we will use the Excel tools Goal Seek and Solver to solve equations.] If the business sells 307 t-shirts or fewer, it will be in the red (losing money). The business begins to make a profit with the sale of the 308 th t-shirt. For cost, revenue, and profit situations, economists refer to the cost of making one more unit as the marginal cost of that unit. Similarly, the additional revenue from selling one more unit is called marginal revenue and the additional profit from making and selling one more unit is called the marginal profit. When cost and quantity have a linear relationship, the marginal cost is constant. This means that the marginal cost does not depend on the quantity produced. It will be the same for 100 units as for 00 units or for 307 units. If you were to calculate C(101) C(100), C(01) C(00), or C(308) C(307) in the t-shirt example you would get the same value each time. It is identical to the slope of the cost function. In other words, if q units have been made then the marginal cost of making q + 1 unit is the variable cost per unit, regardless of the particular value of q. Similar statements apply to linear revenue and profit functions. For linear cost functions, marginal cost equals variable cost per unit Only for linear cost functions is marginal cost a constant. Chapter 5 will help us estimate marginal cost for non-linear cost functions using calculus. Joseph F. Aieta Page 3 Printed 8/14/003

24 Linear Models Part Section 1.4 Page 4 Example 1.4. Comparison of Profit Lines. Suppose CapeCodTees can obtain its supplies more cheaply from another wholesale outfit and thereby reduce its variable cost to $5.50 per shirt. While the quality of the t-shirts is the same, the catch is that the new distributor wants an up-front fee of $00 to cover transportation costs for the season. CapeCodTees would add this $00 to their fixed cost and see how these changes would impact profit and break-even. Their original decision is to stay with the original supplier if they have to raise their unit price above $1.50 in order to be profitable when q = 308. Original Supplier New supplier R(q) = 1.50 q R(q) = 1.50 q (unchanged) C(q) = q C(q) = q P(q) = R(q) C(q) P(q) = R(q) C(q) P(q) = 1.50 q ( q) P(q) = 1.50 q ( q) P(q) = ( ) q 000 P(q) = ( ) q 00 P(q) = 6.50 q 000 P(q) = 7.00 q 00 Solve for q and get q = 000/(6.50) = Profit turns positive with the 308 th t-shirt sold Solve for q and get q = 00/7.00 = Profit turns positive with the 315 th t-shirt sold Under their original criteria, they would stick with the original supplier. Then they start to wonder whether or not they might make more money by the end of the season if they switched to the new supplier. They look back at previous seasons and realize that even in their poorest summer, they never sold fewer than then 500 t-shirts. At q = 500 they compare P(500) = 6.50(500) 000 = $1,50 versus 7.00(500) 00 = $1,300 and decide to go with the new supplier. Applying additional algebra, they find that they are better off with the new supplier if they can sell more than 400 t-shirts. The break-even concept can occur in other contexts that involve linear functions. An individual may have two or more offers and needs to make a decision about which option is optimal. The decision in example is to select the best copy shop. Example You have just received bids from two printing companies for the job of printing copies of a travel brochure for your agency. Speedy Printers charges a fee of $00 plus 75 cents per copy. Quality Printers charges a fee of $150 plus 90 cents per copy a) Plot the cost functions for both print shops on the same coordinate axes with the number of copies on the horizontal axis and the cost on the vertical axis. b) For what range of copies does Quality Printers offer the best deal? c) For what range of copies does Speedy Printers offer the best deal? Joseph F. Aieta Page 4 Printed 8/14/003

25 Linear Models Part Section 1.4 Page 5 Figure contains a cost versus quantity table and graph from the Excel file Speedy vs Quality.xls. By creating a column for the difference between the costs of using Speedy Printers minus the cost of using Quality Printers, the agency can see how much is saved by using a particular print shop rather than the other. The agency can then decide if the savings is worthwhile in light of other considerations, such as convenience. Figure When the difference d = Speedy Quality is positive then choosing Quality Printers saves the agency d dollars. When the difference d = Speedy Quality is negative then choosing Speedy saves the agency d dollars where is the absolute value function. For example, if only 00 copies are printed then choosing Quality will save the agency $0.00 but if 400 copies are printed then choosing Speedy will save the agency $10. The breakpoint occurs between 35 and 350 copies. We will use Goal Seek to find the precise break point as follows: 1. The entries in column A are numerical values from patterned data and cannot contain any formulas. See Tutorials 1 and in Appendix A for the steps to create patterned data with Excel.. Highlight cell D0. 3. From the Tools menu, select Goal Seek. 4. With D0 as the Set cell, you will search for the value that will make D0 zero by changing the numeric value in cell A0. Don t forget to click OK Goal Seek changes cell A0 to We can also find this breakpoint with pencil and paper by solving x = x to get x = 50/0.15 = We conclude that Quality Printers is more cost effective for 333 or fewer copies, but Speedy Printers is the better deal for 334 copies or more. See Tutorial 3 in Appendix A for other applications of Goal Seek. Joseph F. Aieta Page 5 Printed 8/14/003

26 Linear Models Part Section 1.5 Page 6 Iso-value lines in two variables Section 1.5 The origin of the prefix term iso comes from a Greek word meaning same so iso-profit means the same profit, iso-cost means the same cost, iso-dollars means the same amount of money. All iso-value lines for two variables are of the form ax + by = c. For equations in this form, we can quickly determine the two intercepts and the slope. For example, if the iso-profit line is given by 4x + 3y = 480 then the y- intercept is (0, 160), the x-intercept is (10, 0) and the slope is -4/3. Example Iso-profit lines After several successful seasons, the owners of a t-shirt business decide to add caps to their sales line. They observe that some of their customers come into the shop and leave with caps and no t-shirts. Other customers buy t-shirts and no caps, and some customers buy both. In the current season the owners have been able to raise their t-shirt prices without any significant impact on the number of t-shirts sold in a season. Currently they make a profit of $8.00 for each t-shirt sold and $6.50 for each cap sold. In one season they had a gross profit of $080 to the nearest dollar. The equation of this profit line can be written in general form as 8.00x y=,080 where x is the number of t-shirts sold and y is the number of caps sold. In a season with a profit of $3,10.00 the profit line has equation 8.00 x y =3,10. For a profit of $4,160 the profit line is 8.00 x y = 4,160. If we draw the graph of each profit line on the same coordinate axis, as shown in Figure 1.5.1, we observe that each line in this family of profit lines is parallel to each other line iso-profit lines The slope of this family of profit lines is divided by 6.50 which are approximately Each profit line has an equation of the form ax + by = c where c represents profit. The larger the value of c, the further the line is from the origin and the larger the values of the x and y intercepts. For the $080 iso-profit line, the intercepts are (0, 30) and (60, 0). For the $3,10 iso-profit line the intercepts are (0,480) and (390, 0). Similarly the $4,160 iso-profit line has intercepts (0,640) and (50, 0). caps t-shirts Figure Note that the line closest to the origin corresponds to the lowest profit and that the line furthest from the origin corresponds to the lowest profit. Q1. Name two other points on each iso-profit line. A. The points (65, 40) and (130,160) are on the profit line 8.00 x y =,080 The points (65, 400) and (130, 30) are on the profit line 8.00 x y = 3,10 The points (65, 560) and (130, 480) are on the profit line 8.00 x y = 4,160 Joseph F. Aieta Page 6 Printed 8/14/003

27 Linear Models Part Section 1.5 Page 7 Example 1.5. Aunt Agatha has chosen her brother, sister, and a favorite niece to be the heirs of her estate. Rather than have her relatives inherit millions of dollars at the time of her death, she sets up three trust funds that will provide each of them with an annual income. The trustee of Aunt Agatha s estate has been directed to allocate enough money by investing in bonds and money market funds to produce at least $40,000 in annual income for her niece, at least $50,000 in income for her brother, and at least $60,000 in income for her sister. Bond funds are expected to pay eight percent interest and have some risk. Money market funds are expected to return five percent interest and have practically no risk. The trustee is given complete discretion each year as to how much to invest in each category for each relative as long as the total investment produces the desired annual income level. One way the trustee can satisfy the $40,000 income requirement for the niece is to invest $800,000 in money markets and nothing in bonds. Another strategy would be to put $500,000 in bonds and nothing into money markets. A mix of $50,000 in bonds and $400,000 in money markets would also achieve the income goal of $40,000. As the annual income requirement increases to $50,000 for her brother and $60,000 for her sister, the trustee needs to invest more money from Aunt Agatha s estate. Q1. Using the iso-dollar line 0.08x y = 50,000 describe three ways to achieve a $50,000 income? A1. $0 in bonds and $1 million dollars in money markets, 0$ in money markets and $65,000 in bonds, or $400,000 in bonds and $360,000 in money markets. Q. Using the iso-dollar line 0.08x y = 60,000 what are three ways to achieve a $60,000 income? A. $0 in bonds and $1. million in money markets, $0 in money markets and $750,000 in bonds, or $500,000 in bonds and $400,000 in money markets. Figure 1.5. shows the three iso-dollar lines. The line closest to the origin corresponds to an annual income of $40,000. The middle line corresponds to an annual income of $50,000 and the line furthest from the origin corresponds to an annual income of $60,000. The slope of each iso-dollar line in this family is -0.08/0.05 = This can be interpreted as follows: To achieve the same income, each $10,000 increase invested in bonds means a $16,000 decrease for the amount invested in money market funds. money market $1,400,000 $1,00,000 $1,000,000 $800,000 $600,000 $400,000 $00,000 iso-dollar lines $0 $0 $100,000 $00,000 $300,000 $400,000 $500,000 $600,000 $700,000 $800,000 bonds Figure 1.5. Complete Excel Basics Tutorials 3 and 4 in Appendix A before you begin the exercises. Joseph F. Aieta Page 7 Printed 8/14/003

28 Linear Models Activities Linear Models Activities Page 8 Most of the exercises in Linear Activities can be solved using basic algebra. The spreadsheet Break-even Quantity.xls is related to several of these exercises. 1. Approximate cost, revenue and profit lines for a vendor that sells mini-pizzas are displayed in the graph. Use this graph to a) Estimate the fixed costs. b) Estimate the revenue per pizza. c) Estimate the variable cost per pizza. d) Estimate the total cost, revenue, and profit for q = 400. e) Estimate the break-even quantity. f) Estimate the break-even revenue. g) Do your estimates in (a) (f) remain the same if you are told that the cost function contains the points (0, 00) and (150,50) and also that the revenue function contains (500,500)?. A manufacturer has a linear cost function with a fixed cost of $60,000, a variable cost of $ per unit made and sold. The selling price is $5 per unit. a) Write the revenue, cost, and profit functions using q for number of units. b) Compute profit if 5,000 units are made and sold. c) Compute profit if 10,000 units are made and sold. d) Find the break-even quantity. e) Find the break-even dollar volume of sales (revenue at break-even point). f) Construct a break-even graph. Label the cost and revenue lines and the break-even point Joseph F. Aieta Page 8 Printed 8/14/003

29 Linear Models Activities Page 9 3. A manufacturer has a linear cost function for one of its products and has determined that over the next three months it can produce 1,000 units at a total cost of $300,000. This same manufacturer can produce,000 units at a total cost of $400,000. The units sell for $180 each. a) Determine the revenue, cost, and profit functions using q for number of units. b) What is the fixed cost? c) What is the marginal cost? d) Find the break-even quantity. e) What is the break-even dollar volume of sales? 4. For each problem below, assume that monthly cost, revenue, and profit are linear functions of units produced and write formulas for the cost function, the revenue function, and the profit function. Sketch rough graphs of the three lines on the same coordinate axes (Excel graphs are not necessary here). Label the break-even point and write its coordinates on the graph. a) revenue = $8 per unit, variable cost = $8 per unit, monthly fixed cost= $56,000 b) revenue = 85 cents per unit, variable cost = 15 cents per unit, monthly fixed cost =$400 c) revenue = $18 per unit, monthly fixed cost = $98,000, and the company breaks even when 7000 units are sold. 5. Suppose that the manufacturer is a publisher and that the units produced are copies of a particular book. Given that variable cost/unit is $8.75, fixed cost is $5,500, and selling price/unit is $5.00. a) How can you quickly tell whether or not the publisher can make a profit with this book? b) Determine the cost, revenue, and profit if 1000 books are made and sold. c) To the nearest book, what is the break-even quantity (i.e. what is the first book quantity for which revenue minus cost is not a negative value)? d) If selling price and variable cost per unit are unchanged, but fixed cost can be reduced to $4,50, what will be the impact on the break-even quantity? 6. A company has a linear total cost function and has determined that over the next three months it can produce 10,000 units at a total cost of $550,000. This same manufacturer can produce 0,000 units at a total cost of $600,000. The selling price per unit is $5.50. a) Determine the revenue, cost, and profit functions using q for number of units. b) What is the fixed cost? c) What is the marginal cost? d) Find the break-even quantity. e) What is the break-even dollar volume of sales? Joseph F. Aieta Page 9 Printed 8/14/003

30 Linear Models Activities Page Let the variable q represent quantity. Let v stand for the constant cost per unit and let F stand for the fixed cost. a) Write a formula for total cost as a linear function of the variable q. b) Represent the constant for unit selling price by p and write a formula for revenue as a linear function of q. c) Write a simplified formula for profit as a linear function of q. 8. An online distance learning company charges each student $50 to enroll in a one month course. When the company signed up only 400 students in the first month, it went in the red $4,000 (loss of $4,000) During the second month it enrolled 1,00 students and was in the red only $8,000. Assuming monthly revenue, cost and profit are linear functions of the number of students enrolled, find each of the following: a) fixed monthly costs b) variable cost per student c) break-even number of students d) monthly enrollment needed to realize a profit of $5, Tour de Frantic (TdF) is a small bicycle retailer that has monthly fixed costs (rent, utilities, etc.) and variable (marginal) costs that depend upon how many bikes it orders from its supplier. TdF sells each bike for $400. The total cost for ordering 0 bikes in one month is $6,400. The total cost for ordering 5 bikes in one month is $7,800. a) Let x be the number of bikes ordered in a given month. Write expressions for the cost function C(x), revenue function R(x) and profit function P(x) assuming that these functions are linear. b) How many bikes need to be sold so that TdF will break even? c) How many bikes need to be sold so that TdF will make a profit of $,000? d) Graph C(x), R(x), and P(x) with bikes sold on the horizontal axis and dollars on the vertical axis. Identify each function and write the coordinates of the break-even point on the graph. 10. Suppose you have the choice of two taxi companies for a trip across town. Their fare schedules differ. Company A charges $.50 plus 10 cents a mile while Company B charges $1.50 plus 0 cents a mile. a) If your trip is 5 miles which company should you call? b) If your trip is 15 miles which company should you call? c) Graph both fare formulas on the same coordinate axes. d) For what length trip, in miles, will the fare be the same for both companies? What is this fare? Joseph F. Aieta Page 30 Printed 8/14/003

31 Linear Models Activities Page You have received bids from two printing companies for the job of printing copies of an annual report for your organization including delivery. Speedy Printer charges a flat fee of $0 plus 5 cents per page. Quality Printer charges 9 cents per page with a minimum of 50 copies. a) For how many pages does Speedy Printer offer the best deal? b) For how many pages does Quality Printer offer the best deal? c) Plot the cost functions for both printers on the same coordinate axes with the number of pages on the horizontal axis and the cost on the vertical axis. 1. A student plans to order several boxes of ten Read /Writeable computer disks from an online computer store. A Web search leads to three vendors all of which charge the same discount price for the product but have different fees for handling and postage. Company A charges a $.00 handling fee per order plus $0.75 per box. Company B has a handling fee of $4.00 per order plus $0.50 cents per box. Company C charges a flat rate of $1.50 per box. a) If the student plans to order 3 boxes, which company gives the best deal? b) If the student plans to order 5 boxes, which company gives the best deal? c) If the student plans to order 10 boxes, which company gives the best deal? d) Express the handling and postage charges for each on-line vendor in terms of the number of boxes, represented by x, and graph the three lines on the same co-ordinates axes. e) For what value(s) of x does Company A offer the best deal? f) For what value(s) of x does Company B offer the best deal? g) For what value(s) of x does Company C offer the best deal? Joseph F. Aieta Page 31 Printed 8/14/003

32 Linear Models Activities Page Accountant s Perspective on Cost. A company s accountants are less interested in the precise number of units sold. They are more interested in total sales (dollars coming in) and total cost to the company (dollars going out) for bringing in those sales. The accountants have access to data relating the company s revenue, its dollars of sales in a given period, the company s fixed cost, and the total variable amount that the company had to spend in order to bring in that revenue. To represent this graphically, both the horizontal axis and the vertical axis must be measured in dollars. Suppose fixed costs are $70,000 and variable cost, as a percent of sales, is 61%. One way to describe this situation is to say that the manufacturer s contribution to each one dollar of sales is 61 cents. In other words, 61% of each dollar of sales is needed to cover production cost. The fixed cost of $70,000 must be added to variable cost to describe total cost. If x is dollars of sales then C(x) = 0.61 x + 70,000. From the definition of revenue as dollars of sales we see that revenue is sales and write R(x) = x. This is clearly equivalent to R(x) = 1.00x. a) Write the formula for the Profit function. b) Determine the break-even dollar of sales to the nearest dollar. c) Determine the break-even dollar of sales to the nearest hundred dollars 14. For one of its products, the company s total cost is a linear function of sales. The company has determined that in the first few days of production it has a total cost of $36,836 on sales of $15,000. It has also been determined that the total cost on sales of $5,000 is $41,536. Take the accountant s perspective and find: a) the functions for revenue, cost, and profit using x for the sales volume in dollars. b) the variable cost per dollar of sales. c) the fixed cost. d) the variable cost on sales of $7,000? e) the total cost of sales of $7,000. f) the net profit (or loss) before taxes on sales of $80,000? g) the break-even dollar volume of sales. 15. Last year, variable cost for a t-shirt business was $0.40 per dollar of sales and fixed cost was $3,000. This year, variable cost per dollar of sales remains at last year s level but fixed cost has jumped up to $3,600. How much additional revenue, over last year s sales, will be required to break even this year? Draw a rough sketch to illustrate this situation. 16. A couple approaching retirement anticipates a return of 9% for their stock market investments and 4% for their money market investments for the coming year. Draw a family of iso-value lines corresponding to total annual returns of $6000, $1,000, $18,000 and $4,000. Label each line with its total return value and label the coordinates of all intercepts. What is the slope of each of these iso-value lines? Joseph F. Aieta Page 3 Printed 8/14/003

33 Linear Models Activities Page In a particular month, a local juice company is producing two types of juice, mango and papaya. The company can sell all that they In can produce. Each case of mango can be sold for $18.00 and costs $10 to produce Each case of papaya can be sold for $0.00 and costs $14 to produce. a) On the same coordinate axes draw the cost iso-value lines corresponding to total production costs of $1,000, $18,000, $4,000, and $48,000. Plot mango juice production of the horizontal axis and papaya juice production on the vertical axis Label each line with the appropriate cost value and write the coordinates of each intercept. What is the common slope of these lines? b) On a new coordinate axes draw the profit iso-value corresponding to total profits of $100, $400, $3000, and $3600. Label each line with the appropriate profit value and write the coordinates of each intercept. What is the common slope of these lines? 18. Modify the spreadsheet Break-even Quantity.xls to represent the following situation. An electric utility is about to bring a new generating unit into production at a fixed cost of $400,000. It will cost the utility $0.05 (.5 cents) per kilowatt-hour to generate electrical power, which they can sell for $0.075 (7.5 cents) per kwh. Choose an appropriate scale to represent this situation graphically. It will be necessary to make adjustments to the number of decimal places displayed, increment size, and starting value of q. The intersection point of the cost and revenue functions must appear in the final chart. Save this worksheet as kwh.xls. Let q be the number of kilowatt-hours. a) Write simplified formulas for cost, revenue, and profit as functions of q. b) Determine the break-even quantity of kwh in millions. c) To the nearest hundred dollars, what is the break-even dollar volume of sales? (Remember that you can adjust the starting value and the increment). 19. Cruise Control A car travels along a flat highway at a constant speed (cruise control). At point A there were 18 gallons of gas in the tank. At point B, 50 miles from point A, the tank was empty. Let the independent variable be the distance traveled in miles. Let the dependent variable be the gallons remaining in the tank. a) Compute the slope of this linear function and explain what this number is measuring. b) Start with a blank Excel spreadsheet and create a worksheet with cells for entering the original number of gallons in the tank at point A, the total distance traveled. in miles, and the number of gallons of gas remaining in the tank at point B Create a table with fewer than 15 rows showing the gallons remaining in the tank on the vertical axis and the total distance traveled from point A to point B on the horizontal axis. c) Create a scatter plot of the points in the table using the Chart wizard as shown in Tutorial 4 of Appendix A. Joseph F. Aieta Page 33 Printed 8/14/003

34 Linear Programming Part 1 Page 34 Linear Programming Part 1 The Graphical Method Section.1 Linear programming, or LP as it is commonly known, is a special type of optimization problem in which all of the underlying mathematical components are based on linear expressions. The term optimizing is a fancy way of saying finding a best way in the sense that decisions lead to a maximum value or a minimum value. We may want to maximize the annual return from our investment portfolio or minimize the amount of fat in our diet. A manufacturer may want to maximize profit or minimize the total amount of overtime hours. A small airline that operates a shuttle service between several cities, or a large airline that has worldwide operations, may want to maximize the number of passengers per flight or minimize the number of flights needed to service a route. The large airline may also want to minimize the total number of unprofitable routes. A regional or national telecom service may want to minimize such quantities as the total length of cabling in a city or the number of its service vehicles. Airlines and telecoms often deal with the tens of thousands of variables. Problems on this scale require sophisticated computer software and very powerful computers. We will first investigate two variable examples that can be solved with paper and pencil by a graphical method. The special case of just two decision variables enables us to represent relationships between these variables in a familiar, twodimensional setting that we know as the x-y plane. The variables are called decision variables because we are often trying to decide how much or how many achieve our objective. Even these small models with only two-dimensions can illustrate basic mathematical ideas underlying linear optimization problems. To solve LP problems with many decision variables we will utilize Excel spreadsheets and a tool called Solver. This tool uses a method known as the simplex method to solve linear optimization problems. Two articles about Linear Programming applications and the inventor of the simplex method for solving LP problems can be found in the document entitled Dantzig. What computer software cannot do very well is to translate a real situation, given in words, into an algebraic representation that can then be transformed into a number crunching procedure. As you learn how to solve LP problems, keep in mind that the initial formulation stage and the final stage of interpretation of the numerical results are steps that require the most thinking. They are at least as important as learning techniques that produce numerical results. We are not finished when we have obtained numerical results by paper and pencil methods or with a computer. We must return to the original statement of the problem and interpret numerical results within the context of the problem situation. If the results don t make sense in the context of the problem, then a source of potential error could be in the formulation. As pre-requisite skills for the application of the graphical method in LP we need to be able to: graph lines on the same set of axes determine which half-plane corresponds to a linear inequality once its boundary line has been graphed find points of intersection for pairs of boundary lines. identify the region that corresponds to those points in a plane that satisfy an entire set of inequality conditions. To practice these prerequisite skills, see the Systems of Linear Equations and Inequalities Review on page 41. We begin linear programming with a somewhat contrived example that has relatively nice numbers. Joseph F. Aieta Page 34 8/14/003

35 Linear Programming Part 1 Page 35 Example.1.1 Three machines, A, B, and C, are needed in the manufacture of nuts and bolts. To make one pound of nuts requires 5 minutes on machine A, 18 minutes on machine B, and 11 minutes on machine C. To make one pound of bolts requires 10 minutes on machine A, 18 minutes on machine B, and 4 minutes on machine C. On one particular day, machine A will be available for at most 150 minutes, machine B for at most 34 minutes, and machine C for at most 13 minutes. The manufacturer can sell all the nuts and bolts that are produced and make a profit of $4.00 per pound for nuts and $5.00 per pound for bolts. The goal is to determine the optimal mixture of nuts and bolts to manufacture in order to maximize daily profit. Formulation Let x = pounds of nuts and let y = pounds of bolts. The objective is to maximize profit =$4.00x + $5.00y subject to non-negativity and the three time constraints Machine A: 5x + 10y <= 150 Machine B: 18x + 18y <= 34 Machine C: 11x + 4y <= 13 In this problem we ignore other considerations related to profit, such as labor costs or the cost of materials. We focus exclusively on the time needed on machines A, B, and C to produce nuts and bolts and the available time on these machines each day. We want to know what profit levels are possible given that Profit = 4.00x y. $0 profit x y x y = 0 $30 profit x y x y=30 $40 profit x y x y=40 Figure.1.1 $50 profit x y x y=50 $60 profit x y x y=60 Figure.1.1 illustrates production levels that correspond to specific profit levels. Each separate table contains points that lie on the same line. A line whose points all correspond to the same profit level is called an iso-profit line or an iso-value line. The boldface ordered pairs in each table satisfy both of the following conditions: a) each boldface pair satisfies the time constraints for Machines A, B, and C. b) each boldface pair corresponds to the profit level indicated at the top of the table For example, in the leftmost table, a production of 3.00 pounds of nuts and 1.60 pounds of bolts satisfies all constraints since 5(3) + 10(1.6) <= 150, 18(3) + 18(1.6) <= 34, and 11(3) + 4(1.6) <= 13. Profit =4(3.00) + 5(1.60) = 0. In the rightmost table, production of pounds of nuts and 4.00 pounds of bolts satisfies all constraints since 5(10) + 10(4) <= 150, 18(10) + 18(4) <= 34, and 11(10) + 4(4) <= 13. Profit =4(10) + 5(4) =60. The point (1.00,.40) also corresponds to a $60.00 profit but 11(1) + 4(.40) = is greater than the 13 minutes available on machine C, so this point is not feasible. Points in the tables that are not in boldface fail to satisfy at least one of the three Joseph F. Aieta Page 35 8/14/003

36 Linear Programming Part 1 Page 36 constraints. We will soon see graphically that it is not possible to make a $100 profit since there are no feasible points on the $100 profit line. Visualization by Graphing In order to visualize the relationship between profit lines and the three time constraints, we first graph each of the boundary lines in the first quadrant. The two non-negativity constraints, x > 0 and y > 0, are what restrict us to the first quadrant. We then determine whether to shade above or below the boundary line for each inequality. Any point that is not on the boundary line of an inequality can be used as a test point to determine which side of the boundary line to shade. If a test point satisfies an inequality, then all of the points on the same side of the boundary line as the test point will also satisfy that inequality. If a test point does not satisfy an inequality, then we should shade all points on the opposite side of the boundary away from the test point. The origin (0, 0) is not contained on any of the three boundary lines for machines A, B, and C so it is a convenient point to use. [When a formulation produces a boundary line that contains the point (0, 0), such as x -y > 0, then we would select a test point such as (1, 1) that is not on that boundary line.] If we test the point (0, 0) in the inequality 5x +10y < 150, we obtain the true statement < 150 so we shade the region which contains the origin and is below the boundary line 5x+ 10y = 150 If we test the point (0, 0) in the inequality 18x +18y <34, we obtain the true statement < 34 so we shade the region which contains the origin and is below the boundary line? 18x +18y =34 If we test the point (0, 0) in the inequality 11x +4y <13, we obtain the true statement < 13 so we shade the region this contains the origin and is below the boundary line. 11x + 4y = shade below shade below shade below Figure.1. Figure.1.3 Figure.1.4 For a point to be feasible it must satisfy the entire system of inequalities, that is it must simultaneously have non-negative coordinates and satisfy 5x + 10y <= 150, 18x + 18y <= 34, and 11x + 4 y <= 13. For the machine constraints in the nuts and bolts example, we want the set of points that satisfies all constraints (inequalities) simultaneously. This region, called the set of feasible solutions, or the feasible region, is shaded in Figure.1.5. Three corner points on the axes are (0, 15), (0, 0), and (1, 0). To find the remaining corner points, we need to solve a systems of equations. You can review the prerequisite algebra for solving systems in the file TwobyTwoEquations.xls and test your skills with the practice file Solve_Linear_Systems.xls Joseph F. Aieta Page 36 8/14/003

37 Linear Programming Part 1 Page 37 The system of equations 5x + 10y = x + 18y = 34 has solution (6, 1) The system of equations 18x + 18y = 34 11x + 4 y = 13 has the solution (8.57, 9.43) with coordinates rounded to two decimal places. x y Figure.1.5 The point of intersection of the two boundary lines 5x + 10y = x + 4 y = 13 and is above the constraint 18x + 18y < 34 This point does not need to be found. It could not be a corner point since it is not feasible. If an iso-profit line P = 4x + 5y has no points of contact with the feasible region then it is not possible to make a profit of P dollars. Figure.1.5 suggests the location of values of x and y that might maximize profit. Clearly the point (0, 0) minimizes profit so that leaves the remaining four corner points, the edges, and the points in the interior of the feasible region as candidates. By re-introducing the iso-profit lines and superimposing them on the feasible region, we can quickly narrow down our search. We have previously seen that a $60 profit is possible so we continue to increase profit and look for feasible points. The boldface points in the table below are feasible and are on the $70 profit line 4x + 5y = 70 x y Figure.1.6 None of the points on the $100 profit line 4x + 5y = 100 satisfy all three machine constraints. x y In Figure.1.6, the $100 profit line is the dotted iso-value that is furthest from the origin. Points on this line, including the intercepts (0, 0) and (5, 0) are listed in the table to the right of the graph. None of the points on the $100 profit line 4x + 5y = 100 satisfy all three machine constraints. It is not possible to make a profit of $100 since the $100 isoprofit line has no points in common with the feasible region. All iso-profit lines are parallel and profit increases as the iso-profit lines move further away from the origin. Joseph F. Aieta Page 37 8/14/003

38 Linear Programming Part 1 Page 38 Where should we look for the optimal point(s) that correspond to the maximum profit? Imagine for a moment that we have found the maximum profit, which we call MP. The associated iso-profit line would be 4x + 5y = MP. Could any points on the maximum profit line be in the interior of the feasible region? The answer is no and the reasoning is as follows: If an iso-profit line contains points that are inside the feasible region then we could always increase profit by moving the iso-profit line in the direction away from the origin. If we go too far then our iso-profit line will have no points in common with the feasible region. We are left with two possibilities: One possibility is that the maximum profit line has exactly one point in common with the feasible region. That point must be one of the vertices (corners) of the set of feasible points. In such a case we say that the optimal solution is unique. The other possibility is that the maximum profit line 4x + 5y = MP has the same slope as one of the boundary lines for the feasible region. In this case, points on the maximum profit line will coincide with points on a boundary line and optimal solutions will be on an entire edge of the feasible region. Unique Solution versus Alternate Solutions (6, 1) As we move away from the origin and focus on a portion of the feasible region, we see that the last point of contact between the feasible region and the family of iso-profit lines occurs at the point x = 6 and y = 1 The last iso-profit line that is in touch with the feasible region is 4x + 5y = 84 and the point of contact is (6, 1). 4x+5y = 100 (dotted) 4x+5y = 84 Figure.1.7 If a linear programming problem with two decision variables has a unique solution then it occurs at a corner point of the feasible region. Joseph F. Aieta Page 38 8/14/003

39 Linear Programming Part 1 Page 39 Let s change our profit function to reflect a $5.00 profit per pound of nuts sold. The objective function changes from P = 4x + 5y to P = 5x + 5y. Each iso-profit line now has a slope of -1. This is the same slope as the boundary line for the machine B constraint, 18x + 18y = 34. Figure.1.8 shows the $80 profit line which contains interior points and the $100 profit line having no points in common with the feasible region. Maximum profit is between $80 and $ x+5y = 100 (dotted) 10 5x+15y = Figure.1.8 Figure As we move the $80 iso-profit line away from the origin, we see in Figure.1.9 that the $90 profit line goes through the corner points (6, 1) and (8.57, 9.43). If we should move this iso-profit line any further in the direction away from the origin then it will have no contact with the feasible region. All points on the segment with endpoints (6, 1) and (8.57, 9.43) are contained on the boundary line 18x + 18y = 34 and are also on the maximum profit line 5x + 5y = 90. Check that (7, 11) and (8, 10) are two additional points on this edge and on the profit line 5x + 5y = 90. If the solution to a linear programming problem with two decision variables occurs at two corner points then any of the points on the segment (edge) that joins these two corner points will be optimal and correspond to the same value of the objective function. We say that these points are alternative optima. This property of corner points suggests a step by step approach for solving formulated linear programming problems with two decision variables. 1. Graph the constraint boundaries and shade the feasible region which is the set of points that satisfies each of the constraints.. Find all of the corner points of the feasible region. 3. Calculate the value of the objective function at each of the corner points. 4. If the solution occurs at two corner points then all points on that segment (edge) are also optimal solutions. Otherwise the solution occurs at a unique corner point. Joseph F. Aieta Page 39 8/14/003

40 Linear Programming Part 1 Page 40 Suppose that we apply this step by step method to a linear programming problem with the new objective function Profit = x + 5y. Assume that the constraints remain the same. They are non-negativity and (1) 5x +10y < 150 () 18x + 18y < 4 (3) 11x + 4y < 13. Steps 1 and would produce the graph in Figure.1.10 with corner points at (0,15), (6, 1), (8.57, 9.43), (1, 0), (0, 0) Enter the coordinates in a table and complete steps 3 and 4 x y Profit =x+ 5y The solution is unique. The maximum profit is $75.00 which occurs at the corner point (0, 15) on the y axis. Figure.1.10 Q. For this feasible region, find an objective function for which the point (1, 0) on the horizontal axis is the optimal solution. A. We need a fairly steep slope. Profit = 6x + y has a slope of -3. The maximum value of the objective function Profit = 6x + y is 7.0 at (1, 0). x y Profit =6x+ y [In the next section we will learn about Solver, a powerful Excel Add-in, which we will use to solve more complex linear programming problems.] Joseph F. Aieta Page 40 8/14/003

41 Linear Systems Review Page 41 Review Systems of Linear Equations and Inequalities Solve the systems of equations for exercises x - y = 3 x + y = 3.. 4x + 3y = 4 x + 6y = 5 3. x - 3y = 5 6x - 9y = x + 1y = -9 -x - 8y = x - 15y = -3-10x + 5y = 5. Use graph paper for # Using graph paper, graph the solution space ( feasible region) for the system of inequalities x + y > 6 x + 4y < 8 and non-negativity x > 0 y > 0 7. Using graph paper, graph the solution space ( feasible region) for the system of inequalities 4x + 9y > 130 x + 3y < 50 x > 5 Joseph F. Aieta Page 41 8/14/003

42 8. Using graph paper, graph the solution space (feasible region) for the system of inequalities x + y > 0 3x + y < 40 y < 15 Linear Systems Review Page 4 9. Using graph paper, draw the boundary lines, find the coordinates of the corner points, and shade the solution space (feasible region) for the system of inequalities below: 5x + 3y < 10 6x + 8 y < 19 x > 0 y > Graph the feasible region in Quadrant I for the following system of inequalities: x < 700 x + 4 y < 3,00 16 x + 8 y < 9,400 Joseph F. Aieta Page 4 8/14/003

43 Linear Programming Activities 1 Page 43 Linear Programming Activities 1 Open the worksheet Maximize profit in the file Nuts&Bolts1.xls for exercises 1 and. 1. Examine the graph showing the bounded feasible solution set (region) that corresponds to the non-negativity constraints and the constraints for Machine A, Machine B, and Machine C. Change the objective function to Profit = 3.00 x y where x = number of pounds of nuts and y = number of pounds of bolts. Note that the iso-profit line 3.00 x y = 60 has points in common with the feasible region so it is possible to make a profit of $60. The point x = 10 and y = 6 is one of the points on the line 3.00 x y = 60. Is this point (10, 6) feasible? Explain why or why not? a) Name two feasible solutions (ordered pairs) that correspond to a profit of $ b) Is the point (1, 4.8) feasible? Why or why not? c) The point x = 10, y = 10 corresponds to a profit of 3*10 + 5*10 = = 80 dollars. Explain why it is not possible to have a profit of $80 given these constraints and this objective function. d) Use the slider bar to slowly move the iso-profit line away from the origin. What is the only point common to the feasible region and the iso-profit line when 3.00 x y = 78?. Assume that quality control personnel adjust one or more of the machines and the profit for each pound of nuts increases by $1.00 per pound. In Nuts&Bolts1.xls examine the worksheets Maximize 4x+ 5y and Maximize Profit = 5x+ 5y. a) Describe one or more possible ways to obtain a profit of $80 assuming Profit =4.00x+ 5.00y. If a profit of $80 is not feasible then explain why? b) Is it possible to make a profit of $88.00 if the profit function is Profit = 4.00x+ 5.00y? If so, describe how and if not, then explain why not. c) Is it possible to make a profit of $88.00 if Profit = 5.00x+ 5.00y? If so, describe several possible ways and if not, then explain why not. d) What is the maximum profit if Profit = 5.00x+ 5.00y? Describe how this profit can be obtained. Open the worksheet graph in the file Burgers&fries1.xls for exercises Examine the graph showing the unbounded feasible solution set (region) that corresponds to the non-negativity constraints and the constraints for Protein, Carbohydrate, and Calories. Suppose the objective function is fat = 10 x + 18 y where x = number of HB and y = number of FF. a) Is the point (3, 5) feasible? Why or why not? b) Is the point (9, 5/3) feasible? Why or why not? c) Name two feasible solutions (ordered pairs) that correspond to 10 grams of fat. d) What is the slope of the iso-fat line 10 x + 18 y = 10? e) Use the slider bar to slowly move the iso-fat line towards the origin. When grams of fat becomes 90, what point(s) are common to the iso-fat line 10 x + 18 y = 90 and the feasible region. Joseph F. Aieta Page 43 8/14/003

44 Linear Programming Activities 1 Page Suppose the fast food restaurant switches to a cheaper type of meat and the new objective function becomes fat = 18 x + 18 y. All iso-fat lines will now have slope equal to -1. [Note: This new family of iso-cost lines will not be parallel to the dotted line with the double arrows that is shown in the worksheet, so ignore that line] a) Is it possible to get exactly 10 grams of fat and still satisfy the constraints for Protein, Carbohydrate, and Calories? Justify your answer. b) Is it possible to get exactly 150 grams of fat and satisfy the constraints for Protein, Carbohydrate, and Calories? Justify your answer. c) Name one optimal solution to this optimization problem, i.e. what is x, what is y. and what is the minimum amount of fat? d) Name six different feasible solutions to 18 x + 18 y = 144. e) What property of the line 18 x + 18 y = 144 is the reason why there are alternative solutions to this optimization problem instead of just one unique solution? f) Is it feasible to get less than 144 grams of fat with this new objective function? Justify your answer. 5. Suppose the objective function becomes fat = 0 x + 16 y. a) Name one point that minimizes fat and satisfies the constraints for Protein, Carbohydrate, and Calories. b) Are there alternative optimum solutions? Why or why not? Open the file Boxes of CDs.xls for exercises 6 and 7 6. a) Use the step by step graphical method to maximize the profit function Z =.00 x y subject to non-negativity and the four constraints: x + y 4 x + y 10 -x + 3y 0 Note that y 1/3x is equivalent to the constraint -x + 3y 0 x 1 b) Use the graphical method to maximize the objective function Z = 1.5x y subject to non-negativity and the four constraints: x + y 4 x + y 10 -x + 3y 0 x 1 Joseph F. Aieta Page 44 8/14/003

45 Linear Programming Activities 1 Page 45 c) e) Leave the constraints unchanged but change the objective function to Z = 0.50x y. b) What is the slope of the iso-profit line 0.50x y = 1.75? c) Do all iso-profit lines have the same slope? d) What values of x and y produce the maximum value of the new objective function? 7. Change the objective function back to Z =.00 x y in Boxes of CDs.xls and turn your attention to the table in the worksheet Setup for Solver. The values of the decision variables are stored in the blue cells below the labels x and y. These blue cells have been named x and y by first selecting the block of cells J5:K6 and then using the menu commands Insert Name Create Top Row. Since the two cells, J5 and K5 are initially empty, x and y will each have the value zero. Fill in the totals in cells L6, L7, L8, L9, and L10 exactly as instructed below: In this table, the formula in cell L6 is the sum of the products obtained by multiplying the objective function s coefficient of x (in cell J6) times the value of the decision variable x (in cell J5) and then adding the product of the objective function s coefficient of y (in cell K6) times the value of value of the decision variable y (in cell K5). The formula in cell L6 becomes =J6*x+K6*y. Since the initial values of the decision variables are both zero, the value in cell L6 will also be zero. The sum of products formulas in cells L7 through L10 for the left-hand sides of the four inequalities are entered in a similar fashion: =J7*x+K7*y for constraint 1, =J8*x+K8*y for constraint, =J9*x+K9*y for constraint 3, and =J10*x+K10*y for constraint 4. a) Once you have entered these formulas, replace the zero values for x and y to 3 and 6 respectively. What are the five values that appear in cells L6 though L10? b) What cells have been designated to store the values of the right-hand sides of constraints (1) (4) and what are those values? c) What is our objective, i.e. which cell are we trying to optimize? Use the graphical method to determine the optimal solution. Joseph F. Aieta Page 45 8/14/003

46 Linear Programming Activities 1 Page 46 In problems 8-11 below, the symbol θ represents the value of a linear objective function. 8. Find θmin if θ = x + y subject to non-negativity and x+ 3y 1 5x+ y Find θmax if θ = 5x + 8y. subject to non-negativity and x + y 13 x + y x+ y 0 x Find θmax and θmin if θ = 3x + 5y subject to non-negativity and x + 4y 36 x + y 1 x + y Find θmax and θmin if θ = 0.3x - 0.y subject to non-negativity and 0.15x y 15 x + y x 3 1. Graph the following systems of inequalities, find the corner points of the resulting region, and determine any value(s) for x any y that minimizes Z = 10x + 6y. x + 3y 54 4x + 8y 10 x, y Solve the following linear program Minimize Z = 10x + 15y subject to non-negativity and x+ y 10 3x + y 1 -x + 3y 3 Joseph F. Aieta Page 46 8/14/003

47 Linear Programming Activities 1 Page A married couple, who have between $50,000 and $100,000 to invest, has asked for advice on how much to put into stocks and how much to put into a money market. They expect to get a yearly return of 9% on the stocks and 5% on the money market. Because they think stocks are intrinsically risky, they don t want to put more than 5% of their total investment in the stock market. On the other hand, they are attracted to the higher return and have decided to put at least 10% of their total investment into stocks. Their objective is to maximize the annual return on their total investment. a) Determine which of the following investment strategies meet the couple s constraints and for those that do meet their requirements, calculate the yearly return. Stocks Money Market $15,000 $75,000 $0,000 $55,000 $18,000 $80,000 $5,000 $40,000 $9,000 $90,000 b) Given their constraints, what is the largest possible stock market investment for the couple? c) Given their constraints, what is the smallest possible stock market investment for the couple? d) Formulate the problem of maximizing the couple s return algebraically as a linear program. e) Graph the feasible region and find the optimal solution. 15. Your pet pig, Waddles, lives on a diet of French fries and chocolate. The table below contains nutrient and cost information for his diet. You wish to provide him with a diet that contains at least 300 grams of carbohydrates and at most 600 grams of fat and at minimum cost. Rates per cup Fat Carbohydrates Cost Fries 5 gm 10gm $0.50 Chocolate 15 gm 15 gm $.00 a) Suppose Waddles is fed 5 cups of French fries and 10 cups of chocolate, what is his total fat consumption? Carbohydrate consumption? What is the cost? Does this diet meet the nutrient requirements? b) Can Waddles live on a diet of French fries alone? Why or why not? If so, what would the cheapest such diet cost? c) Can Waddles live on a diet of chocolate alone? Why or shy not? If so, what would the cheapest such diet cost? d) Determine a diet for Waddles that meets all of his nutritional requirements and costs less than $ e) Formulate the problem of minimizing the cost of Waddle s diet algebraically as a linear program. f) Graph the feasible region and solve the diet problem. Joseph F. Aieta Page 47 8/14/003

48 Linear Programming Activities 1 Page Java Juice sells two different coffee bean mixtures, House Blend and Deluxe Blend. The house blend is 90% Mocha and 10% Java and sells for $10 a pound. The Deluxe is 80% Mocha and 0% Java and sells for $11 a pound. Assume Java Juice will sell all that it makes. On a particular day, Java Juice has 7 pounds of Mocha and 10 pounds of Java on hand. a) How many pounds of each blend should Java Juice make in order to maximize revenue? b) d) Suppose Java Juice changes its per pound coffee mixture prices. For each of the prices below, determine the optimal product mix and the associated revenue. b) House Blend to sell for $10 and Deluxe to sell for $15 c) House Blend to sell for $10 and Deluxe to sell for $0 d) House Blend to sell for $7 and Deluxe to sell for $11 e) House Blend to sell for $13 and Deluxe to sell for $ Sugar Spectral is a company that manufactures Halloween candy. It has two candy factories, Factory A and Factory B, and two distributors for its product, Distributor 1 and Distributor. In September Distributor 1 needs 8,000 pounds of candy and Distributor needs 7,000 pounds. Factory A has 10,000 pounds ready to ship and Factory B 11,000 pounds. Shipping costs per pound from factories to distributors are given below. Distributor 1 Distributor Factory A $1.00 $0.60 Factory B $1.0 $0.70 a) How much candy should Sugar Spectrals ship from each factory to each distributor if shipping costs are to be kept to a minimum? b) Suppose the price of shipping from Factory A to Distributor 1 doubles. How does this change the optimal solution? 18. The owner of the Sea Wharf Restaurant would like to determine the best way to allocate a monthly advertising budget of $1000, all of which will be split between newspaper and radio advertising. Management has decided that at least 5% of the budget must be spent on each type of media, and that the amount spent on local newspaper advertising must be at least half the amount spent on radio advertising. A marketing consultant has developed an index that measures audience exposure per dollar of advertising on a scale from 0 to 100, with higher values implying greater audience exposure. The value of the index is 50 for a standard advertisement in the local newspaper and 80 for a radio ad. How should management allocate the advertising budget in order to maximize total audience exposure? Joseph F. Aieta Page 48 8/14/003

49 Linear Programming Section. Page 49 Linear Programming Part What is Solver? Section. The Solver in Microsoft s Excel is a powerful optimization and resource allocation tool that can handle complicated situations with dozens of variables. It can help business professionals uncover the best uses of limited resources to maximize quantities, such as profit or return on investment, and to minimize other quantities such as cost or time. There are certain pre-requisites that you should have before you solve LP problems with Solver. First and most important, you must understand the formulation process. This involves transforming an LP problem stated in words into a structured symbolic form. Every formulation contains an identification of the decision variables, a linear combination of these variables that represents the function to be optimized, and linear combinations of the decision variables that model the problem constraints. Once you have the above pre-requisite skills, you will then need to develop a systematic process for embedding the formulated model into a spreadsheet. We will refer to this stage as the setup for Solver. Finally, you will need to know how to obtain numerical results, generate reports from Solver, and interpret Solver output in the context of the original problem. Follow these guidelines in order to learn how to efficiently setup a spreadsheet and apply Solver 1. Study the worked out examples that are related to familiar two variable LP problems.. Complete worksheets that either contain detailed instructions for each stage or have been partially structured for Excel. 3. Start with a blank worksheet and solve new LP problems with two or more variables. Illustrative files of different types can help you learn how to use Solver effectively. File SolverLP.pps About 10 minutes Burgers&Fries.xls All worksheets About 10 minutes Nuts& Bolts.xls Prep for Solver Five minutes or less BoxesofCDs.xls Setup for Solver Five minutes or less Description This is a self-paced PowerPoint tutorial by Prof. Denise Troxell for learning to use Excel s Solver. This presentation shows every detail including menu selections, cell entries, and complete dialog boxes. This is a set of detailed directions included right on the spreadsheet for each stage in the setup and solution of the LP problem Burgers& Fries. This is a partially complete setup for the LP problem Nuts & Bolts. This is a partially complete setup for the LP problem Boxes of CD. Joseph F. Aieta Page 49 8/14/003

50 Linear Programming Section. Page 50 The optimization features of this tool surpass time-consuming paper and pencil methods and simple trial and error approaches. Solver utilizes sophisticated mathematical algorithms and the number-crunching power and speed of a computer. To solve an optimization problem using Solver, you must specify the: target cell that you want to minimize or maximize. changing cells that you want Solver to adjust until a solution is found. constraint cells that must fall within certain limits or equal target values. Solver has extensive on-line help that can assist you with all aspects of using this tool including: finding and installing the Solver Add-in if Solver is not in the Tools menu how to define a problem using Solver how to add, change, or delete constraints in Solver how to complete dialog boxes how to create and interpret reports that summarize the results of a successful solution process how to save new models and load models that have been previously saved. You can use on-line help for other questions such as the difference between linear and nonlinear problems. We will take a relatively simple LP situation, formulate the problem, and then illustrate each of the following stages leading to a solution: Example..1 entering the relevant data, formulas, and cell names, in a structured way, into a blank worksheet entering the relevant parameters and options required to search for an optimal solution with Solver. modifying the model by introducing additional decision variables and additional constraints. Ace Tire Company manufactures two types of tires: Model P (the premium) and Model R (the regular). Model P sells for $95 per tire and costs $85 per tire to make, whereas Model R sells for $50 per tire and costs $4 per tire to make. To make one Model P tire, it requires two hours on Machine A and four hours on Machine B. On the other hand, to make one Model R tire, it takes nine hours on Machine A and three hours on Machine B. Production scheduling indicates that during the coming week Machine A will be available for at most 36 hours and Machine B for at most 4 hours. How many of each tire should the company make in the coming week in order to maximize its profit? What is this maximum profit? The two decision variables in this model are the number of premium tires, x, and the number of regular tires, y. The formulation for this LP problem is given below: Maximize Profit = 10x + 8y subject to non-negativity and the constraints (1) and () for Machine A and for Machine B 1) x + 9y < 36 Machine A time constraint ) 4x + 3y < 4 Machine B time constraint The objective is to find an optimal weekly production plan. This means determining the number of each type of tire to produce each week in order to maximize the total weekly profit. Joseph F. Aieta Page 50 8/14/003

51 Linear Programming Section. Page 51 To prepare an Excel model for Solver, start by entering labels and naming cells as shown below: Type the labels Premium in cell B1 and Regular in cell C1. These labels will become the names for cells B and C by first highlighting the four cells and choosing Insert Name Create Top Row. The cells B and C will contain the values of the decision variables. Type the labels Dir. (direction) in cell E1 and Bounds in cell F1. Type the labels Decision Variables in cell A and Profit Function in cell A3. Type the labels Machine A in cell A4 and Machine B in cell A5. Rows 4 and 5 will contain the constraint information. A B C D E F 1 Premium Regular Totals Dir. Bounds Decision Variables 3 Profit Function 4 Machine A 5 Machine B Figure..1 Cell values or formulas In cell B3 type 10 In cell C3 type 8 Explanation Each premium tire contributes $10 of profit. Each regular tire contributes $8 of profit. In cell D3 type =B3*Premium+C3*Regular Our objective is to maximize the value of this target cell. In cell B4 type In cell C4 type 9 In cell B5 type 4 In cell C5 type 3 Each premium tire requires hours on machine A. Each regular tire requires 9 hours on machine A. Each premium tire requires 4 hours on machine B. Each regular tire requires 3 hours on machine B. In cell D4 type In cell D5 type =B4*Premium+C4*Regular =B5*Premium+C5*Regular This cell contains the left-hand side (LHS) of the constraint for machine A. This cell contains the LHS of the constraint for machine B. In cell F4 type 36 In cell F5 type 4 This cell contains the right-hand side (RHS) of the constraint for machine A. This cell contains the RHS of the constraint for machine B. In cell E4 type the inequality symbol <= and do the same in cell E5 A B C D E F 1 Premium Regular Totals Dir. Bounds Decision Variables 3 Profit Function Machine A 9 <= 36 5 Machine B 4 3 <= 4 Figure.. Joseph F. Aieta Page 51 8/14/003

52 Linear Programming Section. Page 5 Note: Entering symbols for the direction of an inequality are not required and could be omitted since Solver does not use these cells. Indicating the direction < or > is simply a reminder to help us complete a Solver dialog box at a later stage in process. It is often useful to assign names to cells. We can accomplish this by selecting the cells that we want to name and the cells containing the names (in this case B1: C) and then selecting Insert Name Create and directing excel to the location of the names. Notice that the formula for the objective function and the LHS for each of the constraints contain references to the decision variables in B and C. Since cells B and C are currently empty, the values of cells D3, D4, and D5 will be zero. The figure below displays the formulas in column D rather than the values of these cells. A B C D E F 1 Premium Regular Totals Dir. Bounds Decision Variables 3 Profit Function 10 8 =B3*Premium + C3*Regular 4 Machine A 9 =B4*Premium + C4*Regular <= 36 5 Machine B 4 3 =B5*Premium + C5*Regular <= 4 Figure..3 The spreadsheet below serves as a check on our formulas. By placing a 1 in both cells B and C, we obtain the values 18, 11, and 7 in column D. We interpret these numbers as the values of the profit function, the number of machine A hours used, and the number of machine B hours used, respectively. Making exactly one Premium tire and exactly one Regular tire would generate a profit of 18 dollars, using 11 hours on machine A and 7 hours on machine B. Since both machine A and B have considerable available time that has not been used. A B C D E F 1 Premium Regular Totals Dir. Bounds Decision Variables Profit Function Machine A 9 11 <= 36 5 Machine B <= 4 Figure..4 We could use a trial and error approach to test different values of the decision variables and observe the impact on profit. Keep in mind, however, that any pair of values in B and C that produces a total greater than 36 in D4 or a total greater than 4 in D5 must be rejected as not feasible. Is it feasible to make 5 Premium tires and 3 Regular tires? For Machine A we see that (5) +9(3) = = 37 which exceeds 36 so (5, 3) is not a feasible point. Try some other points. When you are finished experimenting by trial and error, restore the worksheet to a state with zeros in B and C A B C D E F 1 Premium Regular Totals Dir. Bounds Decision Variables Profit Function Machine A 9 0 <= 36 5 Machine B <= 4 Figure..5 Joseph F. Aieta Page 5 8/14/003

53 Linear Programming Section. Page 53 Our spreadsheet has been prepared for investigation with the Solver Add-in tool. When we load Solver from the Tools menu, a dialog box entitled Solver Parameters will appear. The first entry to be specified is Set Target Cell. This cell must contain a formula in terms of the decision variables. Solver may make a guess about the location of the target cell. In this case the formula in column D and row 3 is the correct target cell so click on cell D3. Next move to the Equal to row and click on the Max button. This will direct Solver to maximize the weekly profit from the two types of tires. Buttons for Min and Value will be blank since only one of these buttons can be in effect at any one time. Figure..6 Next move to the box labeled By Changing Cells. Clicking on the cells for the two decision variables will cause the entry to appear as $B$:$C$, the absolute address of the range of cells which contain the decision variables. [Note: If Solver is successful at finding an optimal solution then B and C will be the cells into which Solver will store these optimal values.] Next, move to the Subject to the Constraints box. A dotted rectangle will appear in this box. Clicking on the Add button will bring up another dialog box. Cells D4 and D5 of the spreadsheet contain the formula for the left hand sides (LHS) of the machine constraints. These values should be respectively less than or equal to (<=) the right hand sides (RHS) of the boundary values which are stored in cells F4 and F5. Constraints can be added one at a time or several at a time if their direction is the same. Figure..7 Click in the Cell Reference box and highlight cells D4:D5 on the worksheet. You should see the <= sign in the middle box. If not, click on the black down arrow to the right and select the <= sign. This is what actually sets the direction of the constraint. Click in the Constraint box and highlight cells F4:F5, the cells containing the expressions for the Right Hand Side (RHS) of each constraint. Lastly, press the OK button in the dialog box and you will be returned to the Solver Parameters dialog box. This time you will see the constraint highlighted in the Subject to Constraints box. It should read: $D$4:$D$5 <= $F$4:$F$5. Joseph F. Aieta Page 53 8/14/003

54 Linear Programming Section. Page 54 If you make an error, move the cursor over to the Change key and the Change Constraint dialog box will appear. It should look similar to the Add Constraint dialog box and the same instructions apply as before. Move the cursor over to where the change has to be made, make the change, and press OK to return to the Solve Parameters dialog box. Figure..8 Click on the Options button and check the boxes Assume Linear Model and Assume Non-Negative. If you forget to include the nonnegativity constraints, you may obtain wrong results. For now, ignore the rest of the options. Most of these have to do with default settings and solving non-linear optimization problems. Click on the OK button and return to the Solver Parameters dialog box.. Figure..9 Click on the Solve button and Solver will proceed to search for the optimum solution. When Solver has finished, another dialog box will appear indicating whether or not a solution was found. Since the search was successful a solution appears on the worksheet. Figure..10 Joseph F. Aieta Page 54 8/14/003

55 Linear Programming Section. Page 55 For managers, no solution is complete without a report containing an interpretation and a conclusion. The dialog box contains references to the Excel Solver completion reports Answers, Sensitivity, and Limits. As you learn more about linear programming, you will want to generate and interpret Sensitivity and/or Limits reports. For now, we will generate just an Answer report. First highlight Answer in the Reports box and then click on the OK box. The answer report for this two variable problem should look like the output table in Figure..11. M icrosoft Excel 10.0 Answer Report Worksheet: [Book]Sheet1 Report Created: 8//001 4::05 PM Target Cell (Max) Cell Nam e Original Value Final Value $D$3 Profit Function Totals Adjustable C ells Cell Nam e Original Value Final Value $B$ Premium 0 9 $C$ Regular 0 Constraints Cell Nam e Cell Value Form ula Status Slack $D$4 Machine A Totals 36 $D$4<=$F$4 Binding 0 $D$5 Machine B Totals 4 $D$5<=$F$5 Binding 0 Figure..11 Checking the button Keep Solver Solution instructs Solver to display the optimal values of each variable on the spreadsheet. If you choose Restore Original Values, then Solver will restore these variables to their original values. Saving the spreadsheet will also save the Solver parameters in your model along with the spreadsheet data and the Answer Report. Each Answer Report is numbered and identified as a tabbed sheet. New decision variables and/or additional constraints can be introduced in the spreadsheet at any point. For example, suppose Ace Tire decides to make a third model which they refer to as the Discount tire. To make one Model D will require four hours on machine A and four hours on machine B. To improve the quality of their products, Ace Tire must also bring in a new finishing machine; machine C, which can be utilized for at most 30 hours per week. The machine C utilization requirements are as follows: three hours per Premium tire, one-half hour per Discount tire, and two hours per Regular tire. The profit function also changes. The new profit function will be defined by the formula Profit = 8*Premium + 9*Discount + 7*Regular. To add a new decision variable for Discount tires, we must first insert a new column. Highlight the column C containing the data for Regular tires and then select Insert Enter the label Discount in cell C1. Enter the value 9 in cell C3 Enter the value 4 in cell C4 Enter the value 4 in cell C5 To add a new constraint in row 6 for machine C Enter the label Machine C in cell A6. Enter the value 3 in cell B6 Enter the value 0.5 in cell C6 Enter the value in cell D6 Enter the inequality <= in cell F6 Enter the value 30 in cell G6 Joseph F. Aieta Page 55 8/14/003

56 Highlight C1 and C and use Insert Name Create Top row to give C the name Discount The formulas in cells E3:E6 should be modified as follows: Linear Programming Section. Page 56 E3 should contain the formula E4 should contain the formula E5 should contain the formula E6 should contain the formula = B3*Premium+C3*Discount+D3*Regular = B4*Premium+C4*Discount+D4*Regular = B5*Premium+C5*Discount+D5*Regular = B6*Premium+C6*Discount+D6*Regular When the modified worksheet is ready for Solver, it should look like the spreadsheet below A B C D E F G 1 Premium Discount Regular Totals Dir. Bounds Decision Variables Profit Function Machine A <= 36 5 Machine B <= 4 6 Machine C <= 30 Figure..1 Figure..13 shows the parameters of the completed Solver dialog box. Make sure that Linear Model has been checked under Options before you run Solver. Once Solver indicates that a solution has been found, select the Answer Report before you click the OK button. Figure..13 Joseph F. Aieta Page 56 8/14/003

57 Linear Programming Section.3 Page 57 Note that Answer Report in Figure..14 shows the original and final values of the target cell and the changing cells. It also lists the constraints and gives important information about each constraint. Microsoft Excel 10.0 Answer Report Worksheet: [Book]Sheet1 () Report Created: 8//001 4:53: PM Target Cell (Max) Cell Name Original Value Final Value $E$3 Profit Function Totals Adjustable Cells Cell Name Original Value Final Value $B$ Premium 0 3 $C$ Discount $D$ Regular 0 0 Constraints Cell Name Cell Value Formula Status Slack $E$4 Machine A Totals 36 $E$4<=$G$4 Binding 0 $E$5 Machine B Totals 4 $E$5<=$G$5 Binding 0 $E$6 Machine C Totals 1.75 $E$6<=$G$6 Not Binding 17.5 Figure..14 Improving the Efficiency of Building Larger Models Section.3 The Ace Tire worksheet has only three decision variables, three < constraints, and one > constraint (non-negativity). It is not unusual for real - world linear optimization problems to have hundreds (or even thousands) of variables and constraints. Some examples of very large LP models occur in airline reservation systems and the design of telephone switching circuitry. See the document Dantzig.doc. The nutrition model below has nine variables and thirteen constraints. Before we examine this model, we will first learn how to prepare spreadsheets for Solver that uses a more efficient method of entering data and formulas. These modifications will also make it easier for us to read the reports and interpret the results. To understand this new approach we will re-build the entire Ace Tire worksheet starting with a blank worksheet. As you will see, the major difference will be in the way we build the formula for the target cell, E3, and the formulas for the cells, immediately below the target cell, containing totals. Joseph F. Aieta Page 57 8/14/003

58 Linear Programming Section.3 Page 58 Create a blank worksheet and enter the column labels Premium, Discount, and Regular in cells B1, C1, and D1 respectively. Also enter the label bounds in cell G1. Enter the same row labels in column A as shown earlier. Leave cells B:D (B through D) empty for the three decision variables or set these initial values to zero. Enter the coefficients (8, 9, and 7) for the objective function in columns B, C, and D of row 3. Enter the nine coefficients which are to be used in the LHS for the Machine A, B, and C constraints. Enter the three values of the bounds for Machine A, B, and C (These will become the RHS of <= constraints in the Solver parameters). One way to enter the formula for the objective function in cell E3 is =B3*B+C3*C+D3*D Cells E4, E5, and E6 must contain formulas for the LHS of each machine constraint. Using the earlier method, these formulas were expressed as: =B4*B+C4*C+D4*D =B5*B+C5*C+D5*D =B6*B+C6*C+D6*D In each of these four formulas, notice that the values in columns B, C, and D of each row (rows 3, 4, 5, and 6) are multiplied by the corresponding values of the decision variables which are always in cells B, C, and D. These individual products are then added together, as a sum of products. Using Excel s SUMPRODUCT function, the formula for cell E3 can be efficiently implemented as = SUMPRODUCT(B3: D3, $B$:$D$) References to the range of cells B, C, and D with $ signs in front of the column and row position, are called absolute references. This has the effect of insuring that a reference to such a cell will not change if the formula containing the absolute reference is copied to another location. The LHS formulas for cells E4, E5, and E6 can be easily obtained by copying cell E3, which contains the formula = SUMPRODUCT( B3: D3, $B$:$D$), into the three rows directly below E3. One easy way to perform this action is to select cell E3, locate its fill handle in the lower right corner of the cell s border. Then hold the left mouse button down and drag the cursor (in the shape of a crossbar) through the three rows. Once the range E4: E6 has been highlighted, let up on the left mouse button and the filling down will be accomplished. In the spreadsheet below, below, column E shows the formula for the objective function (shaded) and the LHS of the constraints for each machine. Column E shows formulas instead of values A B C D E F G 1 Premium Discount Regular Totals Dir. Bounds Decision Variables Profit Function =SUMPRODUCT(B3:D3,$B$:$D$} 4 Machine A 4 9 =SUMPRODUCT(B4:D4,$B$:$D$) <= 36 5 Machine B =SUMPRODUCT(B5:D5,$B$:$D$) <= 4 6 Machine C =SUMPRODUCT(B6:D6,$B$:$D$) <= 30 Figure.3.1 The efficiencies of building a model in the new way will become more apparent as you investigate the following nutrition/diet problem. Joseph F. Aieta Page 58 8/14/003

59 Linear Programming Section.3 Page 59 Example.3.1 Ten male soccer players are all between 19 and years old and weigh from 154 to 177 pounds. The team intends to eat together at a local McDonald's. Their coach has advised them to avoid carbonated beverages so most of them will drink water or orange juice. From their previous trips to McDonald's they have sampled the food, checked the prices, and obtained nutrition information per serving. Their main concerns are (1) meeting the recommended daily requirements including 100 % of the RDA for Vitamins A and C, Thiamin, Riboflavin, Niacin, Calcium, and Iron () staying within the recommended limits for fat, cholesterol, and sodium (3) minimizing the cost per person and (4) maintaining good relationships with the busy restaurant staff. From the 80 possible choices on the menu they have decided to limit their choices to the nine shown below. They know that no one can order 1.7 hamburgers but they could order 17 and cut them up. Figures.3. shows 1994 unit prices at a local McDonald's and the data in Figure.3.3 was taken from a 1993 McDonald's publication entitled Nutrition Information per Serving. This data has been stored in the file MCDNLD9.XLS HB MF GS SBE W BAP MCOOKIE OJ HCLFYS $0.59 $1.05 $1.89 $1.49 $0.99 $0.85 $0.65 $0.89 $0.95 Hamburger Medium Fries Garden Salad Sausage Biscuit/w Egg Wheaties Baked Apple Pie Figures.3. McDonald's Cookies Orange Juice Hot Caramel Low fat Frozen Yogurt Sundae The recommended limits on daily intake of calories, protein, fat, cholesterol, and sodium for each individual are found on the right side of the table Nutrition Information per Serving in Figure.3.3. Recommended limits: * insignificantly small Nutrition Information per Serving Males 19 - years old 154 to 177 lb menu item HB MF GS SBE W BAP MCOOKIE OJ HCLFYS Calories >= 900 Protein (g) >= 56 Vitamin A (%RDA) 4 * * * 6 >= 100 Vitamin C (%RDA) * 0 * * 10 * >= 100 Thiamin (%RDA) >= 100 Riboflavin (%RDA) 10 * >= 100 Niacin (%RDA) * >= 100 Calcium (%RDA) 10 * 4 10 * 0 >= 100 Iron (%RDA) * >= 100 Fat (g) <= 96 Cholesterol (mg) <= 300 Sodium (mg) <= 3000 Figure.3.3 Current nutritional data from McDonald's can be located at Joseph F. Aieta Page 59 8/14/003

60 Linear Programming Section.3 Page 60 Your task is to build the worksheet and then use Solver to minimize cost. Note that there will be ten > constraints (counting non-negativity) and three < constraints. Load Solver and enter the appropriate parameters. Be sure to check Linear Model in Options and then Solve. A B C D E F G H I J K L menu item HB MF GS SBE W BAP MCOOKIE OJ HCLFYS Totals Dir Bounds Decision Variables Cost Function $0.59 $1.05 $1.89 $1.49 $0.99 $0.85 $0.65 $0.89 $0.95 $9.97 Calories >= 900 Protein (g) >= 56 Vitamin A (%RDA) >= 100 Vitamin C (%RDA) >= 100 Thiamin (%RDA) >= 100 Riboflavin (%RDA) >= 100 Niacin (%RDA) >= 100 Calcium (%RDA) >= 100 Iron (%RDA) >= 100 Fat (g) <= 96 Cholesterol (mg) <= 300 Sodium (mg) <= 3000 Figure.3.4 The target cell and the cells containing the decision variables are shaded as before for ease of identification. As you might expect, the optimal solution, which corresponds to a minimum cost, does not have integer values. Joseph F. Aieta Page 60 8/14/003

61 Linear Programming Activities Page 61 Linear Programming Activities 1. Formulate and Solve XYZ Steel Company manufactures two kinds of wrought-iron rails: Model E (the elegant) and Model D (the distinctive). Model E rails sell for $59 and cost $50 to make, whereas Model D rails sell for $48 and cost $41 to make. To make one Model E rail requires hours on Machine A, 1 hour on Machine B, and 4 hours on Machine C. On the other hand, to make one Model D rail requires 1 hour on A, hours on B, and 5 hours on C. Production scheduling indicates that during the coming week Machine A will be available for at most 30 hours. Machine B for at most 4 hours, and Machine C for at most 7 hours. Find the number of each kind of rail to be made in the coming week in order for the company to maximize its profit. What is this maximum profit? At the maximum, which machines, if any, are not fully utilized?. Formulate and Solve Safety Lock Company makes two kinds of locks: Model SS (the super safe) and Model S (the safe). Each Model SS lock sells for $4 and costs $19 to make, while each Model S lock sells for $18 and costs $15 to make. Each of the locks must be processed on two machines: Model SS requires 3 hours on Machine A and hours on Machine B, whereas Model S requires 7 hours on A and 1 hour on B. During the coming week Machine A will be free for no more than 4 hours and Machine B for no more than 17 hours. Determine the number of each kind of lock to be made in the coming week for the company to maximize its profit. What is this maximum profit? 3. Formulate and Solve. a) A special diet for athletes is to be developed from two foods: Food X and Food Y. The new food is to contain at least 16 milligrams of Vitamin A, at least 0 milligrams of Vitamin B, and at least 1 milligrams of Vitamin C. Each pound of Food X costs $1.50 and contains 1 milligram of Vitamin A, 5 milligrams of Vitamin B, and 1 milligram of Vitamin C. On the other hand, each pound of Food Y costs $.50 and contains milligrams of A, 1 milligram of B, and 1 milligram of C. How many pounds of each food should be used in the mixture in order to meet the preceding requirements at a minimum cost? What is this minimum cost? b) Repeat the previous problem if, in addition, the amount of Food Y in the mixture must be no more than one-half the amount of Food X 4. Formulate and Solve a) ABC Dairy Company wishes to make a new cheese from two of its current cheeses: Cheese X and Cheese Y. The mixture is to weigh no more than four pounds and is to contain at least six ounces of the sharpness ingredient S. Each pound of X costs $4 and contains three ounces of S, whereas each pound of Y costs $1 and contains one ounce of S. How many pounds of each cheese should be used in the mixture in order to meet these requirements at a minimum cost? What is this minimum cost b) Repeat the previous problem if, in addition, the amount of Cheese Y cannot exceed the amount of Cheese X by more than one pound. Joseph F. Aieta Page 61 8/14/003

62 5. Formulate and Solve Linear Programming Activities Page 6 To introduce its new economy automobile to the public, the marketing department of Guarantee Motors, Inc., has decided to sponsor a 90-minute television special featuring the world-famous comedian I. M. Hilarious. From past experience, it is known that to reach a maximum number of viewers, a delicate balance must be maintained between the number of minutes devoted to commercials and the number of minutes that the comedian is on the air. In fact, although Hilarious popularity is such that he attracts 15,000 viewers for every minute that he is on the air, any television program tends to lose about 150 viewers for each minute devoted to commercials. Furthermore, the time devoted to commercials should be no more than 5 percent of the time Hilarious is on the air. On the other hand, the president of Guarantee insists that the commercial time be at least 0 percent of the comedian s time. Under all of these conditions, what is the optimum strategy for allotting the 90 minutes between commercials and Hilarious? What is the maximum number of viewers that Guarantee will reach? 6. Formulate and Solve Strong Steel Company operates two steel mills with different production capacities. Mill I can produce 1,000 tons per day of AAA steel, 3,000 tons per day of AA steel, and 5,000 tons per day of A steel. Mill F can produce,000 tons per day of each grade of steel. The company has made a contract with the construction firm to provide 4,000 tons of AAA steel, 3,000 tons of AA steel, and 40,000 tons of A steel. For each of the following costs, determine the number of days the company should operate each mill in order to meet the terms of the preceding contract most economically, the minimum cost, and also what grade(s) of steel would be over produced: a) The cost of running Mill I is $1,400 per day and Mill F is 1,000 per day. b) The cost of running Mill I is $1,500 per day and Mill F is $3,000 per day. 7. Use Solver to Maximize Z = 3x + 4x + 10x 1 3 subject to non-negativity and x1 + x + 4x3 4 x1 + x + 3x3 40 x + x + x Formulate and Solve Ace Rubber Company manufactures three types of tires: Model P (the premium, Model S (the second line), and Model (the economy). Model P sells for $96 per tire and costs $86 per tire to make; Model S sells for $79 per tire and costs $73 per tire to make; Model E sells for $76 per tire and costs $64 per tire to make. To make one Model P tire, it requires one hour on Machine A, one hour on Machine B, and two hours on Machine C. To make one Model S tire, it takes one hour on Machine A, two hours on Machine B, and one hour on Machine C; to make one Model E tire requires four hours on A, three hours on B, and two hours on C. Production scheduling indicates that during the coming week Machine A will be available for at most 4 hours, Machine B for at most 40 hours, and Machine C for at most 30 hours. How many of each tire should the company make in the coming week in order to maximize its profit? What is this maximum profit? Joseph F. Aieta Page 6 8/14/003

63 Linear Programming Activities Page Formulate and Solve Reliable Sand and Gravel has contracts to transfer solid fill in the next week to each of three cities: Ashland, Bedford and Cambridge. Two hundred loads of fill are required at Ashland, 150 loads of fill at Bedford, and only 60 loads of fill at Cambridge. Reliable has storage pits at Lynn and Quincy, which can supply solid fill at a limited weekly rate. Next week Lynn can supply up to 75 loads, while Quincy can supply up to 175 loads. The delivery cost per truckload depends on the delivery source and destination, and is shown in dollars per load in the following table: Destination $ per load $ per load $ per load Ashland Bedford Cambridge Source Lynn Quincy a) How many truckloads should the company transport from each source to each destination in order to minimize total transportation costs? b) What is the minimum cost? 10. Formulate and Solve Superior Paint Company sells three kinds of exterior finishes: oil-base paint, water-base paint, and stain. In order to keep all its distributors properly supplied, the company must manufacture and store at least three months sales in advance. Oil-base paint costs the company $5.50 per gallon to make, whereas water-base paint costs $4.50 per gallon, and stain costs $3.50 per gallon. The company has determined that the carrying costs for all three finishes are $0.50 per gallon for the average number of gallons in the inventory (the average being typically one-half of the number manufactured). On the basis of past experience, the sales department of the company has projected that its needs for the next three months will be at most 10,000 total gallons but at least 6,000 total gallons. Furthermore, no more than,000 gallons of stain but at least 1,000 gallons of each finish will be needed. In addition, the total number of gallons of paint (oil-base and water base) should be at least 75 percent of the total. How many gallons of each type of finish should the company make? What will be the cost? 11. Formulate and Solve Growth Investment Company is planning a pension fund of $10,000,000 for one of its valued clients. Federal and state regulations require that, for the workers protection, the fund must be made up of stocks, bonds, and a reserve in the form of bank notes or savings accounts. After much investigation, the company has decided upon the following combination: three stocks (S 1, S, and S 3 ) two bond issues (B 1, and B ), and a bank CD, certificate of deposit. Their study shows that the expected yield from S 1 will be 8 percent, S will yield 9 percent, S 3 will yield 7 percent, B 1 will yield 10 percent, B will yield 11 percent, and the CD will yield 6 percent. There are, however, stringent limitations on the investment possibilities written in the regulations on pension funds. First, the amount invested in stocks must be no more than 40 percent of that invested in bonds. Second, a minimum of 5 percent of the total fund must be invested in reserve as a CD. Third, no more than 35 percent of the total fund can be invested in stocks. Finally, no single investment other than a bank CD can constitute more than 30 percent of the total fund. What portfolio will the company recommend for the pension fund? What will be the expected yield of this portfolio? Joseph F. Aieta Page 63 8/14/003

64 More LP More Linear Programming Page KLEAN SOAP: Klean Soap wishes to make a new detergent from two of its current soaps: Soap A and Soap B. The new detergent must contain at least 0 ounces of cleaning ingredient D. Each pound of Soap A costs $0.0 and contains 5 ounces of ingredient D, while each pound of Soap B costs $0.5 and contains 1 ounce of ingredient D. Furthermore, for ecological reasons the amount of Soap A must be no more than that of Soap B. Finally, the new mixture must weigh exactly 1 pounds. How many pounds of each type of soap should be used in the mixture in order to meet the above requirements at a minimum cost? What is this minimum cost?. MKTRES: For a telephone survey, a marketing research group needs to contact at least 150 wives and 10 husbands. It costs $ to make a daytime call and (because of higher labor costs) $4 to make an evening call. The table below lists the results that can be expected. For example, 30% of all daytime calls are answered by a wife, and 30% of all evening calls are answered by a husband. Because of a limited staff, at most two thirds of all phone calls can be evening calls. Determine how to minimize the cost of completing the survey. What is this minimum cost? Phone Survey Data Person Percentage of Percentage of Responding Daytime Calls Evening Calls Wife Husband Others/no answer HARTFORD: Hartford Insurance has just obtained $ million by converting industrial bonds to cash and is now looking for other investment opportunities for these funds. Considering Hartford s current investments, the firm s top financial analyst recommends that all new investments should be made in the oil industry, steel industry, or government bonds. Specifically, the analyst has identified five investment alternatives and projected their annual rates of return. The alternatives and rates of return are shown below. The investment manager has imposed the following investment guidelines: Neither industry should receive more than 50% of the total new investment. Government bonds should be at least 5% of the steel industry investments. The investment in Benson Oil cannot exceed 60% of the total oil industry investment What portfolio recommendation (investments and amounts) should be made for the available $ million? Projected Investment Rate of Return Atlantic Oil 7.3% Benson Oil 10.3% Carnegie Steel 6.3% Draper Steel 7.5% Government bonds 4.5% Joseph F. Aieta Page 64 8/14/003

65 More Linear Programming Page BASEBALL GLOVES: Kelson Sporting Equipment makes two different types of baseball gloves: a regular glove and a catcher's mitt. Over the next planning period the firm has 900 hours of production time available in its cutting and sewing department, 300 hours available in its finishing department, and 100 hours available in its packaging and shipping department. The production time requirements and the profit contribution for each product are as follows: Production Time hours per unit hours per unit hours per unit Cutting Packaging dollars per unit Model & Sewing Finishing & Shipping Profit/Glove Regular glove 1 1/ 1/8 $10 Catcher's mitt 3/ 1/3 1/4 $16 a) How many gloves of each type should Kelson produce to maximize profits b) What is the resulting maximum profit? c) How many hours of production will be scheduled in each department? d) What is the slack time in each department? 5. POWERCO: PowerCo has three electric power plants that supply the power needs of four metropolitan areas. Over the next 4 hours each power plant can supply the energy amounts (in million kilowatt-hours) shown in Table 1. The energy demand during the same period at each metropolitan area (again in million kwh) is given in Table. Finally, the cost (in dollars) of sending a million kwh from each plant to each metropolitan area is given in Table 3. PowerCo wants to find the lowest-cost method for meeting the demand in the four metropolitan areas. Plant Supplies Requirements Metropolitan Area (million kwh) (million kwh) Supply Demand Plant A 40 Area 1 45 Plant B 60 Area 0 Plant C 40 Area 3 30 Area 4 30 Table 1 Table Transmission Costs ($/million kwh) Area 1 Area Area 3 Area 4 Plant A Plant B Plant C Table 3 Joseph F. Aieta Page 65 8/14/003

66 More Linear Programming Page COPYEDITING: The production editor for Rayburn Publishing Company has 000 pages of manuscript that must be copyedited. Because of the short time frame involved, only two copyeditors are available: Alice Mergen and Sue Smith. Alice has 10 days and Sue has 1 days available. Alice can process 100 pages of manuscript per day, Sue can process 150 pages. Rayburn Publishing has developed an index used to measure the overall quality of copyeditors on a scale from 1 (worst) to 10 (best). Alice s quality rating is 9, and Sue s quality rating is 6. In addition, Alice charges $3 per page of copyedited manuscript; Sue charges $ per page. If a budget of $4800 has been allocated for copyediting, how many pages should be assigned to each copyeditor to complete the project with the highest possible quality? 7. CEREALS: Healthtech Foods managers are considering developing a new low-fat snack food. It is to be a blend of two types of cereals, each of which has different fiber, fat, and protein characteristics. The following table shows these nutrition characteristics for one ounce of each type of cereal. Dietary Fiber Fat Protein Cereal (grams) (grams) (grams) A 4 B Note that each ounce of cereal A provides grams of dietary fiber and that each ounce of cereal B provides 1.5 grams of dietary fiber. Thus, if Healthtech were to develop the new product using a mix consisting of 50% cereal B, 1 ounce of the snack food would contain 1.75 grams of dietary fiber. Healthtech s nutrition requirements call for each ounce of the new food to have at least 1.7 grams of dietary fiber, no more than.8 grams of fat, and no more than 3.6 grams of protein. The cost of cereal A is $0.0 per ounce and the cost of cereal B is $0.05 per ounce. a) Determine how much of each cereal is needed to produce 1 ounce of the new food product at the lowest possible cost. b) Which constraints have slacks? Interpret these slacks in the context of this problem. c) If Healthtech markets the new snack food in an 8-ounce package, what is the cost per package? Joseph F. Aieta Page 66 8/14/003

67 Time Value of Money Section 3.1 Page 67 Time Value of Money Section 3.1 Individuals, families, small businesses, and large corporations all make decisions about managing money. Parents and grandparents need to plan wisely for major expenses such as the purchase of a house, the financing of a college education, and the accumulation of wealth for retirement. If you have ever borrowed money, have a credit card, or have any investments or savings, then you already have some experience with personal finance. The mathematics underlying financial formulas has been known for many generations. One difference today is that the computations are much easier to perform due to advances in technology. Other significant changes brought about by information technology are reflected in the rapid flow of information over the Internet. People making decisions about money, in both personal finance and corporate finance, currently have access to vast amounts of financial data and investment information. Not so long ago, making decisions about saving for retirement or borrowing money required going to the local bank and making selections from a very few options. Individuals who needed to borrow money were often presented with charts and tables that left them feeling perplexed and helpless. For millions of workers, preparing for retirement meant participation in the Social Security system. During their working years, individuals would have regular amounts of money automatically deducted from their paychecks. In retirement they would receive their Social Security checks each month. Often the amount of these payments did not completely cover monthly expenses partly because the cost of goods and services had increased over the years. In the first three-quarters of the 0 th century only a small percentage of the working population had built up retirement funds by investing in stocks and bonds. Stock market performance is now of great interest to millions of people. By 1998, stocks represented over half of the financial assets of U.S. portfolios. Participation in ownership of 1 stocks and bonds give people a sense of independence and control over their financial security. Much of the information about publicly traded companies that was formerly available to security brokers working on commission is now widely available to everyone via the Internet. Though the Internet has undergone exponential growth since the mid nineteen nineties, it is still in a state of comparative infancy. We can expect that world-wide access to data on-line will continue to grow rapidly. Before electronic banking, most people paid their bills with cash or by check. It wasn't until the latter part of the 0th century that electronic banking, credit cards, and access to tax-sheltered retirement plans became commonplace. Today the ordinary consumer has easy access to information about credit card rates, mortgage loans, money market funds, government bonds, and opportunities and risks in the stock market. For millions of people, their main source of information on current financial matters are the relatively new Web sites that are maintained by Internet companies such as Yahoo, by financial institutions such as Fidelity, or by companies that have historically had a major presence in print or broadcast media. Information is available on-line 4 hours a day. We may soon reach the point where more people manage their financial accounts electronically than by any other method. Individuals have total access to specialized web-based calculators to answer such questions as What is the best credit card for my needs, Should I refinance my mortgage?" or "How much should I be saving each month for retirement?" 1 Very simply, stocks and bonds can be defined as follows: when you buy stock you gain equity in a company by becoming a part owner. For the risk of owning the stock, you participate in the success of the company by way of periodic payments, called dividends, or by selling the stock at a higher price. When you buy a bond, you have loaned money to the company or to the government. You own a contract for which you are compensated with regular payments, called interest, until the bond matures and the principal is returned to you. Joseph F. Aieta Page 67 8/14/003

68 Time Value of Money Section 3.1 Page 68 The rapid development of computers and computer networks has forever altered the ways in which individuals and corporations gain access to financial data and do business. Whether you are an individual thinking about investing in stocks or bonds, a family member saving for a vacation, a business owner looking for an opportunity, or a manager whose company needs to finance a major purchase, you will make better decisions if you understand the fundamentals of the Time Value of Money (TVM). Take the Cash Now or Accept Monthly Payments? Elisa worked in her high school district s computer center and became very skillful at troubleshooting and making repairs on classroom machines. During her first year at college she listened to her college classmates and her friends at nearby colleges talk about problems they encountered with their computers, especially after the warrantees had expired. Elisa started a business fixing student laptops and renting out loaner machines. Using her contacts in the area, she was able to find someone to work on the machines that she could not fix herself. Elisa was also in demand as a Web site designer during her last two years of college. By the end of her senior year, she had established a strong reputation among students for excellent service. Her business had grown to the point where she needed to train several other students. It is now Elisa s graduation time and she decides to sell the business. She has acquired computer equipment and a van that will be part of the deal. Her asking price is $8,000 but she has no immediate need for this money. Blake, one of her most reliable student employees, is interested in continuing the business but he does not have $8,000. He makes the following offer to Elisa. He will pay her $400 per month for two years starting one month after graduation. Blake knows that Elisa has the opportunity to safely invest money at 6% compounded monthly for the next two years. Elisa will make her decision using what she knows about the time value of money. She does a comparison of accepting $8,000 now versus receiving a stream of $400 payments at the end of each month for two years. By the end of this chapter you will understand how Elisa makes this comparison and how the 6% interest rate figures into her decision. Here is a short list of the types of questions and problem situations that we will be exploring in this chapter: Why is compound interest sometimes called "the eighth wonder of the world"? What is the difference between a nominal interest rate and an effective interest rate? What is meant by the "doubling time" of an investment and how is it related to interest rates? If an individual invests $100 a month into a mutual fund of stocks and bonds and it grows to $10,000 at the end of five years then what is the annual growth rate? If a family knows the largest possible monthly mortgage payment that they can make, then how much can they afford to borrow? How does this depend upon the interest rate? If a corporation needs to repay its bondholders in ten years, then what regular amount should the company put into a fund each month to meet this future obligation? How is this amount affected by the interest rate? Joseph F. Aieta Page 68 8/14/003

69 Time Value of Money Section 3.1 Page 69 Many undergraduate and graduate students have significant educational loans and credit card debt by the time that they finish school. According to a 1998 national Student Loan Survey, excessive debt is the main problem cited by students who drop out of school. Understanding the mathematics of the cost of credit may help borrowers avoid the consequences of building up too much debt. Example Suppose a credit card has an annual rate of 15.00% per year and the minimum monthly payment is either $50 or % of the unpaid balance, whichever is larger. The cardholder has a balance of $3,000 and decides to pay it off by making only the minimum monthly payment. Assuming that the cardholder puts no additional debt on this card, it would about eight and one-half years for this debt to be paid off. The total amount of interest paid over five years would be more than $,000! The first 10 months and the last 10 months of the payment schedule are shown from an Excel table in Figure Figure Joseph F. Aieta Page 69 8/14/003

70 Time Value of Money Section 3.1 Page 70 Simple and compound interest: Lump sum situations We are all familiar with the concept of property that appreciates over time --- oceanfront real estate, the rookie baseball card of a home run champion, an antique automobile, etc. When dealing with financial applications, we must always keep in mind that money itself has the capacity to earn money. The money that an investment earns is called interest. Money earning an annual rate of 8% per year means that each dollar will earn eight cents of interest in one year so one thousand dollars will earn eighty dollars of interest. The total amount of interest earned depends upon the size of the investment, the interest rate, how long the money is invested, and how interest is calculated. In some ways, interest paid by a bank for the use of a depositor s money is analogous to rent paid to a landlord for the use of an apartment. The bank compensates the depositor for the use of his/her money until it is withdrawn. The amount periodically paid by the bank is based upon the size of the deposit, a quoted annual interest rate, and the method that the bank uses to calculate interest. In the rental situation, the renter compensates the landlord for the use of an apartment over a certain agreed upon period of time. The amount that the renter pays the landlord is influenced by a wide variety of factors, both quantitative and non-quantitative. These local factors include the number of rooms, whether or not the apartment is furnished, the desirability of the neighborhood, the proximity to school or work, and the number of people looking for apartments in that neighborhood at a particular time. The interest rates offered by banks are highly influenced by more global conditions, primarily the general state of the economy. The two most common methods for computing interest are simple interest and compound interest. Dollars invested in the present earn interest and grow to a larger value in the future under both simple interest and compound interest. The table and graph in Figure 3.1. show the growth of $1000 over a period of 0 years at an annual rate of 8%. The column labeled Simple shows the growth of $1,000 under simple interest, and the column labeled Compounded shows the growth of $1000 under annual compounding. Joseph F. Aieta Page 70 8/14/003

71 Time Value of Money Section 3.1 Page 71 Table and Graph of Simple and Compound Interest Over a Period of 0 Years from SinglePayment.xls Enter the nominal rate as a decimal or as a percentage followed by the % key. example: 0.08 or 8% entered in the annual rate PV SIMPLE INTEREST blue cell for r will be formatted as 8.00 % 8.00% $1,000 ANNUAL COMPOUNDING Years Simple Compounded Compounded DAILY COMPOUNDING t Interest Annually Daily 0 1, , , $6, , , ,083.8 Simple vs. Compound Interest 1, , , , , ,71. $5, , , , , , , , , , $4, , , , , , , , ,999.00, ,800.00,158.9,5.35 $3, ,880.00,331.64, ,960.00,518.17, ,040.00,719.6,88.89 $,000 14,10.00, , , , , , , , $1,000 17, , , , , ,0.03 $0 19, , , , , , Figure 3.1. In the first few years, the two accounts are not too far apart but, as you can see, there is a significant difference in the simple interest account and the compound interest accounts over an extended period of time. Calculating Simple Interest Consider the values in the simple interest column in Figure What is the pattern to the changing amounts? Each value differs from the amount above it by $80. In other words, this simple interest account grows by $80 each year since $80 is 8% of $1000. The amount at the end of the first year is $1080 = $ $80. Note that only the original $1000 earns interest each year. This is the very essence of simple interest calculations. The $80 simple interest earned in the first year does not earn any interest during the next year or any subsequent years. The amount at the end of the second year is $1160 since 1160 = which is the same as (80) Let s continue to examine the amounts for remaining years as shown in Figure Joseph F. Aieta Page 71 8/14/003

72 Time Value of Money Section 3.1 Page 7 The pattern in the table below reveals that the amount in year t is simply t Time (in years) Amount under Simple Interest = (0.08 ) = (80) 1 = (0.08 ) = (80) = (0.08 ) = (80) 3 = (0.08 ) = (80) 4 = (0.08 ) = (80) 5 = 1400 t = (80) t = 1000( t) Figure Before developing a general formula for the total amount in a simple interest account, we will introduce some common financial terminology and notation. The initial investment (original amount) is often called the principal or present value and is denoted by P or PV. The future value, denoted by F or FV, is the total amount in an account (principal plus any earned interest) at a particular point in time after the initial investment. The annual interest rate or nominal rate is often denoted by r and time in years is given as t. Using this notation PV for principal, FV for Future Value, r for the annual rate as a decimal, and t for the number of years we can write FV = PV + PV r t which is algebraically equivalent to PV (1 + r t ). FUTURE VALUE OF A SIMPLE INTEREST ACCOUNT If PV dollars earns simple interest at an annual interest rate r over a period of t years, then the future value of that account is FV = PV(1 + r t) The amount of interest earned is (principal) ( rate) (time) or I = PV r t This future value formula assumes that a single amount is deposited, no additions other than interest, and no withdrawals are made. Although the interest rate is usually reported as a percentage as in the APR is 4.50%, when the rate r is entered into an ordinary calculator, it is entered in decimal form, as in Joseph F. Aieta Page 7 8/14/003

73 Time Value of Money Section 3.1 Page 73 Example 3.1. A lump sum of $50,000 is deposited into an account that earns simple interest at the annual rate of 6%. a) How much interest is earned in years? b) Find the future value after 5 years and 3 months. c) Are the following statements true or false? Give reasons for your answers. i. The amount of interest earned in 0 years is exactly twice the amount of interest earned in 10 years. ii. Regardless of how many years money is left in this account, the amount of interest earned at a simple interest rate of 6% per year would be three times as much as the amount of interest earned at a simple interest rate of % per year. Solution: a) Using the simple interest formula I = PV r t, we have I =(0,000)(.06)() = $6,000 b) Recall that time is measured in years when using the future value formula. The future value of the $50,000 investment after 5 years and 3 months (5 and 3/1 or 5.5 years ) is FV = 50,000 ( ) = 50,000 ( ) = $65,750 c) i. This is a true statement. The interest earned in 0 years at a simple interest rate of 6% per year is I =(50,000)(0.06)(0) = $60,000 while the interest earned in 10 years is I =(50,000)(0.06)(10) = $30,000 ii. With r = % per year, the simple interest earned in t years is I =(50,000)(0.0) t = 1,000 t and with r = 6% per year, the interest earned is I = (50,000)(0.06) t = 3,000 t = (3)(1,000) t, so this statement is also true. Ordinary versus Exact Methods for Calculating Simple Interest Simple interest is often used for short-term loans measured in days. Because the number of days in a month varies, banks historically adopted certain rules about converting days and months into years before performing calculations. For some transactions, banks are allowed to count a year as twelve 30-day months. The 360-day year is called the ordinary method (also called the Banker s Rule) instead of the exact method that uses 365 days in a regular year (or 366 days in a leap year). Generally, if the type of year is not specified then the exact method with 365 days is assumed. Example A lump sum of $1500 is deposited for 90 days at a simple interest rate of 8% per year. a) What interest is earned if the bank uses the ordinary method? b) How does the amount of interest earned using the exact method with 365 days compare with the ordinary method in part a? Interest amounts should always be rounded off to the nearest cent. Joseph F. Aieta Page 73 8/14/003

74 Time Value of Money Section 3.1 Page 74 Solution: a) Using a 360-day year, we have I = PV r t = (1500)(0.08) 90 = $ b) Assuming that this is not a leap year, the interest would be (1500)( 0.08) 90 $ Rounding off to the nearest cent, the interest is $9.59, an amount that is 41 less than the interest calculated by the ordinary method. Example Bank A borrows $10,000,000 overnight from bank B. They agree upon the ordinary method with an annual rate of 6%. What amount of simple interest will bank B receive from bank A for this one day loan? Solution: $10,000,000 1 (0.06) =$1, using the Banker's Rule. To repay the overnight loan, bank A will pay bank B a 360 total of $10,000,000 + $1, = $10,001, to the nearest cent. Example Two relatives arrange a loan of $10,000 for 00 days at 5% simple interest per year using the exact method. How much does the borrower repay the lender at the end of the 00 days and how much of this is interest? Solution: 00 FV = 10,000 ( ) = 10,000 ( ) or $10,73.97 to the nearest cent. Since $10,000 was the amount 365 borrowed, the portion that is interest = $10, $10,000 = $ Thus far, we have found the amount of simple interest, I, using the formula I = PV r t and the future value, FV, using the formula FV = PV ( 1 + r t). If the values of three of the variables are known, then the value of the remaining variable can be found in either formula with simple algebra and arithmetic. Simple interest calculations can be done with the simplest of calculators (no pun intended) since the only functions necessary are the four arithmetic functions: addition, subtraction, multiplication, and division. Calculating Compound Interest Under compound interest, the interest earned in each period (year, month, quarter, day, and so forth) is based upon the entire amount that has been accumulated up to the beginning of that period. Refer back to the table and graph in Figure 3.1. and observe that in one year $1000 has grown to $1080 for both the simple and compound interest columns. However, the amounts for the second year and all succeeding years are different. According to the table, the accumulated compound interest amount at the end of the second year is $ This is $6.40 more than the corresponding amount of $1660 accumulated under simple interest. Where did this additional amount come from? With compound interest, the interest as well as the principal earns interest. Joseph F. Aieta Page 74 8/14/003

75 Time Value of Money Section 3.1 Page 75 Figure 3.1. shows the growth of $1000 under an 8% annual compound interest rate for 5 years. Each dollar amount has been rounded to the nearest cent. Time (in years) Annual Compound Interest = (0.08) = = (0.08) = (0.08) = (0.08) = (0.08) = Figure3.1.4 Figure 3.1. shows that the original $1,000 grows to $4, under annual compounding at 8% if this process is continued for twenty years. Compare this with the future value of only $600 after 0 years under simple interest of 8%. Such is the power of compounding! The fundamental concept of compounding is that the accumulated value at the end of any one period becomes the basis for computing the interest earned in the next period. This differs from simple interest where interest is earned on only the initial deposit. Under compound interest, the interest earned during any conversion period is based upon the total accumulated value up to that period. Figure shows the pattern in the growth of an amount PV earning annual compound interest. Each year the future value is the sum of the accumulated amount the year before and the interest that this amount earned in one year. Time (in years) Future Value (FV) Under Annual Compound Interest 0 PV = PV(1 + r) 0 1 PV (1 + r) 0 + PV(1 + r) 0 r = PV(1 + r) 0 (1 + r) = PV(1 + r) 1 PV(1 + r) 1 + PV(1 + r) 1 r = PV(1 + r) 1 (1 + r) = PV(1 + r) 3 PV(1 + r) + PV(1 + r) r = PV(1 + r) (1 + r) = PV(1 + r) 3 4 PV(1 + r) 3 + PV(1 + r) 3 r = PV(1 + r) 3 (1 + r) = PV(1 + r) 4 5 PV(1 + r) 4 + PV(1 + r) 4 r = PV(1 + r) 4 (1 + r) = PV(1 + r) 5 t Amount in year t = PV(1 + r) t Figure Joseph F. Aieta Page 75 8/14/003

76 Time Value of Money Section 3.1 Page 76 The total amount in the account at the end of any period is (1 +.08) times the amount in the account at the end of the previous period. To find the amount at the end of period 4 for example, start with the original PV and perform repeated multiplications by (1 +.08) four times. We see that exponentiation is equivalent to repeated multiplication. Multiplying 1000 by 1.08 four times is 1000 (1.08) (1.08) (1.08) (1.08) = 1000 (1.08) 4. Example What is the future value in 5 years of a deposit of $ into an account that pays 3.75% interest compounded annually? Solution: FV= ( ) 5 = to the nearest cent. Since compounding is done only once a year, the formula for the accumulated amount at the end of year t can be expressed as FV= PV (1 + r) t where r is the annual rate and t is the number of years. Compare this exponential expression FV= PV (1 + r) t with the linear expression FV= PV (1 + r t ) for simple interest. Rates quoted by banks and other financial institutions are usually stated as percentages. In financial formulas that contain a variable for the rate, such as r shown above, it is understood to be the decimal form, not the percentage form. When we mean the rate as a percentage we will use the convention % r instead of r. This is the convention that is followed on many financial calculators. Interest paid by banks or charged by credit card companies is compounded more than once a year. The frequency of compounding is usually daily for savings accounts. For credit card debt, the frequency of compounding is usually monthly. If interest is compounded each period (year, month quarter, or day), the pattern illustrated in Figure remains the same if r, the annual rate, is replaced by i, the interest rate for one period. In actual practice, interest rates are generally quoted on an annual basis. Financial institutions use the terms nominal interest rate and Annual Percentage Rate (APR) when they refer to this quoted, advertised annual rate. Rates for one interest period can be calculated by dividing the nominal rate (APR) by the number of interest periods per year. We will use m for the number of conversion periods in one year. Some of the common compounding (conversion) periods and their corresponding interest rates for one period is shown in Figure Joseph F. Aieta Page 76 8/14/003

77 Time Value of Money Section 3.1 Page 77 Frequency of Compounding Period in words Number of Periods in One Year denoted by m Interest Rate for One Period denoted by i annual year 1 r semi-annual half-year quarterly quarter 4 r monthly month 1 1 r r 4 daily (exact year) day 365 daily (ordinary year) day 360 r 365 r 360 Figure For instance, if the annual rate is 6% and interest is compounded quarterly, then there are four interest periods per year, and the interest rate for each period is i = 0.06 = or 1.5%. We would like a general compound interest formula to 4 compute the future value of money when interest is compounded more than once a year. As was the case with simple interest, this future value formula assumes that a single amount is deposited, no additions (other than interest) are made, and no withdrawals occur. The following formula is used for the future value of a lump sum. FUTURE VALUE IN A COMPOUND INTEREST ACCOUNT If PV (the present value) is invested for a total of n periods earning compound interest at an annual (nominal) interest rate, r, that is compounded m times a year over t years then the future value of the account is FV = PV (1 + i) n where i = r m and n = m t When using this formula, the interest rate for one period, i, must be entered as a decimal. The exponent n is the total number of periods and is usually measured in units of years, half-years, quarters, months, or days. As was discussed with simple interest, there is more than one method for computing daily compound interest. The ordinary method uses a 360-day year, and the exact method uses 365 days in a regular year or 366 days in a leap year. Joseph F. Aieta Page 77 8/14/003

78 Time Value of Money Section 3.1 Page 78 Example Future Value of a Lump Sum with Compound Interest. An account is opened with a single deposit of $15,000. The account pays 9% per year compounded monthly. a) How much interest is earned in the first year? b) How much is in the account at the end of two and one-half years? Solution: a) For monthly compounding, the number of periods per year m = 1. Therefore, the rate for one period (one month) is i = 0.09 = The $15 thousand is invested for a total of n = m t = 1(1) = 1 periods. The future value is , = 15,000(1.0075) 1 = 15,000( ) $16, Subtract the original principal of $15,000 to obtain the total amount of interest for the year: $16, $15,000 = $1, b) First, find the total number of periods by converting two and one-half years into months: n = m t = 1 months/year.5 years = 30 months The future value is: , = 15,000(1.0075) 30 = 15,000( ) $18, At the end of.5 years there will be $18, in the account. Notice that in both parts of Example 3.1.7, intermediate calculations were not rounded. Each final answer was rounded to the nearest cent because future value and interest are amounts of money. As the following example shows, you should never round the result of any intermediate calculation in the evaluation of a formula. Joseph F. Aieta Page 78 8/14/003

79 Example Time Value of Money Section 3.1 Page 79 $1 million is deposited in an account paying 6% per year compounded daily. a) Determine how much interest is earned during the first 30 months using the ordinary method. b) If you round the results of all intermediate calculations to 4 decimal places, how large is the resulting error? Solution: a) There are 360 days in a year using the ordinary method, and 30 months =.5 years. The total number of periods is days therefore n = years = 900 days year r 0.06 The interest rate for one period (1 day) is i = = = m The future value is = 1( ) 900 1( ) million dollars or 360 $1,161,819.7 to the nearest cent. b) Performing the same steps as in part a), but rounding each intermediate result to 4 decimal places, we obtain the future value (1+.000) 900 = million dollars = $1,197, These calculations show that you would be in error by $35,380.8! Example For a less extreme example, find the amount of interest earned if $1,000 is invested at a 6% annual rate compounded daily for two years. Solution: The rate for one day is 0.06 i = Multiplying times 365 we obtain 730 days as the number of periods 365 Using the formula FV= 1000 ( /365) 730 without intermediate rounding we would get $1,17.49, which is accurate to the nearest cent. If we had rounded just the quotient to and performed the rest of the 365 calculations, we would obtain $1, This value is off by almost 30 dollars! The lesson 3 here is DO NOT ROUND THE RESULTS OF INTERMEDIATE CALCULATIONS If you do round intermediate results, your final result could vary significantly from the correct value. Can you imagine the bookkeeping errors that would result if banks were not careful about how they handled rounding errors? 3 If you do write down and re-enter the value of i or any other intermediate calculation, then show the maximum number of decimal places that your calculator will display. You can and should avoid manually re-entering values into your calculator by taking advantage of its storage capabilities. Consult the manual for your calculator to learn how to store and recall values. Joseph F. Aieta Page 79 8/14/003

80 Time Value of Money Section 3.1 Page 80 Technology and Notation Inexpensive, non-programmable, scientific calculators have a power key (y x or y^x) as well as keys for the elementary functions such as square root ( ), reciprocal (1/x ), and at least one logarithmic function (ln or log). Financial calculators and graphing calculators are more powerful machines with all of the above features and many more. Financial calculators have special keys with such labels as PV, FV, %i or I/Y, N, Pmt, P/Y, C/Y. As their name suggests, graphing calculators have large displays that are designed to show portions of graphs and tables. Some graphing calculators also have built-in financial functions. The table in Figure shows our notational conventions for lump sum problems. We will add a few more conventions when we study annuities that involve regular periodic payments. Notation r (APR) Definitions nominal rate (quoted rate, annual percentage rate) given as a decimal m frequency of compounding, number of interest (conversion) periods per year i interest rate for one interest period, i= r/m given as a decimal APY effective rate, often called the annual percentage yield or the effective annual rate, usually reported to the nearest hundredth or thousandth of a percentage t time period in years, not necessarily a whole number n or N total number of conversion periods, in t years n = m t PV present value of a lump sum, also known as the principal in lump-sum problems FV future value of a lump sum, also known as the accumulated amount Figure Joseph F. Aieta Page 80 8/14/003

81 Time Value of Money Section 3.1 Page 81 Computer-based spreadsheets, such as Excel, have extensive built-in financial functions and use a notational system that is similar but not exactly the same as what you will find on a dedicated financial calculator. For entering or solving for an interest rate i, Excel uses rate instead of the symbols %i or I/Y that are found on financial calculators. For entering or solving for the number of periods, Excel financial functions use nper instead of n or N. The notational system that we have adopted in Figure reflects a mix of conventions found in textbooks on finance, financial calculators, graphing calculators, and spreadsheet software. This system should be relatively easy for you to follow, regardless of the particular technology that you are using. Once you understand the general principles, you should be able to adjust to variations that you may encounter in books or on computing devices. If your calculator does not have built-in financial functions then be sure to enter rates in decimal form (e.g ) when evaluating financial formulas. If your calculator has keys labeled %i or I/Y, then most likely the rate should be entered as a percentage instead of decimal form. For example, if the rate for one interest period is 6.5% then you would enter 6.5, not When in doubt, check your manual or experiment by using a known formula and comparing results. Unfortunately there is no universally accepted notational system for finance. This is also true for statistics and for other technical disciplines that make heavy use of formulas and symbolic representations. Joseph F. Aieta Page 81 8/14/003

82 TVM Activities 1 TVM Activities 1 Page 8 Give dollar amounts to the nearest cent. For daily compounding, assume an exact year of 365 days unless otherwise specified. 1. Assume that the stated percent is an annual simple interest rate. Find the amount (future value) for each principal (present value) and time period. a.) $1,000; 8 percent; one year. b) $1,000; 9 percent; 6 months. c) $,000; 6 percent; 6 months. d) $000; 1 percent; 18 months e) $5,000; 4 percent; 3 years f) $4,000; 30 percent; years. The simple interest formula FV= PV(1 + r t ) is solved for FV. Algebraically solve for each of the variables PV, r, and t in terms of the other three. a) PV = b) r = c) t = 3. How much was deposited into an account paying 6% simple interest per year for it to be worth $000 at the end of 30 months? 4. In one year an investment grew from $6000 to $6500. This growth is equivalent to what simple interest rate? 5. A credit card holder has owed the credit card company $00 for a month and receives a bill containing an interest charge of $3. Find the simple interest rate. 6. How long does it take an investment which pays 5% simple interest to grow from $4000 to $5000 Joseph F. Aieta Page 8 8/14/003

83 TVM Activities 1 Page How many months will it take at 8 percent simple interest for $,000 to grow to an amount of $, Tom has an investment which pays him 4.5% simple interest per year. If he earns $140 interest in the first six months then what was the original amount of his investment. 9. Compare the growth under simple interest with the growth under annual compounding over a period of years. Start with a principal of $10,000 and a nominal rate of 5%. How much more is earned with annual compounding than with simple interest in 10 years? 10. Given a single deposit of $10,000 and a nominal rate of 5%. Compare the growth under annual compounding with the growth under daily compounding (exact year of 365 days). a) How much more is earned with daily compounding than with annual compounding in 5 years? b) How much more is earned with daily compounding than with annual compounding in 10 years? c) How much more is earned with daily compounding than with annual compounding in 15 years? d) How much more is earned with daily compounding than with annual compounding in 0 years? 11. A bank pays 5.5 percent compounded daily on a certificate of deposit with a maturity of 6 years. Using an exact year of 365 days, what will a deposit of $5,000 today be worth 6 years from now? 1. Paul buys a TV set priced at $500 and 3 months later is expected to pay this amount plus simple interest,. The total bill was $50. Compute the simple interest rate. 13. Find the present value of $1000 receivable years from now if the simple interest rate is 8.5 percent. Joseph F. Aieta Page 83 8/14/003

84 TVM Activities 1 Page Discounted Loans and Effective Simple Interest Rate. A discounted loan is an arrangement in which the amount of interest paid is based upon the amount due rather than on the amount borrowed. The lender gets the interest on the loan in advance and deducts the total amount of interest due from the discounted loan amount. The amount that the borrower pays back when the loan is due is called the maturity value of the loan. The amount that the borrower actually receives is called the proceeds of the loan and consists of the maturity value minus the total interest that was deducted in advance. Compare a standard two-year loan of $900 with a simple interest rate of 5% and a discounted loan of $1000 with a quoted rate discount of 5%. Quoted rate Proceeds (amount received by the borrower) Maturity Value (amount to be repaid in years) standard loan 5% per year $ ( )= $990 discounted loan 5% per year $1000 ( ) = $900 $1000 It is fairly intuitive that the better deal in this case is the standard loan. If we want to make a precise, numerical comparison of the true rates then we need some common yardstick. We could ask Under what annual simple interest rate would $900 grow to $1000 in two years? The effective rate for a discounted loan, re, is the interest rate at which the discounted loan is equivalent to a standard simple interest loan. For this example we need to solve the equation 1000 = 900 (1 + re ) for re. By basic algebra we obtain = r e. Using the four decimal place approximation of for 10/9, this equation becomes = re or re = To the nearest hundredth of a percentage, this effective rate would be reported as 5.56%. The relationship between the discount rate, d, the number of years, t, and the effective rate re, of a discounted loan (as a d percentage) is given by the formula re = 100% 1 d t This formula can be developed using the same algebraic steps given in the numerical example above. See exercise 18. Let FV be the maturity value (instead of 1000), let d be the discount rate (instead of 0.05), let t be the time (instead of ) and then solve FV(1 - d t) (1 + re t) = FV for re. Joseph F. Aieta Page 84 8/14/003

85 TVM Activities 1 Page A borrower has signed a note to pay off a discounted loan of $5000 due in 18 months. Under this arrangement the interest is deducted in advance. The amount that the borrower actually receives is called the proceeds of the loan. Determine the proceeds and the simple effective interest rate for the following: a) bank discount rate 4% per year b) bank discount rate 8% per year c) bank discount rate 1% per year 15. Find the proceeds of a $500, 9-month loan from a bank if the discount rate is 9 percent per year. 16. What discount rate should a lender quote to earn 9 percent effective interest on a 90-day transaction. 17. What discount rate should a lender quote to earn 11 percent effective interest on a 6-month transaction? 18. Solve FV(1 - d t) (1 + re t) = FV for re. Hint : Start by dividing both sides by FV FV(1 - d t) (1 + re t) = FV (1 - d t) (1 + re t) =1. Joseph F. Aieta Page 85 8/14/003

86 Time Value of Money Section 3. Page 86 Effective rate: Compounding more than once a year We have been using the terminology nominal rate, annual rate, and APR for the rate r that financial institutions often advertise or quote. When interest is compounded more than once a year, however, the quoted annual percentage rate is not the interest rate that determines how the money in an account actually grows. When interest is compounded at a frequency greater than once a year, say m times per year, then we must convert the annual interest rate r into an interest rate per period. We will denote this rate per period as i. Calculate i by dividing the annual interest rate by the number of r periods to obtain i =. m How do we compare two annual rates if the number of conversion periods per year is not the same? Example 3..1 How much interest is earned on $1 in one year if the account earns interest at the rate of 9% per year compounded monthly? Solution: 1 The rate for one period is 0.09 i = = The future value at the end of one year is = 1 1 1(1.0075) 1 = The theoretical amount of interest earned in one year on this one-dollar deposit is = If we write this as a percentage rounded to the nearest hundredth, it is 9.38%. The interest earned in one year under a nominal rate of 9% compounded monthly is about 9.38% of the original deposit. In this situation, the bank that compounds interest monthly would report that they offered a 9.00% nominal rate and a 9.38% effective rate. Banks often call this the effective rate or the annual percentage yield (APY). Financial institutions commonly quote both the nominal and the effective rate of interest. The above example leads to the following formula: EFFECTIVE RATE FOR COMPOUND INTEREST For compound interest, the effective interest rate is equivalent to the simple interest earned on one dollar in one year expressed as a percentage. The effective annual rate is also called the annual percentage yield or APY. Given a nominal rate of r with m conversion periods in a year, the effective interest rate is m r 1+ 1 which can be multiplied by 100 and stated as a percentage. m The APY is usually reported as a percent rounded to two, or at most 3, decimal places. Joseph F. Aieta Page 86 Printed 08/14/03

87 Time Value of Money Section 3. Page 87 Notice that for simple interest, the distinction between effective rate and nominal rate is not made since they are always the same. It is also true that for a fixed nominal rate, compounding more frequently increases the effective rate, and compounding less frequently decreases the effective rate. This is the reason that a person should compare effective rates, not nominal rates, when comparing different investment options. Example 3.. a) Use the above formula to calculate the effective rate of 9% compounded daily, assuming an exact year with 365 days. b) If $100,000 is deposited at 9% what would be the difference in actual interest earned in one year between monthly and daily compounding? Solution: 365 a).09 APY = = A bank would report this as 9.4%. For a fixed nominal rate we can 365 see that compounding more frequently does increase the effective rate. b) The difference in actual interest earned in one year would be 100,000 ( ) which is less than $36. The difference on the reported APY between monthly and daily compounding is only about 0.04% ( ) but if your deposit was as large as $100,000 and the bank with daily compounding was convenient for you, then it would make sense to use the bank with the higher effective rate. Example 3..3 a. What is the percentage difference in effective rates between monthly and daily compounding for a nominal rate of 5%? (Assume a 365-day year) b. If a bank advertises a nominal rate of 5% with compounding every hour, then by how much is the effective rate increased over a bank offering daily compounding? Solution: 1 a) For monthly compounding, 0.05 APY = % % b) For daily compounding, 0.05 APY = % % 365 The percentage difference of % % is about 0.01%. In other words this difference in effective rates is about one hundredth of one percent (not one percent). Joseph F. Aieta Page 87 Printed 08/14/03

88 Time Value of Money Section 3. Page 88 When interest is compounded every hour, we have to first calculate m, the number of periods. In this case it is the number of hours in a year which is 365 days 4 hours = 8760 hours. day 8760 The hourly compounding annual percentage yield is % %. In part a) we found 8760 that the APY for daily compounding is about %. Hourly compounding increases the effective rate of the daily compounding by approximately % when the APR is 5%. Note that both of these effective rates would be reported as 5.13% even though hourly compounding would give us a few more pennies of interest on a deposit of $10,000. In the examples we just discussed we noticed that the larger the number of compounding periods, the larger the APY. In fact, the APY is an increasing function of the number of compounding periods in a year. Once we get beyond daily compounding, however, there is only a slight difference in APY as the number of conversion periods per year continues to increase. As we will see, a limiting value is approached as the frequency of compounding gets greater and greater. Example 3..4 Using Effective Rates to Compare Investment Options Ann Uity is the Chief Financial Officer for Dynamic Designs LTD. Her husband, Lum Psum, is a freelance journalist. Ann and Lum are trying to decide which bank to use for their savings account. A check of three banks produces the following information: Bank A quotes a rate of 4.77 % compounded daily (365-day year). Bank B states that they offer an APR of 4.79 % with monthly compounding. Bank C advertises quarterly compounding at a nominal interest rate of 4.81%. Which bank has the highest effective rate? Solution: The effective rates are computed as follows: 365 For Bank A, APY = % = % 4.89% For Bank B, APY = % = % 4.90% 1 4 For Bank C, APY = % = % 4.90% 4 Banks B and C would both report an effective rate of 4.90% (to the nearest hundredth of a percent). However, for computational purposes, Bank C would actually be the best choice -- but not by much. $100,000 invested for one year would grow to $104, in Bank C versus $104, in Bank B, a difference of only 89 cents. Joseph F. Aieta Page 88 Printed 08/14/03

89 Time Value of Money Section 3. Page 89 Example 3..4 Finding the Effective Rate Given One Year of Growth in Dollars Over the course of a year, shares in a mutual fund grow in value from $500 to $ What is the effective growth rate? Solution: Since t = 1, we need to solve for r in the equation 734.5= 500 (1 + r) 1 = r = 500 r r = = = % The mutual fund would report the effective rate as 9.38% to two decimal places or 9.381% to three decimal places. Continuously Compounded Interest We have seen that interest can be compounded using a different number of conversion periods per year. In this section, we explore what is called continuous compounding. Figure 3..1, taken from, SinglePayment.xls, lists the amounts in three accounts each opened with an initial deposit of $1000. Interest is compounded at 8% for each account and the frequencies of compounding are annual, daily, and continuously. Figure 3..1 shows the graphical plots of data in the table. Continuous compounding compared with daily compounding Years nominal rate PV t 8.00% 1,000 Annual Daily Continuous 0 1, , , , , , , , , , ,71. 1, , , , , , , , , , , , , , , , ,999.00,054.7, ,158.9,5.35, ,331.64,410.67, ,518.17,611.4, ,719.6,88.89,89. 14, , , , , , , , , , , , , ,0.03 4, , , , , , , $6,000 $5,000 $4,000 $3,000 $,000 $1,000 $0 Annual Daily Continuous years Figure 3..1 Joseph F. Aieta Page 89 Printed 08/14/03

90 Time Value of Money Section 3. Page 90 It appears at first that Figure 3..1 shows only the graphs of two functions instead of three. Can you explain why the graph seems to be missing one of the curves? Look carefully at the values in the columns labeled Daily and Continuous in the table. Notice that the continuous compounding values are only a few cents higher than the daily compounding values. Consequently, the graphs of the two curves are so close that it is difficult to visually detect any difference between them. For practical purposes continuous compounding does not seem very different from daily compounding. Does this observation hold true for a larger principal (PV) and different nominal rates? What happens if we start with a million dollars? To answer these questions, we need to know more about what is meant by the continuous compounding of interest. Effective Rate for Continuous Compounding Figure 3.. shows the computation of effective rates (written in decimal form to 10 decimal places) given a nominal rate of 8% that is compounded m times per year for selected values of m. It appears that the effective rate increases, but not by very much, as the number of conversion periods per year increases beyond daily compounding. Nominal Rate = 8% Compounding Frequency year) m Effective rate as a decimal semi-annually quarterly monthly daily (exact every hour every minute 55, every second 31,536, Figure 3.. It seems clear that the effective rate gets closer and closer to a specific upper bound as the number of conversion periods gets larger and larger. A graph of the effective rate (expressed in decimal form) as a function of m helps us see what is happening. Joseph F. Aieta Page 90 Printed 08/14/03

91 Time Value of Money Section 3. Page Figure 3..3 shows a graph of the function f ( m) for m between 0 and 00. Notice that f (1) = 0.08, m the nominal rate (as a decimal). You can also verify that f () = , f (365) = , and so forth. = m In the graph of f(m) shown in Figure 3..3, the curve climbs rapidly from to , but then it increases much more slowly. In fact, as m gets larger and larger, the decimal form of the effective rate approaches a limiting value near f(m) m Figure 3..3 As the value of m gets arbitrarily large, the duration of one conversion period becomes arbitrarily small. There is a limiting number which f(m) approaches as the number of conversion periods per year, m, gets larger and larger. The mathematical expression for this limit, e , includes the important constant e, the base of the natural logarithm function. This limit is the annual percentage yield, or effective rate, for continuously compounded interest. EFFECTIVE RATE FOR CONTINUOUS COMPOUNDING When interest is compounded continuously at a nominal rate of %r, the effective interest rate is APY = (e r - 1) 100% In this formula, the nominal rate r is in decimal form. Like other effective rates, this effective rate is usually reported to the nearest hundredth of a percentage. Joseph F. Aieta Page 91 Printed 08/14/03

92 Example 3..5 Effective Rate for Continuous Compounding Time Value of Money Section 3. Page 9 a) Verify that the effective rate is 8.33% for a deposit of $100 earning interest at a nominal rate of 8% compounded continuously. b) What would be the effective rate if the $100 deposited in part a) changes to $3000? Solution: a) To compute the effective rate, find e = and convert to a percent. The effective rate is 8.33%. b) The effective rate, e , expressed as 8.33%, is the same for any lump sum. The formula for the effective interest rate does not depend at all upon the principal PV. A general formula for continuous compounding The fundamental formula for continuous compounding relates future value, present value, nominal rate, and time in years. COMPOUNDED INTEREST FORMULA FOR CONTINUOUS COMPOUNDING When an amount PV (present value) is invested for a total of t years and earns interest that is compounded continuously at nominal rate r, the future value of the amount is FV = PV e r t The nominal rate, r, must be expressed as a decimal when using this formula. Example 3..6 Future Value for Continuous Compounding a) A deposit of $1000 in an account that earns interest at a nominal rate of 4.98% compounded continuously. What is the future value of this account in and ½ years? b) What would be the future value of $1000 in and ½ years if the interest rate is 5.00% compounded daily? Solution: a) FV = 1000 e = $1,13.58 to the nearest cent b) 0.05 FV = = $1, to the nearest cent. 365 Joseph F. Aieta Page 9 Printed 08/14/03

93 TVM Activities Page 93 TVM Activities Assume exact year unless specified otherwise. Give dollar amounts to the nearest cent. 1. A certain mutual fund balance is $5,000 at the end of the year. If the amount in the fund at the beginning of the year was $,000, then what is the effective yield (APY)?. Given a nominal rate of 6.00 percent, find the effective rate of growth for each compounding method in a) f). State the percentage with four (4) instead of two decimal places (example 6.387%) a) compounding is annually. b) compounding is semiannually. c) compounding is quarterly. d) compounding is monthly. e) compounding is weekly. f) compounding is daily. 3. Suppose that you have $100,000 to invest for one year. Find the future value of this deposit at the end of the year if a) bank A offers 6% per year and compounds interest every day for a 365 day year? b) bank B offers 6% per year and compounds interest every hour for a 365 day year? c) Suppose that you can walk to bank A, would you spend $1.00 in gas to drive to bank B? 4. Suppose that you have $10,000 to invest for one year. Bank A offers 6% per year and compounds interest every month. Bank B offers a slightly lower nominal rate but compounds interest every day for a 365 day year. If the interest earned, to the nearest cent, in one year at bank B is identical to bank A then what is the effective rate at bank B to the nearest tenthousandth of a percentage? 5. Suppose that the annual effective rate for a given account is 5.5% per year and that you invest $30,000 in the account. How much money will be in the account at the end of the year? Joseph F. Aieta Page 93 Printed 08/14/03

94 TVM Activities Page You have $10,000 to invest for three years and six months. Bank A pays 1% compounded semiannually, and Bank B pays 1% per year compounded quarterly. a) What is the effective interest rate at Bank A? b) What is the effective interest rate at Bank B? c) To the nearest dollar, how much more can you earn by depositing your money in Bank B? 7. Over a five-year period in the 1980 s a particular mutual fund returned 14.5% per year compounded monthly. At this growth rate how much would a $10,000 investment be worth after five years. 8. A $10,000 investment in a particular mutual fund grew to $3,17 six years later. What was the effective rate of return on this investment? 9. If the nominal rate is 7.5% and interest is compounded continuously, then find the effective rate. 10. What is the effective rate of 7.9 percent per year compounded continuously? 11. Find the future value of $4,000 in 5 years and 8 months if interest is compounded continuously at a nominal rate of 6%. 1. Determine the future value of $100,000 in five years under a nominal rate of 5.90% when interest is a) compounded daily. b) compounded continuously. c) To the nearest dollar, by how much does the answer to part b) exceed the answer to part a)? Joseph F. Aieta Page 94 Printed 08/14/03

95 TVM Activities Page History tells us that Peter Minuit, a Dutch colonial governor of New Netherlands, purchased Manhattan Island in New York from the Indians in 166 for trinkets costing about $4. If the $4 had been invested at 5 percent per year compounded continuously, what would it be worth in the year 000? 14. Imagine an account that paid 100% interest per year compounded continuously. a) If you invested $1 at this rate, what would it grow to in one year? b) If you invested $100,000 at this rate, what would it grow to in one year? 15. Which bank actually offers the better rate? a) Bank A offers 4.40% per year compounded quarterly. b) Bank B offers 4.35% per year compounded continuously During the mid 1980s, annual levels of inflation exceeded 10%. Inflation in the late 1990s dropped to annual levels below 3%. Inflation and taxes, taken together, lower the real return after taxes of an investment in terms of actual purchasing power. We represent the different rates y, rr, roi, and tr as follow: y = annual yield before taxes, rr = annual real return, roi = rate of inflation, tr = tax rate Assuming each rate is in decimal form, then rr + roi + rr roi y = 1 tr which is equivalent to y(1 tr) = rr(1 + roi) + roi For people in a 8% tax bracket to get a real return of 1% on an investment during a period of 4% inflation their pre-tax investment would have to yield 7.00% since y = = Joseph F. Aieta Page 95 Printed 08/14/03

96 TVM Activities Page State all answers to the nearest hundredth of a percentage. 16. For people in a 8% tax bracket to get a real return of 3% per year when inflation is running at 4% per year, what yield would they need on their investment? 17. For people in a 36% tax bracket to get a real return of % per year when inflation is running at 3.0% per year, what yield would they need on their investment? 18. For people in a 15% tax bracket to get a real return of 1% per year when inflation is running at.5% per year, what yield would they need on their investment? 19. What would be the real return for a person in a 8% tax bracket if their investment had a pre-tax yield of 8% during a period when inflation was 4%? [Hint: since rr is the unknown in this case and all other rates are known, start with y(1 tr) = rr(1 + roi) + roi ] 0 The federal government often borrows money by issuing bonds to cover its immediate and long-term debts. A very widely-followed long-term treasury bond has thirty years to maturity from the issue date. If a thirty-year treasury bond in 1999 had an APY of 6% then explain why some investors might prefer a -year bank CD with an APY of only 5.00%. Joseph F. Aieta Page 96 Printed 08/14/03

97 Time Value of Money Section 3.3 Page 97 Finding an unknown present value, rate per period, or number of periods Section 3.3 MATHEMATICAL FUNCTIONS: You can think of a mathematical function as a rule, a formula, a prescription, or a procedure that assigns to each input exactly one output. Mathematical functions are commonly represented by mathematical symbols or formulas. With the advent of calculators and computers, it has become relatively easy to evaluate the majority of mathematical functions that occur in applications. A simple example of a mathematical function is the square root function on a calculator, usually shown as x. If you input the number 9 into x then the output is the number 3. In the rule y = x we think of x as the input and y as the output. Not all numbers are acceptable inputs for a particular function. If you enter the number 9 as an input to x using a standard scientific calculator then you will get probably get some kind of error message instead of a numerical output. To evaluate the square root of positive five with a scientific calculator, we input 5 and the square root key (or vice versa depending on your calculator) and the output will be a decimal close to up to the maximum number of decimal places that the machine can display. NEED FOR LOGARITHMS: To solve problems in the mathematics of finance, we seldom need complicated algorithms or tables to evaluate functions. We do need to understand the properties of certain types of functions and their graphs. Exponential functions are the most appropriate models for compound growth. The exponential function base e is typically shown on a calculator key as e x. The natural logarithmic function, whose base is e, is typically shown on a calculator key as LN(x), ln(x) or just plain ln. If you enter 1 as an input to the function e x then a 9-digit display for the value of e 1 will be Solving equations is closely related to properties of functions. The answer to the question What positive number is the solution to x = 5? is equivalent to the question For what positive value of x is the function f(x) = x - 5 equal to zero? The pair of functions x and x have the property that each function undoes the other. If we select any positive number and square it, we can get back the original number by taking the square root of the result. If we select any positive number and find its square root, we can get back the original number by squaring the result. It is this property that enables us to solve equations of the type x = a or x = b where a and b are positive numbers. The equation x = a has a as a solution and - a as another solution. The equation x = b has b as its solution. Joseph F. Aieta Page 97 Printed 08/14/03

98 Time Value of Money Section 3.3 Page 98 In the mathematics of finance we rarely encounter problems that can be solved by simply squaring a number or by finding a square root. We do, however, encounter situations such as finding the value of x for which the exponential function f(x) = (1.06) x - is equal to zero. This is equivalent to solving the equation (1.06) x =. Computers and calculators that have solver capabilities can produce very good approximations with built-in numerical search algorithms. For this type of equation, however, we do not need to depend on a very high-powered computing device. As we will see, all that is needed is a logarithm function that undoes an exponential function. Identify the e x and the ln keys on your calculator and find e 3. We will refer to the result of raising e to the third power as ans. Next, calculate ln of ans. What do you get? Next, start with any other number and repeat this sequence of steps. What do you get? Repeat this process as many times as you want. Using standard mathematical notation, a symbolic description of what you have observed is that ln( e x ) = x for any real number. This time we start with the natural log function ln and enter a positive number, say Calculate ln(1.05) and call that result ans. Now raise e to the power ans. What is the output? Next, start with some other positive number and repeat this sequence of steps. Repeat this process as many times as you want. Using standard notation, a mathematical description of this pattern is e ln x = x for any positive number. Note that logarithm functions are restricted to positive inputs. To solve exponential equations, any other logarithm function can be used. Another well-known logarithm function is the common log which has a base of 10 instead of e and is displayed as log on many scientific calculators. Find the power key on your calculator (or the key 10 x ), raise 10 to the 3 rd power, and call 10 3 ans. Assuming that you have a key labeled log, calculate log of ans. What do you get? Repeat this process with numbers other than 3. Using standard notation, a mathematical description of the pattern that you observe is log(10 x ) = x for any real number. Next calculate log(1.05) and if necessary store this result as ans. Next raise 10 to the power ans. Now start positive numbers other than 1.05 and repeat this sequence of steps. Using standard notation, a mathematical description of this pattern is 10 log x =x for any positive number. Logarithm and exponential functions undo each other. Natural log Common log ln(e x ) = x log(10 x ) = x for any number e ln x = x 10 log x = x for any positive number Technically speaking, these function pairs are called inverse functions. Joseph F. Aieta Page 98 Printed 08/14/03

99 Time Value of Money Section 3.3 Page 99 The paired exponential and logarithmic functions are pictured below for 0 < x < 5. For the functions on the left, the base is e. For the functions on the right the base is 10. These restricted graphs suggest that the steeper the exponential function the flatter the corresponding logarithm function. e x and ln(x) for 0<x<5 10 x and log(x) for 0<x<5 e to the x natural log 10^x common log Figure More complete graphs of e x and ln(x) illustrate how exponential functiona with a base greater than 1 approach the negative half of the x-axis as x gets further and further to the left of zero. The corresponding logarithm function approaches the negative half of the y-axis as x approaches zero from the right. e^x ln(x) Figure 3.3. Joseph F. Aieta Page 99 Printed 08/14/03

100 Time Value of Money Section 3.3 Page 100 Important algebraic connections between exponential and logarithm functions can also be seen in the table below. Each property of logarithms can be derived from a corresponding property of exponents because of the following definition: If b is any positive base, not equal to one, then log b (y)= x is equivalent to b x = y 3 log 8= 3 is equivalent to 8 and log 0.01= is equivalent to = 10 = In the following table: b > 0 and b 1, m, n, and p are real numbers; M and N are positive real numbers. Properties of Exponents Corresponding Properties of Logarithms b 0 = 1 log b 1 = 0 b 1 = b log b b = 1 b m b n = b m+n log b (M N) = Log b M + log b N b b m n M N m n = b logb = logb M log b N n m m n ( b ) b p = logb M = p logb M If the base b is ten then the logarithm is called the common log. The notation log x is often used as the abbreviation for log 10 x. If the base b is e then log e y = x which is equivalent to e x = y. The logarithm base e is called the natural logarithm The notation ln x is typically used as the abbreviation for log e x. log 10 y = x is equivalent to 10 x = y log e y = x is equivalent to e x = y For solving general exponential equations that occur in the mathematics of finance, the most important property of p logarithms is log b M = p log b M. Translated into words, this property says that the logarithm of a number to a power is the power times the logarithm of the number (independent of the base). Joseph F. Aieta Page 100 Printed 08/14/03

101 Time Value of Money Section 3.3 Page 101 Example Solve for x to four decimal places (1.06) x =. Solution: Take the logarithm of both sides (we will use natural log but we could also have used common log) Apply the power property of logarithms p logb M = p logb M Isolate x by dividing both sides by ln(1.06) ln (1.06) x = ln (). x ln(1.06) =ln() ln( ) x = ln( 106. ) Evaluate with Excel or with a calculator x = Example 3.3. Solve for x a) 10 x = 5 b) e x = 5 Solution: These equations could be solved the same way that we solved (1.06) x = but it is much easy to use the facts that log y = x is equivalent to 10 x = y and log e y = x is equivalent to e x = y 10 a) 10 x = 5 is equivalent to log 10 5 = x. The common log of 5, log (5), is to seven places. b) e x = 5 is equivalent to loge 5 = x. The natural log of 5, ln (5), is to seven places. Example Solve 1 (1.08) x = 6 to four decimal places. Solution: Divide both sides by 1 (1.08) x = 0.5 Take the logarithm of both sides ln ( (1.08) x )= ln (0.5) Apply the power property of logarithms x ln (1.08) =ln (0.5) Isolate x by dividing both sides by ln(1.08) ln( 0.5 ) x = ln( 108. ) Evaluate (to four decimal places) x = Joseph F. Aieta Page 101 Printed 08/14/03

102 Time Value of Money Section 3.3 Page 10 Performing calculations for lump sum problems Financial calculators and spreadsheets have built-in routines that automatically perform the calculations in this section. Graphing calculators have general equation solvers and some models also have built-in financial functions. If you are using a scientific calculator, the following examples illustrate how to use algebra and the future value formulas to solve for unknown values of PV, i or n. If we start with the compound interest formula FV = PV (1 + i) n where FV is future value, PV is present value, i is the interest rate per period, and n is the number of periods then we can algebraically solve for PV by dividing FV by (1 + i) n or by using negative exponents FV PV = (1+ i) n = FV(1+i) -n Example If $0,000 is needed in three years and an account pays 5% interest compounded quarterly then what amount should be deposited into the account now? Solution: i=.05/4 =0.015 $0,000 is an FV that you can withdraw at the end of period 1 PV Think of PV as money that you take out of your pocket and put into the bank. Figure In the time line diagram shown in Figure 3.3.3, FV is 0,000, i is.05/4 = and n is 3 times 4 = We want the unknown present value PV which is given by PV = = 0000(1+.05/4) 1 (1+.05/4) = 0000( ) = 17,30.17 to the nearest cent. [With a scientific calculator it is often easier to raise a number to a negative power and multiply by 0,000 rather than divide 0,000 by a number raised to a positive power.] One interpretation of the final result is that a deposit of $17,30.17 now into an account that pays 5% compounded quarterly will grow to $0,000 three years from now. Another way to describe this situation is to say that $0,000 in three years is worth $17,30.17 in today s dollars under an assumed growth rate of 5% compounded quarterly. Joseph F. Aieta Page 10 Printed 08/14/03

103 Time Value of Money Section 3.3 Page 103 Example For what annual rate will $1000 grow to $000 in 1 years assuming annual compounding? Solution: In this example, the value of FV, PV, and n are known and we want the rate i for one interest period. We need to isolate the variable i in the formula FV = PV (1 + i) n In the timeline shown in Figure 3.3.4, FV is 000, PV is 1000, and n is 1. i is unknown FV= 000 and n = 1 PV = 1000 Figure We want to solve for i in the equation 000 = 1000 (1 + i) 1 step 1) Divide both sides by 1000 to get = (1 + i) 1. step ) Next we "undo" raising an expression to the 1 th power. This can be accomplished by raising each side of the equation to the 1/1 power. 1/1 1 1/1 ( (1 + i ) ) = + i = 1 step 3) Next we subtract 1 from both sides and convert the result to a percentage. i = 1/ 1 1= or 5.95%. FV PV We can apply these same steps to the general formula FV = PV (1 + i) n to obtain i = 1. 1/ n Joseph F. Aieta Page 103 Printed 08/14/03

104 Time Value of Money Section 3.3 Page 104 Comparison shopping for the best rates can be done on-line. Check out the web sites at and Keep in mind that the best credit card rate may not always be the best card for you. You should avoid offers of features that you don't need. 4 Scrutinize the fine print for such terms and conditions as cash advances, how long the introductory rate lasts, and what the penalties are for late payments. The penalty charged for late penalties is often more significant than the annual interest rate itself (to say nothing of the effect on your credit ranking). Example Suppose that your average balance is $3000 and the rate on your current card from Bank A is 15.6% compounded monthly with an APY of %. Over a one-year period you will pay about $503 in interest. Here is a typical new card offer by a bank, let s call it Bank B.: NO Annual Fee: 3.9% Introductory APR for 5 months, then 9.99% Fixed APR for all ONLINE and OFFLINE purchases Assume that the new card also compounds interest monthly. You will save about $503 - $8 = $75 in interest fees over a one year period if you switch to the new card. Suppose that your current credit card company, Bank A, wants to keep your business and will lower their APR to be competitive. As a customer, you might prefer not to go through additional paperwork plus a credit bureau check if Bank A can come close to matching the offer from Bank B. Usually Bank A cannot offer you an introductory rate since you are already a customer. Instead they offer an annual rate, guaranteed for one year that is equivalent to paying annual interest of $50 on an average balance of $3000. What is this new nominal rate (APR) to the nearest hundredth of a percentage? Solution: This APR can be calculated by solving for r in the formula for effective rate. 1 r = 50/3000 = r 1 + = /1 1 r = /1-1 = This is the interest rate for one month. 1 To find the nominal rate, multiply the interest rate for one month by 1 to obtain r = 1( ) = This is reported by Bank A as an APR of 8.03%. 1 Another approach would be to solve = x 1 + for x by using Excel s Solver or Goal-Seek utilities. 1 4 Learning to deal with a future of plastic. A refresher course for students in handling credit cards wisely by Charles A. Jaffe, Globe Staff, 08/3/99 Joseph F. Aieta Page 104 Printed 08/14/03

105 Time Value of Money Section 3.3 Page 105 Example How many years will it take for $000 to grow to $3000 under annual compounding if the nominal rate is 4%? In this example we need logarithms to isolate n, the number of periods. i is 0.04 FV= 3000 and n is unknown PV = 000 Figure Solution: Solve 3000 = 000 (1 +.04) n for n. step 1) Divide both sides by 000 and get 1.5 = (1 +.04) n step ) Take the natural logarithm of both sides and obtain ln(1.5 ) = ln (1 +.04) n step 3) Apply the property of logs that says that the log of a number to a power is that power times the log p of the number. [Symbolically, log a = p log a ]. This gives us ln(1.5 ) = n ln(1 +.04) b b step 4) Now divide both sides above by ln(1.04) to isolate n ln(1.5) n = = ln(1.04) Conclusion: In approximately 10 years and 4 months, $000 will grow to $ Joseph F. Aieta Page 105 Printed 08/14/03

106 Time Value of Money Section 3.3 Page 106 Example Suppose we change Example to monthly compounding. How long will it take for $000 to grow to $3000 if the nominal rate is 4% and compounding is monthly? FV = 3000, PV = 000, and i =.04/1. Note the difference between Examples and In the latter problem, interest is compounded 1 times a year, not once a year. Remember that in solving this problem with a scientific calculator with the formula FV = PV (1 + i) n, we must first calculate the monthly interest rate i. Solution: Solve 3000 = 000 (1 +.04/1) n for n. The value of n that we find will be the unknown number of months. If we want to report the number of years, we would divide n by 1. step 1) Divide both sides of 3000 = 000 (1 +.04/1) n by 000 and get 1.5 = (1 +.04/1) n step ) Take the natural logarithm of both sides [any log function can be used] ln(1.5) = ln(1 +.04/1) n step 3) Apply the property of logs that says that the log of a number to a power is the power times the log of the number. ln( 1.5) = n ln(1 +.04/1 ) step 4) Divide both sides of the above equation by ln(1+.04/1) to isolate n n = ln(1.5) ln( /1 ) ln(1.5) = = months ln( ) This is about years or approximately 10 years and months. Note that this is about two months shorter than the time it would take for the same nominal rate with compounding only once a year. Joseph F. Aieta Page 106 Printed 08/14/03

107 Time Value of Money Section 3.3 Page 107 Doubling Time The ability to quickly estimate the time it takes for an investment to double in value can be very useful in checking the reasonableness of many financial calculations. As we will see, doubling time depends upon rates and not on dollar amounts. It takes the same time for $1000 to grow to $000 as it does for $1 to grow to $ or for $100 to grow to $00. The decimal 0.06 entered for rate r will be formatted as the percentage 6.00% Future value over a 15 year period rate= r r% = 100*r PV 1, Years t 6.00% 8.00% 10.00% $4, , , , $4, % 8.00% 10.00% $3, $3, $, $, $1, $1, $500 years $ Figure The table and the graph in Figure 3.3.6, taken from SinglePayment.xls, illustrate the approximate doubling time for annual compounding given nominal rates of 6%, 8%, and 10%. The number of years required for $100 to grow to $00 is very close to 1 years for a 6% rate, very close to 9 years for an 8% rate, and a little over 7 years for a 10% rate. These estimates can be verified using logarithms. Find the approximate doubling time for a 4% rate and for an 18% rate and then complete the table below for doubling time under annual compounding. See if you can state a simple rule in words that fits the pattern fairly well. Annual Compounding nominal rate as a % 4% 6% 7% 8% 9% 10% 1% 18% doubling time in years? 1? 9? 7+?? If %r is the rate as a percentage and d is the doubling time in years then which is the better model 7 d = -1.5* %r + 1 or d =? % r Joseph F. Aieta Page 107 Printed 08/14/03

108 TVM Activities 3 Page 108 TVM Activities 3 1. What value must x have if a) ln 4 e = x b) c) log3 6 3 = x log4 5 4 = x. Solve for x a) ln x = 0.4. b) ln x = 1.. c) ln x = =-1.1 d) ln x = e) ln (1 + x ) = 0.3 f) ln ( + x) = 1. g) ln (0.5x ) = Write the following in inverse form: Example log 5 15 = 3 is the inverse form of 5 3 = 15 a) log 7 49 =. b) 7 1/3 = 3. c) 64 ½ = 8. d) log 8 = 1/3. e) ln e = 1 f) (1/) 3 = 0.15 g) log N = y. h) log 16 4 = 1/ 4. What sum of money deposited now, in an account paying 8% per year compounded quarterly, will provide just enough money to pay a $1,000 debt due 7 years from now? 5. An account paying interest at 6 % per year compounded semiannually was established 10 years ago. The account balance is currently $8, What was the initial amount when the account was established? Joseph F. Aieta Page 108 Printed 08/14/03

109 TVM Activities 3 Page The population of a town is now 5,000. The town s population has increased at the rate of 6 percent compounded annually for the past five years. Estimate the population 5 years ago. 7. For a given nominal rate, doubling time is the number of years required for $1 to grow to $. Suppose interest is compounded annually and you are given the doubling times in years. a) Complete the table below with estimates of the nominal rate to the nearest whole percentage. annual compounding doubling time in years nominal rate as a % 6 % b) Develop a formula that gives the exact compound growth rate of an investment that doubles in n years. 8. For a given nominal rate, doubling time is the number of years required for $1 to grow to $. Assuming daily compounding, complete the table below with estimates of the doubling time, to the nearest tenth of a year, for nominal rates of 4%, 6%, 7%, 8%, 9%, 10%, 1%, 18%. nominal rate as a % 4% 6% 7% 8% 9% 10% 1% 18% doubling time in years 5.8 (daily compounding) How do these values compare with the pattern that gives doubling time for annual compounding? 9. a) If your money grows at the rate of 6% per year compounded annually then what lump sum should you deposit now in order to have $1,000 in ten years? b) If your money grows at the rate of 6% per year compounded annually then how much should you deposit now in order to have $1,000 in five years? c) Is your answer to part b) double your answer to part a)? Explain why or why not? 10. Suppose that you are considering the following three contracts. Assuming that compounding is annual which contract is worth the most in today s dollars. Justify your answer. a) $500 in 8 years under an agreed upon interest rate of 8% per year. b) $500 in 9 years under an agreed upon interest rate of 7% per year. c) $500 in 10 years under an agreed upon interest rate of 6% per year. 11. Lum Psum buys a three-year CD with $10,000. How much will this CD be worth at the end of years and 180 days if he selects a) Bank A that uses a 360 day year and pays 4.75% interest per year compounded daily? b) Bank B that uses a 360 day year and pays 4.80% interest per year compounded semi-annually? Joseph F. Aieta Page 109 Printed 08/14/03

110 TVM Activities 3 Page Lum made a deposit of $1,000 into a bank that pays 5% per year compounded daily. One year later he moved all of the money in this account from bank A into Bank B. Bank B pays 5.5% per year compounded monthly. If no other deposits are made then how much will he have in savings three years after the initial deposit? 13. Lum is considering an investment in a mutual fund of growth stocks that he expects will have a minimum growth rate of 18% per year compounded annually. a) Under this assumption, how long, to the nearest tenth of a year, will it take for his investment of $10,000 to grow to at least $0,000? b) Another mutual fund has doubled an investor's money over the past 6 years. What was the annual growth rate (to the nearest hundredth of a percent)? c) Lum must be certain of having $10,000 in savings twenty months from now. What amount should he deposit now in the stock market? 14. What is the annual effective rate (APY) on an account that pays 6.0% interest per year, compounded daily? 15. At 9 % per year compounded annually, how many years will it take for $5,000 to grow to $10,000? 16. At 8 % per year compounded annually, how many years will it take for $500 to grow to $1000? 17. Compute the Annual Percentage Yield (APY) on an account paying 5.98% per year compounded daily and on another account paying 6.0% per year compounded semi-annually. a) Which account has a higher APY? b) By how much will the APY on the 6.0% per year account increase if compounding is monthly? 18. INFLATION The compound interest formula FV= PV (1 + i) n can help us understand the impact of inflation on the cost of living. When prices for goods and services rise for extended periods at annual rates higher than or 3 percent, the economic impact on purchasing power can be a hardship for people who are living on fixed incomes. If basic living costs for a retired couple are $3000 monthly and their income is fixed at $3500 then there is some margin for discretionary spending. If, however, basic living costs start to increase at an annual rate of 3.5 % or more, then it will not take too many years before the retirees must draw down savings or make some major adjustments in their standard of living. At a constant 3.5 % annual inflation rate, a purchase that costs 1000 dollars today will cost 1000( ) n dollars n years from now. If we experience 3.5% inflation for four years in a row, then what costs $3000 today will cost about $3443 four years from now. An inflation calculator can be found at Use this inflation calculator to determine the dollar amount today that is equivalent to the purchasing power of $10,000 ten years ago. a) Suppose the cost of renting an apartment increases at an average rate of 8% per year for six years. If an apartment rents for $1800 per month now then how much will it rent for six years later? b) Suppose housing costs in a particular suburb rose at the rate of 10% per year in the decade from 1980 to If a house sold for $50,000 in 1990 then what would it have sold for in 1980? c) (optional) Find the average inflation rate in the U.S. for each of the years from 1978 to Document your sources. d) (optional) Find the average rate of increase of selling prices for single family homes in Massachusetts in the period from 1988 to 1998 Joseph F. Aieta Page 110 Printed 08/14/03

111 TVM Activities 3 Page Sam is looking at the $8,000 sticker price of car. Suppose that the inflation rate over each of the last six years was 1%. How much would a comparable automobile have cost six years ago (nearest dollar)? 0. To simply keep pace with an annual inflation rate of 4%, a person who earns $50,000 per year now will need to earn how much in five years (nearest dollar)? 1. If a vintage sports car sold for $5,000 in 1980 and then sold for $65,000 in 1990, what was the annual percentage increase over those ten years?. If the cost of a first class postage stamp is 34 cents in 000 and we imagine an annual inflation rate of 3% over 100 years, then to the nearest cent, what was the theoretical cost of first class postage in 1900? 3. A painting by a famous artist rose % in value each month since If it sold for $5000 in 1990 then what did it sell for in the same month in 1995? 4. Which investment offers the higher effective rate of return? a) A $0,000 zero coupon bond due in ten years that can be purchased for $11,000 today. b) A long-term certificate of deposit at 5.75% compounded monthly. 5. FV = PV e r t is solved for FV. Start with this equation and solve for a) PV b) r c) t 6. Find the future value of $4,000 in 5 years and 8 months if interest is compounded continuously at a nominal rate of 6%. 7. If the effective rate is 8 % and interest is compounded continuously then find the nominal rate (nearest hundredth of a percent). 8. For a given nominal rate, doubling time is the number of years required for $1 to grow to $. Assuming continuous compounding, complete the table below with estimates of the doubling time, to the nearest tenth of a year, for nominal rates of 4%, 6%, 7%, 8%, 9%, 10%, 1%, 18%. nominal rate as a % 4% 6% 7% 8% 9% 10% 1% 18% doubling time in years 5.8 continuous compounding How do these values compare with doubling time for daily compounding? 9. How much should be invested now at 6% per year compounded continuously if the value of the account in one year must be $5000? 30. How much should be invested now at 6% per year compounded continuously if the value of the account in five years must be $5000? Joseph F. Aieta Page 111 Printed 08/14/03

112 Time Value of Money Section 3.4 Page 11 Annuities: The Future Value of a Stream of Periodic Payments Section 3.4 An annuity is a series of constant payments made over a period of time. Car loans, mortgages, and annual payments to a million-dollar lottery winner are examples of annuities. By now you are very familiar with the basic compound interest formula FV = PV ( 1 + i) n. This formula relates the future value at the end of n interest periods to the single payment, PV that is made at the beginning of the first interest period and to the interest rate for one period. This formula is the foundation of many other compound interest formulas. We will develop a formula for calculating the future value of a series of equal cash flows that are made at the end of each compounding period. Example What is the future value of a series of four $100 payments made at the end of each period for four periods given an interest rate of 5% per period? If no interest (0%) is earned then the sum of these four payments would be simply $400. Since the interest rate per period is 5%, we know that the compounded value of the four payments must have a value greater than $400. First, we will do calculations to find the future value at the end of the fourth period for each of the four individual payments of $100. We will sum these four individual future values to find the future value of the annuity. Interest rate per period i = 0.05 FV is unknown Figure The first payment has earned interest for three periods and is worth at the end of period 4. The second payment has earned interest for two periods and is worth at the end of period 4. The third payment has earned interest for one period and is worth at the end of period 4. The fourth and last payment has earned no interest and is worth at the end of period 4. The total value of these four amounts at the end of period 4 is = Joseph F. Aieta Page 11 Printed 08/14/03

113 Time Value of Money Section 3.4 Page 113 The table in Figure 3.4. shows how each $100 payment grows. The last payment, made at the end of period 4, earns no interest. start end of period end of period end of period end of period terms expressed as =PV (1 + i ) n 1 st payment =100 (1+0.05) 3 nd payment =100 (1+0.05) 3 rd payment =100 (1+0.05) 1 4 th payment =100 (1+0.05) 0 =100 (1) Figure 3.4. sum is The sum is obviously the same whether we write the four terms in descending order or ascending order. In ascending order the sum is (1+0.05) (1+0.05) (1+0.05) 3. We can see that each successive term is ( ) times the previous term. This fits a well-known mathematical pattern, called a geometric series, in which any two consecutive terms have a common ratio. In descending order this sum is expressed in equation (1) below: (1) S = (1 +.05) (1 +.05) (1 +.05) +100 We will multiply each side of equation (1) by (1+.05) and then subtract equation (1) from the result S (1+.05) = 100 (1 +.05) (1 +.05) (1 +.05) (1 +.05) S = -100 (1 +.05) -100 (1 +.05) -100 (1 +.05) S (1+.05) 1 S = 100 (1 +.05) S (1+.05) 1 S = (1 +.05) = 100 [ (1.05) 4 1] S ( ) = S (0.05) = 100 [ (1 +.05) 4 1] + This is equivalent to 100 [ (1 +.05) 4 1] We solve this last equation for S and get S =.05. In the general case with parameters Pmt, i, and n the future value can be expressed with summation notation as 5 FV = n k = 1 Pmt n k ( 1+ i) has the sum FV = (1 + i) Pmt i n 1. 5 Summation notation is a compact way to indicate a sum of terms that fit a pattern. To expand this sum, substitute k = 1,, 3,..., n and write a plus sign between each term. FV = n k = 1 Pmt n k ( 1+ i) = Pmt (1+ i) n-1 + Pmt (1+ i) n- + Pmt (1+ i) n Pmt (1+ i) 0 Joseph F. Aieta Page 113 Printed 08/14/03

114 Time Value of Money Section 3.4 Page 114 Example 3.4. Figure was created with a worksheet in the file Annuties.xls. Regular payments of $000 are made into an account at the end of each period for 1 periods. Interest is compounded at a rate of 5% per period. At the end of period 1 the accumulated value is $31, All but the very last payment accumulates interest for some length of time. The very first payment accumulates the most interest since it is earning interest over the longest time interval. This first $000 deposit made at the end of period one is worth $ at the end of period 1. Interest rate Periods per period $,000? N i PMT FV 5.00% %i The sum of these 1 amounts is the future value of an annuity with 1 periodic payments of $,000 made at the end of each period at a rate of 5.00% per period. Figure We can confirm the formula for the future value of annuity by substituting.05 for i, 1 for n, and 000 for Pmt to obtain 1 (1 +.05) 1 FV = 000 = $31,834.5 to the nearest cent..05 Joseph F. Aieta Page 114 Printed 08/14/03

115 Time Value of Money Section 3.4 Page 115 Future value of an ordinary annuity If regular payments of size Pmt are made into an account at the end of each period for n periods and the interest rate is i per period then the future value of the account at the end of the last period, will be n ( 1 + i) 1 FV = Pmt i An annuity in which the payment intervals coincide with the interest conversion periods is called a simple annuity. If payments are made at the end of the period then it is called an ordinary annuity. If payments are made at the beginning of each period then the final amount would be worth slightly more. This type of annuity with payments at the beginning of a period is called an annuity due. Unless otherwise stated, assume that the examples and exercises refer to simple, ordinary annuities. A special type of annuity problem is called a sinking fund in which an account is established to accumulate funds on a regular basis in order to meet some future obligation. Example A corporation has issued ten year bonds totaling $10 million in order to finance the construction of new manufacturing plants. To redeem the bonds 10 years from now, the company will transfer a regular amount to a reserve fund at the end of each 6-month period. How much should be transferred on a semi-annual basis if the account earns 7 percent per year compounded semiannually and payments are made at the end of each period? i = 0.07/ = n = 0 Pmt Pmt Pmt Pmt Pmt Pmt Pmt Pmt Pmt Pmt Pmt Pmt is unknown Figure The sum of the terms in figure can be expressed with the formula for the future value of an ordinary annuity, n (1 + i) 1 FV = Pmt. i 0 ( ) 1 In this example FV is 10,000,000, i = 0.035, and n is 0 so we have 10,000,000 = Pmt To solve for Pmt with a scientific calculator, we first compute the value of the expression ( ) 1 and then divide 10,000,000 by the resulting This is equivalent to multiplying 10,000,000 by the reciprocal of 0 ( ) 1. The regular semi-annual payment is $353, to the nearest cent Joseph F. Aieta Page 115 Printed 08/14/03

116 Time Value of Money Section 3.4 Page 116 Important points to keep in mind about the FUTURE VALUE OF A SIMPLE, ORDINARY ANNUITY (1) Each cash flow or payment occurs at the end of the period. () The future value of the annuity is the equivalent lump sum at the time of the last payment. (3) The number of payments per year is the same as the frequency of compounding per year. In subsequent coursework related to the mathematics of finance you will learn to solve problems in which the number of payments per year is not necessarily the same as the frequency of compounding per year. Using financial calculators A sound problem solving strategy that is independent of the computational tool that you select is to hold off pressing keys to perform calculations until you fully understand the type of problem that you are dealing with. You should read the problem several times and classify it into an identifiable category or try to relate it to a familiar problem. Typical notation used by financial calculators is shown below. common keys N I/Y function to enter the total number of periods to enter the annual interest rate as a percentage, P/Y C/Y PV Pmt FV to enter the number of payments per year to enter the number of conversion periods per year (in this chapter, we assume that P/Y and C/Y are the same) to enter the present value to enter the periodic payment (0 for lump sum problems) to enter the future value Many specialized, interactive, Web-based calculators have been programmed to accept and report only positive values. This is also the case for the financial calculator emulated in Annuties.xls. When using either Excel or a financial calculator such as the TI-BA II plus, TI-83, or the HP 10-B, the direction of cash flow must be taken into account. From the depositor s perspective, money that is paid by making regular deposits to an individual savings account or by a company that is building up a sinking fund to replace equipment is considered as a negative cash flow. From this perspective, values should be entered as negative numbers. A positive number can represent cash that is received, such as a dividend check or money received from a bank loan. From the perspective of an individual saver, would represent a deposit of $1,000 from the saver s pocket to the bank and from this perspective a $000 withdrawal or loan from the bank would be We will generally look at cash flows from the perspective of an individual or entity that is depositing and withdrawing money, rather than the perspective of a bank that is receiving or paying out money. It is important to be aware of the conventions adopted by the particular computational device you are using. In the case of financial calculators, these conventions can be found in manuals. In some cases, the calculator manufacturer maintains a Website with pages devoted to various calculator models.. Spreadsheet conventions can be found in manuals and other books. Excel s on-line help and its function wizard are convenient resources that are readily available. Joseph F. Aieta Page 116 Printed 08/14/03

117 Time Value of Money Section 3.4 Page 117 Example Ann Uity is the Chief Financial Officer for Babson Designs LTD. Ann's company is a customer of a bank that pays 3.90% per year compounded monthly. Her company plans to replace some graphic design equipment in five years at an estimated cost of $80,000. Ann sets up a fund with a bank into which the company makes monthly payments for five years at the end of each month. Each payment earns 3.9 % per year compounded monthly. What is the amount of the monthly payment (cash outflow) that the company must make? i=.039/1 FV = $80,000 at the end of period 60 Pmt Pmt Pmt Pmt Pmt Pmt Pmt Pmt Pmt Pmt Pmt Figure We will solve this problem using three different methods (1) Using a formula and a scientific calculator with keys for storage and for raising a number to a power. () With fincalc, a generic financial calculator (3) With a spreadsheet. Method (1) formula and scientific calculator (1+ i) In the formula FV= Pmt i n 1 the known values are FV = 80,000, i = 0.039/1, n = (1)(5) =60 60 The unknow ( /1) 1 Pmt. Solve = for 80,000 Pmt 0.039/1 Pmt. To avoid repeated data entry and rounding errors, first calculate i = 0.039/1 and store the result. Add 1 to i and raise the result to the power 60. Subtract 1 and divide the result by the stored value of i. The display should read if your calculator displays nine places. Since division is multiplication by the reciprocal, you can use the 1/x key to find the reciprocal of which is to nine decimal places. Multiply the value in the display by 80,000 to get to the nearest cent. The size of the regular monthly payment that the company must make is $ Joseph F. Aieta Page 117 Printed 08/14/03

118 Time Value of Money Section 3.4 Page 118 Method () fincalc, generic financial calculator You should clear your calculator of all stored values between problems and enter new values, one at a time, followed by the appropriate key. In Example enter 60 into N, enter 3.9 as the annual rate in percentage form, enter 1 as the number of periods per year (which is also the number of payments per year) and enter in FV. In this future value problem you will need to check that 0 has been stored in PV since no money is in the account before the deposits begin. An x is displayed under PMT simply to identify that this is the unknown value which fincalc will calculate N I/Y P/Y PV PMT FV CPT PMT x Based upon the calculator s cash flow conventions, we interpret the result to signify that cash outflows of $ per month for five years are equivalent to a future cash inflow of $80,000 at the end of year five, assuming that money grows at the rate of 3.9% compounded monthly. Method (3) Spreadsheet Excel has an extensive collection of financial functions including PMT for periodic payment. The value (output) of the PMT function is calculated b y the function =PMT(rate, nper, pv, fv, type) and depends upon the parameters (also called arguments) that are entered in the indicated order, separated by commas. [Note: do not use commas as separators in numerical values, i.e. enter eighty thousand as rather than 80,000] In contrast to financial calculators that store the nominal rate in one location and the periods per year in another location, the rate in the Excel PMT function is the rate for one period. It can be easily calculated as the nominal rate divided by the number of compounding periods per year. In the current example, rate can be entered in Excel either as 3.9%/1 or as 0.039/1. If you mistakenly enter 3.9/1 without the % sign then Excel will calculate 3.5 percent instead of 0.35 percent. The number of periods, nper, is 60= (5)(1), the number of years times the number of periods per year. The first payment is made at the end of the first period. The example involves the future value of an annuity and no initial deposit has been made so the present value, pv, is zero. The future value of $80,000 is a cash outflow so future value, fv, is entered as Since this is an ordinary annuity with payments at the end of each period, enter 0 for type or leave that argument blank. You also have the option of letting Excel s function wizard guide you through the process as shown in Figures 3.4.6, 3.47, and Start by opening the function wizard by clicking on the f x icon. [Your Excel screen may look slightly different] Figures Joseph F. Aieta Page 118 Printed 08/14/03

119 Time Value of Money Section 3.4 Page 119 Click on the Financial function category and then scroll down until you find PMT. Click on the PMT function and Excel will open the dialog box for you to supply values for each of the stated parameters as shown in Figure Figure Enter 3.9%/1 for the interest rate per period, enter 5*1 or 60 for the number of periods, enter 0 for present value, enter for the future value, and enter 0 for the type. Figure The Excel function =PMT(0.039/1,5*1,0,80000,0) returns the value -1,09.71 to the nearest cent. The zero for the last parameter indicates the default assumption that payments are made at the end of the period (ordinary annuity). From the perspective of the depositor, a cash outflow of $1,09.71 at the end of each month for 60 months into an account paying 3.9% compounded monthly is equivalent to a cash inflow of $80,000 at the end of the 60 th month. For payments at the beginning of the period (annuity due), the value of the parameter for Type must be 1 and cannot be omitted. The formula =PMT (0.039/1, 60, 0, 80000, 1) for an annuity due would return -1,05.80, a slightly smaller cash outflow since the initial payment has had one additional period to earn interest. Joseph F. Aieta Page 119 Printed 08/14/03

120 Time Value of Money Section 3.4 Page 10 Example Let s return to a problem that was posed earlier. "How many years will it take for $000 to grow to $3000 if the nominal rate is 4% under annual compounding?" We will illustrate how this problem can be solved with a financial calculator and with Excel. fincalc: From the perspective of the depositor, $000 is a negative cash flow and $3000 is a positive cash flow. We enter -000 in PV, 3000 in FV, 4 as the percentage in I/Y, and 1 as the number of periods per year in P/Y. This problem does not involve periodic payments. It is a lump sum problem so we should make sure that zero is stored in PMT N I/Y P/Y PV PMT FV CP T N x Excel: We can use the financial function =NPER(rate, pmt, pv, fv, type). Since there are no periodic payments we enter 0 for pmt and ignore type. We can enter the formula directly into Excel as =NPER(0.04,0,-000,3000) or we can use the function wizard to guide us. In either case, Excel returns From the statement of the problem, we need to keep in mind that 4% is an annual rate and the time units are in years. Since times 1 is approximately 4, we can report years as approximately 10 years and 4 months. Example You are offered the alternative of $4,000 today or receiving a guaranteed $6,700 at the end of 10 years. If you accept the $4,000 today, your plan is to purchase a ten-year certificate of deposit at a bank that compounds interest daily. Which alternative would you prefer a) if the nominal interest rate on the bank certificate is 5%? b) if the nominal interest rate on the bank certificate is 6.5%? One possible approach is to compare amounts at the end of 10 years: fincalc a) assuming a 5% annual rate N I/Y P/Y PV PMT FV CPT FV 3, ,000 0 x 6, Excel: a) Use the financial function FV in Excel with parameters (rate, nper, pmt, pv, type). This problem does not involve periodic payments. It is a lump sum problem so we should make sure that zero is stored in PMT. The result from the Excel formula =FV(5%/365,3650,0,-4000) is also 6, $6, is less than the guaranteed $6,700 in ten years. If the nominal rate is 5% then the best option is to receive the guaranteed $6, at the end of 10 years. fincalc: b) assuming a 6.5% annual rate N I/Y P/Y PV PMT FV CPT FV 3, ,000 0 x 7,47.58 Excel: b) The output from Excel s future value function =FV (6.5%/365, 3650, 0,-4000) is also 7,47.58 $7,47.58 is more than the guaranteed $6,700 in ten years so the better option is to accept the $4, today and deposit it in the bank paying 6.5% per year compounded daily. Joseph F. Aieta Page 10 Printed 08/14/03

121 Time Value of Money Section 3.4 Page 11 Present Value approach comparing amounts in today s dollars: Making comparisons in today s dollars rather than imagining future dollars is a preferable way to solve these problems. We make the comparison in today s dollars by calculating the present value of $ in ten years under each rate. fincalc: a) If the interest rate is 5% compounded daily then the present value of $6,700 is $ which is more than the $ today. This means that the guaranteed $6,700 in 10 years is the better option. N I/Y P/Y PV PMT FV CPT PV 3, x Excel: a) Use the financial function PV in Excel with parameters (rate, nper, pmt, fv). Since there are no periodic payments we enter 0 for pmt and ignore type. =PV(5%/365,3650,0,6700) returns fincalc: b) If the assumed interest rate is 6.5% compounded daily then the present value of $6,700 is $3, which is less than $ making the preferred option to be the $4, today? N I/Y P/Y PV PMT FV CPT PV 3, x , Excel: b) Use the financial function PV in Excel =PV(6.5%/365,3650,0,6700) returns -3, You will encounter many situations in which the approach of discounting back to the present and comparing present values is the preferred approach. Joseph F. Aieta Page 11 Printed 08/14/03

122 Time Value Money Activities 4 Page 1 TVM Activities Open the file Annuities.xls and go to the worksheet FVORDANN (future value of an ordinary annuity). This worksheet shows a stream of twelve payments, each made at the end of the period. Set the payment size to $,000 and the interest rate per period to 5.00% as shown in Figure The first $,000 payment in Figure has grown to what amount at the end of period 5?. The first $,000 payment in Figure has grown to what amount at the end of period 1? 3. The fifth $,000 payment in Figure (i.e. the payment made at the end of period 5) has grown to what amount at the end of period 1? 4. If exactly five payments had been made, instead of twelve, then what would be the total value of the account at the end of the fifth period? 5. When Cathy was born, her parents decided to deposit $500 every 6 months thereafter for 15 years in an account earning 6 percent compounded semiannually. How much will be in the account after last deposit is made? 6. What amount should be deposited at the end of each 6-month period into a sinking fund earning 6 percent compounded semiannually if the amount in the fund after 15 years is to be $75,000? 7. In order to accumulate $15,000 for a down payment on a home 8 years from now, the Jones s are going to deposit a sum of money at the end of each 6-month period in an account earning 8 percent compounded semiannually. What should be the amount of each deposit? 8. Hitech Corporation has issued $10 million worth of bonds to obtain money now for expanding its corporate activities. To redeem the bonds, which fall due in 30 years, Hitech will transfer an amount to a reserve fund at the end of each month. How much should be transferred if the account earns 6 percent compounded monthly? Joseph F. Aieta Page 1 Printed 08/14/03

123 Time Value Money Activities 4 Page A college alumni club has decided to establish a scholarship fund with the first grants to be made five years from now. What should be placed in the fund at the end of each year in order to have $80,000 in 5 years if interest on the account is earned at the rate of 8 percent compounded annually? 10. Ann's company plans to replace some graphic design equipment in five years at an estimated cost of $80,000. Ann sets up a fund at a bank into which the company makes monthly payments for five years at the end of each month. Each payment earns 5.9 % compounded monthly. What is the amount of the monthly payment needed in order to meet this goal? 11. A self-employed consultant puts $10,000 at the end of each year into a Keogh retirement plan. This money is invested in a fund that pays 8% compounded annually. What will the consultant have in this account at the end of 0 years? 1. A 30-year-old self-employed person is planning a retirement program that will enable her to retire at age 55. How much should she set aside each month in an annuity paying 8%, compounded monthly, in order to have $500,000 in 5 years? 13. How much should you deposit each week into an annuity paying 6 % interest, compounded weekly if you want to have $50,000 in this account at the end of 5 years? 14. A family deposits $00 per month into an account that pays 6.00% interest per year, compounded monthly, for the next 10 years. For the next ten years they make no additional deposits. They make no withdrawals and the interest rate remains at 6% throughout. a) How much will be in the account at the end of the twentieth year years? b) If they had kept making the $00 deposits each month for all 0 years what would be in their account? 15. Paul receives a $4,000 bonus and puts it into an account that pays 9% compounded monthly. He also adds a portion of his monthly paycheck into the same account at the end of each month for five years. He plans to use this account to make a down payment on a house. If he makes payments of $300 a month into this account then what will he have for a down payment at the end of the 5th year? Joseph F. Aieta Page 13 Printed 08/14/03

124 Time Value Money Activities 4 Page A family wants to have $8000 in four years to take a vacation. What should they put into their vacation fund every month if it pays 4% compounded monthly? 17. Lum and Ann are both 39 and ½ years old and each of them puts $000 into a Roth IRA at the end of the year for 0 years. They invest in mutual funds that they expect to grow at the annual rate of 10%. What do they estimate will be in their combined IRA accounts when they turn 59 and ½? 18. In five years an electronics company must replace equipment that they expect will cost $80,000. What quarterly payment should they make in order to have the $80,000 if they can earn 8% compounded quarterly on their payments? 19. Annual costs at a certain private college are expected to be $40,000 eight years from now. The grandparents of a ten-year-old decide to create a college account into which they will make monthly payments. This account will earn 6% compounded monthly. How much will they need to put into this account so that it accumulates to $40,000 in eight years? 0. How much should you deposit each week into an annuity paying 8% interest, compounded weekly if you would like to have a balance of $100,000 after 10 years? Joseph F. Aieta Page 14 Printed 08/14/03

125 Time Value of Money Section 3.5 Page 15 Annuities: The present value of a stream of periodic payments Section 3.5 Example Imagine that you, along with hundreds of others, have applied to a philanthropic organization for a college scholarship. Suppose you are awarded $1,000 a year for four years starting one year from now. The organization does not offer you the option of receiving a single check now. What amount of money should they set aside in order to be certain that they can send you $1,000 each year for all four years? Clearly this will depend on what return that they can get on their money in a guaranteed investment. Suppose that they can earn 8% compounded annually. i = 0.08 $1000 $1000 $1000 $1000 PV is unknown Figure Figure 3.5. shows the present value of each payment made in Figure If we solve FV = PV(1 +i) n for PV we get PV = FV(1 +i ) -n. All values have been rounded to nearest cent. period 1 period period 3 period 4 PV of 1 st payment = 1000(1+.08) -1 = PV of nd payment = 1000(1+.08) - = PV of 3 rd payment = 1000(1+.08) -3 = PV of 4 th payment = 1000(1+.08) -4 = sum is 3,31.13 Figure 3.5. At this interest rate, the organization needs to put aside about $3300 in order to make the $1,000 payments at the end of each year for four years. The first payment has been earning interest for one year and has a present value of $95.93 The second payment has been earning interest for two years and has a present value of $ The third payment has been earning interest for three years and has a present value of $ The fourth payment has been earning interest for four years and has a present value of $ The present value of all four payments is $3,31.13 As you might expect from our earlier work with the future value of an ordinary annuity, we can derive a formula for the present value of an ordinary annuity. Joseph F. Aieta Page 15 Printed 08/14/03

126 Time Value of Money Section 3.5 Page 16 One way to derive the formula for the present value of an ordinary annuity is shown below. We start with the basic n n compound interest lump sum formula PV = FV ( 1+ i) and replace FV with the formula (1 + i) 1 Pmt from the i future value of an annuity. PV = FV ) n ( 1+ i = (1 + i) Pmt i n 1 (1 + i) n n n (1 + i) (1 + i) 1(1 + i) = Pmt i n this is equivalent to 0 (1 + i) (1 + i) Pmt i n 1 (1 + i) = Pmt i n. Present value of an ordinary annuity When regular payments of size Pmt are made at the end of each period for n periods and the interest rate is i per period then the formula 1 (1 + i) PV = Pmt i gives the present value of this stream of payments. We say that PV is the value of all of these payments in today s dollars (discounted to the present). n It is important to keep in mind that the use of this formula requires that the payments per year are the same as the number of compounding (or conversion) periods per year and that the rate per period must be entered in decimal form. Example 3.5. A corporation has borrowed money at an annual rate 18% per year and agrees to repay this debt with a stream of $8,000 payments at the end of each month for 1 months. To find the amount of the loan, each payment can be discounted or brought back in time to the present. Since the annual rate is 18%, the assumed interest rate per month is 18% divided by 1 or 1.5% per month. PV i = or 1.5% per month Figure We will see that this stream of payments is worth $87,60.04 in present day dollars. Joseph F. Aieta Page 16 Printed 08/14/03

127 Figure was produced with a worksheet in the file Annuities.xls. Time Value of Money Section 3.5 Page 17 Interest rate Periods per period $8,000 N i PMT 1.50% %i Present value < < < < < < < < < < < ,60.04 The sum of these 1 amounts is the present value of an ordinary annuity with 1 equal periodic payments of $8,000 made at the end of each period at a growth rate of 1.50% per period. Figure Checking the sum of these present day dollars with the formula 1 ( ) PV = = 87,60.04 PV 1 (1 + i) = Pmt i n we get Joseph F. Aieta Page 17 Printed 08/14/03

128 Time Value of Money Section 3.5 Page 18 Question and Answer Q1. What is the present value of the $8,000 payment that will be made at the end of month 5? A1. $ Q. What is the present value of the $8,000 payment that will be made at the end of period 1? A. $ Q3. What is the present value of just the first five payments? A = $38, This can also be obtained by formula as PV 1 ( ) = = As a check to A3, apply the formula 1 (1 + i) PV = PMT i n to question 3 "What is the present value of the first 5 1 ( ) five payments?" The sum of these terms has the value PV = 8000 = The reason for the difference of one cent is the fact that the five values in Q3 have each been rounded to the nearest cent. When these five rounded values are added we get $38, The more accurate value is $38, Excel can display two decimal place accuracy but it calculates results to many more significant digits. Example A car loan of $30,000 at an annual rate of 9% is to be repaid by making equal payments at the end of each month for 3 years. The table and graph in Figure shows a schedule for paying off this car loan with regular payments that include both interest due and partial repayments of the principal. The portion of each payment applied as interest is in the nd column and the portion applied to reduce the principal is in the 3 rd column. At the end of month 36 the unpaid balance is zero. This is called an amortization table. The Car Loan worksheet that produced Figure is included in the file Annuities.xls. Joseph F. Aieta Page 18 Printed 08/14/03

129 Amount Borrowed: $30,000 Time Value of Money Section 3.5 Page 19 Annual Interest Rate: 9.00% Monthly rate: 0.75% Number of Years: 3 Months: 36 Monthly Payment: $ Payment Paid in Paid on Unpaid Number Interest Principal Balance 30, , , , , , , , , ,38.76 year , , , , , , , , , , , , , , , , , , , year Interest Principal , , , , , , , , , , , , $4, year 3 Figure For this three-year car loan notice that a much larger portion of each month's payment is used to reduce the principal (unpaid balance) rather than paid to the bank as interest due. Joseph F. Aieta Page 19 Printed 08/14/03

130 Time Value of Money Section 3.5 Page 130 Since we are borrowing $30,000 today, PV is 30,000. Over a three-year period there are 36 monthly payment periods so n = 36. The annual interest rate is 9% but interest is compounded monthly so the value of i is.09/1 = (or 0.75% as a percentage). These tables can be produced efficiently with Excel and with less likelihood of making an error compared to using a calculator. If you are careful, it is possible to determine the value of Pmt in the equation calculator that has no built-in financial functions ( /1) = Pmt 0.09 /1 with a Since 0.09/1, the monthly rate as a decimal occurs twice (first in the numerator and again as a divisor), it would be wise to copy 0.09/1 into your calculator s storage. Keep in mind that it is never a good idea to round an interest rate and then to re-enter a rounded value into your calculator. All scientific calculators have a key to raise a number to a power (y x, x y, or ^ ), a key for changing the sign of a number ( ± or +/- ), and another key to calculate the reciprocal of a number, 1/x. Normally, intermediate results can be displayed by pressing the = key. The following sequence of steps uses the formula n 1 (1 + i) PV = PMT and solves for Pmt in a systematic way that avoids most common errors. i Steps (Use the = key to display intermediate results) Use the division key to get the value of 0.09/1 in your display Store this value into memory Use the addition key to add Press the power key Enter 36 and press the change sign key -36 Press = Press the change sign key Use the addition key to add Press the division key Recall the value of 0.09/1 that is stored in memory Press = To divide by a number is the same as multiplying by its reciprocal so press the reciprocal key Press the multiplication key and enter If the interest rate i were the unknown instead of Pmt then we could not use a similar approach to finding a solution since there is no closed-form algebraic expression to isolate i in the equation 1 (1 + i) PV = Pmt i instead is to use an Excel built-in financial function for approximating the unknown rate. n. What we will do Joseph F. Aieta Page 130 Printed 08/14/03

131 Time Value of Money Section 3.5 Page 131 Example Home Loan: A 9% loan for $150,000 is to be repaid over 30 years by making equal payments at the end of each month. Examine the first 1 months of the amortization table in Figure produced by Annuities.xls Amount Borrowed: $150,000 Annual Interest Rate: 9.00% Monthly Rate: 0.75% Number of Years: 30 Months: 360 Monthly Payment: $1,06.93 Payment Paid in Paid on Unpaid Number Interest Principal Balance 150, , , , , , , , , , , , , , , , , , , , , , , , ,975.0 $1,400 $1,00 $1,000 $800 $600 $400 $00 $0 The First year Interest Principal month Figure Questions and Answers Q1. In the first month, why is the amount of interest paid equal to $1,15 and the amount of the unpaid balance equal to $149,918.07? A1. The borrower had the use of the entire $150,000 for one month at the monthly interest rate of 0.09/1 = so interest due in the 1 st month is (150,000)(0.0075) = $1,15. The difference between the monthly payment and the interest due $ $1,15 = $81.93 and this amount can be applied to reduce the principal. The unpaid balance becomes $150, $81.93 = $149, Q. How are the amounts in the second month computed? A. During the nd month the borrower had the use of $149, at the monthly interest rate of 0.09/1 = so interest due in the nd month is (149,918.07)(0.0075) = $1, The amount that can be applied to reduce the principal is the monthly payment due minus the interest due which is $ $1,14.39 = $8.55. The new unpaid balance becomes $149, $8.55= $149,835.5 Joseph F. Aieta Page 131 Printed 08/14/03

132 Time Value of Money Section 3.5 Page 13 Q3. Is it possible to compute the unpaid balance after 1 months without doing all the computations for the previous months? A3. At the end of month 1 there are still 348 payments to be made. This stream of 348 remaining payments forms an annuity. If we use Pmt = $ (rounded to the nearest cent), i =0.09/1 = , n = (1)(0)=40, and the ( /1) present value formula then we would obtain PV = = $148, /1 If we don t round Pmt to the nearest cent and use six decimal places in the calculated vale Pmt $ then we would obtain ( /1) = $148, /1 Amortization tables often contain slight discrepancies due to rounding. Banks frequently make adjustments in the last payment. Q4. Prior to the closing date on the loan the interest rate drops to 8.50% from 9.00%. Over the 30 year term of the loan, approximately how much less will be paid in total interest under the lower rate than under the higher rate? A4. The monthly payment drops to $ with the lower rate of 8.50% so the total interest paid over the life of the loan is approximately ( )(360) - 150,000 = 415, ,000 = $65,13.8. At the 9% rate the total amount of interest paid over the life of the loan is (1,06.93)(360) - 150, = 434, ,000 = $84, The difference between these two cumulative interest amounts is $19,81.5. While these calculations do tell you something about the impact of a 0.5% rate change in a 30 year mortgage, the dollar amounts themselves, $84, and $65,13.8, do not give a truly meaningful measure of the cost of each loan. To be meaningful, the cost of a loan should be measured in today s dollars and reflect an assumed rate of inflation over the term of the loan. An inflation factor should be used to discount the monthly interest payments back to the present day. If we were to assume a 5% inflation rate over the life of the loan then a more accurate measure of the cost of the 9% loan in today s dollars would be approximately $77,700. The cost of the 8.5% loan in today s dollars would be approximately $67,594. This is a difference of about $10,000 in today s dollars. The calculations underlying these more realistic loan costs and their relationship to different inflation scenarios are beyond the content of this course. Joseph F. Aieta Page 13 Printed 08/14/03

133 Time Value of Money Section 3.5 Page 133 Interest rates since 197 for a 30-year conventional mortgage reached a peak in 1981 as shown in Figure How do mortgage rates today compare with 1981? AVERAGE ANNUAL INTEREST RATE ON 30-YEAR CONVENTIONAL MORTGAGES: m onth avg. rate percent year Figure average Jan. Feb. Mar Apr May Jun Jul Aug Sep Oct Nov Dec percentage Example Ann and Lum are considering a $00,000 thirty-year mortgage for a new house. They currently estimate that they can afford to make $1, monthly payments at the end of each month. If rates move up to 8.75% then what is the largest 30-year loan they can afford to repay with monthly payments of $1,500.00? Solution: Using the PV function =PV(8.75%/1,360,-1500,0,0) in Excel returns $190, to the nearest cent. Note that the monthly payment is a cash outflow and has been entered as the negative number Percent source average over 1 months in 1981 was % based upon data from 30-Year Conventional Mortgage Rate Average Contract Rate on Commitments for Fixed-Rate First Mortgages: Reprinted with from the Federal Home Loan Mortgage Corporation. Joseph F. Aieta Page 133 Printed 08/14/03

134 Time Value of Money Section 3.5 Page 134 If Ann and Lum want to know the nominal rate that would enable them to borrow $00,000 by making $ monthly payments then they would need to first approximate the value of i in the equation (1 + i) 00,000 = 1500 and then multiply by 1 since i is a monthly rate. The answer is approximately 8.3%. i You should be aware of the fact that this result was not obtained using basic algebraic techniques and ordinary calculators. It can be obtained with most financial calculators, with Excel using Solver, or with Excel s financial functions. The Excel function RATE has parameters (nper, pmt, pv, fv, type, guess). If we use Excel to solve for the interest rate in this annuity problem, we need to provide a guess for the monthly interest rate. Entering a guess is accomplished by scrolling down in the dialog box for the RATE function. We will enter 1% as our guess. The number of periods, 360, has scrolled off the top of the dialog box. Excel uses a numerical algorithm to improve on the initial guess and returns %. Since this is the monthly rate, we need to multiply by 1 to get the annual rate of 8.30% Figure The one line formula =1*RATE(360,-1500,00000,0,0,1%) also gives us 8.30%. Joseph F. Aieta Page 134 Printed 08/14/03

135 Time Value of Money Section 3.5 Page 135 Example Elisa was accepted to graduate school and was ready to sell her successful campus computer business. Her asking price was $8,000 but she had no immediate need for this much money. Blake was one of her undergraduate employees but he did not have access to $8,000. He made Elisa the following offer. He would pay her $400 per month for two years starting one month after graduation. Both Elisa and Blake had the opportunity to safely invest money at 6% compounded monthly for the next two years. Assuming that she has confidence in Blake, should Elisa accept Blake s offer? Solution: You should first complete a time- line diagram and then ask yourself What is the value today of monthly payments of $400 starting one month from now if the time value of money is 6% compounded monthly? Applying Excel s PV function with the parameter values =PV(6%/1,4,-400,0,0) we find that Excel returns $ The present value of these $400 payments discounted to the present is significantly higher than the asking price of $8,000 so Elisa decides to accept Blake s offer. Joseph F. Aieta Page 135 Printed 08/14/03

136 Time Value of Money Section 3.5 Page 136 Multi-step problems Example A couple puts $000 each into a Roth IRA at the end of each year until they both retire 0 years later. At the end of their first year of retirement, they start making the first of 10 withdrawals. After their tenth withdrawal, the amount in this retirement account is zero. Assume that money in this account can earn 8% per year compounded annually over the entire 30-year period, and then what will be the size of each of the 10 annual withdrawals? Solution: 0 ( ) 1 The build-up stage accumulates to a future value of 4000 = $183, at the time of the 0 th 0.08 payment. The $183, amount then becomes the present value of an annuity that can fund a stream of ten equal 10 n 1 (1 + i) 1 ( ) payments. Substitute in the formula PV = PMT and then solve $183, = Pmt for i 0.08 Pmt to get $7, This means that the savings plan will provide payments of $7,79.53 each year for ten years. Portions of a spreadsheet for solving this multi-step problem are shown in Figure with values and again in Figure displaying formulas. The first few rows correspond to the pay-in stage and the last few rows correspond to the pay-out stage. Notice how the negative of the future value of the pay-in stage is copied into the pay-out stage as a present value. Since this present value is entered as a negative number, Excel calculates the annual payment and reports it as a positive cash flow. Figure Figure Joseph F. Aieta Page 136 Printed 08/14/03

137 Time Value of Money Section 3.5 Page 137 Example Nancy has just celebrated her 40th birthday and plans to retire on her 65th birthday. She is planning a savings program to supplement her Social Security. She plans to make equal annual payments into this account starting on her 41st birthday and ending on her 65th birthday. This plan must provide her with an income of $10,000 per year for the period that starts on her 66th birthday and ends on her 85th birthday. Assume that her investments can grow at the annual rate of 8% for the entire 45-year period. What equal annual amounts must she deposit in this savings account to achieve her retirement income goal? We start by drawing a time line from age 40 to age 85. Figure This problem can also be broken down into two separate but related problems. The two separate problems are linked by the amount accumulated at the time of her 65th birthday. This amount will fund the stream of 0 annual payments of $10,000. The interest rate assumption during the pay-in stage and the pay-out stage is assumed to be 8% per year compounded annually. When you use Excel (as opposed to using the formulas in which all amounts are positive), you must the direction of cash flow in mind. STEP ONE (pay-out) We need to determine the amount of money (PV) that Nancy should have in her account at age 65 in order to be able to make the twenty periodic withdrawals of $10,000. Since this is the present value of an ordinary annuity we enter the values 0.08,0, and for the parameters rate, nper, and pmt and enter 0 for fv and type. The Excel function =PV(0.08,0,10000,0,0) returns STEP TWO (pay-in) We need to determine the payment size such that a stream of these periodic payments will build up in the future to produce the value that she will need to accumulate by age 65. In stage one Excel returned the For stage we need to change the sign of the $98, amount from negative in the pay-out stage to positive when it becomes the FV of the pay-in stage. Change the sign of the present value of the pay-out stage when it becomes the future value of the pay-in stage. The Excel formula =PMT(0.08,5,0, ,0) returns Interpretation: Nancy must make annual payments of $1,343 into this retirement account for 5 years to fund the 0 withdrawals of $10,000 that she will start making one year after her 65th birthday. Joseph F. Aieta Page 137 Printed 08/14/03

138 Time Value of Money Section 3.5 Page 138 Figure gives us a visual representation of Nancy s cash account balance each year over the 45-year period. These amounts are based upon an assumption of consistent 8% growth. Changing the assumption about the time value of money to 6% or to 4% would have a major impact on the cash account balance. 5 years of payments and 0 years of withdrawals $10,000 $100,000 $80,000 $60,000 $40,000 $0,000 $0 0 ($0,000) years Figure If the assumed interest rate for both phases drops from 8% to 6% then at age 65 Nancy would need to accumulate $114,699.0 to make withdrawals of $10,000 each at the end of each year for 0 years. From Excel, =PV(0.06,0,10000,0,0) is and =PMT(0.06,5,0, ,0) is If Nancy starts at the end of her 40 th year and wants to accumulate enough to withdraw $10,000 per year starting at age 66 then she would need to make 5 annual payments of $ assuming a growth rate of only 6%. The difference between the annual payments under the 8% growth rate versus the 6% growth rate is almost $750. Joseph F. Aieta Page 138 Printed 08/14/03

139 Time Value of Money Activities 5 Page 139 TVM Activities 5 1. Is the $30,000 in the car loan example, Figure 6.5.5, a future value or a present value?. In the car loan example, Figure 6.5.5, the interest due in the first month is equal to $5.00. Why? 3. In any amortization schedule the interest due in each successive month is less than the previous month. Why? 4. In Figures and 6.5.6, the height of all of the stacked bars remains constant. Why? 5. A company borrows $100,000 at 1 percent compounded semiannually. The debt is amortized by making equal payments at the end of each 6 months for 7 years. a) How many payments will there be? b) What is the interest rate per period? c) Find the amount of each semi-annual payment. d) How much of the first payment is interest due? e) By how much is the balance reduced in the first half year? f) How much of the second payment is interest due? g) By how much is the balance reduced in the second half-year? 6. Ann borrowed $16,000 at 9 % compounded monthly to buy a car. The debt is to be discharged by equal payments at the end of each month for 3 years. a) How many payments will there be? b) What is the interest rate per period? c) Find the monthly payment. d) How much of the first payment is for interest? e) By how much does it reduce the balance owed? 7. Marilyn will make 0 equal semiannual deposits to an account earning 8 percent compounded semiannually. Then, after the last deposit, she will use the amount in the account to establish an ordinary annuity earning 6 percent compounded annually which will provide her with $10,000 at the end of each year for 5 years. How much should Marilyn s semiannual deposit be? Joseph F. Aieta Page 139 Printed 08/14/03

140 Time Value of Money Activities 5 Page Assume that you have just purchased a $10,000 annuity that yields 9.5% interest per year, compounded weekly. a) How much can you withdraw every week from the annuity so that it will run out of money in exactly four years? b) What is the total amount of money that you will withdraw from the annuity? 9. Ann decides to borrow $0,000 for a new car. Payments are made at the end of each month. She can get a 36-month car loan at a nominal rate of 7.5% or a 48-month car loan at a nominal rate of 8.0%. a) What is the amount of the monthly payment on the 36-month loan? b) What is the amount of the monthly payment on the 48-month loan? c) What is the difference in the total interest paid over the life of each of the two loans? 10. Ann and Lum were planning to borrow $175,000 for a house but they experience some unexpected change in their combined incomes. They are now looking to buy a less expensive house. They can afford to take out a twenty-year $150,000 mortgage at a rate of 8.5%. a) What is the size of the monthly payment? b) If interest rates fell to 7.5% could they make the payments on $175,000 loan with a term of 5 years? 11. The Smiths have taken out a $80,000, 10-year home equity loan in order to pay expected tuition bills for their three teenagers. Their interest rate is currently 7.75 percent compounded monthly. a) What is the size of their monthly payment? b) How much more interest do they pay over the life of the loan if they spread it out over 15 years instead of 10? 1. The board of directors of a company has voted to establish a fund that will provide a retiring executive with an income of $0,000 at the end of each quarter for 10 years. The fund will be invested in company bonds, which earn 1 percent interest compounded quarterly. Find the amount that should be invested to provide these payments. 13. Paul and Cathy have $160,000 left to pay on their mortgage and are considering refinancing over 5, 0, or 15 years. a) If the 5 year rate is 7.5% then what is the size of the monthly payment and what is the total amount of interest that they will pay over the 5 year period? b) If the 0 year rate is 7.5% then what is the size of the monthly payment and what is the total amount of interest that they will pay over the 0 year period? c) If the 15 year rate is 7.00% then what is the size of the monthly payment and what is the total amount of interest that they will pay over the 15 year period? Joseph F. Aieta Page 140 Printed 08/14/03

141 Time Value of Money Activities 5 Page Aunt Agatha left $500,000 to her niece Ann as an inheritance. Ann decides to live aboard and travel. She invests the $500,000 in an account paying interest compounded monthly. She makes equal monthly withdrawals at the end of the month until the account is depleted after 5 years. a) If the nominal interest rate is 7% then what size monthly withdrawals can Ann make? b) If the nominal interest rate is 3.5% then what size monthly withdrawals can she make? 15. Cindy would like to take a 3-month world tour. One plan is to save the money by making monthly deposits into an account which pays 5 percent interest compounded monthly and take the tour in four years when the cost of the trip is estimated to be $7000. An alternative is to borrow $6000 and take the trip now. Her loan would be for four years at the rate of 8.5% compounded monthly. a) What would she need to contribute monthly for 4 years in order to have the $7000 four years from now? b) What would be the size of her monthly payments if she borrows $6000 and takes the trip now? c) What are some of the pro and cons of each plan? 16. Harry added a $4,500 Jacuzzi to his master bathroom. He could not get a conventional home improvement loan so he borrowed the money from a friend of a friend and signed a promissory note with a maturity date 3 years from the transaction date at an agreed upon 1 percent compounded monthly. The note contained a clause that the loan could be sold after one year. Harry learned at the end of the first year that the loan had indeed been sold to an international financial organization. Harry had made payments for one year and at that point the financial organization demanded payment in full of the unpaid balance. What was the amount demanded? Suppose that Harry had made no payments and was planning to pay off the entire amount due, $ , at the end of the third year. At the end of one year, the new owners of his note demanded payment in full but they were willing to accept the present value of the $ at the rate of 9% compounded monthly. What was the amount that they would accept? 17. Several states sell million-dollar lottery tickets every week. Rather than pay the winner in one lump sum, the state is only obligated to pay the winner $50,000 per year for 0 years. Suppose that the time value of money is 6% for the entire twenty-year period. a) An investment group offers you the present value of the 0 payments of $50,000 assuming a 6% annual interest rate in return for your lottery ticket. How much will you receive now and how does this amount compare with one million dollars? b) If you took the money received in part a and invested the lump sum in an account paying 6% compounded annually then in what year would you become a millionaire? c) If you received payments of $50,000 a year and invested each payment in an account paying 6% compounded annually then in what year would you become a millionaire? Joseph F. Aieta Page 141 Printed 08/14/03

142 Time Value of Money Activities 5 Page A 45 year-old person is planning to retire at the age of 65. How much should this person set aside each month into a retirement account in order to withdraw $000 per month for 5 years starting one month after retiring. Assume that the retirement account pays 6% compounded monthly over the entire 45-year period? 19. Mr. Grady was just about to take out a home mortgage of $100,000 for 0 years at the annual rate of 8%. The monthly payments would have been $ A competitive bank offered him a 30-year mortgage at 7.75% that has monthly payments of Mr. Grady went with the second bank because he assumed that a lower monthly payment and a lower interest rate would be a better bargain. Assuming that he could afford the higher payment, evaluate Mr. Grady s decision to go with the bank offering the 30 year loan at 7.75% 0. In Example 6 a couple put $000 each into a Roth IRA at the end of each year until they both retired 0 years later. At the end of their first year of retirement, they started making the first of ten withdrawals. After their tenth withdrawal the amount in this retirement account is zero. a) Suppose that money in this account earns 10% during the 0-year build-up stage but then the interest rate changes to 6% during the 10-year pay-out stage. What will be the size of each of the 10 annual withdrawals? b) Suppose that money in this account earns 6% during the 0-year build-up stage but then the interest rate changes to 10% during the 10-year payout stage. What will be the size of each of the 10 annual withdrawals? 1. During a 3-year period (beginning of end of 199), when his business was prospering, Jack was able to deposit $1,000 at the end of each month into an account earning an APR of 6 percent, compounded monthly. The business slackened and Jack could not continue the deposits. Moreover, the nominal interest rate on his accumulated deposits fell to 4 percent compounded quarterly and remained at this level for years (beginning of end of 1994). At the end of this period Jack decided to exhaust the account by withdrawing equal amounts at the end of every month for 3 years (beginning of end of 1997). Over the time of the withdrawals, the APR was 5 percent compounded monthly. How much did Jack withdraw each month? Joseph F. Aieta Page 14 Printed 08/14/03

143 Time Value of Money Activities 5 Page 143. a) Find an interactive calculator on a financial web site and verify that the monthly payment for a loan of $180,000 at an annual rate of 7.75% for 30 years is $1,89.54 a) Identify the source of the on-line calculator that you have found and use this calculator to verify your answers to any three homework problems. b) What other types of calculators are available at this web site? Sample: Calculate Loan Amount: 180,000 dollars Interest Rate: 7.75 % Term: 30 years Payment $ 1,89.54 Calculator Disclosure from a Website Calculators are provided as simple tools. Information obtained from use of calculators should be discussed with a qualified professional as this tool can only illustrate trends and may not be accurate with regard to your particular circumstances. The calculations are not guaranteed to be accurate. This information is provided by a 3rd party unaffiliated with Wingspan Bank. Wingspan Bank does not warrant or guarantee the reliability of this information E-LOAN, Inc. All rights reserved 3. Suppose that you have a balance of $1000 on your credit card and the current rate is 15.99% compounded monthly. You make only the minimum monthly payments of $50 at the end of each month. The table shows only the first three months. Beginning Interest Amount Principal Ending Balance Due Paid Paid Balance Month BB ID AP PP EB 1 $1, $13.33 $50.00 $36.68 $ $ $1.84 $50.00 $37.16 $ $96.16 $1.34 $50.00 $37.66 $ Assume that these payments remain at $50 per month. Verify with Excel that it will take until month 3 for the ending balance to be less than or equal to $ a) Show that the amount paid in interest before the debt is paid off is about $171. Joseph F. Aieta Page 143 Printed 08/14/03

144 Time Value of Money Activities 5 Page 144 b) Describe what happens in the first three months if the original credit card balance is $000 instead of $1000 and only the minimum monthly payment of $50 is made at the end of each month? Beginning Interest Amount Principal Ending Balance Due Paid Paid Balance Month BB ID Pmt PP EB 1 $, A company earns interest on the $15,000 that it deposits in a sinking fund at the end of each year for 10 years. This fund grows to $00,000 after the tenth payment is made. The unknown is the annual interest rate. a) What is the rate if payments are made at the end of the period? b) Look up annuities due in your calculator manual or in Excel s on-line help. What is the annual interest rate if payments are made at the beginning of the period? 5. Complete the following table to show the monthly amortization breakdown of equal monthly payments made for two years on a $ loan taken out at 9% interest. a) Determine the size of the monthly payment. b) Determine the unpaid balance at the end of month 6. Month Balance at the beginning of the month Paid in Interest Paid on the Principal Balance at the end of the month 6. Ann has been making monthly payments of $400 at the end of each month into her 401-K retirement plan. The mutual fund account has grown to $18,000 at the end of three years. What was her annual rate of return (APR) over this period? 7. Suppose you deposited $100 at the beginning of 000 into an account that pays 6% annual interest. You repeat this every year (deposit $100 at the beginning of each year) until the beginning of 030. How much money will you have accumulated at the end of 030? Joseph F. Aieta Page 144 Printed 08/14/03

145 Time Value of Money Activities 5 Page A family is thinking about taking out a $10,000 twenty-year mortgage for a new house. They can afford to make $1000 monthly payments at the end of each month. Long term rates are currently 8.5 %, compounded monthly, and are expected to fall. a) At what nominal rate (APR) will the family be able to afford the new house? b) If rates stay at 8.5% then what is the largest twenty-year loan the family can repay with monthly payments of $1000? 9. Suppose you own a campus franchise and you are going to sell the business. You ask for $5,000 cash today. You could invest your money at 8% per year compounded monthly. The prospective owner makes a counter offer of $150 at the end of each month for 36 months a) Should you accept the offer? Why or why not? b) What monthly payments by the buyer would give you an equivalent return? 30. How much would you be willing to pay today for an annuity which will generate payments of $100 per year for 1 years, if the first payment will be received 5 years from today? Your opportunity cost of capital is 15%. Round your answer to the nearest dollar. 31. A contractor can buy a piece of property for $30,0000 now or by making four annual payments of $9,000. If money can be invested at the rate of 9% compounded annually then which option is better? Justify your conclusion. 3. An investment of $40,000 provides payments of $4000 at the end of each year for 15 years. If interest is compounded annually then what is the annual interest rate? 33. In the middle of a four year period interest rates on two-year certificates of deposit declined from 6% compounded daily to 4%. Suppose that you invested $10,000 in the 6% CD and at the end of the second year "rolled" the entire amount into the 4% CD. 34. Santina purchased a computer for $1450 and charged it to her Visa card which gives her a 9.9% rate. On the same day, she does a balance transfer of $000 to Visa from another card that charged her 16.9%. Visa requires her to pay 13.9% on the transfer balance and treats the two balances as two separate accounts. How much interest will she owe after one month? 35. Ann has been making monthly payments of $350 at the end of each month into her 401-K retirement plan. The mutual fund account has grown to $15,000 at the end of three years. What was her annual rate of return over this period? Joseph F. Aieta Page 145 Printed 08/14/03

146 36. Paul and Cathy pay $700 each month on a 7.50% ten-year home equity loan. a) How much did they borrow? b) What is the unpaid balance at the end of the 5th year? Time Value of Money Activities 5 Page 146 c) Starting with month 61, they make an extra payment of $100 each month directly to the principal in order to reduce the balance of the loan. If they are able to keep this up until the loan has been repaid then in what month would the unpaid balance be below $700? 37. Bill plans to retire eight years from now. His retirement plan is partly built around an account that will provide him with monthly income for ten years. Bill wants to be able to withdraw $4,000 at the end of each month, starting at the end of his first month in retirement. Money in this account grows at the rate of 7% compounded monthly for the entire 18 year period. a) What is the size of the monthly payments that Bill must make for the next eight years in order to have a sufficient amount to fund this retirement plan? b) Repeat part a) with the additional information that Bill makes a $100,000 lump sum deposit into this account now. 38. For what nominal interest rate will $1,000 grow to $4,000 over a period of 18 years if interest is compounded 365 times per year? (Nearest hundredth of a percentage) 39. Amanda has been named an heir by her aunt. Her inheritance will be in the form of $00 each month for twenty years with the first payment to be made one month from now. If money is worth 5% compounded monthly, then what is this inheritance worth in today's dollars? 40. During a 3-year period when his business was prospering. Jack was able to deposit $1,000 at the end of each month in an account earning 1 percent, compounded monthly. The business slackened, and Jack could not continue the deposits. Moreover, the interest rate on his accumulated deposits fell to 8 percent compounded quarterly and remained at this level for 10 years, at which time Jack decided to exhaust the account in five years by withdrawing equal amounts every 6 months. The interest rate remained at 8 percent compounded semiannually over the time of the withdrawals Joseph F. Aieta Page 146 Printed 08/14/03

147 Functions and Trendlines Section 4.1 Page 147. Functions and Trend Lines: Fitting Curves to Data Section 4.1 Figure A graph may not reveal the exact values that are in tables or are obtained from formulas but it has the advantage of showing the general features of a function in a very efficient way. A graph can quickly reveal important features such as where the function may be increasing or decreasing, where it may reach a peak (local maximum) or a valley (local minimum), and whether or not it comes to a set of observed data points. Mathematical functions are often used to represent real phenomena. Studying the behavior of functions to represent quantitative situations can enhance understanding of business, economics, and the physical world. Today s technological tools have made it practical for us to implement standard function-fitting techniques using fast and accurate number-crunching algorithms. Mathematical modeling frequently requires fitting a curve to real world data. Typically, we enter the data into a spreadsheet (or graphing calculator) and then generate a scatter plot of the discrete points. We then inspect the scatter plot and search for a line or curve that is a good fit to the data. While an eyeball test can help us decide if a model comes close to the points, choosing a good fit to the data takes much more than guesswork. Extensive familiarity with the properties of mathematical functions and experience with the situational context contributes greatly to the art and a science of model building. Real-world data from which models are constructed are often less complete than we would like. Models generally reflect shortcomings in the amount and accuracy of the data that is collected by observation. Building a mathematical model of a business situation or physical phenomena usually requires focusing only on certain aspects of the given situation. A model always represents a simplification of real phenomenon. Details are excluded if they are considered to be irrelevant or too complex to be included in the model. Despite all of these limitations, a good mathematical model can capture enough of the essence of a business situation to provide useful insights into the structure and solution of many real-world problems and suggest conclusions that are helpful to management. Joseph F. Aieta page 147 8/14/003

148 Functions and Trendlines Section 4.1 Page 148 Fitting a Linear Function by the Method of Least Squares Every linear function is completely defined if the slope of the line and one point are known. Lines are relatively easy to work with both algebraically and graphically. If the graph of the plotted points suggests a linear pattern then a linear function is often selected to model the relationship between variables. A few examples in which linear functions appear in business, finance, or economics are linear depreciation, sales growth, federal and state taxes, energy consumption, consumer price changes, cost, revenue, and profit functions. We are already familiar with manufacturing situations in which the total cost of making a product can be separated into fixed cost and variable cost. Fixed cost is sometimes referred to as start-up cost or overhead, and may include rent, utilities, insurance, and equipment that is purchased or leased. These cost components are independent of the number of units made. Variable cost does depend directly on how many units are made. This component may include the cost of raw materials, labor, and other expenses that increase as the number of units increases. In the context of cost functions, we would interpret the linear formula C(X) = 6.00X to signify that the variable cost per unit is $6.00 and the fixed cost is $000. In the following example, a t-shirt business is looking at data they have collected. The company is trying to analyze how total cost is related to quantity. Example Encouraged by their success as part of a campus-based business, a small group of Babson students has been selling t- shirts during the summer months. They live near a resort community that is visited by thousands of tourists every summer. They have located a small building that they can rent for a very reasonable fee. Some equipment, such as a press and computers, has already been paid for but they do have to pay rent and other expenses that are independent of the number of t-shirts that they imprint. They order blank t-shirts and imprint them with different logos, slogans, and images. In early June they start buying and imprinting blank t-shirts in batches of 50 per week. By the start of the tourist season they are ready to begin selling. They plan to divide up the profits and do not include their labor as a cost. For the past two summers they have kept accurate records of their total expenses. Using Excel, they plot total cost versus the number of t-shirts that they buy and imprint. The data has followed a linear pattern in each of the past two summers. They are now looking at cost data from the third summer in order to make some projections about their fourth summer. The owners have recorded their total cost from ten weeks of imprinting t-shirts. Quantity is in the top row and the corresponding observed cost is in the second row of Figure X (units) Y (dollars) Figure 4.1. Joseph F. Aieta page 148 8/14/003

149 Figure shows a scatter plot of the data obtained with Excel s Chart Wizard. Functions and Trendlines Section 4.1 Page 149 $6,000 $5,000 $4,000 $3,000 $,000 $1,000 $ units Figure Based upon the scatter plot in Figure 4.1. and on previous history, the business decides that a linear function would be a reasonably good fit to the observed data. Obviously, the data points do not belong to a straight line but there is a general linear pattern and the owners would like to find the particular slope and intercept of that line that best fits the data. In their first summer, there were an unusual number of rainy days and they were able to sell only 380 t- shirts. They are curious to know the predicted y coordinate on the fitted line if x = 380. The following definition for the line of best fit is the one that they studied in their quantitative coursework at Babson. The line of best fit has the property that the sum of the squares of the vertical distances between the observed points (x, y) and the corresponding points on the fitted line (X, Y) is as small as possible. Suppose that Y = 6.00 X is suggested as a line that comes close to the observed data. For X = 100 the fitted Y value would be 6.00(100) = 600. For x = 100 the observed cost to the t-shirt business is actually $501. The vertical difference of the observed Y minus the fitted Y is called an error or a residual. In the case of x =100, the vertical error is = -99. If we find the vertical error for each of the other points, we see from Figure that some errors are positive and others are negative. If we were to simply add the errors then negative errors would cancel positive errors and the resulting sum would not tell us very much. X observed Y - fitted Y Figure Joseph F. Aieta page 149 8/14/003

150 Functions and Trendlines Section 4.1 Page 150 If we square the errors as shown in Figure 4.1.5, we will get a positive sum. The one exception would be a perfect fit (no errors) in which all of the observed points lie on the line. The sum of the squares of the errors would be zero in this unusual case. X (observed Y - fitted Y) Figure Figure shows the observed points and the fitted points on the line Y = 6X The vertical segments connect the observed points to the fitted points for each input 50,100, 150, up to 500 in steps of 50. $6,000 $5,000 $4,000 $3,000 $,000 $1,000 $ units Figure slope 6.00 intercept,000 number of observed t-shirts cost X Y fitted Y error error^ SSE = 1,307, Figure In Figure the fitted line is Y = 6.00 X and the Sum of Squares of the Errors, SSE, is 1,307,570. slope 6.50 intercept 1,800 number of observed t-shirts cost * X Y fitted Y error error^ SSE = 1,38, Figure In Figure the fitted line is Y = 6.50 X and the Sum of Squares of the Errors, SSE, is 1,38,545 Joseph F. Aieta page 150 8/14/003

151 Functions and Trendlines Section 4.1 Page 151 Open the Excel file FitBestLine.xls and find the worksheet sliders. Move the sliders and observe changes in SSE as you change the slope and vertical intercept.. $6,000 $5,000 $4,000 $3,000 $,000 $1,000 $ units Figure Go the worksheet Set up for Solver in the Excel workbook FitBestLine.xls In this worksheet the slope and intercept have been estimated to be 6.00 and 1900, respectively. The Sum of the Squares of the Errors, SSE, is currently 1,168,770. We are going to use the powerful Excel Add-in, called Solver, to minimize SSE by adjusting the two cells that contain the parameters for the slope and intercept. Figure a shows a blank dialog box and b shows the completed dialog box. Start by Invoking Solver from the Tools menu. This brings up a dialog box, named Solver Parameters Figure a Joseph F. Aieta page 151 8/14/003

152 Functions and Trendlines Section 4.1 Page 15 Activate Set Target Cell and click on F18, the cell containing the value of SSE Indicate that the target cell is to be minimized by clicking the Min button Select the two cells containing the slope and the intercept of the fitted line as the Changing Cells. In this case they are in C and C3 Click the Solve button Figure b Solver changes the worksheet values of the slope and intercept to and respectively and also displays the minimum SSE of 1,164, The line of best fit is reported as Y = 6.05 X where the slope (variable cost per unit) is rounded to the nearest cent and the intercept (fixed cost) is rounded to the nearest dollar. Independent of how many t-shirts have already been imprinted, each additional t-shirt contributes $6.05 to cost. We substitute 380 for X in the equation for this fitted line and estimate that the total cost of making 380 t-shirts is Y = 6.05 (380) = $4,166 to the nearest dollar. Excel has a built-in utility, called Add Trendline, for finding and displaying the line of best fit by the method of least squares. Joseph F. Aieta page 15 8/14/003

153 Function and Trendlines Section 4. Page 153 Excel s Trendline Section 4. Start by creating a scatter plot for the observed data below and then follow the steps described above Figure 4..1 through Figure X (units) Y (dollars) Activate the graph by clicking near the border of the plot area. Select the observed points in the scatter plot (do not be concerned if a few points do not appear to be selected). Figure 4..1 Figure 4.. Right-click and choose Add Trendline click Type and select Linear Click Options and select Display equation on chart Figure 4..3 Figure 4..4 By right-clicking on Format Trendline you can change certain features of the Trendline such as style, thickness (weight), and color. Joseph F. Aieta page 153 8/14/003

154 Function and Trendlines Section 4. Page 154 $6,000 y = 6.05x $5,000 $4,000 $3,000 $,000 $1,000 units $ Figure 4..5 Notice we have rounded the coefficient for the slope to 6.05 (to the nearest cent) and the coefficient for the intercept to 1867 (nearest dollar). These Trendline results agree with the coefficients found earlier by minimizing the sum of the squares of the errors using Solver shown below Figure [Note: If you should select Display R-squared value on chart, then you can think of R as a measure of the goodness of fit. An R value of 1 represents a perfect fit with no error. The closer R is to 1, the stronger the linear relationship between the observed values of x and y. You will learn much more about R when you study linear regression in statistics.] Non-Linear Curve Fitting Excel s Trendline can also be used to fit other types of functions to X-Y data by the method of least squares. Let s imagine that the cost to make some product is given by the ten ordered pairs (units, total cost) in Figure First, create a scatter plot for the data points. It should become clear that total cost is increasing at a steeper rate as q increases. The business that produces this product decides to fit a quadratic function rather than a linear function as the more appropriate mathematical model. The company is interested in the cost of making 18 units. units total cost Figure 4..6 Activate the graph by clicking near the border of the plot area as shown in Figure Next, select the observed points in the scatter plot as shown in Figure 4..8 Joseph F. Aieta page 154 8/14/003

155 Function and Trendlines Section 4. Page 155 Figure 4..7 Right-click and choose Add Trendline, Click Type,and select Polynomial of Order Figure 4..8 Click Options and select Display equation on chart Figure 4..9 Figure Use Format Trendline to change the thickness and/or color of the Trendline as shown in Figure $7,000 $6,000 $5,000 $4,000 $3,000 $,000 $1,000 $0 y = 0.108x x units Figure Joseph F. Aieta page 155 8/14/003

156 Function and Trendlines Section 4. Page 156 For x = 18, the fitted value of Y is (0.108)(18 ) (5.97)(18) = $3, Based upon the quadratic model produced by Excel s Trendline, the company would estimate a cost of about $3,337 to make 18 units. It is important to recognize that we are always talking about partial fits. This means that only a portion of the curve is a good fit to the actual data which is given over a finite interval. We should not expect a model to provide useful information by estimating very far beyond the interval that contains the observed data. For example, we may have some measure of confidence in projecting a cost near $6664 when x =75 but it would be very unwise for us to project a cost for x = 375. As you become familiar with several types of models and different kinds of problem situations, you will soon realize that is not always obvious which type of function to select when you are working with real world data. Always start with a scatter plot. It will often be necessary to adjust and align the scales on the horizontal and vertical axis. For example if your sales data is given in Table 4..1 then the horizontal and vertical scales should probably be adjusted as shown in Table before you try to fit a line or a curve to the data. year Sales in $ 4,560,000 5,385,000 5,600,400 5,84,000 6,057,000 6,69,400 6,58,000 6,944,000 7,78,000 Figure 4..1 years since Sales in millions of $ Figure You should always examine the scatter plot carefully and ask yourself which type of function might give a smooth fit. If the pattern of the data changes dramatically then you should be prepared to break up the data into pieces as we did for piecewise linear functions. Guidelines and Observations Computational software always has limitations and often has flaws. Excel is no exception. Even though Excel may be the most popular tool for mathematical modeling in business, many people who use this computer software are not well informed about its flaws, particularly in statistics. In curve fitting the coefficients of an equation reported by Trendline may be incorrect due to Excel's choice of algorithm or because the coefficients of an equation are not reported with sufficient accuracy. Models can be very sensitive to small changes in the coefficients. One way to check an equation generated by Trendline is to plot the points that satisfy the equation and visually check whether or not the resulting graph is a reasonably good fit to the observed data. Finding a good fit for the data is an important start but it is equally important to provide an interpretation that explains the connection between the generated results (algebraic or numeric) and the context of the original problem situation. You will be expected to use Excel s Trendline in the exercises. Keep in mind that this introductory material contains only the very basic concepts and mechanics of curve fitting. To pursue this topic in further depth requires knowledge of statistics and higher-level mathematics. Joseph F. Aieta page 156 8/14/003

157 Functions and Trendlines Activities 1 Page 157 Functions and Trendlines Activities1 1. a) Given the five series of points below, with x- values 0, 1,, 3, 4, 5, 6, 7, 8, and 9 in the first column, determine which of the corresponding scatter plots could be approximated reasonably close by a linear model. b) For each series that can be closely approximated by a linear function, write the equation of the line of best fit based upon the method of least squares. c) For those data series that do follow a linear pattern what would be the output value (to the nearest tenth) that corresponds to an input value of x = 10? series 5 series 4 series 3 series series 1 x The price of movie tickets since 1970 in a particular town is given in the table below. a) Develop a linear model for the price of a ticket from 1970 to 000. b) Use the model to find the ticket price in 199. This is called interpolating. c) Use the model to predict the ticket price in 00. This is called extrapolating beyond the observed data. What assumptions are you making when you make predictions for years that come before or after the observed data? actual adjusted ticket prices year year in dollars Joseph F. Aieta page 157 8/14/003

158 3. First Class Postal Rates in June 000. Source: Adapted from the United States Postal Service rate calculator for Letters, Flats, and Parcels. Functions and Trendlines Activities 1 Page 158 [Note: pieces less than 3 ounces in cells with gray background are at a different rate than pieces over 3 ounces] Bring the following first class postage data into Excel. Weights are in ounces and postage is in dollars. Weight in Weight in ounces not single ounces not exceeding piece exceeding pre-sorted a) Without drawing a scatter plot, how could you tell that the data in the first two columns would be a perfect fit to a straight line? b) Write a formula for the postage of a single piece in terms of its weight up to 13 ounces. c) Find the line of best fit for first class postage of pre-sorted mail up to 13 ounces. d) (optional) Update the table beyond June 000 and comment on your observations. 4. a ) Use Trendline to find the best quadratic fit to the monthly profit data on round trip flights between two cities. ticket price in $ profit in thousands of $ 3,080 3,50 3,760 3,80 3,700 3,380 b) If this trend continues as ticket prices increase then at what ticket price would the airline begin to see a loss? Joseph F. Aieta page 158 8/14/003

159 5. Numbers of AIDS cases in taken from the 1991 Statistical Abstract. Functions and Trendlines Activities 1 Page 159 Coded Number of Year Cases year X Y , , , , , , , ,633 a) Fit a linear or a quadratic function to the cumulative number of reported AIDS cases and justify your choice. b) What number would your model predict in 1991 if you extrapolated from the observed data? 6. U. S. Spending on Defense from 1985 to 1989 adjusted Expenditures year year in billions of $ In the second half of the 1980s the U.S. spent increasing amounts on national defense. Bring the data from the table into Excel and create a scatter plot for the data points in columns and 3. a) Find the equation of the line of best fit (regression line) and plot this Trendline on your graph. b) In plain English, explain the meaning of the coefficient of x that is shown in the Trendline equation. c) If the actual military expenditures were 99.3 billion in 1990 and 71.6 billion in 1995 how do these numbers compare with the trend established in the late 1980s? d) What historical events changed the trend of military spending that occurred in the late 1980s? e) (optional) Update the table for the years 1990 through 000 and comment on your observations. Joseph F. Aieta page 159 8/14/003

160 7. The population of California in the nineteen nineties is given in the table below. Source: population Year Coded Year in millions a) Develop a linear model for California s population over this period. Functions and Trendlines Activities 1 Page 160 b) In what years were the observed data points above the line of best fit and in what years were the observed data points below the line of best fit? c) Predict the population in 000. What assumptions are you making when you make predictions for years that come before or after the observed data? d) Go to the Web and choose a state whose first letter is the first letter in your first or last name. Perform an analysis similar to a) c) for California. If a good linear fit is not possible then explain why or state what function family you think would better fit the points. 8. X = Options Ordered Y = Delivery time in days a) Let X = the number of different options ordered by customers on a popular new car and let Y = the number of days required for delivery by a particular car dealership. Develop a linear model for fitted Y values. b) Give a plain English, in-context interpretation for the slope of this fitted line. c) If this line could be extrapolated back to X =0 how would you interpret the fitted Y value in words? d) Interpolate for X = 10 and for X = 15 Joseph F. Aieta page 160 8/14/003

161 Functions and Trendlines Activities 1 Page Swimmers World published an article in 199 that related a swimmer s age to the average time of swimming 100 meters for swimmers of that age. Age in years Time in seconds to swim 100 m a) How can you tell just by looking at the table that a linear model would not be a good choice? b) Create a scatter plot and then select a Trendline type that fits the data very closely. c) Interpolate to estimate the average time for a 5-year-old swimmer to swim 100 meters. 10. The Consumer Price Index (CPI) for all items from 1980 to 1998 is given in the table below in constant 1984 dollars. Source: Year Adjusted CPI a) Find the line of best fit for the CPI of all items b) Go to the Web and the slope of the fitted line for just medical care in The U.S. between 1980 and Compare the medical care slope with the slope in part a). Joseph F. Aieta page 161 8/14/003

162 Functions and Trendlines Activities 1 Page A retailer of a certain building product observes the following relationship between the units sold q and the corresponding price, p, in dollars. Economists use the convention of putting price on the vertical axis and quantity demanded on the horizontal axis. The graph that relates them is called the price-demand function. Quantity Price per unit $19.00 $14.00 $1.50 $1.00 $11.50 $11.00 $8.00 The table indicates that a demand level of 800 units corresponds to a unit price of $1.00. Another way of looking at this price-demand data point is that consumers are willing to buy 800 units at a price of $1.00. a) Find the line of best fit for the price-demand function. b) Use the fitted line and the fact that revenue is price times quantity [R(q) = p*q] to create the revenue function for values of q starting at q = 100 up to q =1500 in steps of 100. c) Give an estimate (to the nearest hundred units) of the quantity where revenue is a maximum. d) At a unit price of $7.90 the supplier is willing to supply 400 items and at a price of $9.00 the supplier is willing to supply 1700 units. Calculate the price-supply function based upon these two points. Find the equilibrium point, i.e. the point where the supply and demand functions intersect. e) g) Suppose a supplier has a variable cost of $6.80 and a fixed cost of $700. e) At a price of $7.90, where is the break-even point? f) What is the quantity demanded at this price of $7.90? g) Calculate the supplier s profit at this $7.90 price level. 1. A credit card research alert contained a title stating "Credit Card Spending Experiencing Slowest Holiday Growth In Decade". Adjusted year * Holiday credit card debt Year X in billions of $ $ $ $ $ $ $ $ $ $ $ $ $11.40 * (VISA, MasterCard, American Express and Discover) Source: a) Create a new column in the table that tests the claim made in the above title. b) Show graphically that a quadratic fit to the data is slightly better than a linear fit. c) Give a predicted value for 00 using the linear approximation. d) Give a predicted value for 00 using the quadratic approximation. Joseph F. Aieta page 16 8/14/003

163 Functions and Trendlines Activities 1 Page The table below shows the dividends paid to Pepsico shareholders from 1989 to 000. Fit an appropriate function Y =f(x) to this data where X is the number of years since 1989 and ranges from 0 to 11. Adjusted year Year X Dividends per share Source: a) What is the predicted value from the fitted line for X = 6 (1995) to the nearest cent and what was the actual amount? b) What is the predicted value from the fitted line for X = 11 (000) to the nearest cent and what was the actual amount? c) Calculate the predicted value for 001 and compare it with the actual dividend for that year. 14. Adjusted year Number of females female % of labor force population of age Year X (thousands) 16 and over , , , , , , , The data in the table are related to women in the civilian labor force at the end of each decade from 1940 to 000. Create two scatter plots with the adjusted years since 1940 on the horizontal axis as follows: The first plot is for the number of women in the labor force versus the adjusted years since 1940 and the second plot is for the female percentage of labor force population of age 16 and over. Fit a quadratic function N(X) where X is the number of years since 1940, to the scatter plot of number data and fit a quadratic function P(X) to the scatter plot of the percentage data. Use these fitted curves to answer a) - d) a) For 1975 estimate the number of women in the labor force (to the nearest thousand) and the female % of the labor force population aged 16 and over (to the nearest tenth of a percentage point) b) Repeat part a) for the year c) Use your quadratic model to predict the number of women in the labor force in 001 (to the nearest thousand). Find the actual value and comment on the difference. Joseph F. Aieta page 163 8/14/003

164 Functions and Trendlines Activities 1 Page Age 65 and older Year X (in millions) * * * The table shows the number of Americans of age 65 and older (in millions) at the end of each decade 1980, 1990, and 000. Source for : The table also contains *projections for the next three decades (based upon sophisticated statistical analysis of demographical and actuarial data). Projections:TIME magazine, August 30, 1999,page.5 a) Find a quadratic function that fits the data in the table for X = 0 to 50. b) When is this function increasing most rapidly? c) Would you be confident about using the resulting quadratic fit to estimate the number of Americans age 65 and older in 1985 and 1995? Explain. d) Find out how many Americans were actually 65 or older in 1985 and 1995 (nearest million) 16. DVD and VCR unit sales The sale of DVD players for home entertainment was launched in March In that initial year, the average price of a DVD player was $481. In 001 the average unit price had dropped to somewhere between $150 and $00. For the same period, 1997 through 001, the average price of a VCR tape deck dropped from $157 to about $70. adj. year (in thousands) (in thousands) Year X DVD players VCR decks a) Use the link to find and copy the data that completes the tables for VCR and DVD player sales from 1997 to 001. (Note: slightly different DVD player sales data can be found at ) b) On the same coordinate axes, create scatter plots of the DVD and VCR data obtained from the Consumer Electronics Association and determine if either data set can be closely approximated by a fitted line. c) If you can find a good linear fit to the VCR data and/or the DVD data then give a concise interpretation for the slope of that fitted line. Determine what the line would predict for 00 in thousands of units. d) As a consumer, explain what you think is happening from an historical and technological perspective. e) (optional) Describe some other relatively recent phenomena in electronics that also has a pattern of rapid growth. Identify Websites that provide supportive data. Joseph F. Aieta page 164 8/14/003

165 Functions and Trendlines Activities 1 Page U.S. Sales of Compact Discs (net after returns). Adj. year (millions) Year X CDs Fit a line to a scatter plot of sales of compact discs (in millions) versus the adjusted years where X = 0 is the number of years since a) In what years are actual sales below this fitted line? b) What are the predicted CD sales for the year 001 (X = 11) according to the fitted line? c) Investigate the Website and determine whether or not the unit sales of Singles also exhibit a linear trend from 1990 to 1999? Explain. Compiled from two sources: Recording Industry Association of America Year End Statistics Recording Industry Association of America. 000 Yearend Statistics Joseph F. Aieta page 165 8/14/003

166 18. The data show the enrollments in private colleges in 1980, 1985, and Source: Functions and Trendlines Activities 1 Page 166 Adj. year enrollments Year X in thousands a) What linear model is the best fit to this data? b) What number of students does the model predict for 000 c) What did the National Center for Educational Statistics (NCES) predict for 000? d) Check the Web and determine if public college data also suggest a linear trend over the same time period. Joseph F. Aieta page 166 8/14/003

167 Functions and Trendlines Section 4.3 Page 167 Exponential, Logistic, and Cubic Models Section 4.3 Linear functions have the property that the same amount is added to the output variable for each unit change in the input variable. This ratio: change in output divided by change in input is the constant slope of the line. For linear cost functions we refer to this constant rate of change as the variable cost per unit. For linear deprecation we think of the slope as the constant amount by which the output variable decreases for each unit of time (usually years). If a set of data follows a linear pattern we can describe trends in a very efficient way. For example, suppose the input variable is the number of square feet in a Boston apartment on the waterfront and the output variable is the monthly rental fee. If the relationship between these variables is approximately linear, we might report the trend in the following way: For each additional 100 square feet we can expect to pay an additional $75 per month. Describing trends in models that are not linear is not so straightforward. Certain types of non-linear functions have special characteristics that we will learn to recognize. Quadratic and cubic polynomials may be more familiar to you than other types of non-linear function. Next to linear functions, however, the most important mathematical models for analyzing trends in business, financial, and economic data are based upon exponential functions. Exponential functions do possess a type of constant change but it is not additive. For exponential functions we observe a pattern of multiplication of the output variable by a constant factor for each unit change in the input variable. This is reflected in applications such as the growth of money under compound interest. The future value of a savings account at the end of a time period is a constant times the amount at the beginning of that time period. For the growth of money or for population growth, this constant is referred to as the growth factor. If an exponential model is given by f(t)= a b t where t is some unit of time, the growth factor, or base, is b. We determine this constant percentage change by calculating (b -1)100%. Suppose that an initial investment of $1000 earns 6% interest compounded annually. The future value of this investment t years later is given by FV(t) = 1000(1.06) t. The constant annual percentage change in the balance of this account is (1.06 1)100% = (0.06)100% = 6%. Each year, the balance of the compound interest account increases by 6% of the previous year. The balance of an account paying 6% simple interest would increase each year by a constant amount of $60. Exponential growth involves a constant percentage change while linear growth involves a constant amount of change. In linear depreciation, linear functions can depict a constant decrease in amount per unit of time. Similarly, exponential functions can represent a constant percentage decrease over time. We use the general term growth just as we used the term profit. We recognize that growth does not always correspond to an increase. When profit is negative we generally talk about a loss rather than a negative profit. Examples that can be modeled by decreasing exponential functions include depreciation in automobiles, decline in populations, and radioactive decay in a material that loses its level of radioactivity over time. Suppose an automobile purchased for $3,000 loses 0% of its value each year. At the end of the year its value is = 0.80 times its value at the beginning of that year or $3,000 (0.80) = $5,600. After four years the undepreciated value (residual value) of the car would be 3,000(0.80)(0.80)(0.80)(0.80) = 3,000(0.80) 4 which is about $13,100. Suppose the population of a town followed a trend given by the function P(t) =(8,000) t where t is the number years after 1990 then the constant percentage change is ( ) 100% = (-0.036) 100*%= -3.6%. The population of this town is decreasing at the rate of 3.6% per year. Growth curves are commonly given in the form f(t)= a e r t where a is the value at time = zero and r is some constant. The model for continuous compounding, P e r t, is a well known function in this exponential family. The future value in t years of $1000 invested at 6% compounded continuously is given by the formula FV = 1000 e 0.06 t. Joseph F. Aieta page 167 8/14/003

168 x = x e = From the exponent law m n ( m ) n we know that r t ( r ) t written as the base b for a growth function in the form f(t) = a b t. Functions and Trendlines Section 4.3 Page 168 e and since If a population model is given by P(t) = 70.6 e t then P(t) = ( ) t since 0.018t t t e = e = In this example, the base is our growth factor. ( ) r e is some constant, it can be Any exponential function given by f(t)= a e r t can be re-expressed in another base b as f(t) = a b t where b = e r Example Consider the population of Mexico from 198 to When we fit an exponential function to the points in the scatter plot using Excel s Trendline, we obtain y = 70.6 e 0.018x Since we know that e t = ( e ) t = t, we can express this population function as P(t) = ( ) t. CIEMAX-WEFA Mexican Economic Outlook Service (October 1994) Population Growth: Mexico Population of Mexico Adjusted population Exponential Year Year (in millions) Fitted Y in miliions y = 70.6e 0.018x years since 198 Figure The constant percentage change during this period is ( )100% =(.01837)100% = 1.837%. Mexico s population grew at a rate of approximately 1.84% per year. Over this same time period the population of the United States grew at a rate that was about half the growth rate of Mexico. Q1. The actual population of Mexico in 1995 was reported to be approximately 94 million. What population was projected by the exponential function P(t) = ( ) t A1. This Trendline would project the value ( ) 13 = million or about 5 million under the reported number. Joseph F. Aieta page 168 8/14/003

169 Example 4.3. An exponential curve that is increasing at a decreasing rate Functions and Trendlines Section 4.3 Page 169 The graph on the right is a graphical representation of the temperature of a glass of cold iced tea that was left on a kitchen counter on a warm afternoon. Fahrenheit Assume that the input variable is hours and the output variable is degrees Fahrenheit. The horizontal line above the curve is not part of the function but simply indicates its limiting value. In this physical example, assume that the iced tea at time zero is 45 degrees and that the limiting value of room temperature is 75 degrees Fahrenheit. Label the figure. Figure 4.3. hours Example Complete the Exercise Suppose we start with a cup of hot coffee at 10 degrees Fahrenheit and let it cool to room temperature. Draw a rough sketch of what you think the graph of temperature versus time would look like. Indicate a scale for each axis assuming the unit for time is hours and the unit on the vertical axis is degrees Fahrenheit. Fahrenheit Figure Joseph F. Aieta page 169 8/14/003

170 Functions and Trendlines Section 4.3 Page 170 Example The scale on the vertical axis in Figure ranges from 0 to 50% indicating that portion of the target audience that responds to an advertising campaign by buying the product within t days. We can estimate from the graph that 5% of the target audience responded sometime near 100 days. From the graph in Figure 4.3.4, we see that the response function increases rapidly and then levels off beyond 300 days. In the first 100 days, less than 30% of the target audience of 00,000 radio listeners had responded. The original plan for the length of the campaign is to be under one year. Each positive response contributes an average of $0 to company profit (before advertising costs are taken into account). Since the marketing campaign has fixed and variable costs, the marketing team plans to examine cost and revenue projections to make a recommendation on how long to run the advertising campaign. Figure Questions.01 a) If the formula for this response function is r( t) = 0.40(1 e 0 t ) then what percentage of the target audience has responded within 80 days?.01 b) If the formula for this response function is r( t) = 0.40(1 e 0 t ) then what percentage of the target audience has responded within 150 days? c) If fixed costs are $400,000 and variable costs are $1,800 per day then what are total costs for a 00-day advertising campaign? d) Based upon this model, estimate the anticipated cumulative contribution to profit before advertising (nearest thousand dollars) after 00 days? Answers about 5% about 33% $760,000 $1,455,000 e) Is net profit increasing or decreasing between 100 days and 140 days? increasing f) Is net profit increasing or decreasing between 40 days and 300 days? decreasing g) Build a table with columns for the time in days, the portion of the market responding within t days, the cumulative contribution to profit before advertising as a function of t, the total cost of advertising as a function of t, and the net profit after advertising costs are subtracted. Use the column headings t, pmr(t), R(t), C(t), and P(t). Start at t= 0 and add increments of 5 days. Insert t as the name for this range of cells. Show the formulas entered in B, C, D, and E Use this table to check the answers to the questions above and show the graph from t = 0 to t =350 for profit after advertising. Joseph F. Aieta page 170 8/14/003

171 Functions and Trendlines Section 4.3 Page 171 PROFIT AS A FUNCTION IF TIME Figure Logistic growth is the term given to phenomena in which something grows slowly at first, then grows more rapidly, and finally levels off to a limiting value. Consider unit sales in a city over a period of a few months after some new audio CD or video DVD is released by a popular artist or the growth of an animal population after the animals have been placed into a restricted breeding area such as a wildlife reserve. Graphs depicting logistic growth are often described as having S shaped behavior as seen in Figure In the CD or DVD sales example, sales may grow quickly after the word gets around and then will level off when enthusiasm wanes and most of the potential consumers have had ample opportunity to purchase the product. In the wildlife reserve example the population can be expected to grow rapidly (assuming that there are no predators) and then level out when the population approaches the maximum number that the environment can support. Other examples of logistic growth as functions of time include the number of people infected by a contagious biological virus or the number of computers on a campus infected by a computer virus Figure A limiting value for the spread of a virus that affects humans may be approached when an entire community has been exposed. The limiting value of the computer virus may be related to the speed of dissemination of alerts and the availability of anti virus software. Joseph F. Aieta page 171 8/14/003

172 Functions and Trendlines Section 4.3 Page 17 There are several mathematical formulas for functions that represent logistic growth. Two widely used logistic formulas are A (1) Nt () = t C + b () () L Nt= Bt 1 + Ae Time t is often the independent variable and N(t) is the corresponding number at time t. In formula (1) the exponential base is represented by the parameter b. In formula () the number e is the base of the natural logarithm function, which is why formula () is sometimes called the natural logistic curve. Example Suppose that a deer population grows at the rate of introduced Nt= () t where t is the number of years since the herd was Questions Answers (to the nearest whole number) a) How many deer were initially on the reserve? 40 b) Calculate N(5) and explain what it means in this context. Expect 140 deer 5 years after the herd is introduced c) Based upon this model, how many deer will be present after 10 years? 371 d) Based upon this model, how many deer will be present after 15 years? 639 e) What is the average increase in deer per year from the 5 th year to the 10 th year? 46/year f) What is the average increase in deer per year from the 10 th year to the 15 th year? 54/year g) What is the percentage increase from the 5 th year to the 10 th year? 166% h) What is the percentage increase from the 10 th year to the 15 th year? 7% i) Will the population ever reach 850? No Joseph F. Aieta page 17 8/14/003

173 Functions and Trendlines Section 4.3 Page 173 Example A few students in the same dormitory show symptoms of a contagious virus at 1 noon. Suppose that the number of students in the dormitory infected t hours later is given by the formula 80 Nt () = 0.7t 1 0e where t = 0 corresponds to 1:00 noon Number of Students Infected hours since noon Figure Questions Answers a) What is the limiting number of students infected? 80 b) About how long does it take to reach the limiting value? About 1 hours c) At what time is the rate of infection the steepest? About 5:00 PM d) How does the shape of the curve up to the hour of steepest increase compare with the shape of the curve after the hour of steepest increase? Increasing rapidly until a point near 5 hours when the rate of increase starts declining and eventually levels off Joseph F. Aieta page 173 8/14/003

174 Functions and Trendlines Section 4.3 Page 174 Example A marketing consultant for a large established automobile dealership has been asked to examine financial data for all showroom locations across the state. The primary objective is to determine the best allocation of advertising dollars that are allocated to a particular popular model. In some years the company spends as little as $5,000 on advertising and offers discounts as a way to bring customers into the showroom. In most years the advertising budget is several hundred thousand dollars. Figure 4.3.8, the first spreadsheet on the left, is from a year in which the company spent $300,000 on advertising. In that year the base unit sales not attributed to advertising was 1,00 with an additional 14 units attributed to advertising. The dealer pays $16,000 for each unit and the average selling price for this model is $0,000. Fixed costs for the dealer are $1,50,000. The sales force earns a 5% commission on all sales. Profit before taxes is total revenue minus the sum of the following costs: what the dealer paid for the cars, the total commissions paid, advertising costs, and the remaining cost of overhead. Figure Figure Figure was obtained by fitting a logistic curve through data points in which the input variable is advertising dollars (in thousands) and the output variable is additional sales. It is obvious from this table that spending more than 350 thousand dollars on advertising will not contribute to the bottom line. In fact, excessive advertising can substantially lower profits. L L A slight variation on formula () Nt () = is Nt () = where A = e (-k) Bt 1 + Ae + ( Bt k) 1 e + If we let x be units of advertising (in thousands of dollars) then the logistic curve f(x) for incremental sales due to advertising can be given by the formula ( Bx k ) $ in thousands advert_cost $300.0 base_sales 1,800 adv_sales 194 tot_sales 1,994 selling_price $0.0 unit_cost $16.0 cmsn_rate 5.00% tot_revenue $39,880 cogs $31,904 gross_margin $7,976 tot_cmsns $1,994 fixed_costs $1,50 pre-tax_profit $4,43 advertising additional costs in sales from thousands $ advertising x L f( x) = 1 + e + where L = 00, B = 0.05, and k = -4. The graph of the logistic curve f(x) for additional sales from advertising and for profit P(x) as a function of advertising dollars are shown below in Figure and Figure Joseph F. Aieta page 174 8/14/003

175 Functions and Trendlines Section 4.3 Page Additional sales from advertising 4,500 Total Profit vs. Advertising Costs unit sales above base thousands of advertising $ Figure profit in thousands of $ 4,450 4,400 4,350 4,300 4,50 4,00 4,150 thousands of advertising $ 4, Figure Develop a spreadsheet for additional sales due to advertising and for pre-tax profit as a function of advertising dollars. Verify that f(300) = 194 and that P(300) = 4,43. Joseph F. Aieta page 175 8/14/003

176 Example Functions and Trendlines Section 4.3 Page 176 Open the file Cable TV-subscribers.xls containing the following data and carefully examine each of the worksheets year thousands 4,500 9,800 16,000 1,000 5,000 30,000 3,000 37,500 41,100 44,000 year thousands 47,500 50,000 51,000 53,000 55,000 57,000 58,000 60,80 64,050 64,170 Source: Warren Publishing, Inc. Television & Cable Fact book, annual Title: Cable Television-Systems and Subscribers: 1970 to 1998 Source: Statistical Abstract of the United States: NA, 1999 Since Excel presently has no logistic Trendline, we have used Excel s Solver in Cable TV-subscribers.xls to find a good fit to the data by the least squares method. According to the least squares criterion, a good logistic fit to these data points is given by () where L = 68.3, A = 1.71, B = Nt L = 1 + Ae Bt Cable Television Subscribers: 1970 to 1998 S(t) =L/(1+A*EXP(-B*t) L A B Subscribers in millions (4.500 represents 4,500,000) As of adjusted Observed Projected Jan. 1 year Subscribers Subscribers errors Cable Television Subscribers t (in millions) (in millions) errors squared years since SSE = in thousands Figure The year of most rapid increase (inflection point) occurs somewhere between 1985 and 1990 Joseph F. Aieta page 176 8/14/003

177 Functions and Trendlines Section 4.3 Page 177 Example The scatter plot of GDP (Gross Domestic Product) from Statistical Abstracts 1993 suggests a point of inflection somewhere near 1983 (year 15). U.S. GDP from 1969 to 1991 Actual Year: Coded yr X GDP in billions of $ Actual Year: Coded yr X GDP in billions of $ Q1. What are the coefficients of that cubic polynomial that best fits the data given in the table above? A1. y = -0.17x x x Q. To the nearest billion, what would be the projected GDP for 199 if this trend had continued? A. 899 billion billins of $ years since 1969 Figure Plotting any function of a single variable General purpose utilities are widely available to make it easy for you to investigate functions of a single variable given the rule for y = f(x) as a closed form expression. You already know how to start with a blank worksheet and produce a scatter plot from a table. The utility Plot&Trace.xls allows you to stay within the spreadsheet environment and efficiently determine the general shape of a graph and its important characteristics. You can zoom in and out and identify key points on the graph, such as x-intercepts (roots) and points where the function reaches relative (local) high points or low points. We will illustrate how to use this utility 3 with the function f(x) = x 0x + 10 Start by typing =x^3-0*x+10 <Enter> into the cell directly below the macro button new_f. A macro is simply a sequence of pre-recorded Excel commands. These steps are executed automatically when the macro is run. Figure Joseph F. Aieta page 177 8/14/003

178 Functions and Trendlines Section 4.3 Page 178 When you press the macro button new_f, the y-values and the graph are automatically updated. If you want a different view, you can adjust the starting value for x (xmin) and our choice for the x increment. By changing the values of xmin and xincr you can perform the equivalent of zooming in or zooming out on portions of the graph. You have indirect control over the maximum value of x. If you choose xmin to be - and the increment to be 0.15 then the x coordinate of the fiftieth point will be 4.5. x max y min y max guess bounds target f(x) = x^3-0*x , , a f(a) Figure If Excel should display the message ERROR IN FORMULA then press OK and make your correction. The most common omissions are the = sign at the beginning of the function definition and the asterisk symbol for multiplication. Observe that a local minimum was found at the point whose coordinates are approximately (,58, -4.47). The target cell initially contained the value of y corresponding to the initial guess, which was x =. We then used Solver to minimize y (our target) by adjusting the guess. We added a constraint that looks for an optimal value to the left of 4. This cubic function also has a local maximum but it is not visible in Figure since it is to the left of x = -. In order to see this relative maximum, we can change xmin to -4 and enter an initial guess of -3. To complete the setup for Solver we also change the bounds to -. We then ask Solver to find the coordinates of the point to the left of x = - that corresponds to a local maximum. The coordinates of the local maximum turn out to be approximately (-,58, 44.47). Joseph F. Aieta page 178 8/14/003

179 Functions and Trendlines Activities Page 179 Functions and Trendlines Activities 1. Use the file GenQuad.xls to check your answers to a) f) Quadratics have coefficients a, b, and c, with the leading coefficient a not equal to zero. If a = 0 then the graph would be a line and not a parabola). a) f) Match the quadratic function described on the left with the equation y = ax + bx + c on the right. Write the Equation Properties Equation a) Vertex at (1,4.5) and one root at (.5,0) i y=1x -4x+3 b) Vertex at (,-1) and one root at (3,0) ii y=-x +4x+.5 c) Roots at (-1,0), (5,0) and y intercept at (0,5) iii y=x +4x d) Vertex at (,0) and y intercept at (0,-4) iv y=-1x +4x e) Contains the points (0,0) and (1,3) v y=-1x +4x+5 f) Roots at (-,0), (0,0) and vertex at (-1,-)) vi y=-1x +4x-4 g) If a relatively small change in a coefficient can have a significant impact on the graph of a function, we say that the function is highly sensitive to that coefficient. Quadratic functions are most sensitive to which coefficient? h) Which coefficient a, b, or c, is the most directly related to shifting the graph up or down? i) Which coefficient is the most directly related to shifting the graph to the left or to the right? j) Explain how the type of curvature of a quadratic function, concave up or concave down, can be predicted from its coefficients. Give an example. k) Explain how x-intercepts, generally referred to as the roots of the function, can be determined from the coefficients a, b, and c. Give an example. Joseph F. Aieta page 179 8/14/003

180 Functions and Trendlines Activities Page 180. Use the file Gen_Cubic.xls to answer a) f) a) f) Cubic polynomials have coefficients a, b, c, and d with the leading coefficient a not equal to zero. Match: the cubic function described on the left with the equation y = ax 3 + bx + cx + d on the right. Write the Equation Properties Cubics a) Low point at x = -0.56, high point at x = 3.56 i y=-x 3 +0 b) Never increases. ii y=x x +3x+ c) Low point at x = 0.5, high point at x = 1 iii y=-1x 3 +6 x -5x+5 d) Never decreases. iv y=1x 3-3 x +3x-1 e) Low point at x = 0.47, high point at x = 3.58 v y=-x 3 +9 x +5x-15 f) k) Answer always, sometimes, or never Circle one f) A cubic function has exactly one local maximum. always sometimes never g) A cubic function has exactly one local minimum. always sometimes never h) A cubic function has exactly one root. always sometimes never i) A cubic function increases from left to right. always sometimes never j) A cubic function decreases from left to right. always sometimes never k) A cubic function has at least one point where it is neither increasing nor decreasing. always sometimes never 3. The general cubic polynomial is y = ax 3 + bx + cx + d a) The graph of a cubic is most sensitive to which coefficient? b) The graph of a cubic is shifted up or down by changing which coefficient? c) d) Without consulting any notes, write definitions for each of the following in your own words: c) high point d) low point Joseph F. Aieta page 180 8/14/003

181 Functions and Trendlines Activities Page Make the appropriate entries and adjustments to the utility Plot&Trace.xls in order to obtain a graph and determine the characteristics for each of the following polynomials. 4. f ( x) = -.1 x x =-.1*x^ *x with Excel for 0 < x < 9 Let xmin = 0 and xincr = 0.5 Display two decimal places Sketch the graph. a) f() =? b) f(5 ) =? c) For what value of x is f(x) a maximum and what is that maximum? d) For what value of x is f(x) a minimum and what is that minimum? Joseph F. Aieta page 181 8/14/003

182 Functions and Trendlines Activities Page f ( x) = x x + x + 5 for -3 < x < 3 Display two decimal places 3 Sketch the graph. Evaluate y to the nearest hundredth a) f(-3) =? b) f(3) =? c) For what value(s) of x is f(x) =0 d) For what value of x is f(x) a local (relative) maximum and what is the maximum? e) For what value of x is f(x) a local (relative) minimum and what is the minimum? Joseph F. Aieta page 18 8/14/003

183 Functions and Trendlines Activities Page Cellular Phone Subscribers are given in the table for the years 1985 to 000 based on data form the Cellular Telecommunications Industry Association Source: adjusted Total Year t Subscribers a) What does a scatter plot of this data suggest about the plausibility of a good linear fit? b) Compare a quadratic fit to an exponential fit. Which curve is a better fit for the early years and which curve is a better fit for the later years? c) If you were to use the quadratic fit, what would you predict for 001? d) If you were to use the exponential fit, what would you predict for 001? e) Find the actual number of cell phone subscribers in 001 according to the Semi-Annual Wireless Industry Survey of the Cellular Telecommunications & Internet Association 7. a) Fit a cubic function to the data that relates the amounts spent on pollution control in the U. S. from 1983 to Source: Statistical Abstracts, Year Adjusted year Amount in billions of $ b) What do you observe about the curvature of the data that suggests a possible cubic fit? c) (optional) Obtain updated data for and explain some of the risks associated with extrapolating from a fitted cubic function. Joseph F. Aieta page 183 8/14/003

184 8. Recall that e x, the exponential function base e, is EXP(x) in Excel. Functions and Trendlines Activities Page 184 a) Open the file GenExpntl.xls and move the sliders to produce the graph of the concave response function 0.05t r( t) = 0.40(1 e ) b) Open the file GenLogistic.xls and move the sliders to create the graph of the logistic response function 0.40 r( t) = 0. 05t 1+ 5e c) Compare the two graphs. In what ways are the two curves similar and it what ways are they different? For the response functions rt () L(1 e Kt L = ) and rt () = time t is the input variable. Kt 1 + Ae d) e) Assume that the parameters L, A and K are all positive numbers. Describe the significance of each parameter in terms of the impact on the graphs when these values increase or decrease. d) Concave response function: L K e) Logistic response function: L K A Joseph F. Aieta page 184 8/14/003

185 Functions and Trendlines Activities Page Population growth State of Massachusetts Est. Change 4/90-7/98 Number 8 yr % Annzd % 130,707.17% 0.6% Apr-90 6,016,45 Jul-98 6,147,13 Source: The estimated population in the state of Massachusetts in July 1, 1998 compared with April 1,1990 shows an increase of 130,707. This was a percentage increase of.17% over 8.5 years which is equivalent to an annualized percentage change of 0.6%. Massachusetts Institute for Social and Economic Research Box University of Massachusetts Amherst a) Show how the annualized rate is obtained by solving the equation (1 + r) 8.5 = b) If this annual percentage increase continued for five years then what would be the projected population of the state in 003? From 1990 to 1998 the regions in the state of Massachusetts with the highest population growth rates were Cape Cod and the islands of Martha s Vineyard and Nantucket. c) Bring the following table into Excel. Nantucket year adjusted year Census Bureau estimates (April 1) ,01 (July 1) ,036 (July 1) ,10 (July 1) ,08 (July 1) ,419 (July 1) ,793 (July 1) ,043 (July 1) ,301 (July 1) ,489 (July 1) ,844 d) Determine the line that best fits the Nantucket data and give an interpretation of the slope. e) Suppose that this linear trend were to continue for five years, project the population of Nantucket in 003. f) Determine the percentage growth rate and the annualized percentage growth rate from April 1, 1990 to July g) If this annual percentage increase were to continue for five years then what would be the projected population of Nantucket in five years? h) If this annual percentage increase were to continue indefinitely then how long after 1990 would it take for Nantucket to double in population? i) Explain some of the pros and cons of using these models for extrapolation to the year 00. Joseph F. Aieta page 185 8/14/003

186 Functions and Trendlines Activities Page Automobile Depreciation Age of the car Residual Year in years Value $16, $13, $11, $9, $7, $6, $5, $4, $3, $, $,00 The table gives the market value (to the nearest hundred dollars) from 1989 up to 1999 of a particular Toyota sports model in good condition. It had a sale price of $16,100 in 1989 and has less than 100,000 miles in a) How can you determine that a linear fit would not be appropriate just by examining the table? x b) Find an exponential model to fit the data and then write its equation in the form y = a b where x is the years since 1989, b is the growth or decay factor, and a is a constant. Since the residual value is declining with time, find the value of b as a decimal in the interval 0.00 < b < c) Linear depreciation may be applicable to evaluating equipment for corporate tax purposes but it does not generally apply to calculating the value of a used car. Explain why. 11. Health Club Revenues: Cubic Fit adjusted year in millions dates 1993 = 0 of dollars 01/01/ /01/ /01/ /01/ /01/ /01/ /01/ /01/ /01/ /01/ The table reports annual health club dues twice a year starting in January 1993 and ending in July a) Fit a cubic curve to this data and use it to project the revenues in January of b) Explain why neither a linear nor quadratic model would be good choices for fitting this data. c) Discuss the limitations of the cubic fit relative to the purpose of extrapolation to the year Joseph F. Aieta page 186 8/14/003

187 1. Health Club Revenues: Logistic Fit An examination of the health club data in the previous problem would support the following curve as a reasonably good logistic fit to the revenue function Rt () = t e Functions and Trendlines Activities Page 187 adjusted year in millions 1993 = 0 of dollars 01/01/ /01/ /01/ /01/ /01/ /01/ /01/ /01/ /01/ /01/ a) Show the scatter plot of the points and the graph of this logistic function R(t) on the same axes. b) Find the exact coefficients for the best logistic fit by the least squares criterion. c) If you have to choose between fitting a cubic or fitting a logistic curve, then which is your choice and why? d) If you choose the logistic models then what are you accepting as a limiting value for revenue? 13. Wal-Mart Sales from 1976 to 1991 The first billion dollar year in sales, $1.48 billion, was in Source: Observed Year Adjusted Sales x y , , , , , , , , , , , , ,887 a) Of the function families (linear, quadratic, cubic, exponential, or logistic) decide which type of curve is your choice to fit the Wal-Mart sales data from 1976 to The input variable, x, is years since 1976 and y is sales in millions. Explain the reasoning that supports your choice. b) Use your fitted curve to project sales for 199 and c) (optional) Obtain Wal-Mart sales data from 199 to 1998 and comment on how well the model in part a) predicted sales. Joseph F. Aieta page 187 8/14/003

188 Measuring Change Section 5.1 Page 188 Derivative Concepts: Marginal Analysis Section 5.1 A derivative is a measure of the rate at which one quantity changes with respect to another quantity. The temperature at :00 PM is increasing at the rate of 3 degrees per hour At a level of production of 00 units, cost is increasing at the rate of $10 per unit At the end of year five the residual value of an automobile was decreasing at the rate of $100 per year If you have driven an automobile then you have some firsthand experience with derivative concepts. Suppose distance traveled is measured in miles and time is measured in hours. When the speedometer reads 40 we say that we are traveling at a speed of 40 miles per hour at that instant. If we travel a total distance of 40 miles in one hour we say that our average speed is 40 miles per hour. Example Distance versus time The vertical axes in the scatter plots below indicate miles from the start of the trip. The horizontal axis represents a one hour time interval separated into 6 minute intervals (every one tenth of an hour). Figure reveals a stop at the 6 th mile. The rates after travel is resumed are greater than the rates before the stop. miles Distance as a function of time hours Figure miles Cruise Control hours Figure 5.1. If cruise control had been set at 40 mph for the entire trip with no stops then our scatter plot would look more like Figure Joseph F. Aieta page 188 8/14/003

189 Measuring Change Section 5.1 Page 189 Figure 5.1.3a shows distances recorded in one minute intervals. Corresponding points on the graph have been connected by a smooth curve in Figure 5.1.3b. 45 miles traveled versus hours 45 miles traveled versus hours miles 5 0 miles hours hours Figure a Figure b Example 5.1. Temperature versus time The table and graph in Figure show temperatures that were recorded every hour from 6:00 AM to 6:00 PM on a spring day in New England. The time scale has been adjusted so that noon = 0, six hours before noon at 6:00 AM corresponds to -6, and six hours after noon corresponds to positive 6. Time Fahrenheit t adjusted t F 6:00 AM :00 AM :00 AM :00 AM :00 AM :00 AM :00 noon :00 PM 1 65 :00 PM 65 3:00 PM :00 PM :00 PM :00 PM 6 47 degrees fahrenheit Temperature from 6 AM to 6 PM AM noon = 0 6 PM Figure The table and graph show a general warming trend between 6:00 AM and 1:00 PM followed by a cooling trend from about :00 PM to 6:00 PM. If we examine some two-hour intervals we see that the average rate of change in temperature between 6:00 AM and 8:00 AM was = = ( 6) On average, the temperature was rising at a rate of 10 degrees per hour. Joseph F. Aieta page 189 8/14/003

190 Measuring Change Section 5.1 Page 190 Over the two-hour time interval between :00 PM and 4:00 PM the average rate of change in temperature was = = 3. On average, the temperature was falling at a rate of 3 degrees per hour over this interval. If we 4 examine average rates of change over shorter and shorter time intervals, we can approximate how fast temperatures were rising or falling at particular points in time. Figure 5.1.5a Figure 5.1.5b One advantage of having a smooth curve, as shown in Figure 5.1.5b, is that it enables us to explore the concept of instantaneous rate of change. We will give a graphical description of this concept before we do any calculations. Associated with each point on this smooth curve is a line called a tangent line. The slope of a curve at a point is the slope of the tangent line to the curve at that point. The slope of the curve at a point tells us how rapidly the curve is increasing or decreasing at that point. At some point on the graph pictured in Figure the curve is neither increasing nor decreasing. This occurs at some point between 1:00 and :00 PM. The tangent line at that point is a horizontal line (slope = 0) Joseph F. Aieta page 190 8/14/003

191 Measuring Change Section 5.1 Page 191 If we focus our attention on the point (-3, 51) in Figure we might ask What is the instantaneous rate of change of temperature at 9:00 AM? Imagine that we have placed a transparent ruler with its edge through the point in question. We will use our geometric intuition to draw an initial candidate for a tangent line and then step back and determine if we can make improvements. Temperature 6 AM to 6 PM degrees fahrenheit noon = 0 Figure Line 1 crosses the vertical axis near 75 and is actually too steep to be the tangent line at (-3, 51). Line crosses the vertical axis near 63 and is not steep enough to be the tangent line at (-3, 51). We carefully draw line 3 with our transparent ruler so that it just kisses the curve at (-3, 51). Line 3, between lines 1 and, is our best approximation to the tangent line at the point (-3, 51). Since the point (4, 100) appears to be another point on line 3 the slope of this line is = = 7. 4 ( 3) 7 In the context of temperature and time, we say that the temperature is increasing at the rate of approximately 7 degrees per hour at 9:00 AM. Joseph F. Aieta page 191 8/14/003

192 Measuring Change Section 5.1 Page 19 The Excel file EstimateSlopes.xls contains the graphs of functions whose graphs are also smooth curves. Each worksheet includes a collection of lines, with a portion of the tangent line at certain points. You are asked to estimate the slope of the corresponding tangent line at each of these points. Question: Estimate the slope of the tangent line at each of these five points: a) Point A (-, -3) b) Point B (0, -1) c) Point C (1, -1.5) d) Point D (, -1) e) Point E (3, 0.5) Solution: a) -3 b) -1 c) 0 d) 1 e) #REF! 10 y h *x^+*x+3 =0.5*x^-x Figure What is special about the point (1, -1.5)? Estimating the Slope of a Tangent Line At this stage we have an idea of what a tangent line is geometrically but we do not have a way of calculating the numerical slope of a tangent line from the equation that defines the function. Going back to Figures 5.1 and and the automobile trip that covered 40 miles in one hour, shorter time intervals of six minutes or one minute gave us a more refined idea of the relationship between distance traveled as a function of time. If we knew precisely how far the car had traveled in an interval of one or two seconds then we could get a very good approximation to the number displayed on the speedometer. If we knew the formula for a function, y = f(x), and could calculate average rates of change over smaller and smaller intervals then we could get reasonably good approximations to the instantaneous rate of change at a point on the curve. We will use Excel to perform the calculations for a numerical approach that sneaks up on the slope of the tangent line. Joseph F. Aieta page 19 8/14/003

193 Numerical Method 1 The Difference Quotient Consider the simplest quadratic function, f(x) = x We sneak up on the slope of the tangent line to f(x) = x at the point P(1,1) by choosing a nearby point Q (, 4) one horizontal unit to the right of P on the curve. As we move the point Q closer and closer to point P, the secant lines from P to Q get closer and closer to the tangent line at point P as point Q moves closer to P. The slope of the secant line through (1, 1) and (, 4) is the average rate of change of the function over the interval from x = 1 to x =. You could think of the points on the positive side of the x axis as the number of seconds after the instant x = 0 and points on the positive y axis as distances in meters. Measuring Change Section 5.1 Page Q (, 4) P 1, Figure In the one-second interval between x = 1 and x = the change in y is 4-1 = 3 meters so the slope is = = The average rate of change in this one second interval is 3 meters per second. Next, we consider what is happening between x = 1 and x = 1.1, one-tenth of a second after x = 1, by calculating the average rate of change over the interval [1, 1.1]. The horizontal difference h between the points is now h = 0.1. The second secant line, with h=0.1 is through the points (1,1) and (1.1, 1.1 ) Slope = = = We continue to calculate the average rate of change of the function over intervals [1, 1+h] as we reduce the size of h by a factor of ten. The third secant line, with h=0.01 is through the points (1,1) and (1.01,1.01 ) Slope = = = For each value of h, where h 0, the slope of the corresponding secant line is ( 1+ h) 1 h For a general function, y = f(x), picture a point P on the curve with coordinates (a, f(a)). Another point Q on the curve y = f(x) is h units away measured by its horizontal distance from point P. The first coordinate of point Q is a + h so its second coordinate must be f(a+h). The slope of the secant line through points P (a, f(a)) and Q (a+h, f(a+h)) is the ratio f ( a + h) f ( a) = a + h a f ( a + h) f ( a). This expression is referred to as the h difference quotient or the forward difference quotient. Joseph F. Aieta page 193 8/14/003

194 Measuring Change Section 5.1 Page 194 In the spreadsheet in Figure 5.1.9a column F shows successive values of f ( a + h) f ( a) for the function f(x) = x h at the point (1, 1 ). Figure 5.1.9a shows numerical values and Figure 5.1.9b shows the Excel formulas that produced those values. The formula in cell B3 under the heading f(a) instructs Excel to square the value in cell A3 to its immediate left. The values of h in column C are decreasing by a factor of ten from 1 to Column D contains the x coordinate of the point that is h units to the right of x = a. When the formula in column B, with its relative cell reference, is copied to column E, Excel automatically calculates the square of the value of the cell to its immediate left. Finally the difference quotient is calculated in column F. The table shows a series of five secant lines. It could easily be extended down to display several more rows containing successively smaller values of h including , , , etc. Values Formulas Figure 5.1.9a Figure 5.1.9b It seems reasonable that the slope of the tangent line to f(x) = x at the point (1, 1) could be. By changing the value of a in column A we can produce a similar series of secant slopes for other points on the curve. Use this numeric approach shown in Figure to estimate the slope of the tangent line to f(x) = x at the points below. Point (-, 4) (-1, 1) (, 4) Estimated Slope of the tangent line For each point, it appears that the slope of the tangent line is two times the x coordinate of the point. What would this pattern suggest for the slope of the tangent line at the points below? , , 4 4, 4 3 3, 9 Point (-3, 9) (0, 0) (3, 9) Estimated Slope of the tangent line -1, 1 1 1, Figure Joseph F. Aieta page 194 8/14/003

195 Measuring Change Section 5.1 Page 195 Algebraic Definition of the Slope of a Tangent line: The First Derivative The difference quotient f ( a + h) f ( a) gives the average rate of change of a function over a closed interval h [a, a+h ]. The numerator f ( a + h) f ( a) is the change in the output variable of the function and the denominator h is the change (a + h) a in the input variable of the function. As we have seen, the slope of the tangent line at a point (a, f(a)) can be closely approximated by the difference quotient f ( a + h) f ( a) for small positive values of h. The exact slope of a tangent line at an arbitrary point on a function y = f(x) h is found by a limit process. We are now ready for the formal definition of the first derivative as the slope of the tangent line at a point (a, f(a)) The slope of the tangent line to f(x) at the point x = a, is f (a) = lim h 0 f ( a + h) h f ( a) Let s examine the terminology and notation of this definition. The term lim is an abbreviation for limit. This limit is called the derivative of f(x) at x = a, abbreviated as f (a) and pronounced f prime of a. We read the expression f ( a + h) f ( a) lim as the limit of the difference quotient as h approaches zero. The limit concept was first suggested h 0 h in the tables of Figure We observed how numbers in one column seemed to converge to a certain value as the values in the column containing h became closer and closer to zero. While the properties of limits can be formally investigated, an intuitive grasp of the concept will be sufficient for business applications. Finding the derivative of a function f(x) is called differentiating. As we shall see, differentiating a function f(x) produces a new function f (x) called the slope function or the derived function. From our investigation of the squaring function f(x) = x pattern the slopes of tangent lines to selected points suggest a clear x f (x) or slope at x Conjecture: For the squaring function f(x) = x the slope at any value of x is found by multiplying the x coordinate by. In other words, the slope of the tangent line to the quadratic x at an arbitrary point (x, f(x)) is given by x. In derivative notation, if f(x) = x then f ' (x) = x. We can justify this claim algebraically by simplifying the difference quotient until we can determine the limit as h approaches zero. ( x + h) x h in consecutive steps Joseph F. Aieta page 195 8/14/003

196 Measuring Change Section 5.1 Page 196 ( x + h) lim h 0 h x x + xh + h lim h 0 h xh + h lim h 0 h (x + h) h lim h 0 h lim(x + h) = x 0 h x Write the difference quotient for the general quadratic and expand it. Collect like terms and see if it is possible to determine the value of this expression as h approaches zero. Since x is fixed we see that both the numerator and the denominator are approaching zero as h approaches 0. At this stage the algebra needed is factoring. We factor h from both terms in the numerator and then recognize that we can divide top and bottom by h since h is never zero. As h gets closer and closer to zero, the limiting value of x+h is x. Therefore ( x + h) lim h 0 h x = x We can think of the term h in x+h as an error term; it is a measure of the disagreement between an approximation to the slope, obtained numerically with a specific h, and the exact slope. To calculate the derivative of f(x) = x 3 we again start with the limit definition f (x) = f ( x + h) f ( x) lim h 0 h and continue to simplify until the limit as h approaches zero becomes apparent. Keep in mind that x is fixed and only h changes by becoming smaller and smaller. f ( x + h) f ( x) lim h 0 h ( x + h) lim h 0 h ( x lim h 0 x lim h x 3 + 3x h + 3xh h + 3x h + 3xh h 3x h + 3xh lim h 0 h h(3x lim h 0 lim (3x h 0 + h 3 + 3xh + h ) h + 3xh + h ) h ) x + h 3 x 3 Limit definition of the derivative The function f in words says raise the input value to the third power Expand the expression (x+h) 3 Algebraic Simplification Algebraic Simplification Factor h from all terms in the numerator h h = 1 Therefore ( x + h) lim h 0 h 3 x 3 = 3x The terms 3xh and h both approach zero as h approaches zero In this case the expression 3x + 3xh + h differs from the limit 3x by the error term 3xh + h which depends upon both x and h. Joseph F. Aieta page 196 8/14/003

197 Measuring Change Section 5.1 Page 197 The Derivative is Not Always Defined Not every curve has a first derivative at all points but the majority of the functions that we will encounter in business applications do have this property. One example of a function that is defined at all points but does not have a derivative at all points is 7 f ( x) = x 3 shown in Figure f(x)=x^(/3) Figure For all other points on this function, other than x = 0, the derivative exists and is given by the formula 1 3 f ( x) = x or. At the point (0.0) this curve actually comes to sharp point, called a cusp, and does not have a x derivative. For the Excel graph on the left, a higher density of points near zero would be needed to capture the sharpness of the cusp. If we try to calculate f using the formula above, we get which is undefined. Thus we say that the function 3 f ( x) = x 0 is not differentiable at the point where x = 0. This is true for all power functions x n where 0 < n< 1. 7 Excel cannot raise a negative number to a fractional power directly. Since the results for this function are positive, we get around this limitation by defining the piece to the left of the origin as =(ABS(x))^(/3). Joseph F. Aieta page 197 8/14/003

198 Marginal Analysis Measuring Change Section 5.1 Page 198 Marginal analysis is the study of how sensitive the value of an output variable is to small changes in an input variable (or variables). For functions of a single variable, the input variable is often referred to as the independent variable (or predictor) and the output variable as the dependent variable (or response). In the context of manufacturing, the input variable is the number of units produced and the output variable is the corresponding cost in dollars. Geometrically, marginal cost is related to the slope of the cost function. In the simplest case of a linear cost function we have seen that the constant slope of the line is the variable cost per unit. In the case of a non-linear cost function, marginal cost has two commonly used interpretations. Consider the following example of a quadratic cost function: Example A company determines that the daily cost in dollars to produce and ship q units of one of its products can be described as Fixed cost F plus A times the number of units plus S times the square of the number of units. F, A, and S are parameters of the function, sometimes referred to as coefficients. Assume that the parameters A and S are related to production and shipping. In practice the values of these parameters are often statistically determined in the context of the production process of a particular item. In a particular week, the company determines that the costs associated with a particular product are approximately F = 500, A=.00 and S = 0.0 so the quadratic cost function becomes C(q) = q q Suppose that 50 units have been produced and we want to know how much more it will cost to produce one additional unit. The additional cost of producing the 51 st unit is $.40 as shown by the following calculation. ( ) ( ) = =.40 This leads to the first definition of marginal cost. Marginal cost at a production level of q units is the additional cost of making one more unit. At a production level of q units, the marginal cost of making one more unit is C(q+1) - C(q) For the cost function C(q) = q q we can see from Figure that the marginal cost of producing the 31 st unit is $754.0 minus $ which is $14.0 and the marginal cost of producing the 101 st unit is $74.0 minus $ which is $4.0. $6,000 Cost as a Function of Quantity $5,000 $4,000 $3,000 $,000 $1,000 quantity $ Figure Joseph F. Aieta page 198 8/14/003

199 Measuring Change Section 5.1 Page 199 Let s apply the numerical difference quotient approach to the cost function C(q) = q q at the point P (50, 1100). f(a+h) - f(a) a f(a) h a + h f(a+h) h Figure Observe in Figure that as h gets smaller and smaller, the slopes of successive secant lines are becoming closer and closer to.00. Once you create your own table in Excel, check other points on the cost function by simply changing the value of the first coordinate a in the first column. For example, change the value 50 in the first column to 49 and observe that the difference quotient is approaching It is possible to calculate the exact slope of a tangent line to this quadratic function at (50, 1100), or at any other point, directly from the rule that defines the function. The method for doing this is called differential calculus. The creation of the calculus of differentials in the latter part of the 17th century came about as scientists searched for a method to describe rates of change of non-linear functions related to motion. There was some controversy about who first invented the calculus. The German philosopher and mathematician Gottfried Leibniz claimed to have developed his solution sooner than the English scholar Sir Isaac Newton but the prestigious Royal Society declared that Isaac Newton had actually discovered the calculus first. The body of their work, developed independently, revolved around the idea of measuring the instantaneous rate of change, known today as the first derivative. The notation that Leibniz developed for describing derivatives is still used today. Before we introduce notation for derivatives or investigate any of the formal aspects of differential calculus, we further explore the intuitive, geometric properties of a tangent line to a curve at a point. One way of thinking about a tangent line is the best linear approximation to the curve in the neighborhood of the point. This description is based upon the notion that the tangent line at a point closely follows the curve in the vicinity of the point. This relationship between the curve and a tangent line at a point is not so obvious if we back away (zoom out) and look at a large portion of a graph, as shown on the left in Figure As we zoom in on the point P, this relationship between the curve and the tangent line becomes more apparent. Before zooming in Zoom at magnification X Zoom at magnification 4X Figure Joseph F. Aieta page 199 8/14/003

200 Measuring Change Section 5.1 Page 00 For the cost function C(q) = q q at the point P (50, 1100) the first graph on the left side in Figure shows the tangent line taken from an Excel graph at normal viewing (size 100%). Using Excel to zoom-in by 00% and then 400%, we are looking at smaller and smaller regions containing the point P. We can observe that the cost curve appears to become straight. We cannot easily distinguish the tangent line at point P from the function itself, although we know it is curved by zooming out. Functions that we will encounter in business situations will normally have this property which is known as local-linearity. Before zooming in Zoom at magnification X Zoom at magnification 4X Figure We have calculated the marginal cost of making the 51 st unit as C(51) C(50) = $.0. Figure examines the cost function in a relatively small interval around q = 50 and the graph of the tangent line to the curve at (50, 1100). In the figure it is hard to tell the difference between the tangent line to the curve at (50, 1100) and the curve itself. $1,140 $1,10 $1,100 $1,080 $1,060 C(q) = q q 51, , , $1,040 $1,00 quantity $1, Figure Joseph F. Aieta page 00 8/14/003

201 Measuring Change Section 5.1 Page 01 Once we learn to calculate the slope of a curve at a point (instantaneous rate of change) we will be able to show that the exact slope of the tangent line to the cost function C(q) = q q at the point (50, 1100) is.00. We interpret this slope as follows: The cost function is increasing at the rate of.00 dollars per unit at a production level of 50 units. The application of calculus to economics is a recent phenomenon compared with the application of calculus to physics and astronomy. Economists often use point marginal cost instead of the additional cost of making one more item. Certain types of analysis are easier to perform and describe with the language and notation of calculus than they are with the finite difference formula C(q+1) C(q). The context will generally determine whether a reference to the term marginal cost refers to the cost of making one additional item or to the slope of the curve at a particular level of production. To avoid possible ambiguity, we will refer to the latter as the instantaneous rate of change or point marginal cost. Exact formulas for the slope of a tangent line can be derived with the definition of the derivative as a limit and properties of limits. Fortunately, certain patterns emerge that allow us to state derivatives for entire classes of functions in a very compact way. The family of power functions can be expressed as f(x) = x n where n is a constant. Keep in mind that the first derivative, f ' (x), measures the slope of the graph of the function at x. If n = 0 we have f(x) = x 0 =1. The graph of f(x)=1 is a horizontal line that cuts the vertical axis at 1. Its slope is zero. Any horizontal line f(x)=b where b is a constant has a slope of zero. The derivative of a constant is zero. [In a business context, the observation that a graph of a company s sales over several quarters looks like a horizontal line is often described as sales were flat.] If n = 1 we have f(x)=x=1x whose derivative is 1 since f ( x + h) f ( x) x + h x h lim = lim = lim = 1 h 0 h h 0 h h 0 h For the linear function g(x) = ½ x in Figure the slope of the line is ½ = 0.5. The slope of the linear function f(x) is also ½ = 0.5. Figure The value of the y-intercept b for the function f(x) is 4 The graph of f(x) is related to the graph of g(x) by a vertical translation of the entire curve g(x) 4 units up. A vertical translation, up or down, has no effect on the slope of a function. For the general linear function f(x) = mx + b, the derivative at any point is simply m, the slope of the line. Figure The derivative of a linear function is a constant If n = we have f(x) = x and we have shown that f ' (x )= x The derivative of a quadratic function is a linear function If n = 3 we have f(x) = x 3 and we have shown that f ' (x )= 3x The derivative of a cubic function is a quadratic function Joseph F. Aieta page 01 8/14/003

202 Measuring Change Section 5.1 Page 0 The shortcut formula, known as the simple power rule, applies to all powers including negative and fractional powers. If f(x) = x n where n is some constant, then f' (x )= n x n-1 In words, bring down the power and multiply by x raised to the power n minus 1 Recall the definition x n 1 = regarding negative exponents. For n = -1 we have the reciprocal function n x 1 *Challenge Problem: Use the limit definition to show that the first derivative of is f ( x) = 1x = x x 1 1 f ( x) = x 1 =. x Here are several important derivative rules stated both in words and symbolically in two forms. When we write the derivative of x 3 with respect to x as d 3 x = 3x, think of d as a one-piece symbol, not as a fraction. It is simply an dx dx abbreviation for the phrase the derivative with respect to x. The derivative of constant times a function is the constant times the derivative of the function, Bring down the power multiply by the constant In prime notation ( k u) = k u. In differential notation d dx k u = k du dx We can take derivatives term by term. The derivative of a sum is the sum of the derivatives In prime notation ( u v) u v d + = du + dv dx dx dx + = + In differential notation ( u v) The derivative of a difference is the difference of the derivatives. In prime notation ( u v) = u v. In differential notation d ( u v) = du dv dx dx dx For a one page listing of the derivative rules that we will need for the business applications in this course see Derivative Rules.doc. Joseph F. Aieta page 0 8/14/003

203 Measuring Change Section 5.1 Page 03 Examples through illustrate the above derivative rules together with some algebraic simplification. Example x x f ( x) = + x find f ' (x ) 1 1 f ( x) = 3 x x + 1 = x x Example f ( x) = 3x + x find f '(4) f(x) can be written as 3x+ 48 x f ( x) = 3 48 x = 3 x 48 f ( 4) = = 3 = 3 3 = 0 4 Example f ( x) = 6x x find f '(8) 1 3 f ( x) = 6 x 3 4 f (8) = x = 4x x = = = = x x 3 Note in examples and 5.1.7, when we wanted the derivative at the points x = 4 and x = 8 respectively, we first take the derivative with respect to x before replacing x by the values 4 or 8. Example Given the square root function 1 f ( x) = x = x, state the rule for f ' (x ) Then find the slopes of the tangent lines to f ( x) = x at x = 0.01, 9, and 100. The derivative d d 1 1 x = x = 1 = x x dx dx which can be written without negative exponents as 1 = 1. 1 x x At x = 0.01, the slope of the tangent line is = = = (0.1) x 1 As x increases, the denominator of the expression also increases and the slope of the tangent line to f ( x) = x decreases. 1 x f ' (9) = 1 1 = and f ' (100) = = Joseph F. Aieta page 03 8/14/003

204 Measuring Change Section 5.1 Page 04 The first derivative f (x) is a standardized measure of the rate of change of the function f (x). From the derivative rules, we know that the derivative of any quadratic function is a linear function. The quadratic whose graph is shown in Figure is f(x) = -x + 8x and its first derivative is f (x) = x+8 At the point P (3, 13) on the curve, we see a portion of the tangent line. This tangent line has slope. Points with x coordinate 3 are indicated on both the graph of the function itself and on the corresponding first derivative (slope) function. At x = 3 the function is increasing and the slope at the point x = 3 is positive. Notice that the function y = f (x) is increasing at points where its derivative is positive and is decreasing at points where its derivative is negative. For 5 < x < 7, the quadratic is clearly decreasing and the corresponding points on the derived function f (x) are below the x-axis. Figure If the derivative is zero at a point then the function is neither increasing nor decreasing at that point. Q1. What is the x-value that makes the first derivative zero? A1. f (x) = x+8 = 0 at the point x = 4 Q. Describe the behavior of the curve f itself and the behavior its first derivative f ' in the neighborhood of the point (4, f(4)). A. To the immediate left of x = 4 the curve y = f (x) is increasing and its first derivative f is positive. To the immediate right of x = 4 the curve y = f (x) is deceasing and its first derivative f is negative. Joseph F. Aieta page 04 8/14/003

205 Measuring Change Section 5.1 Page 05 To examine the graphical relationship between quadratic functions and their linear first derivatives, open the file DerivQuadratic.xls,, and move the slider bars that control the coefficients a, b, and c as shown in Figure Quadratic and Derivative f(x) = -4*x^ + 10*x x y y ' turning point a (1.50, 6.50 ) accuracy to b 10 0 decimal places c f ' (x) = -8*x Figure Joseph F. Aieta page 05 8/14/003

206 Measuring Change Section 5.1 Page 06 When working with cost C(q), revenue R(q), or profit functions P(q) of quantity, economists find it convenient to refer to point (or instantaneous) marginal cost, point marginal revenue, and point marginal profit as dc dr dp,, and respectively. If cost C(q) = q q and revenue R(q) = 40 q then profit dq dq dq dc dr dp P(q) =R(q)-C(q)= q +38q 500 and = q, = 40, and = q dq dq dq At a production level and sales level of 100 units, verify that dc dr is 4, is 40, and dq dq dp is -. dq What is the significance in a business context of the fact that point marginal profit is negative at q = 100? Average Cost In addition to understanding marginal cost at a particular production level, manufacturers are also interested in the average cost of making a particular number of units. These are very different concepts. Marginal cost focuses on the behavior of the cost function at a specific point, such as the 50 th unit. Average cost looks at an interval of production, say from 0 to 50 units. Average cost is given by C ( q), the total cost divided by the number of units produced. q Average cost is widely used in economics. When cost is a linear function, C(q) = v q + F, total cost equals variable cost plus fixed cost. Recall that fixed cost, sometimes referred to as overhead, includes such expenses as the cost of rent, utilities, insurance, and security. The fixed component of cost does not vary with the number of units made but variable costs, such as the cost of materials, do vary with the number of units made. When a cost function is linear, average cost will steadily decrease as we increase the quantity of manufactured goods, since we are spreading the fixed cost over more and more units. For the linear cost function C(q) = 00q + 60,000 where q is quantity and C(q) is dollars, the average cost of making 1000 units is C (1000) 60,000 = = 60 dollars per unit while the average cost of making 000 units is C (000) 460,000 = = 30 dollars per unit Average cost is not so predictable for non-linear cost functions. Let s return to the quadratic cost function first introduced in example A company determines that the daily cost in dollars to produce and ship q units of one of its products can be described as Fixed cost F plus A times the number of units plus S times the square of the number of units. F, A, and S are parameters of the function, sometimes referred to as coefficients. The parameters A and S may be related to production and shipping. In a particular week, the company determines that the costs associated with a particular product are approximately F = 500, A=.00 and S = 0.0. The quadratic cost function for the week is becomes C(q) = q q Joseph F. Aieta page 06 8/14/003

207 Measuring Change Section 5.1 Page 07 C (30) (30) If 30 units are made then average cost is = = = dollars per unit rounded to the nearest cent. If 100 units are made then average cost is C(100) (100) = = = dollars per unit The table and graphs in Figure were obtained from the worksheet average cost in the file QuadraticCost.xls. The company plans to produce between 0 and 150 units in any given week. The graph of the average cost function decreases for small values of q, reaches a minimum value around q = 50 and then increases as q increases. Average cost is $7.00 dollars per unit at q = 5 and again at q = 100. For values of q to the left of 50, the primary reason why the average cost is relatively high is because the fixed cost is being divided by a relatively small number of units. As q increases beyond 50, the cost function increases from left to right but not at a constant rate. The average cost function is not linear but approaches a line as q gets larger and larger. [Note: In order for the slider bar to work properly, the Calculation Option in Excel Tools must be set to Automatic.] Figure Joseph F. Aieta page 07 8/14/003

208 Measuring Change Section 5.1 Page 08 Numeric Method The Symmetric Difference Quotient Computational tools with a fraction of the power of Excel were not available until the twentieth century. Analytic tools, including those available to Newton and Leibniz evolved over thousands of years. Intellectual contributions, such as theirs, enabled mathematicians and scientists to make enormous strides that led to many of the problem solving tools that we take for granted today. If we are interested in approximating numerical derivatives with computational efficiency, we can use an approach that gives more accurate results for not much more work. It is the basis for how many graphical calculators, such as the TI -83, approximate slopes of tangent lines. Instead of connecting point P with coordinates (a, f(a) ) and a nearby point Q with coordinates (a+h, f(a+h)), take one point to the left of P with coordinates (a-h, f(a-h) ) and another point to the right of P with coordinates (a+h, f(a+h)). The x-coordinate of point P is symmetrically located between x = a-h and x = a+h. The average rate of change over the interval [a-h, a+h] becomes f ( a + h) f ( a h). The numerator is called the symmetric difference and h the expression is called the symmetric difference quotient. Figure shows eight decimal place approximations to the slope of the tangent line at (50, 1100) on the cost function C(q) = q q using the symmetric difference quotient. f(a+h) -f(a-h) a h a-h f(a-h) a + h f(a+h) h Figure Quadratic functions are rather special as we see in this example where the symmetric difference quotient is identical to the exact derivative. The extent to which the symmetric difference quotient gives better results than the ordinary difference quotient (also called the forward difference quotient) depends upon the type of function and on the point in question. Let s compare the forward difference quotient with the symmetric difference quotient at the point (, 9) on the cubic function f(x) =x 3 + x - 4x+ 5. Here the values of h decrease by a factor of (1, ½, ¼, etc. ) so that we can see what is happening on the graph. If we start with h =1 and examine successive secant lines where each successive value of h is smaller by half than the previous h. For the 11 th secant line h is which is about one-one thousandth. The ordinary difference quotient gives us to eight decimal places as an approximation to the slope of the tangent line. For this same value of h near 0.001, the symmetric difference quotient gives which is much closer to the limiting value of 1. secant f(a+h) -f(a) f(a+h) -f(a-h) number a f(a) h a-h f(a-h) a+h f(a+h) h h Figure Joseph F. Aieta page 08 8/14/003

209 Measuring Change Section 5.1 Page 09 Example The graphs of the cubic function f(x) = x 3 + x - 4x + 5 in Figures and show the first few secant lines for each of the numerical methods of approximating the slope of the tangent line at x =. Forward Difference Quotient f(a+h) -f(a) h Symmetric Difference Quotient f(a+h) -f(a-h) h Figure Figure These graphs provide some insight as to why the symmetric difference quotient converges more rapidly. If we had used the sequence 1, 0.1, 0.001, instead of 1, ½, ¼, we would not be able to distinguish the second secant line from the third secant line in Figures and Joseph F. Aieta page 09 8/14/003

210 Measuring Change Activities 1 Page 10 Measuring Change Activities 1 1. The graph on the right represents miles traveled by an automobile as a function of the passage of time in minutes. a) d) What point or intervals of time correspond to the following events: a) The auto is miles from the starting point of the trip. b) The auto comes to a brief stop. c) The speed of the auto is decreasing. d) The driver sets cruise control. e) What was the average speed over the 6 minute time interval? miles minutes. Draw a rough sketch of a smooth curve that relates highway distance traveled versus time over a 30-minute interval.. The scale on the vertical axis does not need to be precise, but the rough sketch should tell the following story: The car moved slowly up the entrance ramp and then quickly accelerated until it reached a cruising speed of 60 mph. It stayed at that speed for the next 10 minutes. Heavy traffic was then encountered and everyone slowed down to about 30 mph for the next 10 minutes. The car was stopped for about five minutes. The heavy traffic had cleared and the speed of the car was brought up to 50 mph. Near the end of the trip, the car hit 70 mph as it passed several trucks that were in the right lane. The car slowed down to take its exit and then traveled to its final destination at a steady speed of 30mph. 3. Inflation rates year Adj. year percent 7.00 inflation percentage years since a) Which two year period had the highest average change (positive or negative) in the rate of inflation? b) Which two year period had the lowest average change (positive or negative) in the rate of inflation? Joseph F. Aieta page 10 8/14/003

211 Measuring Change Activities 1 Page actual clock time 6:00 AM 7:00 AM 8:00 AM 9:00 AM 10:00 AM 11:00 AM 1:00 noon 1:00 PM :00 PM 3:00 PM 4:00 PM 5:00 PM 6:00 PM Temp. degrees F From the temperature table above, calculate the average rate of change in temperature between 9:00 AM and a) 10:00 AM b) 11:00 AM c) 1:00 PM From the temperature table, calculate the average rate of change in temperature between 1:00 PM and d) :00 PM e) 4:00 PM f) 6:00 PM 5. Use a ruler to draw a tangent line to the curve for each of the indicated times and then complete the statements a) d). Temperature 6 AM to 6 PM degrees fahrenheit noon = 0 a) At 8:00 AM the temperature is at the rate of per hour. b) At 10:00 AM the temperature is at the rate of per hour. c) At 4:00 PM the temperature is at the rate of per hour. d) At 6:00 PM the temperature is at the rate of per hour. Joseph F. Aieta page 11 8/14/003

212 Measuring Change Activities 1 Page 1 6. Assume that a function f(x) has a first derivative at all points. Give two descriptions of f '(x), the first derivative of the function at a point P (x, y) a) geometric description b) limit definition c) if f is any linear function f(x) = mx + b, show that the average rate change over any interval [x 1, x ] is the same as the instantaneous rate of change at any point For the cost function given by C( x) = x + find the average rate of change over x the interval [50, 50+h] where h = This value will be very close to the point marginal cost at x = 50. Use the difference quotient C ( x+ h) C ( x) h with accuracy to eight decimal places. 8. Suppose that a company s annual revenues from 1994 through 000 can be modeled by the function R(t) = -0.1 t t over the interval 0 < t < 6. where R(t) is revenue in billions of dollars and t = 0 represents January 1994 a) Use the ordinary difference quotient to estimate the average rate of change of R(t) over the period January 1996 through January Interpret your answer. b) Estimate the instantaneous rate of change of R(t) in January Interpret your answer. 9. Suppose the demand price function for a new brand of footwear is given by the function on the right. Dq is the quantity demanded at price p where p is in dollars. 5,000,000 Dq( p) = p Find Dq(100) and estimate the point marginal cost at p =100. Interpret your answers. 10. a) Explain why we cannot substitute 0 for h in the limit definition. b) Use the limit definition to show that the derivative of a constant is zero. f '( x) = lim h 0 f ( x + h) h f ( x) Joseph F. Aieta page 1 8/14/003

213 11-1. Recall that the first derivative f (x) is defined as follows: Measuring Change Activities 1 Page 13 f (x ) = lim h 0 f ( x + h) h f ( x) Question 11 is about the reciprocal function, y =1/x. Question 1 is about the natural log function. At x = 0.5, approximations for the first derivative have been entered to three decimal places. Note that each function is undefined at a point (or collection of points) and for these points the first derivative does not exist. a) For other points in the tables, use the difference quotient to approximate the derivatives at the given points with accuracy to three-decimal places. b) Make conjectures about the derivative rules for these functions 11. Reciprocal function a) Complete the second and third columns of the table with accuracy to three (3) decimal places. b) Make a conjecture? If f (x) 1 f ( x) = x = then x y =1/x Estimate f'(x) undefined undefined Natural Log function a) Complete the second and third columns of the table with accuracy to three (3) decimal places. b) Make a conjecture? If f ( x) LN( x) = then f (x) = x f(x)=ln(x) Estimate f'(x) -.0 undefined undefined -1.5 undefined undefined -1.0 undefined undefined 0.0 undefined undefined Joseph F. Aieta page 13 8/14/003

214 Measuring Change Activities 1 Page Given the function 1 f ( x) = x + x a) Determine the slope of the tangent line to the graph of this function at x = shown to the right. b) What is the slope of the tangent line at x = 1? c) Write the equation of the tangent line at x = Justify the following derivative rules using the limit definition a) If f ( x) = = x then f ( x) = 1x = x b) If f(x) = x 4 then f '(x) = 4 x 3 x f ( x) = lim h 0 f ( x + h) f ( x) h 15. Estimate the derivative of the function f(x) = e x at the point (0, 1) using the difference quotient with h = Recall that the built-in exponential function for e x in Excel is =EXP(x). 16. If the total dollar cost of making x pounds of material is C(x) = x x then what is the point marginal cost of making the 10 th pound? 17. Find the slope of the tangent line to the curve 1 f ( x) = 3x + at x = x State the first derivative for each function. Derivative rules that you will need are given below in both differential d notation and f notation dx Power rule Constant times a function Sum and difference d dx x n = n x n 1 d dx k u = ( u+ v) = + ( u v) k du dx d du dv dx dx dx d = du dv dx dx dx f ( x) = x n, f ( x) = n ( k u) = k u ( u+ v) = u + v ( u v) = u v Joseph F. Aieta page 14 8/14/003

215 Measuring Change Activities 1 Page ( ) 5 f x = x 19. ( ) f x = x ( ) f x = x ( ) f x = x x. f 1 = + x ( x) x y = x x 4. 1 ( ) = 3 g x x x 5. 1 h ( x) = + x 3 x 6. s( x) = x + 1 x 7. t x x x 3 x 1 ( x) = = t ( x) = x 0 x 8. d dt ( at 3 4at) where a is a constant 9. find dv dr given V = 4 π r f ( x) = a x + b x + cx + dx + e Joseph F. Aieta page 15 8/14/003

216 Measuring Change Activities 1 Page After you find the derivative, find the slope of the tangent line to the function at the given point f ( x) = 3x + at x = x 3. f ( x) = x at x = f 3 ( x) = x at x = First find the derivative and then find each value of x such that the tangent line at (x, f(x)) is horizontal. If there are no stationary points, state none. 34. y = x + 3x y = x + x f ( x) = 3x + x 37. The average cost per unit is the total cost divided by the number of units: If the cost function is C(q) = q q then the expression for average cost can be simplified as q. q a) What is the average cost if 40 units are produced? C ( q) where q 0. q q + 0.0q q 500 b) Determine the value of q, to the nearest whole number, that minimizes the average cost A( q) = q. q 38. Use the symmetric difference quotient f ( a + h) f ( a h) with h = to approximate the slope of the tangent line to h x the function ( ) = at the point a = 0.5. f x x e Joseph F. Aieta page 16 8/14/003

217 Measuring Change Section 5. Page 17 Derivative Concepts: Rules, Properties, and Tests Section 5. Let us re-examine quadratic cost functions of the form C (q) = F + A q +S q. The input variable, q, represents the quantity of units produced and the output variable, C (q), is measured in dollars. At a particular point in time, a manufacturer determines that the cost function for one of its products is C(q) = q + 0.0q. This new product is in high demand and consumers are currently willing to pay $40 per unit so revenue is the linear function R(q) = 40q. This week the company will produce a minimum of 0 units and a maximum of 150 units. Since the company can sell all of the items that it produces, management is interested in finding out if there is an ideal production level. The tables and graphs in QuadraticCost.xls, illustrate how cost changes as quantity increases. The rate of change is clearly not a constant. At a production level of 50 units, the marginal cost of making one more unit is C(51) C(50) which is or.0. At a production level of 100 units, the marginal cost of making one more unit is C(101) C(100) which is = 4.0. A good approximation to the marginal cost of making one more item at any level of production, can be calculated quickly with the first derivative C (q) = q. The slope of the linear revenue function R(q) = 40q is 40. To obtain the formula for the profit function, subtract cost from revenue P(q) = 40q ( q + 0.0q which simplifies to P(q)= 38.00q q. The marginal profit, P(q+1) P(q), gained by the manufacture and sale of one more item is closely approximated by the derivative of the profit function P' (q). Figure 5..1 shows a table of the cost, revenue, and profit functions together with their associated derivatives. It reveals several interesting patterns and relationships. Figure 5..1 Joseph F. Aieta page 17 8/14/003

218 Measuring Change Section 5. Page 18 Note that P' (q) = R ' (q) C '(q) and the formula for point marginal profit is P ' (q) = q. At q = 100, the marginal profit gained by the manufacture and sale of the 101 st item is = -.0. The point marginal profit is P' (100) = (100) = How can we interpret this negative marginal profit at q = 100? The graph of the three functions in Figure 5.. gives us a clearer picture of what is happening. At the point q = 100, the profit function is decreasing at the rate of $.00 per unit. The profit function has actually reached a maximum value at some point to the left of q = 100. $5,500 Cost, Revenue, and Profit $4, , 4000 $3,500 $, , $1, , $500 -$ quantity Figure 5.. P'(q) gives the slope of the tangent line to the quadratic profit function. From the table in Figure 5..1 we see that P '(q) is zero at q = 95. Without a table, we could find this point by solving the equation P ' (q) = q =0. What we are finding is a value of q on the graph of P(q) = 38.00q q whose tangent line is horizontal. This stationary point must be a local maximum since the quadratic function opens down. Another way to justify our conclusion that this point is a local maximum is to observe that P (q) is positive just to the left of q = 95 and P'(q) is negative just to the right of q = 95. The profit function changes from increasing to decreasing in the neighborhood of the point (95, 1305). At the point of maximum profit the rate of change of the revenue function is the same as the rate of change of the cost function. At the point of maximum profit, point marginal revenue is equal to point marginal cost. Stated symbolically, R' (q) = C' (q) at the point where profit is a maximum. Joseph F. Aieta page 18 8/14/003

219 Measuring Change Section 5. Page 19 The graph in Figure 5..3 is taken from the worksheet MC = MR in the file QuadraticCost.xls. Cost, Revenue, and Profit $5,500 F A S quantity cost dc dq quantity revenue revf $,500 40q dr dq $4,500 $3,500 $1, q q ^ costf 95, q + 0.q ^ 95, prftf 38 95, q q ^ q -0.q ^ quantity profit dp dq 0.00 $500 -$ quantity Figure 5..3 For the specific parameters shown in Figures 5..1, 5.., and 5.5.3, the steps involved in finding the optimal quantity that maximizes profit are shown on the left. Once we have found the optimal q, we can substitute for q in the profit function to determine the maximum profit in dollars. The parameters that define cost and revenue functions are rarely static over time. For example, a drop in consumer demand might require the company to lower the unit price below $40. Production costs can also change due to such factors as higher (or lower) fixed cost, cost of materials, labor, etc. Rather than rework the problem each time the value of the parameters change, it would be more efficient if we could describe the optimal value of q explicitly in terms of the parameters F, A, S, and p. Since the derivative P'(q) of the quadratic profit function is a linear function, it is possible to state the optimal solution in terms of the parameters using basic algebra as shown below in the middle column. Joseph F. Aieta page 19 8/14/003

220 Measuring Change Section 5. Page 0 Specific case P(q) = R(q) C(q) P(q) = 40.00q ( q + 0.0q ) P(q) =40.00q q -0.0q P(q) = 38.00q q P '(q) = = q P '(q) = 0 solve q = = 0.40 q so q = = 95 and P(95) = General case P(q) is R(q) C(q) P(q) = p q (F+ A q + S q ) P(q) = p q F A q - S q P(q) =(p-a) q F S q P '(q) = (p-a) - S q P '(q) = 0 solve (p-a) - S q = 0 (p-a) = S q so q = p A S Definition Substitution Algebra Algebra Derivative rules Find roots of first derivative Algebra Algebra and Arithmetic Once we have found the optimal q, we can substitute for q in the profit function to determine the maximum profit in dollars. The worksheet sliders & maximum profit in QuadraticCost.xls, shows cost, revenue, and profit curves that correspond to values of parameters controlled by the sliders. In Figure 5..4 the parameters of the cost function are F = 550, A = 3.50, S = For the revenue function, the selling price p is dollars per unit. According to our general solution, the optimal value of q occurs at p A q = = = S 0.15 In Figure 5..4, this optimal value of q has been rounded down to 88 units and the optimal point (88, 60) is shown on the profit curve. Joseph F. Aieta page 0 8/14/003

221 Measuring Change Section 5. Page F A S p F A S p Cost Revenue Profit $5, $4,500 $3,500 $,500 $300 $500 F $0.50 $1.00 A $10.00 $0.00 B $0.05 $ $1,500 88, 60 $ $500 quantity Figure 5..4 Note that the point marginal profit for q = 88 is a small positive number and the point marginal profit for q = 89 is a small negative number. We apply Excel s Goal Seek or Solver to drive the marginal profit to zero by changing the quantity q. This confirms that the value of q that makes the first derivative zero is Marginal Marginal Marginal q Cost Revenue Profit Cost C' (q) Revenue R'(q) Profit P'(q) Before Goal Seek or Solver After Goal Seek or Solver Figure 5..5 Joseph F. Aieta page 1 8/14/003

222 Measuring Change Section 5. Page Properties of Quadratic Functions: All quadratic functions are of the form f(x) = ax + bx + c. In Excel s Trendline, quadratic functions are in the category polynomials of order. Figure 5..6 A quadratic function has exactly one high point at a peak or one low point in a valley. At the turning point, or vertex of the parabola, the tangent line to the quadratic is horizontal. In Figure 5..6 the graph of the quadratic function f(x) = -1x been positive then the parabola would be concave up. - x + is concave down. If the leading coefficient a, had In the Excel file GenQuad.xls. explore the relationships of the parameters a, b, c and the graph of the quadratic function by moving the sliders and observing changes in the graph s shape, location, and turning point. Joseph F. Aieta page 8/14/003

223 Measuring Change Section 5. Page 3 Example 5..1 A company s revenue for a particular product may reflect a relationship between price and the level of consumer demand. Consider the pricing of popular movie DVDs by video retailers. Many customers will make a purchase if they happen to be in the store and the price is in line with other DVDs. There is a group of customers who want to be among the first to own the DVD of a box office success. They pay little attention to price. Others want to wait until after the initial rush is over and the price has dropped. Then there are the bargain hunters who want to eventually own the DVD but are motivated to shop carefully on price. A retailer collects and analyzes the following data on unit price and consumer demand. Clearly lower consumer demand is associated with higher prices and higher consumer demand corresponds to lower prices. price quantity $16.00 $14.50 $15.00 $18.00 $0.00 $.00 $5.00 The scatter plot in Figure 5..6 suggests a linear trend for price versus quantity. Economists have adopted the convention of plotting price on the vertical axis. The price-demand equation from Excel s Trendline is p = q The revenue function price times quantity becomes q p 1000 $ $14.50 $30.00 $5.00 y = x R(q) = p q = ( q ) q R(q) = q q 100 $ $ $ $ $5.00 price $0.00 $15.00 $10.00 $5.00 Assume that the wholesale cost is $6.50 per DVD. $ quantity Figure 5..7 Profit = revenue minus cost P(q) = q q q P(q) = q q. Since the leading coefficient of this quadratic profit function is negative, we can expect that its graph will be a parabola that opens down. The peak value of the profit function will occur at a point where the first derivative of the profit function is zero. Point marginal profit P (q) = q P'(q) = 0 when q = which suggests that maximum profit occurs near 900 units which corresponds to a unit price of approximately $ $1,000 $10,000 $8,000 $6,000 $4,000 $,000 $0 Profit as a function of quantity y = x x demand quantity Figure 5..8 Joseph F. Aieta page 3 8/14/003

224 Measuring Change Section 5. Page 4 Properties of cubic polynomials: All cubic functions are of the form f(x) = ax 3 + bx + cx + d. In Excel s Trendline, cubic functions are in the category polynomials of order 3. All cubic polynomials have an inflection point which separates the curve into two parts as follows: the curve is increasing (decreasing) rapidly on one part and then increases (decreases) less rapidly on the other part. This type of inflection point is the point of most rapid increase (decrease). The other possibility is that the curve flattens out (becomes stationary) and is neither increasing nor decreasing at the inflection point. On one side of an inflection point the curve is said to be concave up (down) and on the other side the curve is said to be concave down (up). For the cubic f(x) = -1.6x 3 + 9x + 5x 15 the curve is changing from concave up to concave down around the inflection point (1.875, ), as shown in Figure a -10 b 10-5 c 5-0 d y = -1.6*x^3 + 9*x^+ 5*x guess value of y Cubic functions always have at least one real root (x-intercept) and may have up to three real roots. Roots of a cubic are not always easy to find algebraically. If a cubic changes from increasing to decreasing then it will have a high point called a relative (local) maximum. If a cubic changes from decreasing to increasing then it will have a low point which is called a relative (local) minimum. Cubics may be constantly increasing or constantly decreasing and may have no local maximum or minimum points. For example the cubic f(x) =x 3 +x +x+1 is constantly increasing from left to right. x y inflection point Figure 5..9 If the leading coefficient is negative then the curve would be concave down on one side and concave up on the other side of the inflection point. For a general review of the properties of cubic functions, see the Excel file GenCubic.xls. For polynomial functions of degree 4, 5, or higher, the situation gets more complex. Consider the curve in Figure 5..10, which is the graph of a fifth degree polynomial. Joseph F. Aieta page 4 8/14/003

225 Measuring Change Section 5. Page 5 Each of the named points A through G in the closed interval [-4, 8] has some distinctive property. The point A at x = -4 is a left hand endpoint and the point G at x = 8 is a right hand endpoint. D E -4.0, 8.0 A F C B G 8.0, Figure Sub-interval or critical point Properties of the function f(x) Properties of the derived function f '(x) between points A and B the function is decreasing the first derivative is negative at point B the function is stationary the first derivative is zero between points B and D the function is increasing the first derivative is positive at point C the function changes concavity the first derivative is positive at point D the function is stationary the first derivative is zero between points D and F the function is decreasing the first derivative is negative at point E the function changes concavity the first derivative is negative at point F the function is stationary and the the first derivative is zero function changes concavity between points F and G the function is decreasing the first derivative is negative A stationary point has the property that the curve is neither increasing nor decreasing at that point. The first derivative f '(x) is zero at such a point. The stationary point B in Figure is a relative (or local) minimum. Nearby points on either side are all above point B. Point B is not the absolute lowest point on the curve in the interval [-4, 8]. The point G has the property that it is the absolute (or global) minimum. In the interval [-4, 8] no point on the graph of the function is lower than point G. The stationary point D is called a relative (or local) maximum since nearby points on either side are all below point D. The endpoint A is an endpoint maximum but not an absolute maximum. Points immediately to the right of point A are below it but point D is the absolute maximum in this interval. Points C and E are both ordinary inflection points. Point C is a point of most rapid increase. The curve changes from concave up on one side of C to concave down on the other side. Point E is a point of most rapid decrease. The curve changes from concave down on one side of E to concave up on the other side. Since point F has the property that it is both a stationary point and an inflection point, we call it a stationary inflection point. Joseph F. Aieta page 5 8/14/003

226 Measuring Change Section 5. Page 6 Finding stationary points involves solving the equation f ' (x)= 0. This is also referred to as finding the roots of the first derivative. These roots are candidates for relative maxima or relative minima. As we have seen with point F in Figure 5..10, a stationary point may be an inflection point and be neither a relative minimum nor a relative maximum. The second derivative will be introduced later to help us in locating inflection points. Example 5.. Consider some polynomial function defined over the interval [-1, 4] that contains the points shown in Figure 5..11, Based only the table and the values of the first derivative, describe the properties of the graph in words. Include the approximate locations of any relative maxima or minima (but not inflection points). Since we are told that this is a polynomial function, its graph is a smooth curve without any breaks, cusps, or holes. Conjectures: At the left-hand endpoint (-1, -1.4), the function has a global minimum since no other point has a y coordinate lower than The function seems to be increasing from (-1, -1.4), to some point just to the right of x= 0.5. At this point the function has a relative maximum. The first derivative changes from positive at x = 0.5 (curve is increasing) to negative at x = 0.50 (curve is decreasing). To the right of the relative minimum the function increases until the point (1, 5). This point may be a relative minimum since the first derivative changes from negative just to the left of x = 1 (curve is decreasing) to positive to the right of x = 1 (curve is increasing). At a point somewhere between x =.00 and.5, the curve will have another relative maximum. The first derivative is positive at x =.0 (curve is increasing) and then changes to negative at x =.5 (curve is decreasing). Another relative minimum is near (3, 5) since the first derivative changes from negative to the left of x = 3 (curve is decreasing) to positive to the right of x = 3 (curve is increasing). x function 1 st derivative The curve continues to increase between this relative minimum up to x = 4. The endpoint (4, 10.5) may be a global maximum. Figure Joseph F. Aieta page 6 8/14/003

227 Measuring Change Section 5. Page 7 Figure 5..1 shows a sketch of the polynomial function containing the points shown in Figure Figure 5..1 Example 5..3 This time we begin with a cubic whose equation is given as f(x) = 4x 3-4x +x. Determine the coordinates of all relative maxima or minima for this function in the interval [-1, ]. We start with a quick sketch of the graph created with Excel.. x f (x) cubic y = 4x^3-4x^ + x Figure A quick inspection of Figure would suggest that this cubic never decreases from left to right. Without a compelling reason to investigate further by zooming-in, we might conclude that there are no turning points in this interval. Joseph F. Aieta page 7 8/14/003

228 Measuring Change Section 5. Page 8 An application of differential calculus will provide us with an irrefutable answer to the question Are there any maxima or minima? To calculate the first derivative of any function whose terms contain only power functions and constants use the simple power rule: d n n 1 x = n x dx d the rule for a constant times a function: dx k u = k du dx the sum and difference rules: d ( u+ v) = du + dv and d ( u v) = du dv dx dx dx dx dx dx d. The roots of the first derivative can be found by the quadratic formula. dx Solving 1x 8x + 1= 0 we obtain the two roots x = or 4 4 Since 3 ( 4x - 4x + x) = 1x 8x + 1 The simplified roots are 8 4 = or =. 4 If the solution of f'(x) = 0 by the quadratic formula had produced a negative number under the radical sign then we would know that there were no stationary points and the cubic function could not have any relative maxima or minima. Figure shows a zoomed - in portion of the graph in an interval that contains the points at x=1/6 and x = ½. Example Figure A leasing company has developed a model for their annual expenditures on rental machinery that they lease for more than one year. a) They would like to know how long to keep the equipment in order to minimize annual expenditures. b) They are interested in how fast annual expenditures are increasing in year six, since several important customers have expressed a preference to lease in six-year intervals. The company has developed a modeling function of the form R p C( x) = + a x The input variable, x, is the age of the machines in years, R is the replacement cost, a, is the average x cost of repairs in the first year, and the power p is a constant that reflects the increased annual repair cost as the machines age. For one particular type of machine, the company has determined that the replacement cost is R = $4,000, yearly repairs in the first year are about $100, and the aging factor is For this type of machine the annual cost function is 4000 C( x) = x x This problem could be approached at several levels of detail. If the company wants just a quick answer to part a), then they could create the table and graph shown in Figure to read the nearest year that corresponds to a low point on the annual cost curve. With calculus, they can obtain a more precise value of the optimal x and also can get an answer to part b). A more ambitious goal would be to try and solve the general problem in terms of the parameters R, a, and p. We will aim our solution at the middle level The first derivative of C( x) = x is C ( x) = + 150x x Joseph F. Aieta page 8 8/14/003 x

229 Measuring Change Section 5. Page 9 C' (x) = -R/x^+p*a*x^(p-1) Because of the algebraic order of operations, exponentiation is performed before division in -R/x^ so () aren t needed around x^ but () are needed in x^(p-1) otherwise x would be raised to the p power and then 1 would be subtracted from the result. years x C(x) C '(x) R a p Annual Expenditures $4,500 $4,000 $3,500 $3,000 $,500 $,000 $1,500 $1,000 $500 $ age in years Figure The table in Figure shows that annual expenditures are increasing at the rate of $56 per year when the equipment is 6 years old. Annual expenditures seem to reach a low near x = 4. An algebraic approach to finding the stationary point near x = 4 is to solve the equation C '(x) = + 150x = 0 x x x 150x = 150x 150x x.5 = = x x = 4000 = = x = A numerical approach would be to apply Goal Seek or Solver to the fourth row and drive C'(x) in the third column to zero by changing the value of x. x C(x) C'(x) x C(x) C'(x) Before Goal Seek After Goal Seek The minimum annual expenditure of approximately $1800 occurs after 3 years and roughly 8 months Joseph F. Aieta page 9 8/14/003

230 Measuring Change Section 5. Page 30 Exponential and Logarithmic Functions How can we measure the rate of change of an exponential function? For exponential functions of the form b x, what type of function is the first derivative, or slope function? We will look for patterns in graphs produced by numerical methods. [A rigorous justification of the derivative rules for exponential and logarithmic functions requires limit theory which is beyond the scope of our work.] A common MISTAKE is to think that the derivative of b x is x b x-1. This doesn t work since the simple power rule does not apply to functions of the form b x where the base b is a constant and the exponent is the variable. The simple power rule applies only to power functions which are of the form x n where the variable x is in the base and the exponent is a constant. The simple power rule does not apply to exponential functions. We will first consider the two exponential functions f(x) = x and g(x) = 3 x and estimate the slopes of their respective tangent lines at the points (0.5, 0.5 ) and (0.5, ). The tables in Figures and show the difference quotient with h = 1, 0.001, and The column containing the difference quotient displays nine decimal places. d x Estimate at x = 0. 5 dx Difference f(x)= x Quotient f(a+h) -f(a) a f(a) h a+h f(a+h) h Figure d x Estimate 3 at x = 0. 5 dx g(x)=3 x Difference Quotient g(a+h) -g(a) a g(a) h a+h g(a+h) h Figure Joseph F. Aieta page 30 8/14/003

231 Measuring Change Section 5. Page 31 To explore conjectures about the derivatives f functions, open the Excel file tangents.xls. This file uses the symmetric difference quotient with h= for computational efficiency and gives a very close approximation to the actual derivatives. In this utility file the function is entered once and the path of the derivative is shown in the bottom half of the screen. Figure It would appear from Figure that the derivative of any exponential function is also an exponential function. For base the slope function is below the function x. For base 3 the slope function is slightly above the function 3 x. For some base between.5 and 3, there is a unique exponential function whose rate of change is identical to the function itself. You may have guessed that the base for this unique exponential function is e, discovered by the mathematician Leonard Euler. The number e is also known as the base of the natural logarithm function. When we apply the limit definition f ( x + h) f ( x) lim to the function f(x) = e x x+ h x x h x x h e e e e e e ( e 1) we have lim = lim = lim. Since x is h 0 h h 0 h h 0 h h 0 h x h h e ( e 1) x ( e 1) ( e h 1) fixed, lim = e lim and we need to investigate the ratio as h approaches zero. Recall that in h 0 h h 0 h h Excel notation e x is =EXP(x) The table to the right suggests that e h is getting closer and closer to 1+h as h approaches zero. If we accept that ( e h 1) 1+ h 1 h = = 1 h h h as h approaches zero then we can conclude h x ( e 1) x e lim = e h 0 h h EXP(h) 1+h Joseph F. Aieta page 31 8/14/003

232 Measuring Change Section 5. Page 3 For any exponential function f(x) = b x, where b is a positive number other than one d e dx x is e x d x x and b is b ln(b) dx Example 5..4 d d (i) x x is ln() (ii) 3 x x d is 3 ln(3) (iii) ( 1.04) x x is 1.04 ln(1.04) dx dx dx From (i) and (ii) we see that the exact value of d x 0.5 at x = 0.5 is ln() dx = to nine decimal places. The exact value of d 3 x 0.5 at x = 0.5 is 3 ln(3) dx. When we use Excel to evaluate ln(3 ) to nine places, we get Compare this with from the difference quotient in Figure with h = If an investment was originally worth $10,000 and is growing at 6% per year compounded monthly (or 0.5 % per month) then each month it is worth 0.5% more than it was one month ago. At the end of month 30 it is worth $11, and at the end of month 31 it is worth $11, Exactly how fast is the investment increasing at the particular point in time t =30 where t is measured in months. The derivative of 10,000( ) t is 10,000 ( ) t ln(1.005) At month 30 this account is increasing at the rate of 10,000 ( ) 30 ln(1.005) dollars per month or roughly $57.93 per month. At points after t = 30 the instantaneous rate of change will be higher. The Natural Logarithm The function y = ln(x) is defined only for positive values of x. We estimate the derivative at the point ( 4, ln(4) ) to be 0.5. Recall that the natural log is =LN(x) in Excel. In Figure we observe that the function itself increases steeply for values 0 < x < 1 and continues to increase but at a slower rate for values of x to the right of x = 1. In other words, the first derivative of ln(x) is very large for x near zero and then becomes smaller as x gets larger and larger. These observations are consistent with the fact that the derivative of the natural log function is the reciprocal function. At point P ( 4.00, ) slope is 0.50 f(x) =LN(x) 3 function and its derivative , , Figure d x) is 1 dx x ln( for all values of x > 0 Joseph F. Aieta page 3 8/14/003

233 Measuring Change Section 5. Page 33 You may recall that the functions e x and ln(x) are inverse functions and that their graphs are mirror images over the line y = x. In the exercises following section 5.4 you will be asked for an algebraic argument supporting the derivative rule above. The function f(x) = e x is concave up at all points and its inverse function g(x) = ln(x) is concave down at all points. For functions that do change concavity, we will need a strategy for locating points of inflection. Derivative Tests Figure 5..0 contains a table of values for the function f(x) = 3x 4-4x 3-5x + 6 and its first derivative f '(x) = 1x 3-1x - 10x =x (6x - 6x 5). One root of the first derivative is at x = 0 but finding the other roots requires more work. Column BColumn D x f(x) f'(x) Row 6 R o w Sign change in f (x) indicates a stationary point in the subinterval [-0.65, ] Another stationary point is located at x =0.The first derivative f (x) is positive to the left of x =0 and f (x) is negative to the right of x = 0. Sign change in f (x) indicates another stationary point in the sub-interval [1.5, 1.65] x f(x) f'(x) f' <0 local Minima Solver f' > f' >0 local Maxima f' < f' <0 local Minima Solver f' > Figure 5..0 To find stationary points with Solver we first locate an interval that contains a root. We see that one root is between and so we use Solver to drive the target cell D6 to zero by adjusting the changing cell B6 that contains the initial guess of Solver returns the value which can be displayed to higher levels of accuracy as desired. Joseph F. Aieta page 33 8/14/003

234 Measuring Change Section 5. Page 34 The dialog box in Figure 5..1 illustrates how we found the root at x = The target cell (D6) contains a value evaluated by the formula 1x 3-1x - 10x where x is the value contained in the changing cell (B6) Figure 5..1 We see from the table in Figure 5..0 that there is another sign change between x = and x = We use Solver again, this time with cell D as the target cell and B as the changing cell to find the second stationary point at x = Figure 5.. We could have used Goal Seek instead of Solver if we needed only accuracy to three decimal places. It is also possible to find the other roots of the first derivative by factoring 1x 3-1x - 10x as x (6x - 6x 5) and then applying the quadratic formula to solve 6x - 6x 5 = 0. Joseph F. Aieta page 34 8/14/003

235 Measuring Change Section 5. Page 35 Concavity The first derivative of function of the fourth degree polynomial f(x) = 3x 4-4x 3-5x + 6 is the third degree polynomial f '(x) = 1x 3-1x - 10x. This new function f '(x) also has a derivative. We refer to the derivative of the derivative of f as the second derivative of the function f, denoted by f ''(x), and pronounced f double prime of x. The first derivative f '(x) = 1x 3-1x - 10x gives us information for locating stationary points of the original function f(x) 3x 4-4x 3-5x + 6. The second derivative f ''(x) = 36x -4x-10 tells us about the rate of change of the first derivative and gives us some additional information about the function itself. x f(x) f'(x) f''(x) This sign change in f (x) indicates the existence of a point at which the first derivative changes from increasing to decreasing. This sign change in f (x) indicates the existence of a point at which the first derivative changes from deceasing to increasing. Figure 5..3 A curve will be concave down at a point if its second derivative is negative at that point. A curve will be concave up at a point if its second derivative is positive at that point. A point at which a curve changes concavity is called a point of inflection. Candidates for inflection points are roots of f ''(x) = 0. From the table in Figure 5..3, we verify that x = and x = are indeed inflection points for the function f(x) = 3x 4-4x 3-5x + 6. The second derivative changes from positive to negative in the neighborhood of x = The second derivative changes from negative to positive in the neighborhood of x = Joseph F. Aieta page 35 8/14/003

236 Measuring Change Section 5. Page 36 At this stage, we have detailed information about the function f in the interval [-1, ]. We first solve f (x) = 0 and f ''(x) = 0, then we check for sign changes to determine the locations of possible maxima, minima, and inflection points. We conclude that there could be no other maxima, minima, or inflection points for the fourth degree polynomial f(x) = 3x 4-4x 3-5x + 6 since there can be at most three real roots of the cubic equation f (x) = 0 and at most two real roots of the quadratic equation f ''(x) = 0. Example 5..5 f(x) = 3x^4-4x^3-5x^ Figure 5..4 A dynamic illustration of the relationship between a function g, its derivative g', and its second derivative g'', is found in the file Numerical_vs_Symbolic.xls. The thick function shown in Figure 5..5 is the cubic polynomial g(x) = x 3 18x + 60x 6. The functions g, g', and g'' are each defined in the Visual Basic Editor. As x increases from left to right in the interval [, 10], we observe the position of the point on the function itself, the corresponding point on its first derivative, and the corresponding point on its second derivative Function g(x) g = x ^ 3-18 * x ^ + 60 * x - 6 End Function Function g_prime(x) g_prime = 3 * x ^ - 36 * x + 60 End Function Function g_double_prime(x) g_double_prime = 6 * x - 36 End Function Point P ( 6.00, ) slope is second derivative is Figure 5..5 Joseph F. Aieta page 36 8/14/003

237 Measuring Change Section 5. Page 37 We see that the function f increases from x = - to x = and then begins to decrease. The slope function, f ', is at a low point at x = 6. The second derivative, f '', is changing from negative on the left side of x = 6 to positive on the right side of x = 6. To the immediate right of x = 6, the original curve is still decreasing but not so rapidly. The inflection point at (6, -78) has the property that it is the point of most rapid decrease of the function f. At the point at, or very near, x = 10 the curve has a relative minimum. To the right of that point the curve begins to increase again. Example 5..6 For the function f(x) = x e x +3x in the closed interval [-, 4] find any turning points or inflection points. Use the formulas for the first and second derivatives, f '(x = 4x e x +3 and f '' (x) =4 e x, to create a table with columns for x, f(x), f '(x),and f '' (x). Look for changes in sign of the first derivative to locate possible local maxima or local maxima. Look for changes in sign of the second derivative to locate possible inflection points. Values of x that make the first derivative or the second derivative zero can be found to desired levels of accuracy using Solver. x f(x) f ' (x) f '' (x) x f(x) f ' (x) f '' (x) First derivative local sign change Minimum Second derivative point of sign change inflection local First derivative Maximum sign change Figure 5..6 Figure 5..7 The sign changes, framed in boxes in Figure 5..6, locate the intervals that contain roots of f '(x) or f '' (x). After applying Solver in Figure 5..7 we see that there are stationary points at x = and x =.593. We conclude that (-0.615, -1,69) is a local minimum since the first derivative changes from negative to positive in the vicinity of x = We also know that the curve is concave up at this point since the second derivative is positive. The point (.593, 7.856) is a local maximum since the first derivative changes from positive to negative in the vicinity of x=.593. We also know that the curve is concave down at this point since the second derivative is negative. The point (1.386, 4.003) is an ordinary inflection point where the curve changes from concave up to concave down. The second derivative changes from positive to negative in the vicinity of x = This inflection point marks the point of most rapid increase of the function f(x). Joseph F. Aieta page 37 8/14/003

238 Measuring Change Section 5. Page 38 A sketch of the function f(x) = x e x +3x is shown in Figure f(x) = *x^-exp(x)+3*x Figure 5..8 Joseph F. Aieta page 38 8/14/003

239 Measuring Change Activities Measuring Change Activities Page a) Find the exact location of all the relative and absolute maxima and minima. b) Find the coordinates of any inflection points. 1. g( x) = x 3 1x over the interval [-4,4] ) = with no restrictions on t g ( t t t + t g( x) = x 4 x for x > 0 f ( x x x 4. ) = ln ( ) for x >0 5. t g( t) = e t over the interval [-1,1] 6. 3 f ( x) = x 6x + 1x 3 over the interval [-,4] 7. f (x) = (6-x) (6-x)x for 0 < x < 3 8. Sketch a rough graph of a function f with domain the set of all real numbers, with the following properties: f is not linear f and has no relative maxima or minima 9. The cost in thousands of dollars for airing x television commercials during a Super Bowl game is given by C( x) = x 0.00x a) Find the point marginal cost function and use it to estimate how fast the cost is going up when x = 4. Compare this with the exact additional cost of airing the fifth commercial. b) Find the average cost function, and evaluate the rate of change of average cost when x = 4. What does the answer tell you? 10. Suppose that P(x) represents the profit on the sale of x videocassettes. You know that P(1000) = 3000 and P (1000) = -3. What do these values tell you about the profit function at x = 1000? Joseph F. Aieta page 39 8/14/003

240 Measuring Change Activities Page The percentage of U.S. car sales that were compacts has been obtained for the following years: 1980, 1985, 1990, 1991, 199, 1993, 1994, 1995, 1996, 1997, 1998, and 1999 The 4 th degree polynomial P(x ) = x x x x is a reasonably good fit to the data where x is years since 1980 and P(x) is market share given as a percentage. (Source: United States. Department of Transportation, Bureau of Transportation Statistics. National Transportation Statistics a) Use the graph to give a dp rough estimate for dx when x = 15 (1995). b) Use the equation for P(x) to determine P (15). c) Interpret your answer to part b. Percentage Market Share of Compacts years since The cost of controlling emissions at a firm goes up rapidly as the amount of emissions reduced goes up. In other words, if 90% of the emissions have already been controlled, then bringing 5% more under control will cost considerably more than controlling 5% more if only 60% of the emissions had been brought under control. 4 3 Suppose C ( q) = q is a model for the cost function where q is the reduction in emissions ( in pounds of pollutant per day) and C(q) is the daily cost (in dollars) of this reduction. a) If a firm is currently reducing its emissions by 64 pounds each day, what is the point marginal cost of reducing emissions further? [This is quicker to compute than the additional cost of reducing emissions by one more pound, C(65) C(64) which is $3,08.30 ] b) Government clean-air subsidies to the firm are based on the formula S(q) = 4000 q where q is again the reduction in emissions (in pounds per day) and S(q) is the subsidy (in dollars). At what reduction level does the marginal cost surpass the marginal subsidy? Joseph F. Aieta page 40 8/14/003

241 Measuring Change Activities Page Before the Alaskan pipeline was built,, there was speculation as to whether it might be more economical to transport the oil by large tankers, The following cost equation was estimated by the National Academy of Sciences: C( t) = where C(t) is the cost in dollars of transporting a barrel of oil 1000 nautical miles and t is t t the size of an oil tanker in deadweight tons. a) How much would it cost to transport a barrel of oil 1000 nautical miles in a tanker that weighed 1000 tons? b) By how much is this cost increasing or decreasing as the weight of the tanker increases beyond 1000 tons? 14. Explain the difference between average cost and marginal cost a) in terms of mathematical definitions in words and symbols b) in graphical terms 15. The population of a country was 4,000,000 at the start of 1990 and was doubling every 10 years. How fast was it growing per year at the start of 1995? ( to the nearest thousand people per year). 16. If $10,000 is invested in a savings account yielding 4% per year, compounded semiannually, how fast is the balance growing after 3 years 17. If f(x) = e x then what is its derived function f (x)? 18. Annual expenditures at private colleges in constant ( dollars) can be modeled by the cubic function E(x) = 3.0x x x where x is the number of years since 1970 and E(x) is in millions of dollars. a) How fast were expenditures increasing or decreasing in 1980? b) In what year from 1970 to 000 were expenditures growing most rapidly? Source: National Center for Education Statistics Joseph F. Aieta page 41 8/14/003

242 Measuring Change Activities Page Find the coordinates of all local and endpoint optimum points, and all stationary inflection points f ( x) = 6x x 9x. no restrictions on x 0. f 3 1 ( x) = x 1x. x > f ( x) = 4x + x > 0 x. 3 f ( x) = x 9x + 7x on [ 1,6] f ( x) = x 4x + 30 on [,4]. 4. Given the function C x R x a x p ( ) = + where x is positive and R, a, and p are positive constants with p > 1, find any stationary points in terms of the parameters R, a, and p. If this function has any stationary point(s) can you determine what type (maximum, minimum, or stationary inflection) without knowing specific values for the parameters R, a, and p? Joseph F. Aieta page 4 8/14/003

243 Measuring Change Section 5.3 Page 43 Non-Linear Optimization Section 5.3 The process of formulating and solving a non-linear optimization problem has certain parallels with the process of solving linear optimization problems, also known as linear programming. The table below is a deliberate over-simplification designed to compare and contrast a familiar problem solving strategy (on the left) with strategies (on the right) that you will employ and refine as you gain experience with non-linear optimization. The first and foremost stage in solving quantitative problems is the development of a mathematical model. This requires a translation from words and/or diagrams into an organized algebraic representation of the underlying quantitative relationships embedded in the statement of the problem. Linear Programming Formulation Non-Linear Optimization Two or more variables Objective function is linear Constraints: Linear inequalities or linear equations Graphical Method for the two decision variable case Simplex method implemented with the Solver Add-in to Excel Identification of the maximum or minimum value of the objective function and the corresponding values of the decision variables. Identify the decision variable or variables Determine the objective function Determine constraints or conditions Solution Paper and pencil Computer based Interpretation of numerical results Reducible to a single variable 8 Objective function is non-linear Conditions: Interval confining the decision variable and/or a relationship between two decision variables that enables one or more decision variables to be expressed in terms of a single variable. Algebraic methods using calculus finds exact solutions Search (guess and check) algorithms to find numerical results implemented with the Solver Add-in to Excel. Identification of the optimal value of the objective function and the corresponding value of the decision variable and other variables that depend on the decision variable. Once a non-linear optimization problem is correctly formulated, algebra and calculus may be applied to obtain an exact solution or Excel s Solver may be applied to obtain a numerical solution to a reasonable degree of accuracy. Solver can be used to verify results obtained with calculus or vice versa. Just as the graphical method for linear optimization can provide certain insights (such as alternate optima), that are not always apparent from Solver s reports, the calculus solution to a non-linear optimization problem can provide valuable insights about the relationship between an optimal solution and the problem parameters. Regardless of whether the numerical results are obtained algebraically or generated from a computerbased algorithm, it is important to give a concise interpretation of the numerical results in the context of the given problem. 8 The calculus of several variables is very important in applications but our work will be confined to the calculus of a single variable. Joseph F. Aieta page 43 8/14/003

244 Measuring Change Section 5.3 Page 44 Surprising as it might seem, a formulated quantitative problem that leads to a recognizable equation, may not have a solution that can be described in algebraic terms. For example, it is true that all solutions of an equation of the form ax + bx + c = 0 can be described by a formula (known as the quadratic formula), but there is no formula at all, simple or complex, for the roots of a general fifth degree polynomial. We will not encounter any fifth degree polynomials in the following applications but we will occasionally encounter an equation whose solution cannot be described by a closedform algebraic expression. Even when a problem can be solved entirely with algebra, its solution may require a level of manipulative skill that is not universal among students. To be fair, we should point out that, under certain circumstances, computer-based numerical algorithms can also fail to find an optimal solution even though an optimal solution can be shown to exist. The following problem is an example of a classical non-linear maximization problem. We will approach this problem three ways: (1) with Solver and no graph, () with a graph and Solver, and finally (3) with calculus and no graph. Example A flat sheet of material with dimensions 8.5 by 11 inches is to be shaped into an open box by cutting identical squares from each of the four corners and then folding up the sides x x x x x x x x x Figure Let the input variable x represent the side of the cut-out squares. The height of the open box will be equal to x and the other two dimensions will be length = 11- x and width = 8.5 x. The objective is to find the length, width, and height of that particular box that has the greatest volume. [Imagine if you will that you are constructing an open box that is to be filled with gold dust and then given to you as a gift]. Formulation: Maximize Volume V(x) = (11 x) (8.5 x) x subject to 0 < x < 4.5 Joseph F. Aieta page 44 8/14/003

245 Preparing the Spreadsheet for Solver Measuring Change Section 5.3 Page 45 Start with a blank worksheet and enter the labels, values, and formulas as shown in the first rows below. The value of 1.0 in cell C has been entered as an initial guess and 4.5 is the current upper bound in our search. Use Insert Name Create Top Row to name the cells in row using the labels in row 1. Formulas, rather than values, are shown in columns A, B, and E of Figure 5.3. and corresponding values are shown in Figure A B C D E 1 L W x bounds Volume =11-*x =8.5-*x =L*W*x Figure L W x bounds Volume Load Excel Solver Figure Complete the dialog box as shown in Figure Unlike linear programming, this is not a Linear Model. Figure Click on Solve. After applying Solver, we obtain the results shown in Figure A B C D E 1 L W x bounds Volume Interpretation: Figure The maximum volume of the open box is about cubic inches which corresponds to a box with dimensions 7.89 by 5.39 by with accuracy to four decimal places. Joseph F. Aieta page 45 8/14/003

246 Measuring Change Section 5.3 Page 46 Using the Graphing Utility Plot&Trace.xls Open the file and enter the cubic function = (11 *x) * (8.5 *x) *x into the blue cell under the button new_f. Press <Enter> and then press the macro button new_f. Assume that 1 has been entered as the initial guess. x min x incr x max y min y max guess bounds target x new_f f(x) = (11-*x)*(8.5-*x)*x , , a f(a) Invoke Solver and complete the dialog box in order to maximize the target cell by changing the value in the changing cell guess Add the constraint that sets an upper bound of 4.5 on the search for an optimal solution. Figure Do not make any changes in Options. Click on Solve to obtain the values that are shown in Figure Numbers are displayed here to 4 decimal places but can be changed to display higher or lower levels of accuracy. Figure Joseph F. Aieta page 46 8/14/003

247 Calculus Approach Measuring Change Section 5.3 Page 47 Calculate the first derivative of the volume function. Candidates for maxima/minima are stationary points that can be found by setting the first derivative equal to zero and solving for x. We can justify our final conclusion by referring to the graph or by applying a test that examines the sign of the first derivative on each side of a stationary point. V(x) = (11 x)(8.5 x)x =[(11)(8.5) -17x - x + 4x ] x = 93.5x 39x +4x 3 = 4x 3 39 x x V (x) = 1 x - 78x Now solve V (x) = 0 1x - 78x = 0. Formulate the volume function in terms of x as length, 11 x, times width, 8.5 x, times height x. Distribute the product of the three linear factors to obtain a cubic polynomial in the form V(x) = ax 3 + bx +c x + d where a = 4, b = -39, c = 93.5 and d = 0 Apply the simple power rule and other derivative rules to obtain the first derivative V (x) which is a quadratic of the form Apply the quadratic formula: 78 ± x = (1)(93.5) If ax + bx +c = 0 then b ± x = b 4 a c a x = or x = To four decimal places x = or x = V (1) = = 7.5 > 0 and V () = = = < 0 V(1.5854) is approximately Reject the second root since it exceeds 4.5 (it also happens to be a local minimum). Check the first derivative to the left (x = 1) and to the right (x = ) of the root x = We conclude that the volume function has a local maximum at x = The optimal point on the volume function is (1.5854, ). This example was introduced as a classical non-linear maximization problem. It is classical in the sense that just about every textbook on differential calculus includes this type of problem. It is part of a group of classical problems in a category that we will call Container Problems. In the open box problem, the condition that had to be satisfied was that x, the side of the cut-out square could not exceed twice the width (shorter dimension) of the flat sheet. Other problems in this category often include conditions that specify that a certain area or volume must remain constant. We will investigate some other container problems later but first we turn our attention to a managerial problem related operations. Example 5.3. Optimal Batch Size (also known as the Economic Order Quantity or simply EOQ) This problem focuses on two types of cost that do not include the cost of materials. The first cost is associated with the setup for production. The second cost is related to the storage of items that are periodically produced in batches and ultimately purchased by customers at a steady rate over the course of a year. Joseph F. Aieta page 47 8/14/003

248 Measuring Change Section 5.3 Page 48 Applied Electronics operates a versatile assembly plant that can produce a variety of electronic devices. Rather than dedicate their production line to any one product for a long period of time, they periodically produce batches of a particular component and ship each batch to a storage facility. We will assume that the assembly plant operates 360 days per year. The annual demand, D, for one particular component is 7,000 units per year. Each production run has an associated setup cost, S, of $3,000 per batch. The holding cost, H, of storing one item in inventory is currently $.50 per year. Note that the cost of materials or manufacturing costs other than setup costs and inventory costs are not under consideration in this business model. Consumption of the component is fairly uniform throughout the year so one option is to produce the items in batches of 6000 which would mean 1 batches per year at an annual setup cost of =$ 36,000. Another option would be to produce batches of 10,000 which would correspond to 7. batches per year or one batch every 50 days with annual setup cost of $1,600. You can think of the 7. batches per year as the average number of batches per year over several years so 7. batches per year is equivalent to 36 batches over a period of 5 years. [Averaging over the long run is a common strategy for computing costs over short time periods.] Calculating the holding cost, or inventory cost, also requires an averaging concept. There are several ways to determine holding cost over a one year period (just as there are several ways to calculate annual depreciation ---the simplest of which is straight-line depreciation. The simplest method assumes that the items in a batch are sold at a uniform rate throughout the year. In the case of batches of 6,000 for example, the number in storage declines from 6,000 to 0 at a constant rate over each 30 day period as shown in Figure We interpret the graph in Figure to indicate that Applied Electronics delivers the first batch of 6,000 units and then delivers the next batch 30 days later just in time to replenish the supply. On average, storage is 3,000 units over this time period, half way between 6,000 and 0. This situation is the same for each production run so we will assume an average of 3,000 units in inventory over the course of an entire year. units day Figure Assuming an annual storage cost of $.50 per item, the annual storage cost is estimated to be.50 3,000 = $7,500. Figure illustrates a larger batch size of 18,000 items. On average, there are 18,000/ = 9,000 units in inventory over the course of a year. At a cost of $.50 per item, the annual inventory cost would be: = $,500. Why would the company consider paying $,500 for storage when it could pay $7,500 with a batch size of 6,000 items? units day Figure The answer of course is that the company is interested in minimizing the total cost of holding and setup. Joseph F. Aieta page 48 8/14/003

249 Measuring Change Section 5.3 Page 49 For a batch size of 18,000 there are only 4 setups per year at a cost of $3000 per setup or $1,000 in setup costs per year. The total cost of $1,000 plus $,500 is $34,500 for batches of size 18,000. Total annual cost for batches of size of 6000 would be $36,000 for setup plus $7,500 for inventory or a total of $43,500. Smaller batch sizes lead to lower holding cost but higher setup cost. Larger batch sizes, on the other hand lead to higher holding cost but lower setup cost. In this trade-off between holding cost and setup cost, the company managers want to know how to minimize total cost. Figure is taken from the file OptimalBatchSize.xls. A quick reading of the table for D = 7,000 units, S = $3000 per batch and H = $.50 per item, tells us that the optimal batch size is near 14,000 units. D = 7,000 S=3000 H=.50 Number Annual Avg. number Annual Setup Batch of Setup of units in Inventory plus Size Batches Cost inventory Cost Holding x D/x (S*D)/x x/ (H*x)/ (S*D)/x+(Hx)/, , , , , , , , , , , , , , , , , , , , Figure Use Solver to minimize setup plus holding cost by adjusting batch size (as we did in Example to maximize volume). Guess Number Annual Ordering Average number Annual Setup Batch of Setup of units in Inventory plus Size Batches Cost inventory Cost Holding 14, ,49 7,000 17,500 $3,99 Before Solver Interpretation: 13, ,43 6,573 16,43 $3,863 After Solver The company should schedule about 5.5 production runs per (11 runs every two years) in order to attain a minimum setup plus holding cost close to $33,000. Joseph F. Aieta page 49 8/14/003

250 Measuring Change Section 5.3 Page 50 Applied Electronics actually has several regular customers whose demand for electronic components range from 50,000 to 00,000 per year. Setup costs for the products that these customers demand range from $000 to $15,000 per setup. Annual holding cost per item ranges from $0.5 to $5.00. The company would like to be able to vary the parameters D, for annual demand, S for setup cost, and H for holding cost and immediately read off the optimal batch size. In the case where D= 150,000, S=$9300, and H = $4.50, the optimal batch size is 4,900 and the corresponding minimum cost is $11,049 to the nearest dollar. Figure is also taken from OptimalBatchSize.xls and includes a graph of the total cost as a function of batch size. 50K 00K D Annual Demand 150,000 K 15K S Setup per batch $9, H Inventory per unit $4.50 optimal batch size is 4,900 minimum cost is $11,049 $350,000 Total cost vs batch size $300,000 $50,000 $00,000 $150,000 $100,000 $50,000 batch size $0 0 10,000 0,000 30,000 40,000 50,000 60,000 70,000 80,000 90,000 Figure Joseph F. Aieta page 50 8/14/003

251 Measuring Change Section 5.3 Page 51 Calculus solution: Annual Demand is D, Setup Cost is S, and Holding cost is H. The input variable for batch size is represented by x and the output C(x) is in dollars. In the general case, the batch size that minimizes cost can be found with calculus as follows: C( x) = D x S + H x Setup cost is number of setups, x D, times S and holding cost is average inventory x times H D S H x C( x) = + = D S x x C (x) 1 H + x H D S H = D S x + = + x ) Solve C (x = 0 for x where x is positive D S H + x = 0, equivalent to D S H = x D S = H x D S, equivalent to x = H D S x = ± reject the negative root since x > 0 H Algebra Simplification Find the first derivative of the cost function using d x dx n = nx n 1 and other derivative rules Find candidates for maxima and minima Algebra Simplification Algebra Simplification Algebra Simplification The optimal batch size occurs at x = D S H DS The second derivative C ( x) = is positive 3 x for all positive values of x. We know that we have a local minimum since the curve is concave up at all positive values of x. Checking the results from Figure , when D= 150,000, S=$9300, and H = $4.50 Optimal batch size is 150, =4, or 4,900 to the nearest unit and ,000 4,900 Minimum cost is C (4,900 ) = = $11,049 to the nearest dollar. 4,900 Joseph F. Aieta page 51 8/14/003

252 Measuring Change Section 5.3 Page 5 Container City Container City s customers provide specifications for the containers that they order. The majority of these containers are rectangular boxes and cylinders. Jack and Jill are working at Container City while they are on break from business school. Their job is to describe the precise dimensions of containers that meet the customer s specifications. They then forward this information to the appropriate company subsidiary, depending upon what type of material is required in the production of the container. Typical container problems involve optimizing the area, volume, or construction cost related to some two or three dimensional geometric shape such as a rectangular box or cylinder. A natural first step in solving these classical geometric problems is to draw a picture and label it with as few variables as possible. Jack volunteers to solve problems with Excel and Jill elects to solve problems with calculus. Both of them have agreed to check the other person s results. Since many orders that require materials of different quality and cost, Jack and Jill are expected to figure out how to minimize Container City s cost. In the area where Jack and Jill do their work, the following formulas for geometric shapes are posted on the walls. a) Area of a rectangle with dimensions L and W is L W b) Perimeter of a rectangle is L + W c) Surface area for a rectangular closed box with dimensions H, W, and L for Height, Width, and Length is W H + L W + L H d) Volume of a rectangular box with dimensions H, W, and L is L W H e) Surface area of a circle is π radius where the constant π is approximately f) Perimeter of a circle (circumference) is π radius g) Surface area for the side of a cylinder is π radius height h) Surface area for the entire cylinder (side, top & bottom) is π radius + π radius height i) Volume of a cylinder is π radius height Example Container City Closed Box Problem Design a closed box with a square top and bottom. The volume of the box must be 50 cubic inches. Material for the sides costs $1.50 per square inch. The top and bottom material must be stronger and will cost $.00 per square inch. What are the dimensions of the box that will minimize the total cost of materials? L x x Figure Before Jack starts preparing his spreadsheet for Solver, he draws the figure above with the labels shown as the first step in the formulation --- The decision variable(s) are apparently L and x. In the next stage of the formulation, Jack needs to clearly understand the difference between a problem constraint and an expression for the objective function. The problem contains a volume constraint which can be stated algebraically as x x L = 50. The objective function is about minimizing total cost so Jack needs to think about how to express total cost in terms of variables and parameters. Total cost is the cost of the four sides plus the cost of the top and bottom cost of the four sides plus cost of the top and bottom 4 x L (cost per square inch of side material) plus x x (cost per square inch of material for the top & bottom) Joseph F. Aieta page 5 8/14/003

253 Measuring Change Section 5.3 Page 53 In designing a spreadsheet, Jack decides to create named cells for the values of the three parameters, volume, cost of side material and cost of material for the top and bottom. Jack creates another cell that will contain a guess for that value of x that minimizes total cost. His plan is to use Solver with the value of total cost as the target cell and the cell containing a guess for x as the changing cell. The other cells contain corresponding values of the length of the box, the cost of the sides, and the cost of the top and bottom. [If Jack planned on creating a table by copying formulas then he would make sure that certain cell references were absolute (fixed) references rather than relative references] Figure In B7, Jack needs to describe length L in terms of the volume in C1 and the guess in A7. The number in cell B7 is the value in C1 divided by the square of the value in A7. This comes from the volume constraint which is x x L = 50 or L = 50/x in equivalent algebraic terms. Since the cost of the sides is 4 x L (cost per square inch of side material), the value in C7 is = 4*A7*L*C Since the cost of the top and bottom is x x (cost per square inch of material for top & bottom), the value in D7 is =*A7*A7*B3 Finally the value in E7 is simply C7 + D7. The value in C7 depends directly on the value in A7 as do the values in B7, C7, and D7. Jack confirms that the formulas in these three cells are correct by performing the arithmetic 50/5 = 50/5 = 10 4(5)(10)(1.50) = 300 and (5)(5)(.00) = 100. Total cost is = 400. Can Jack discover dimensions of the box that correspond to a lower cost? Instead of playing guess and check manually, Jack uses Solver and finds cost of cost of total guess L sides top&bottom cost $6.07 $ $ Figure One of the first things that Jill does is to write the objective function Cost = x L +.00 x as 50 C(x) = x +.00 x which she simplifies to C(x) = x x x Complete the analysis with differential calculus. Joseph F. Aieta page 53 8/14/003

254 Measuring Change Section 5.3 Page 54 Example Container City Cylinder Problem Design a cylindrical container that will contain one gallon of liquid (approximately 31 cubic inches). The unit cost of the materials for the side is the same as the cost of materials for the circular top and bottom so finding the least expensive cylinder that satisfies the given condition is equivalent to finding the dimensions that minimize surface area Figure Diameter = r Jill decides that she will solve a more general problem that can be used with orders having different specifications for the volume of the container. This way she will make substitutions for Volume V if the specifications call for cylindrical containers that must hold a half-gallon or a quart. Jill s restated problem: If a cylindrical can must hold V cubic inches, then what are the dimensions of the can with minimum surface area. Recall the fact that surface area for the entire cylinder (side, top & bottom) is π r + π r h where r is the radius of the circular base and h is the height of the cylinder Jill will construct a function representing the total surface area in terms of the parameters V, for volume, and the input variable for the radius of the circular base of the cylinder. The sequence of steps that Jill adopts in her solution is shown below: Draw and label a figure Choose decision variables. Figure with labels r and h for radius and height Let r be the radius of the circular base of the cylinder Obtain an expression for the height h of the cylinder in terms of the volume V and radius r. Since V = π r h it follows that V h = π r Define a function A(r) for the total surface area of surface area = π r + π r h the cylinder in terms of the parameter V and the variable r. V V A( r) = π r + π r = π r + π r r Determine the first derivative of A(r). V A ( r) = 4π r r Find the value of r that minimizes A(r). V V 4π r = 0 is equivalent to 4π r = r r Verify that this value of r corresponds to a local minimum. equivalent to equivalent to 3 4π r = V 3 V V r = = 4 π π 3 so V r = π 4 V A ( r) = 4π + for all positive values of r so the curve 3 r 3 for A(r) is concave up and V r = is a local minimum. π 1 1 Joseph F. Aieta page 54 8/14/003

255 Measuring Change Section 5.3 Page 55 For the parameter value V =31 cubic inches the optimal value of r will be 31 r = =3.35 inches so the diameter of π the cylinder is inches and the height is 31 h = = Jack verifies this result with a straightforward π 3.35 spreadsheet analysis. Is this just a coincidence or is the diameter and height always the same for the cylinder with minimum surface area and fixed volume? 1 3 Example Container City UPS problem United Parcel Service charges extra for the delivery of packages whose length plus girth exceeds 108 inches. A customer wants to ship food products in rectangular boxes with square ends that meet the UPS requirement for lowest shipping rate. These boxes are to be made out of perforated material and will require the largest possible surface area in order to get as much ventilation as possible. The length of the box is L, and its girth is 4x. Find the values of x and L that will maximize surface area. L Girth in this case is 4*x L+4*x=108 L=108-4*x x x Figure Again, Jack puts a guess in the cell for x and constructs other cells that will all depend on the guess. The surface area of the four sides is 4 x L and the surface are for the square ends is x x The formula in B5 is =108-4*A5 The formula in C5 is =4*A5*B5 The formula in D5 is =*A5*A5 The formula in E5 is =C5 + D5 After we apply Solver to minimize E5 by adjusting A5, row 5 displays Figure The dimensions of the box with maximum surface area of square inches are by by 46.9 Jill recognizes that the graph of the quadratic surface area function A(x) = 4x (108-4x) + x is a parabola that opens down so any stationary point will be a local maximum. A(x) = 4x (108-4x) + x A(x) = -16 x + x + 43x A(x) = -14 x + 43x A' (x) = -8x +43. Now solve A' (x) = 0 to find stationary points -8x +43 = 0 x = 43/8 Check: The second derivative is A'' (x) = -8 < 0 which confirms that the stationary point is a relative minimum. Jill concludes that x = 43/8 = is the optimal value of x corresponding to a box with maximum surface area. Joseph F. Aieta page 55 8/14/003

256 Measuring Change Section 5.3 Page 56 Steps for Solving Non-Linear Optimization Problems Step 1: Step : Step 3: Step 4: Step 5: Step 6: Identify what is to be maximized or minimized (output) and the changing quantity or quantities (input). Sketch and label a picture if the problem has a geometric context. Build a function for the quantity that is to be maximized or minimized. If calculus will be used in step 6 then rewrite the function in terms of one input variable. Identify the appropriate interval for the input variable. Apply an optimization technique to determine the maximum or minimum of the final function and the corresponding optimal value of the input variable. Interpret numerical results in order to answer the original questions posed in the context of the problem. Joseph F. Aieta page 56 8/14/003

257 Measuring Change Activities 3 Measuring Change Activities 3 Page 57 Non-Linear Optimization 1. If we use 100 linear feet of fence to enclose a rectangular plot of land, we need x feet for one pair of sides (or x feet for one of these sides), leaving x feet for the other pair of sides (or 50- x feet for one of these sides). The rectangle is then x feet by (50 - x) feet and its area, A(x), is Ax ( ) = x( 50 x) = 50x x a) Find the value of x that maximizes the area of the plot. b) Prove that your answer to (a) is a maximum c) What are the dimensions of the rectangle with maximum-area? d) What is the maximum area?. Modify the Container City UPS problem such that each end of the rectangular box is not a square but a rectangle with one edge that is twice as long as the other edge. The three dimensions of the box in this case are x by x by L and the girth of the box becomes (x+x) = 6x. The UPS requirement for lowest shipping rate is that length plus girth must be less than or equal to 108 inches. Find the values of x and L that will maximize surface area and qualify for the lowest shipping rate. 3. A manufacturer can fill orders for up to 800 units per day of a particular piece of office equipment. For each day that the plant is in operation, the fixed costs are $3,600 and the variable cost of the product components is $180 per unit. The manufacturer has been selling the product to a chain of office supply retail stores at a base price of $70 per unit. To boost sales, the manufacturer is considering a sales incentive discount that would lower the unit price by $50 for every 100 units that the chain buys. In other words, if q units are ordered then the unit price will be dropped by 0.50q dollars. The revenue function is defined by R(q) =q (70-0.5q), which is a quadratic function of q. The manufacturer's objective is to maximize daily profit. a) What is the first break-even quantity? b) At the lower break-even point is the marginal profit positive or negative? c) What is the second break-even point? d) At the second break-even point is the marginal profit positive or negative? e) Between the first and second break-even points there is a point where the derivative of the profit function is zero. What is that point? Joseph F. Aieta page 57 8/14/003

258 Measuring Change Activities 3 Page In many sampling surveys only a small number n of a large population of people are interviewed. Published survey results state what proportion, p, of the n people interviewed answered yes to a particular question. Inherently, because only a fraction of the population is interviewed, the proportion p is expected to be in error. Statisticians measure this error by V(p), where V stands for variance, and p V( p)= n p n. Thus variance is a function of p where sample size n is some constant.. a) Find the value of p that maximizes V(p); that is, the proportion that leads to the largest expected error. b) Justify that your answer to part a) is indeed a maximum. c) If n =10, find V(0.5) and V(0.1). 5. A rectangular warehouse must have 400 square feet of floor area. The warehouse has four exterior walls. It also has one interior wall that divides the warehouse into two rectangular rooms. The total cost of the walls will vary depending on the outside dimensions of the warehouse. The cost of building exterior walls is $15 per linear foot and the cost of building interior walls is $80 per linear foot. The objective is to satisfy this floor area condition and complete the wall construction for minimum total cost. x x interior wall y interior wall y x interior wall parallel to horizontal dimension x interior wall parallel to vertical dimension a) Use Excel to find the dimensions that will minimize total cost of building the walls. b) Modify your spreadsheet in a) and show that the optimal dimensions do not depend on whether you formulate the problem using the figure on the left or the figure on the right. c) Parameterize the problem using A for the required floor area, ext for the cost per linear foot of exterior walls, and int for the cost per linear foot of the interior wall. Write a function of one variable (x or y) that describes total cost in terms of A, ext, and int, and your choice of x or y for the input variable. [You will need to write an expression of one of the variables in terms of the other]. d) Use calculus to describe the optimal dimensions in terms of A, ext, and int 6. A cylindrical storage tank is to contain V = 16,000π cubic feet (about 400,000 gallons). The cost of the tank is proportional to its area, so building a tank for minimal cost will correspond to building the tank which has the minimum surface area. Find the dimensions, r and h, of the tank with the minimal surface area. Joseph F. Aieta page 58 8/14/003

259 Measuring Change Activities 3 Page Jill has constructed the cost function C(x) = x +.00 x x Container City Problem 1 for building a rectangular box with different materials. in order to check Jack s results in Jill simplifies the cost function as C(x) = x. Complete the analysis using calculus. x 8. A temporary holding tank for industrial waste is a cubical container of side x feet (surface area = 6x square feet and volume = x3 cubic feet). The cost to build this container is $ per square foot and the container must be decontaminated and replaced at the end of a year. The cost to decontaminate the interior of the container is $0.50 per square foot. Each time the container fills with industrial waste it must be emptied and its interior of 6x square feet must be decontaminated. The manufacturing process generates 4,096 cubic feet of waste per year. If the side of the container is 16 feet then it could hold all of the waste produced in one year at a total annual cost of $ If the side of the container is 4 feet then it could be built for only $19 but would require 64 decontaminations costing $307 for a total annual cost of $364. Suppose the cost to build the container is b dollars per square foot, the cost to decontaminate is d dollars per square foot, and the annual amount of industrial waste is W cubic feet. What are the parameterized expressions for: a) the cost of constructing the container b) the number of decontaminations per year? c) the yearly decontamination cost? d) the yearly sum of the container cost plus the decontamination cost? e) the container dimension, x that minimizes the total annual cost? f) What is the optimum container dimension for the parameter values of b, d, and W given in the problem statement? 9. Farmers use a certain plant food costing $4 per ounce to help them in growing oranges. It is estimated that when x ounces of the food are used on an acre of orange grove, the farmer is able to get ln(4x + 5) crates of oranges from that acre. If the farmer can sell the oranges at $0 per crate, how many ounces of plant food should be used per acre to maximize the net value of the orange crop per acre? Note: ln is the natural log function 10. Suppose the annual retailer demand for a product is 60 units and the fixed cost of a purchase order is $400. The unit base price is ordinarily $5,500 but the wholesaler offers a discount on the unit price to $5,400 when ten or more units are ordered. Carrying costs are estimated at $100 per unit. We will assume that an average of Q/ units is held in inventory where Q is the order size. Total annual cost is base price*demand + annual ordering cost + annual inventory cost a) What is the theoretical total annual cost if the size of the regular order is Q=5 b) What is the theoretical total annual cost if the size of the regular order is Q=0? c) What is the annual cost if exactly four orders are placed d) If exactly two orders are placed each year then how close is the cost to the theoretical minimum cost? Joseph F. Aieta page 59 8/14/003

260 Measuring Change Activities 3 Page Solve by a spreadsheet analysis. C By the Pythagorean Theorem AC=1 mile CD=sqrt(1^+x^) A D B x x AB = Jack is living on an island (point C), one mile away from the nearest point on the shore (point A). He goes to visit Jill who lives on the mainland three miles down the coast (point B). Jim can get to the shore in his fishing boat and then jog to Jill s house. Jack can run 6 miles per hour and he can travel 4 miles per hour in his fishing boat. a) Where on the shore should Jack land if he wants to make the trip as fast as possible? b) How long would it take Jack to make this fastest trip? c) If Jack decided to go all the way to Jill s house by boat, how much longer would his trip take than the shortest possible trip? 1. Green Buckets, Inc. has an order for cylindrical plastic buckets that hold 5000 cubic centimeters. The buckets are open-topped cylinders. The company wants to know what bucket dimensions will satisfy the volume condition and use the least amount of plastic 13. The Chocolate Box Co. is going to make open-topped boxes out of 6 by 16 rectangular sheets of cardboard by cutting squares out of the corners and folding up the sides. What is the largest volume box it can make this way? 14. A warehouse is to have an interior wall that separates the warehouse into two rectangular rooms. The cost of building exterior walls is $15 per linear foot and the cost of building interior walls is $80 per linear foot. The amount of money allocated to build the walls is $49,500. a) What are the dimensions of the warehouse with the largest total floor area that can be built under this cost constraint? b) What is the maximum floor area to the nearest whole number of square feet? c) Suppose the larger rectangular room must have twice the area of the smaller room. How does this new constraint impact the solution to parts a) and b)? x interior wall x y Joseph F. Aieta page 60 8/14/003

261 Measuring Change Activities 3 Page A box with a square bottom and no top is to be made to contain a volume of 500 cubic inches. What should be the dimensions of the box if its surface area is to be a minimum? What is this minimum surface area? 16. A bank has purchased a rectangular piece of property with dimensions 00 ft. by 100 ft. The bank officials intend to develop the property as a parking lot. The town, in which the bank is located, has a green space requirement for any commercial property that faces a public walkway. The current regulation is that at least 10% of the property must be set aside as green space. The bank officials decide to put the green space at the corners of the lot and enclose each rectangular piece with a fence on three sides as shown in the figure below. The cost of fencing is $18 per linear foot and the cost of paving is $.50 per square foot. sidewalk y y x x width length a) What is the minimum total cost of fencing the green space and paving the rest of the lot and what are the corresponding dimensions of the unpaved areas at the corners of the lot? b) Find the minimum cost if the town requires 15% of the lot to be green space. 17. In their surveys, the Bureau of Justice Statistics Correctional Surveys differentiates between prison and jail as follows: Prison - confinement in a State or Federal correctional facility to serve a sentence of more than 1 year, although in some jurisdictions the length of sentence which results in prison confinement is longer. Reference date is December 31 of the given year. Jail - confinement in a local jail while pending trial, awaiting sentencing, serving a sentence that is usually less than 1 year, or awaiting transfer to other facilities after conviction. Reference date is June 30 of the given year. Joseph F. Aieta page 61 8/14/003

262 Measuring Change Activities 3 Page The values in the table below are from the survey completed on 08/7/01 adj. year Year Xx in Prison , , , , , , , , , , , , , , , ,078, ,17, ,176, ,4, ,87, ,31,354 a) Create a spreadsheet with the three columns shown and an additional column for the cubic function that best fits the data from 1980 to 000 (X=0 to 0) b) Add columns for the first and second derivative of this cubic. c) How fast was the prison population changing in 1995? d) Approximately when was the prison population growing most rapidly? Source: Bureau of Justice Statistics Correctional Surveys (The Annual Probation Survey, National Prisoner Statistics, Survey of Jails, and The Annual Parole Survey.) 18. The table below shows the average U.S. household credit card debt for 1990 (X= 0) through 001(X=11) Source: year X Y $,985 $3,3 $3,444 $3,601 $4,811 $5,83 $6,487 $6,900 $7,188 $7,564 $8,13 $8,367 a) In what year was the percentage increase the lowest? b) Find the equation for the cubic function C(X) that best fits the data where X is the number of years after 1990 and C(X) is in dollars. c) Calculate the first and second derivatives C' and C'' for the fitted cubic. d) What was the approximate rate of change for average household credit card debt in 199? e) In which year did the average credit card debt increase most rapidly? Joseph F. Aieta page 6 8/14/003

263 Measuring Change Activities 3 Page Data in the table below have been compiled from tariff and trade data from the U.S. Department of Commerce, the U.S. Treasury, and the U.S. International Trade Commission. U.S. Imports of Toys, Games, and Dolls imports in year X thousands of $ ,110, ,896, ,445, ,041, ,53, ,06, ,970, ,094, ,385,764 a) Create a scatter plot of imports versus adjusted year X, where X is the number of years since 199. b) Explain why a quadratic function would not be a very good choice as a model for this data. c) Fit a cubic polynomial to the data and determine the first coordinate of the observed point that is furthest from the fitted curve. d) In what year was the value of these imports increasing most rapidly? e) Can we predict the year in which the value of these imports will drop below ten billion dollars? Explain The polynomial D(t) = t t t t is the fourth degree function that best fits the data on total farm business debt from 1980 to 000 where t is the number of years after 1980 and D(t) is in millions of dollars a) Evaluate D(10) and compare it with the actual value in b) What was the actual change in total farm business debt from 1985 to 1986? c) Use Excel s Goal Seek or Solver to find the coordinates of each point of inflection on the function D(t) in the interval [0,1] and give an interpretation of each value. Joseph F. Aieta page 63 8/14/003

264 Measuring Change Section 5.4 Page 64 Composite Derivative Rules and More Non-Linear Optimization Section 5.4 In order to solve non-linear optimization problems with calculus, we must be able to find the derivatives of a wide variety of functions. It also helps to be proficient with manipulative algebra. The additional derivative rules that we will need for problem solving are rules for dealing with functions that are built up by composing two (or more) functions in a particular way. We have seen situations in which one function produces output that then becomes the input to another function. Suppose one function has a rule that says take any input number, double it, and then subtract 1.and the rule for a second function says take any input number and raise it to the third power. We will start with x = 3 and perform the first function to get (3) -1 = 5. We then feed this result to the second function to get 5 3 = 15. The first function can be described symbolically as u(x) = x 1 and the second function as v(x) = x 3. The result of first performing function u and then performing function v is a new function f with the rule f(x) = (x - 1) 3. The simple power rule tells us that the derivative of x 3 (x - 1) 3. is dv x dx 3x 3 = but we want the derivative of a function to a power, namely What applies here is the function power rule. In differential notation, the function power rule says d dx u n = n u n 1 du dx where u is some function of x. This is an extension of the simple power rule in that we start by writing the power times the function raised to the power minus 1 and then we must multiply by the derivative of the inside function. To find the derivative of (x - 1) 3, the inside function is u(x) = x 1 and the derivative of this inside function is du d = ( x 1) = dx dx. The derivative of u(x) du du becomes 3 u(x) = 3 u( x). Substituting x 1 for u dx dx d 3 du we have ( x 1) = 3(x 1) = 3(x 1) = 6(x 1) dx dx Since the expression (x - 1) 3 can be expanded algebraically, we can convince ourselves that this rule is consistent with the earlier derivative rules. (x - 1) 3 is equivalent to (x - 1) (x - 1) which is equivalent to (4x -4x+1) (x-1) = 8x 3-8x + x (4x -4x+1) = 8x 3-8x + x 4x + 4x-1 = 8x 3-1x 3 + 6x -1 and ( 8x 1x + 6x 1) = 4x 4x + 6 derivative rules. d dx from our familiar How does this compare with the derivative 6(x-1) obtained by the function power rule? Expanding 6(x-1) we get 6( 4x - 4x+1) = 4x - 4x+6 and we see that these rules produce identical results. A rigorous mathematical proof of the function power rule requires limit theory. Joseph F. Aieta page 64 8/14/003

265 Measuring Change Section 5.4 Page 65 Here are two more applications of the function power rule followed by algebraic simplification of the result. Example Let f(x) = ( 6x + 3). 3 Here the outside function says raise the quantity to the 3/ power. The quantity is the inside function du u(x) = 6x+3 and the derivative of the inside function = 6 so dx d ( 6x + 3) = (6x + 3) 6 = 6(6x + 3) = 9(6x + 3) dx Example Let f(x) = = 6 ( x + 4x) Here the inside function is u(x) = x + 4x. The outside function is the reciprocal x + 4x function 1 f ( x) = and 6 is just a constant, x d dx 6( x + 4x) 1 = 6 ( 1)( x + 4x) (x + 4) 6(x + 4) which can be written without negative exponents as ( x + 4x) We now have the necessary calculus tools to tackle the following pipeline problem. A pipeline that carries both natural gas and coaxial cable is to be extended from the main line starting at point A and ending at point B located on an island. Point C is directly landward 3 miles from the island. The distance from point A to C is 8 miles. A contractor bids on the job and quotes a cost of $1. million per mile for laying pipeline under water and $750,000 or $0.75 million for laying the pipeline along the mainland. The municipalities involved have budgeted $9 million for the completed job. B BC =3 miles By the Pythagorean Theorem BD=sqrt(3^+x^) C D A x x CA = Figure The straight route from A to C and then from C to B would cost 8(0.75) +3(1.) = 9.6 million dollars which is over budget. Is it possible to come in under budget? At what point D should the pipeline leave the mainland in order to minimize total cost? We need to express total cost in terms of the decision variable x in Figure Total cost (in millions$) = 0.75 AD + 1. DB. The distance CD is x. To express the distance DB in terms of x, we apply the Pythagorean Theorem to the right triangle with vertices BCD. The underwater distance, BD, is the hypotenuse of the right triangle with legs BC and CD. In words the Pythagorean Theorem says that in a right triangle the sum of the squares of the legs equals the square of the hypotenuse. BC + CD = BD so 3 + (x) = BD. Consequently the distance BD is 9 + x. Joseph F. Aieta page 65 8/14/003

266 Measuring Change Section 5.4 Page 66 The total cost of the pipeline in millions of dollars can be expressed as = 0.75(8- x) + 1. Since 9 + x = a a for positive values of a, cost as a function of x is can be written as 0.75(8- x) + 1. a = ( 9 + x The function ( 9 + x ) 0.5 is the composite of two functions, the inside function u(x) = (9 + x ) and the outside function that says raise the inside function to the ½ power (take its square root). ) 0.5 d By the function power rule (9 + x ) dx x (9 + x ) = x(9 + x ) 0.5 = 0.5(9 + x ) 0.5 x equivalent to In millions of dollars, C(x) = 0.75(8 - x) + 1. ( 9 + x ) 0.5 = x + 1. ( 9 + x and the first derivative of C(x) is (x) = x(9 + x ) x (9 + x ) C 0.5 ) = x Solve C (x) = 0 to find candidates for turning points. The equation = (9 + x ) is equivalent to 1.x (9 + x ) = x + x algebra = 0.75 (9 ) multiply both sides by ) x (9 ) ( ) = + x = x divide both sides by x x = (9 + ) 0.5 simplify ( 9 + x 0.5 ( 1.6x ) [(9 + x ) ] = (9 + x ).56x + x = square both sides = 9 simplify.56x x = 9 subtract x from both sides 1.56x = so x = and x = ± = ± Since x can t be negative, the distance CD is about.40 miles. C'() = and C'(3) = Since the first derivative is negative the left of.40 and positive to the right of.40, this point corresponds to a local minimum. The distance DA is about miles and the distance BD is about miles. The minimum cost is approximately 0.75(5.598) + 1.(3.843) or million dollars. Joseph F. Aieta page 66 8/14/003

267 Measuring Change Section 5.4 Page 67 B By the Pythagorean Theorem BC =3 miles BD=sqrt(3^+x^) C D A x x CA = We can build the spreadsheet shown in Figure 5.4. to check this result. The value 3.00 in A7 is our initial guess After Solver Figure 5.4. The formulas in row 7 are We used Solver to minimize cell F7 by adjusting the value in cell A7. Joseph F. Aieta page 67 8/14/003

268 Measuring Change Section 5.4 Page 68 Derivative of e to a function power The next application will require us to take the derivative of base e raised to a power given by a function. d u u du The derivative rule that then applies is e = e dt dt [In many applications the decision variable is time so we will make u a function of t in the first two examples] Example Find d dt e d t+5 e dt t+ 5 = e t+ 5 du dt = e t+ 5 () = e t+ 5 Example d -0.07t Find (1- e ) dt d -0.07t d d -0.07t -0.07t (1- e ) = 1 - (-e ) = -e ( 0.07) = 0.07 e dt dt dt -0.07t Example ,x Find stationary points for the function f(x) = e 3x + 50 and then classify the point(s) that you find as a relative minimum, a relative maximum, or a stationary inflection point. d dx ( e 0.x 3x + 50) = e 0.x = 0.e 3 0. x 0.e 3 = 0 0.x 0.e = 3 0.x 3 e = = x ln( e ) = ln(15) 0.x ln() e = ln(15) 0.x = ln(15) ln(15) x = = x (0.) 3 by the composite rule algebra d dx e u u du = e dx set the first derivative equal to zero and solve for x. take the natural logarithm of both sides Apply laws of logarithms and the definition ln(e) =1 algebra accuracy to 8 decimal places 0.e 0.e 0.(13) 0.(14) 3 3 = 0.e 3 3 = 0.e therefore x = must be a local minimum calculate the first derivative at x =13 and at x = 14 f ' is negative to the left of and positive to the right of Joseph F. Aieta page 68 8/14/003

269 Measuring Change Section 5.4 Page 69 Advertising Campaign A company ready to roll out a new product estimates that the target audience for its advertising campaign is 300,000 people. On average, purchase by an individual responding to the advertising will generate a $1 contribution to profit. Fixed costs for the entire promotional campaign are expected to be $5,000 and variable costs are estimated to be $,000 per day. The response function that estimates the proportion of the target market that responds positively within t days, is given by: pmr=r(t) = 0.0 (1-e t ) a) To the nearest day, how long should the advertising campaign continue to maximize cumulative profit? b) What proportion of the potential audience has responded when the campaign is halted? c) What is the maximum profit? Notation for the parameters: npc appr pmr fcc vcpd is the size of the market which is the number of potential customers, is the average profit per response, is the proportion of the market (target audience) that responds and buys the product if the campaign lasts t da This is actually a function of t, sometimes written as r(t) is the fixed cost of the entire campaign is the variable cost per day. FORMULATION In general: Profit = Revenue Cost = pmr npc appr - ( fcc + vcpd t) In this specific case: Profit(t) = 0.0 (1-e t ) 300, ( 5, t) The objective is to maximize the profit function The key to this problem is to understand the role of the response function, which we have called pmr, shown in Figure The output of this function at any value of t is the portion of the market that responds by making a purchase within t days. This response function has the properties that pmr increases as t increases but the rate of increase is less and less as the number for increased days. The upper limit of pmr is 0.0 which follows from the fact that e t is a positive number that approaches zero as t gets larger and larger. t pmr pmr=r(t) = 0.0 * (1-e -0.07*t ) days Figure Joseph F. Aieta page 69 8/14/003

270 Measuring Change Section 5.4 Page 70 Since 300,000 is the size of the target audience and $1 is the average profit per response, the cumulative contribution to profit from customer purchasing over time is the constant 300,000 1 times the response function. Consequently the graph of the function describing cash coming in, which we will call revenue, has the same shape as the graph of the pmr function. As t gets larger, the marginal revenue gets smaller and smaller. The cost of running the campaign is a linear function of time with variable advertising cost of $000 per day. Marginal cost is always 000 and at some point in time, marginal cost will exceed marginal revenue and profit will begin to decline. The company would like to know the optimal value of t that corresponds to a maximum profit. Solver Approach We start with a guess that the optimal number of days to run the campaign is 100. Assume that each cell in the second row has been given the name in the cell above it. The second row shows the values of each cell based on the initial guess of 100 for t and the on the results of computational formulas for pmr, Revenue, Cost, and Profit. t pmr npc appr fcc vcpd Revenue Cost Profit , ,000, ,343 5, ,343 The formulas for pmr, Revenue, Cost, and cumulative Profit are indicated below. 100 =0.0*(1-EXP(-0.07*t)) 300, ,000,000 =pmr*npc*appr =fcc + vcpd*t =Revenue- Cost When we search for the maximum profit by adjusting the number of days with Solver, we obtain the results below , ,000, ,49 117, ,33 Solver indicates that running the campaign for about 46 days will result in cumulative profit of about $574,000. Calculus Approach This method requires us to take the derivative of base e raised to a linear function. d dt e u e du dt u = where u is a function of t du In this case u(t) = -0.07t so = dt Profit(t) = 0.0 (1-e t ) 300, ( 5, t). Rearrange and simplify to obtain Profit (t) = ,000 1 (1-e t ) - 5, t = 70,000 (1-e t ) - 5, t Taking the first derivative Profit'(t) = 70,000 (0 -e t -0.07) which simplifies to Profit'(t) = 70,000 (-0.07 ) (-e t ) -000 = 50,400 e t Joseph F. Aieta page 70 8/14/003

271 Measuring Change Section 5.4 Page 71 Candidates for turning points are the solutions to the equation Profit'(t) = 0. 50,400 e t =0 50,400 e t = 000. e t = 000/50,400 Divide both sides by 50,400 LN (e t ) = LN(000/50,400). Take the natural log of both sides To isolate t we use a property of log functions The log of a number to a power is the power times the log of the number t LN (e) = LN(000/50,400) Finally, t is the quotient of LN(000/50,400) divided by which is This number agrees with our result from Solver. We can convince ourselves using derivatives or graphs that this stationary point is indeed a local minimum and not a local maximum nor a stationary inflection point. For a generalization of this problem that displays tables and graphs, see the Excel file ExponentialResponse.xls as shown in Figure pmr(t) = L*(1-EXP(b*t)) upper limit coefficient audience avg. profit fixed cost variable cost K 1 mill 1 0 0K 00K 0K 40K , ,000,000 L b npc appr fcc vcpd Advertising Campaign $1,000, % Rev 800,000Cost 8.00 Profit 15, #NAME? #NAME? $800,000 #NAME? #NAME? #NAME? $600,000 #NAME? #NAME? #NAME? 157, $431,77 #NAME? $400,000 #NAME? #NAME? #NAME? #NAME? $00,000 #NAME? #NAME? #NAME? #NAME? $0 #NAME? #NAME? #NAME? days -$00,000 #NAME? #NAME? #NAME? #NAME? Optimal number of days to maximize profit is 157 #NAME? Maximum Profit is $431,77 Figure Joseph F. Aieta page 71 8/14/003

272 Quadratic Cost Function Revisited Measuring Change Section 5.4 Page 7 Recall the example where the cost function is the form C(q) = F + A q +S q where the input variable q represents produced and sold in a certain time period. The cost of making q units, C(q), and the revenue from selling q units, R(q), are measured in dollars. The three parameters are F for fixed cost, A for variable cost per unit related to the cost of materials and S related to shipping cost. The revenue function is of the form R(q) = p q, where the parameter p is the price per unit. Profit is revenue minus cost. We examined the specific case C(q) = q q and R(q) = 40q which gave us the profit function P(q) = 40q ( q q ) that simplifies to P(q) = 38q q 500. Using point marginal revenue, point marginal cost, and point marginal profit, we were able to show that a maximum profit of $1305 occurs at q = 95. New Cost Function Now imagine that the situation has changed because of a recasting of production costs. An analysis of the production process suggests a lower fixed cost but reveals another variable cost that is directly proportional to the square root of quantity. According to this new analysis, fixed cost is actually $00 but the variable cost for making the product is more accurately described as.00q q. The adjusted cost function for this product is C ( q) q profit function R(q) C(q) is P(q) = 40q ( q + 0 q q ) = 38q 00 0 q 0.0q. = + q + q. Marketing has no reason to change the sales price of $40 per unit so the new We create an Excel table in Figure for cost, revenue, and profit functions and their corresponding marginal functions. Figure In the framed rectangle we observe that marginal cost is approximately the same as marginal revenue at q = 9. Marginal profit changes from a positive value of 0.16 at q = 9 to a negative value of -0.4 at q = 93. Joseph F. Aieta page 7 8/14/003

273 Measuring Change Section 5.4 Page 73 If we find a stationary point by solving P ' (q)= 0, we could justify that it is a local minimum by confirming that marginal profit is positive to the left and negative to the right of the root. If we start with initial guess at q = 90 and apply Goal Seek to find a value of q that corresponds to P'(q) = 0 we obtain as shown in Figure If we apply Solver then we can achieve accuracy to additional decimal places. Rev-Cost Marginal Marginal Marginal q Cost Revenue =Profit Cost Revenue Profit No Algebraic Solution Figure Some interesting things happen when we try to find a positive root of the first derivative algebraically. 0.5 Since P(q) = 38q 00 0 q 0.0q the first derivative is P '(q)= 38 10q 0. 40q = q 0. 5 q We can algebraically transform the equation q = 0 until we get a cubic equation q q = q q = algebra 0.5 q ( ) 0.5 q = q q q 1444q 30.4q = q 100 simplify q 3 square both sides = 100 multiply both sides by q (assumes that q 0) 0.16q q q 100 = 0 algebra Since this cubic equation is not factorable there is no obvious algebraic solution. We do get a value very close to zero by substituting the value found with Solver, q = , into this cubic equation q q q 100 If we decide to use Goal Seek or Solver to find roots of the transformed equation 0.16q 30.4q q 100 = 0 then we must be careful about our initial guess. If we enter 100 as the initial guess then Excel finds a root at which is not a root of38 10q 0. 40q. The graph of the cubic in Figure contains a portion of the cubic function in the interval [9, 98]. The graph shows one root near 97.5 and another near 9.5. Squaring both sides has introduced what is called an extraneous root. If an equation has more than one root then Excel s root finding algorithms generally find the root closest to the initial guess Figure Joseph F. Aieta page 73 8/14/003

274 Measuring Change Activities.4 Page 74 Measuring Change Activities 4 More Non-Linear Optimization 1-8) Composite rules for derivatives function power rule exponential base e exponential base b natural log d prime notation differential notation dx n ( u ) = n n 1 d u u dx u n = n u n 1 du dx u u d ( e ) = e u dx e u = e u du dx u u d ( b ) = b l n( b) u ( ) dx b b n b du u = u l dx 1 [ ln ( u) ] u u u d = = ( ) u dx l n u = 1 du u dx 1. f ( x) = ( x 1 / x) a) f (x) = b) Find f () to two decimal places 5. f ( x ) = 4 x + x 3 a) f (x) = b) b) Find f (1) to two decimal places. 0.x+ 3 f ( x) = 4 e a) f (x) = b) Find f (1/) to two decimal places 6. ( x 1 f ( x ) = e ) a) f (x) = b) Find f (-1/) to two decimal places 3. 3 / f ( x) = (8x 3). a) f (x) = b) Find f () to two decimal places 3 / 7. f ( x) = ( x + 9). a) f (x) = b) Find f (4) to five (5) decimal places 4. f ( x ) 4 = x 3. a) f (x) = b) Find f ( 0) to two decimal places 8. x 9 f ( x) =, x 3 x 3 a) f (x) = b) Find f ( ) to two decimal places Joseph F. Aieta page 74 8/14/003

275 9. a d) Find the first and second derivatives of each of the following functions. Simplify if possible. a) f(x) = 5 x b) f(x) = 3e x c) f(x) = 10 e 0.3x-6 d) f(x) =1,000(1.06) x Measuring Change Activities.4 Page Find the slope of the line tangent to the following curve at the point where x = f ( x) = e 0.1x + x f(x) = 00,000(0)(0.40) (1- e -0.01x ) - (1880x ) a) f (x) = b) Solve for x if f (x) = 0 1. Find the derivatives of the following functions: a) f(x) = 3ln(x). b) f(x) = ln(x+3). c) f(x) = ln(x +x). 13. A company, ready to roll out a new product, estimates that the target audience for its advertising campaign is 800,000 people. On average, a purchase by an individual responding to the advertising will generate a $6.00 contribution to profit. Fixed costs for the entire promotional campaign are expected to be $50,000 and variable costs are estimated to be $3,000 per day. The marketing department has looked at historical data and has developed a response function that estimates the proportion of the market that responds, pmr, within t days, to be pmr = r(t) = 0.15 (1-e t ) a) According to this response function, what is the most optimistic estimate of the number of people who will respond to this advertising campaign? b) Write a simplified expression for net profit P(t) as a function of the number of days, t, that the advertising campaign is run. c) The company is thinking about running the campaign for at least 00 days. Why is that not a good idea? d) Using a spreadsheet or calculus, determine the optimal number of days to run the campaign and the corresponding profit. Joseph F. Aieta page 75 8/14/003

276 Measuring Change Activities.4 Page Open the workbook ExponentialResponse.xls and find the worksheet with sliders. Check your answer to the previous problem by moving the sliders to display the values of the parameters below. Reflect on each question below and then make a conjecture. Use the slider bars to help you determine whether or not your conjecture is correct. If using the sliders leads you to a conclusion different from your conjecture then try to understand why your conjecture was incorrect. upper limit coefficient audience avg. profit fixed cost variable cost , ,000 3,000 a) e) Select one always sometimes never a) As the average profit per response increases from a low of $5.00 to a high of $10.00 and all the other parameters are unchanged, then the maximum profit will increase. always sometimes never b) If the fixed cost increases from $50,000 to $100,000 and all the other parameters are unchanged then the optimal number of days will decrease. always sometimes never c) If variable cost per day decreases from $3,000 to $500 all the other parameters are unchanged, then the optimal number of days and maximum profit will increase. always sometimes never d) If the other parameter values below remain fixed and the coefficient b in the response function L* (1-e b t ) increases from to then the optimal number of days will increase. always sometimes never upper limit coefficient audience avg. profit fixed cost variable cost , ,000 3,000 e) If the parameter values above remain fixed and the coefficient b in the response function L* (1-e b t ) increases from to then the maximum profit will increase. always sometimes never 15 16) First formulate each optimization problem. Be sure to clearly identify the variables. Solve in two ways: (1) with calculus and algebra () with Excel s Solver. 15. Newprod Company estimates the total potential number of customers for a new a product is 1,000,000. It plans to operate a promotional campaign to sell the product and uses the response function 0.01t 0.01t r( t) = e which is equivalent to r( t) = 0.5(1 e ) as a measure of the proportion of total customer potential responding to the promotion after it has been in operation for t days. On the average, one response generates $5 to profit. Campaign costs consist of a fixed cost of $15,000 plus a variable cost of $1,000 per day of operation. a) How long should the campaign continue if profit (revenue minus cost) is to be maximized? b) Compute the maximum profit. Joseph F. Aieta page 76 8/14/003

277 Measuring Change Activities.4 Page An oil deposit contains 1,000,000 barrels of oil, which, after being pumped from the deposit, yields revenue of $1.00 per barrel. Operating costs are $345,600 per year. The proportion of the deposit that will have been pumped out after t years of pumping is 0. 16t 0.16t e = 0.9(1 e ) a) How long should pumping be continued in order to maximize profit? b) Compute the maximum profit. c) Answer the previous problem if revenue per barrel is $ Suppose that the Revenue and Cost functions, R(t) = 4e 0.3 t and C(t) = 1.5e 0.4 t, are in millions of dollars and that the input variable t is time in years. For what value of t is the profit function maximized? [ Hint: to solve an equation of the type a e bt = c e dt start by dividing both sides by c e dt ] 18. You are planning your first three-day charitable fundraiser and are trying to figure out what attendance fee to charge. Your objective is to maximize the net amount of money raised for the charity. The entertainers have agreed to waive their fees but fixed costs are estimated at $000. Also, the caterer will provide the building and light refreshments for 5% of the total revenue (revenue = number who attend times ticket price). You do some research on similar fund raisers and obtain the following data: Price p $10 $0 $30 $40 Attendance bp You decide to approximate the data with an exponential function of the form A( p) = a e where a is some whole number and b is a constant with two decimal places. The input variable is price p. a) Create a spreadsheet with columns for price p, approximate number of attendees predicted A(p), gross revenue, total cost, and the net difference. Include cells for fixed cost and the percentage of the revenue given to the caterer. b) Determine what ticket price (nearest dollar) will maximize the net amount for the charity and state the corresponding net (nearest dollar). c) Will the optimal ticket price change if you find that fixed cost is actually $3000? Explain graphically or algebraically. d) Does the optimal ticket price change if the caterers do not charge anything at all for their services? Joseph F. Aieta page 77 8/14/003

278 Measuring Change Activities.4 Page A natural gas pipe is to be laid from the main gas line G to the Island Airport located at a point A on the shoreline of an island. The island is 3 miles from the mainland and the main gas line is 6 miles away from a point B that is directly landward of the island. A BC =3 miles x x B P G BG = a) Suppose land costs are million dollars per mile and cost of running the pipeline in the water is 3.5 times as expensive. What would be the total cost of running the pipeline from G to B and then from B to A? b) If the cost of the pipeline is 3.5 times as great in the water as on the land then at what distance GP should the pipe leave the mainland shore in order to minimize total cost? c) If the cost of the pipeline is.5 times as great in the water as on the land then at what distance GP should the pipe leave the mainland shore in order to minimize total cost? d) If the cost of the pipeline is only 1.5 times as great in the water as on the land then at what distance GP should the pipe leave the mainland shore in order to minimize total cost? e) Use calculus to explain why it is unnecessary to know the actual land cost in parts b) through d) to determine the optimal location of point P that minimizes total cost. 0. Use the composite rule d dx e u( x) = e u( x) du dx give an algebraic argument supporting the derivative rule and the fact that e x and ln(x) are inverse functions to d ln( x) is 1 for all values of x > 0 dx x Joseph F. Aieta page 78 8/14/003

279 Appendix A Page 79 EXCEL BASICS FOR QTM1300 In the business world, Excel is the most prevalent tool for quantitative analysis. By completing these tutorials, you will extend your repertoire of problem-solving strategies. These tutorials are not intended to provide a comprehensive examination of spreadsheet software but to give you a quick start on the path of building your own Excel spreadsheets to model quantitative situations. The tutorials have been designed so that each one can be completed in ten to fifteen minutes. Each individual tutorial is in a problem solving context related to a main topic in a first course on quantitative methods for business. Overview of Tutorials Tutorial 1 involves entering labels and values into cells and editing and formatting cell contents. Tutorial involves entering and copying formulas containing both relative and absolute cell references. Tutorial 3 involves break-even analysis for cost, revenue, and profit functions. Tutorial 4 involves creating charts from data in tables and modifying the appearance of these charts. Objectives By the end of the first tutorial, students should be able to: 1. enter text and values in cells. use the Formatting toolbar and the Format menu to modify how the contents of a cell are displayed 3. adjust column widths and row heights 4. enter a simple formula 5. create an automated series by extending a pattern based upon two or more cells By the end of the second tutorial, students should be able to: 1. recognize the need for both relative and/or absolute cell references in formulas. switch back and fourth from the normal view of worksheet values to a view of worksheet formulas 3. copy formulas to fill a table 4. use the SUM function built-in to Excel By the end of the third tutorial, students should be able to: 1. name individual cells and name a range of cells. enter formulas using names 3. interpret tabular data 4. apply Goal Seek By the end of the fourth tutorial, students should be able to: 1. select data in a worksheet to be represented by a graph. create an initial graph using the Chart Wizard 3. activate, manipulate, and modify various regions and features of a graph 4. change function parameters and observe adjustments to the graph For each tutorial you will work with a corresponding Excel spreadsheet. The files for tutorials one through four can be found in the Website and also in a folder on Babson s K: drive with the path K:\Courses\QTM1300\ExcelTutorials. Joseph F. Aieta Page 79 Printed 8/14/003

280 Appendix A Tutorial 1 Page 80 Credit Card Payments Tutorials 1 and Kristin has credit card debt with a current balance of $300. She has decided not to make any additional charges on this card and to pay $00 at the end of each month to reduce the balance. Once her unpaid balance drops below $00, she will add an extra amount to her last payment to pay off the loan by reducing the balance to zero. Kristin decides to create a spreadsheet with months in consecutive rows. Separate columns will contain the balance at the beginning of the month, the amount of interest due, the size of the monthly payment, the amount that can be used to reduce the unpaid balance of the loan, and the balance at the end of the month. The credit card company specifies that the amount of interest that must be paid to avoid penalties is one month s interest applied to the balance of the loan at the beginning of that month. For example, suppose the balance at the beginning of a month is $1000 and the interest rate is 1% per year then the interest due for that month is (1000)(0.1/1) = 1000 ( 0.01) = 10 dollars. After the interest due each month is subtracted from the monthly payment, the remaining amount is applied to reduce the unpaid balance of the loan. Kristin s credit card company states a nominal interest rate of % per year. Consequently the monthly interest rate is % divided by 1. Assuming that she continues to make payments each month, Kristin would like answers to the following questions: What will be the size of her last payment? What is the total amount of interest that she will pay the credit card company? If she can transfer this $300 debt to a credit card that charges an annual rate of 9.99% then what will she save in interest? If you have some familiarity with Excel, you may be able to skip the explanations and illustrations and work directly with the four Excel files listed below. These files contain multiple worksheets designed specifically for learning or reviewing purposes. If you have not already saved these files locally, then do so now. CreditCardPart1.xls CreditCardPart.xls BreakEvenTable.xls BreakEvenGraph.xls The credit card problem can best be solved by building an Excel worksheet. Open the Excel file CreditCardPart1.xls in a new window. Toolbars Excel provides toolbars to help users work more efficiently. The Standard and Formatting toolbars are normally visible below the main menu bar. If either of these toolbars is not visible when you open Excel then go the View menu at the top, click Toolbars and then activate Standard and Formatting from the list. To get from one worksheet to the next, click on the tabs found at the bottom of the CreditCardPart1 workbook. Etc. Joseph F. Aieta Page 80 Printed 8/14/003

281 Appendix A Tutorial 1 Page 81 The font size has been pre-set to Ariel 8 in order to conserve space on a laptop screen. Individual worksheets contain a short set of instructions given as red text on a yellow background. Complete instructions directly on the worksheet and then compare your results with the next worksheet in the series until you reach the end of the tutorial. Entering and formatting labels and values Tutorial 1 Enter Labels Start by opening the worksheet entitled In cell A1 enter the label annual rate and press <Enter>. In cell A enter monthly rate and press <Enter>. In cell A3 enter initial balance and press <Enter>. In cell A4 enter monthly payment and press <Enter>. At the end of steps that involve entering labels, values, or formulas, it will be implicitly assumed that the <Enter> key is pressed. Notice that the text in cells A3 and A4 spills over into cells A4 and B4. This can be easily corrected by moving the cursor to the vertical divider between columns A and B. Double-click when the cursor turns into a black cross with a double-headed horizontal arrow. Enter Values Open the next worksheet entitled Excel allows percentages to be entered in percentage form or as decimal equivalents. In cell B1 enter followed by the percent key %. Excel computes with this number as Now for your first formula. The monthly interest rate is the annual rate divided by 1. If you click in cell B and type B1, division sign /, 1, then what appears in cell B is the ordinary text B1/1, which is not what you want. Every Excel formula must begin with the = sign. You can correct the above error directly by simply double-clicking in cell B and editing the contents of the cell by inserting the = sign in the front part of the formula. Joseph F. Aieta Page 81 Printed 8/14/003

282 Appendix A Tutorial 1 Page 8 There are many ways to correct mistakes in Excel. In general, when you make an error you can press the Undo button in the standard toolbar and start over. Experimentation is a very good way to learn how to use new software and the Undo key makes it easy to recover from mistakes. Undo is probably the most popular button in Excel. In cell B3 enter 300, the initial amount owed to the credit card company. In B4, enter 00, the dollar amount that Kristin intends to pay at the end of each month. Using % and $ Formats To see what each button does in the Formatting toolbar, move the cursor (white arrow) very slowly over each button icon. To display the monthly rate in percentage format, select the cell B and click, the Percent style button from the Formatting toolbar. The cell B initially displays 1 %, which is only correct to the nearest whole percentage. You will change the formatting of this cell so that it displays a percentage with two decimal places. Click the Increase Decimal button twice to display two decimal places. If you ever need to decrease the number of decimal places you can click the Decrease Decimal button. Formatting changes can also be accomplished with selections from the main menu. For example, to format cells B3 and B4 as dollar amounts to the nearest cent, highlight these two cells, click Format in the top menu, click Cells, and choose Currency. Click OK. The default is two decimal places which you will accept. The cursor has the white arrow shape, as shown next to the Undo button above, when it is outside the cells in a worksheet,. The cursor changes to a white cross when you place it over worksheet cells. When you click in the formula bar (directly above the worksheet) the cursor will change to a vertical insertion bar. If you move the cursor to the lower right corner of an active cell, it will change to a black cross. This black cross will be very important later when you start copying and filling blocks of cells. Joseph F. Aieta Page 8 Printed 8/14/003

283 Appendix A Tutorial 1 Page 83 Months by Name A cell range is made up of one or more cells that form a rectangular group. Cell ranges often contain data that fits into a pattern, Excel makes it easy to extend a pattern based upon two or more cells Enter the label Month in cell A6 and the names September and October in cells A7 and A8, enter Select both cells A7 and A8 and move the cursor to the lower right corner of cell A8. When the black cross appears, drag it down to cell A18.. Months by Number Delete the range of cells A7:A18 and then enter the numbers 1 and in cells A7 and A8. When you copy this pattern down to cell A18, the months will be numbered 1 through 1. To improve the appearance of the spreadsheet, select the range of cells A6 through A18 and center the contents by clicking on, the Center button in the Formatting toolbar. Joseph F. Aieta Page 83 Printed 8/14/003

284 Appendix A Tutorial 1 Page 84 Column Headings Text labels are often used as column headings. Kristin wants to see the status of her credit card account at the beginning and end of each month. In cell B6 enter the label Beginning Balance. In cell C6 enter the label Interest Due. Do not be concerned that the cells in row 6 do not display the complete labels. In cell D6 enter the label Monthly Payment. In cell E6 enter the label Principal Reduction. Finally, in cell F6 enter the label Ending Balance. Wrap Text Instead of using wider columns to display the five column headings in cells B6 through F6, you will use a text formatting feature called wrapping text. It works as follows: Highlight the five cells B6 through F6, go to Format in the main menu then Cells then Alignment and check the control box for Wrap Text. Then click OK. Notice that only the first part of each heading is displayed. To see the complete headings, increase the height of row 6 as follows: Click on row 6 and move the cursor on the horizontal border between rows 6 and 7 until the cursor turns into a black cross with a double-headed vertical arrow. Drag the cursor down until both parts of each column heading are displayed. Center columns B through F. Joseph F. Aieta Page 84 Printed 8/14/003

285 Appendix A Tutorial 1 Page 85 Bottom Border under Labels To emphasize the separation of headings and values, select cells A6 through F6, find the drop down menu next to the Borders button in the Formatting toolbar, and choose Thick Bottom Border from the borders palette. Click anywhere below row 7 in orders to see this border. Next, go to the worksheet and compare it to your work thus far. You will complete this table in tutorial. Joseph F. Aieta Page 85 Printed 8/14/003

286 Appendix A Tutorial Page 86 Writing and copying formulas with relative and absolute cell references Tutorial Open the Excel file CreditCardPart.xls in a new window and go to the first worksheet. After you have completed the instructions contained on a worksheet, compare your results with the next worksheet in the series until you reach the end of the tutorial Note: If any column contains a string of pound signs #, it simply means that you need to widen that column. First put the cursor on the vertical divider between columns. Double-click when the cursor becomes a black cross with a double-headed horizontal arrow. Month 1 Formulas Before Kristin makes her first monthly payment, her beginning balance is $300. Since this amount is contained in cell B3, you simply enter = B3 in cell B7. For the amount of interest due, the credit card company charges one month s interest on the beginning balance for that month. This is the result of multiplying the beginning balance by the monthly interest rate. Using the point and click language of Excel, select cell C7, type =, click cell B7 (the cell to the left containing the beginning balance), type the multiplication symbol *, click cell B which contains the annual rate divided by 1, and finally click the <Enter> key. Kristin s regular monthly payment of $00 in cell B4 can be copied into month 1 by selecting cell D7 and entering = B4. Out of the $00 monthly payment, (300) (.1599/1)=30.65 dollars is paid to the credit card company as interest. Consequently $00 -$30.65 = $ can be used to reduce the unpaid balance at the end of the month. In cell E7 enter =D7- C7. The ending balance for month 1 is calculated by selecting F7 and entering the formula =B7 E7. Once you are finished with month 1, you can turn your attention to month. Before you select a command from a menu or start typing, it is generally a good idea to keep track of where your cursor is located. Make sure that your cursor is in the cell where you want a formula to be evaluated before you start building the formula. Joseph F. Aieta page 86 Printed 8/14/003

287 Appendix A Tutorial Page 87 Month Formulas After month 1, the unpaid balance at the beginning of a month will be the ending balance in the previous month. For month, click on cell B8 and type the formula =F7. Recall that the credit card company charges one month s interest on the unpaid balance. Kristin has a new balance of $, Consequently, the amount of interest due in the second month will be less than the interest that was due in the first month on the full $300. In cell C8, type = and then click cell B8 which contains the beginning balance for the second month, type the asterisk * for multiplication, and then click cell B, which contains the monthly interest rate. Don t forget the <Enter> key. Double-click in cell C8 and observe that the cells used in the formula are displayed with color-coded borders. Click anywhere outside of a cell to restore the normal view of the worksheet. Since the monthly payment remains at $00, type =B4 in cell D8. To obtain the amount that can be used to reduce the balance of the loan, you subtract the interest due from the monthly payment. In cell E8 enter =D8 - C8. As before, the ending balance is the principal reduction subtracted from the beginning balance for that month. In cell F8, enter the formula =B8 E8. Joseph F. Aieta page 87 Printed 8/14/003

288 Appendix A Tutorial Page 88 Formula View As a worksheet grows in complexity, detecting errors is often made easier by viewing all of the formulas contained in the worksheet. The quickest way to switch back and forth between the normal display of values in the worksheet and the display of worksheet formulas is to hold down the <Ctrl> key while pressing the backwards quote key ` (located in the upper left of the keyboard below the Tab key). Since the columns generally expand wider than necessary to accommodate formulas, you will want to adjust the column widths. By inspecting the formulas in rows 7 and 8, you can see that the formulas for month are very similar to the formulas for month 1 with the exception of the beginning balance. The initial balance for month 1 is obtained by copying the contents of cell B3 into cell B7. The beginning balance for month refers back to the ending balance of month 1 so you copy the contents of cell F7 into cell B8. With the cursor anchored in cell B8 you are telling Excel to copy the contents in a cell which is four cells to its right and one row above. This is an example of what is called a relative reference. Looking ahead to months 3 through 1, you will find the beginning balance for each successive row (month) by referring to the cell containing the ending balance in the row (month) above it. For another example of relative referencing, examine the cells E7 and E8 which define the principal reduction for months 1 and. The formula in cell E7 tells Excel to subtract the value found two cells to the left of E7 from the value found one cell to the left of E7. The formula in cell E8 tells Excel to subtract the value found two cells to the left of E8 from the value found one cell to the left of E8. Similarly, you compute the ending balances in column F by subtracting the value found one cell to the left from the value found four cells to the left. By again holding down the <Ctrl> key and pressing ` (the backwards quote) you can switch (toggle) back to the normal view displaying worksheet values. You may need to widen some of the columns. Joseph F. Aieta page 88 Printed 8/14/003

289 Appendix A Tutorial Page 89 Month 3 Manually entering formulas for every month isn t very efficient. You should be able to copy formulas in month and obtain correct results for months 3 through 1. The C column for interest due contains references to both the beginning balance in B7 and the monthly interest rate in B. Column D refers to the single cell B4. At this point you will attempt to compute the results for month 3 by copying the formulas in the block of cells B8 through F8 down to B9 through F9. Select the range of cells from B8 to F8. Move the cursor to the lower right corner of cell F8 until it becomes a black cross. Drag down to month 3 in row 9 and release. Widen any column that contains # signs. A beginning balance of $1, for month 3 shown in cell B9 makes sense but none of the values to the right makes any sense. What went wrong? Need Absolute References Delete the contents of cells B9 through F9. Before you can get correct results from copying the formulas in month, you must first indicate that the reference to the monthly interest rate should remain fixed to cell B. Similarly, the reference to the monthly payment should be fixed to cell B4. Toggle back to formula view by holding down the <Ctrl> key and pressing ~ and click in cell C8. Observe that =B8*B appears in the formula bar. Move the cursor to the right of the asterisk in the formula bar and press the function key <F4> exactly once. Dollar signs now appear in front of the B and in front of the. The appearance of $B$ in the formula =B8*$B$ indicates that the reference to cell B will not change when this cell is copied. This is called an absolute reference to the cell B. In cell C8 the reference to cell B8 remains a relative reference. Also change the reference in cell D8 for month to the absolute reference $B$4 by clicking once on cell D8, putting the cursor over the cell address in the formula bar, and pressing the <F4> key once. Joseph F. Aieta page 89 Printed 8/14/003

290 Appendix A Tutorial Page 90 Excel uses a "select, then do" design. One way to select a range of cells is to select the cell at one end of the range, hold down the <Shift> key, and then select the cell at the other end. You can also select a range of contiguous cells by left-clicking on the first cell, dragging the mouse, and releasing the mouse on the last cell. The cursor should remain a white cross throughout the selection process. Correct Table Fill Select the range of cells from B8 to F8. Move the cursor to the lower right corner of cell F8 until it becomes a black cross. Drag down to month 1 in row 18 and release the mouse. Switch back from a formula view to a normal worksheet displaying values by again holding the <Ctrl> key and pressing `, the backwards quote. Now the values make sense! The unpaid balance at the end of month 1 is $11.0. Entering and copying formulas with absolute and relative references gives Excel its power! The values in months 1 and look the same as they did when the references were all relative. The difference of the absolute references in month becomes clear when you fill down from month to month 1. To see worksheet formulas at any stage simply toggle on the formula view and then adjust column widths. Joseph F. Aieta page 90 Printed 8/14/003

291 Appendix A Tutorial Page 91 The Last Payment Copy down the block of cells A18:F18 to A19:F19. The negative ending balance in cell F19 tells you that Kristin has overpaid. To fix this error enter the formula = B19+C19 in cell D19. Kristin needs to make a final payment of $ which is the balance of $11.0 at the end of month 1 plus the $1.49 of interest due on the unpaid balance. Observe that cells E19 and F19 adjust automatically and that the unpaid balance at the end of month 13 is now zero. Select cells A19 through F19 and put a border below these cells by clicking the drop down menu next to the Borders button on the Formatting toolbar and choosing Thick Bottom Border. Excel has several hundred built-in functions. You can locate functions by clicking on the fx button to the left of the formula bar which opens the Insert Function dialog box. Total Interest Paid There are many ways that you could add up the total interest paid in months 1 through 13. The SUM function is used so often that it has its own button, pronounced sigma, on the Standard toolbar. Click in cell C0, which is directly below the monthly interest payments, and then click the summation button. The built-in SUM function appears followed by C7:C19 in parentheses. Press <Enter>. The range in parentheses indicates the arguments used by the function. By the time that Kristen pays off the loan she will have paid cumulative credit card interest of $ Joseph F. Aieta page 91 Printed 8/14/003

292 Appendix A Tutorial Page 9 Total Interest at a Lower Rate Formulas that contain cell references automatically calculate new results when the contents of any cell are changed. Change the annual rate in cell B1 to 9.99% and observe that the total amount of interest drops to $17.8. Use your worksheet to explore other What if? scenarios. Save your work on your own computer with the Save As option in the File menu. Joseph F. Aieta page 9 Printed 8/14/003

293 Appendix A Tutorial 3 Page 93 Break-even Analysis Tutorials 3 and 4 The Babson Press is considering the publication of a paperback book about the history of personal computers. The text is to be used in an elective course with an expected enrollment of 70 students per year. You have been hired to create an interactive spreadsheet containing tables for the cost, revenue, and profit functions as the number of books printed ranges from 0 to You will design the spreadsheet in such a way that any change to the parameters for variable cost per book, fixed cost, or selling price per book will automatically be reflected in the table. All cost information is expressed in dollars to the nearest cent. The parameters that you will use are v for the variable cost per book, F for the fixed cost, and p for the selling price per book. The estimated values of these parameters for the book under consideration are v = 10.00, F = , and p = Symbolically, the cost and revenue functions are expressed as C(q) = 10.00q and R(q) = 16.00q where q is the quantity, or number of books printed. Obtain the expression for the profit function from the definition 'Profit equals Revenue minus Cost or P(q) = R(q) - C(q) in function notation. When you have finished building the worksheet, examine the question, "Under what conditions should the Babson press go ahead with the publication of this book?" Naming cells and ranges, building tables, and using Goal Seek Tutorial 3 Open the Excel file BreakEvenTable.xls in a new window and go to the worksheet following the introduction. After you have completed the instructions contained on a worksheet, compare your results with the next worksheet in the series until you reach the end of the tutorial. Labels and Values In Column A, text has been entered in three cells and the column has been widened to accommodate the text. in cell B1 and enter the label v Press the <Enter> key in cell B enter the label F or the down arrow key in cell B3 enter the label p after each entry in cell C1 enter 10 in cell C enter 3000 in cell C3 enter 16 Display the values in cells C1, C, and C3 to decimal places by selecting the three cells and clicking Increase Decimal icon, until each of the values displays two decimal places, the Make the backgrounds of each of these three cells a Light turquoise. While the cells are selected, find, the Fill Color icon. Click on the drop down menu, and select the color in the fifth column of the bottom row. Put borders around each of these three cells by clicking on the drop down menu next to the Borders icon, pull down the selection of borders, and select All Borders Center the cells in the block B1:C3 by selecting the six cells and clicking on, the Center icon. Joseph F. Aieta page 93 Printed 8/14/003

294 Appendix A Tutorial 3 Page 94 Name individual cells You now associate the three labels in cells B1 through B3 with the three parameter values in cells C1 through C3 Highlight the six cells B1 through C3. Find Insert in the main menu and select Name then Create. Excel will guess that you want to use the names in the Left Column and you will accept that suggestion by clicking OK. Enter the description quantity in cell E1. When you enter the label q in cell E, hit the delete key to remove all letters to the right of the letter q. In cell E3, enter the starting value 0 and in cell E4 enter 50 for the next value of q. Fill a Pattern Continue the pattern of increasing q in steps of 50 until you reach the value 1000 as follows: Select the two cells E3 and E4. While these two cells are selected, move the cursor to the lower right corner of cell E4. When it changes to a black cross + drag down to cell E3. Activate column E and then click on the Center icon in the Formatting toolbar. Name a Range Associate the name q with the range of cells E3 through E3 as follows: Select cells E through E3. Click on the Insert icon in the Standard toolbar Select the option sequence Name Create Top Row To see the current list of names for any worksheet, click on the drop down arrow next to the Name Box located in the upper left of the Excel window just under the Formatting toolbar and above column A. When you click on any one of the names in the Name Box, the corresponding cell(s) will be highlighted on the worksheet. Associations of names and cells can be removed or modified by replacing existing definitions. You are now ready to create tables for the cost, revenue, and profit functions. By naming cells, as you have done, you can enter formulas in Excel that actually look like the formulas you might see in a math textbook. Joseph F. Aieta page 94 Printed 8/14/003

295 Appendix A Tutorial 3 Page 95 Cost, Revenue, and Profit In cells F, G, and H, enter the labels: Cost, Revenue, and Profit. In cell H1 enter the descriptive text Revenue-Cost. Put a heavy border at the bottom of cells E through H by selecting the four cells and choosing Thick Bottom Border from the drop down menu next to the Borders icon. Reminder: Press the <Enter> key after entering each formula. In cell F3 you can type =v*q + F. Alternatively, you can type =, click in cell C1, type *, click in cell E3, type +, and click in cell C. If you choose to click in cell E3 instead of typing q, the formula will display the cell address instead of the name q. Similarly, in cell G3 you can either type =p*q or you can type =, click in C3, type the asterisk *, and click in cell E3. In cell H3 you can either type =G3-F3 or use the point and click approach. In row 3, display each of the dollar values of cost, revenue, and profit with two decimal places using the Increase Decimal button. Copying Formulas with Named Cells Naming a cell has the effect of making a reference to that cell absolute. When a formula containing a named cell is copied, the reference to the named cell remains fixed and does not depend upon the relative locations of cells. Here are two ways to copy the formulas for cost, revenue, and profit. (1) Click and drag approach: First select cells F3 through H3, move the cursor to the lower right corner of H3 until it changes to a black cross +, and then drag down to row 3. () Double-click on the black cross: Values of quantity in cells E3 to E3 form a border to the left of the cells being filled. This allows you to simply double click on the black cross + and the fill will continue until Excel comes to the first blank cell in column E. Highlight the block of cells F1:H3 and Center the data in columns F, G, and H as you have done before. Joseph F. Aieta page 95 Printed 8/14/003

296 Appendix A Tutorial 3 Page 96 Interpret the spreadsheet For the parameters v = 10.00, F = , and p = 16.00, you can locate the break-even quantity and the break-even dollars of sales just by inspecting the table. When quantity is 500 books you see that Revenue =Cost and Profit = 0. The break-even dollar of sales is $8,000. If the Babson Press can sell more than 500 books they will be able to cover their expenses and start to make a profit. Interpretation is an important part of the process and should be made explicit. The company expects a concise conclusion in plain English. A good consultant will not simply hand in a spreadsheet model to company management and leave them the task of interpreting the model. Change Parameter Values Make sure that you are on the worksheet Change parameter values. Clearly, you cannot expect to always find a zero in the Profit column. When you aren t so lucky you will need to use other techniques for finding the value of q that corresponds to P(q) = 0. Replace the value in cell C1 with 9.5. This might represent a situation in which the variable cost has dropped by 75 cents due to less expensive materials or improvements in the binding process. Assume for now that fixed cost remains at $ and the selling price is also unchanged at $ What would you expect to be true about the new break-even point? Basic mathematical intuition tells you that the new break-even point should be under 500 books. Always try to anticipate relative magnitudes before you change one or more of the problem parameters. You can begin searching for the break-even point by starting at q = 300 instead of q = 0 and increasing q in steps of 10 instead of steps of 50. Fill in the pattern 300, 310, 30 etc. stopping in cell E3 as you did before. Joseph F. Aieta page 96 Printed 8/14/003

297 Apply Goal Seek Appendix A Tutorial 3 Page 97 As you can see from the updated worksheet, the break-even quantity is between 440 and 450 books. If 440 books are sold then profit is $30.00 (losing money) and if 450 books are sold then profit is $37.50 (making money). Find a value of q that makes P(q) = 0. You could narrow the search by starting at q = 440 and increasing q in steps of 1 but there is another way. From the Tools menu select Goal Seek and bring up a dialog window with three boxes. In the first box enter H17 as the Set cell Enter 0 in the middle box In the third box enter E17 the changing cell Press Ok Excel s Goal Seek is an efficient and easy-to-use search algorithm. In general, Goal Seek can be useful in solving equations provided that you start relatively close to the solution and you don t need accuracy beyond three decimal places. Interpretation if the variable cost per book drops to $9.5 then the solution to P(q) = 0 is books which means that the Babson Press starts making money with the 445 th book that they sell. Joseph F. Aieta page 97 Printed 8/14/003

298 Appendix A Tutorial 4 Page 98 Creating and Modifying Charts Tutorial 4 Initial Plot Open the Excel file BreakEvenGraph.xls in a new window and go to the first worksheet after the Introduction. You will start this tutorial with the table that was built in the previous tutorial. After you have completed the instructions contained on a worksheet, compare your results with the next worksheet in the series until you reach the end of the tutorial. Select the column headings and the numerical data in the block of cells E to cell H3. Find and open, the Chart Wizard from the Standard toolbar at the top Click on XY Scatter from Standard Types and choose the Chart sub-type in the nd row and nd column Scatter with data points connected by smooth lines without markers Click Finish (enhancements can be added later) Cost Revenue Prof it Resize and Move Activate the chart by clicking inside the chart until you see the eight black handles on the edges. Then click on the chart and drag it up and to the left into its new location. Use the handles to resize the chart so that it fits in the rectangular outline covering cells A4 to D17. Joseph F. Aieta Page 98 Printed 8/14/003

299 Appendix A Tutorial 4 Page 99 Place Legend at the Top A legend is created when column labels are selected with the data. For the initial plot, the legend is placed to the right of the graph by default. You will place the legend above the graph. Right - click on the Legend box and choose Format Legend Cost Revenue Prof it Open the Placement menu and click Top. Click OK in order to complete the instruction Format the vertical scale To change the display on the vertical scale to a currency format, start by right-clicking anywhere on the vertical axis. When you see the handles on the top and bottom of the vertical axis click on Format axis, select the Number tab select Currency with 0 Decimal places $0,000 $15,000 $10,000 $5,000 Cost Revenue Prof it Click OK. $ $5,000 Format the profit line To change the color of the profit function, right-click anywhere on the yellow profit line to activate points on the line. Choose Format Data Series. $0,000 $15,000 Cost Revenue Prof it Display the Patterns tab and then the drop down menu next to Color. Select the color in the 4 th column and nd row of the palette in order to change the color of the profit line from yellow to green. $10,000 $5,000 $ $5,000 Click OK. Joseph F. Aieta Page 99 Printed 8/14/003

300 Format Plot Area Background To change the background of the plot area, right-click on the sides of the gray plot area until you see a border and handles. Choose Format Plot Area Select the pale yellow color in the 3 rd column and nd row from the bottom of the color palette. Click OK. $0,000 $15,000 $10,000 $5,000 Appendix A Tutorial 4 Page 300 Cost Revenue Prof it Manipulate the Chart $ $5,000 Activate the entire chart so that it displays the eight black handles. Drag the lower right handle straight down so that the chart fills the space from cell A4 to cell D3. Change Parameters You can ask any number of what-if questions by simply modifying the values of the parameters. You will change the values in each of the three cells with a blue background and also change the values of q in column E. With each change, the table and graph are updated automatically. $10,000 $100,000 $80,000 Cost Revenue Prof it Change the variable cost v to $3.75 Change the fixed cost F to $0,000 Change the selling price p to $5.00 Increase the step size of quantity q to 1000 as follows: Change cell E4 to 1000, select both cells E3 and E4, and move the cursor to the lower right of E4. Drag the black cross down as before or double click when the cursor changes to a black cross in the lower right corner of cell E4. $60,000 $40,000 $0,000 $ $0,000 -$40,000 Interpretation The last worksheet highlights the new break-even point at q = 16,000 units. If you move the cursor (white arrow) slowly along the graph of the cost, revenue, or profit functions, you should see coordinates of points appear inside a box on the chart. For example, the point where the green profit line crosses the horizontal axis will appear with the coordinates shown on the right. Joseph F. Aieta Page 300 Printed 8/14/003

301 Appendix B Linear Models Answers Appendix B: Answer Linear Activities 1 Page a) indicates that the residual value decreases by $6500 each year b) $60,000 is the initial cost of the equipment c) $47,000. a) -7,50 indicates that the residual value decreases by $7,50 each year b) $6,000 is the initial cost of the equipment c) V(t) = 6,000 7,50 t d) $1, a) At end of year, book value = $5800 b) ( )/5 = 1600 V(t) = t 4. a) 7 years b) Answers will vary. Most people will use paper and pencil. 5. a) $0,000 b) about $14,000 c) depreciates by $1,500 each year d) V(t) = 0, t 6. a) Slope is -15/300 = 0.05 ( 5 per copy) V(x) = x b) V(00) =5.00 c) 60 copies 0 = 0,000-1,500t t = 0,000/1500 = years or about 13 years and 4 months 7. y 40 = 3( x 10) or y = 3 x y = 3x y = 0.5x a) = 16 dollars increase b) 10 = 1 dollars increase c) slope = 1/16 = ¾ or 0.75 x = sales and y = selling expense line is y 10 = 0.75 (x-14) or y = 0.75x a) y = 0.80x b) E = 0.10V + 50 slope = commission which is 10% of sales 1. a) 500 b) 90,000 c) 1500 d) 130,0000 [$40,000 more than b) 13. a) If u =number of units then u 0 u 0 C( u) = 9.60u +.806( u 0) u > 0 b) cost =$3.70 avg. cost/unit = $1.58 per unit c) cost =$53.78, avg. cost/unit = $1.08 per unit 14. a) $1, b) $ 35,67 e) Solve 1, ( x 1,000) = 3, 900 AGI is $30, d) 75 units Joseph F. Aieta Page 30 Printed 8/14/003

302 Appendix B: Answer Linear Activities Page c) answers will vary. Answers Linear Activities 1. a) $00 b) $1 c) variable cost=1/3=33 cents per unit d) cost=$ revenue=$400 profit=$66.67 e) q = 300 f) $300. a) c(q)=60,000+q r(q)=5q p(q)=3q b) $15,000 c) -$30,000 d) $0,000 e) $100, a) C(q)-= 100q R(q) = 180 q, C(q) = 100 q +00,000 P(q) = 80 q 00,000 b) $00,000 c) 100 dollars per unit d) q =500 units e) $450, a) R(q) = 8q C(q) = 8q + 56,000 P(q) = R(q) C(q) = 0q 56,000 BE point at (800, 78400) b) R(q)=0.85q C(q)= q, P(q)=0.70q-400 BE point at (6000, 5100) c) C(0) = 98,000 R(q)=18q and R(7000) = 16,000 slope (0, 98000) (7000, 16000) is 4 C(q) = 4q and P(q) = 14q BE is (7000,16000) 5. a) Revenue is increasing faster than variable cost b) C(q)=8.75q+5500, R(q)=5q, P(q)=16.5q-5500 C(1000) = $14,50, R(1000) = $5,000, P(1000) = $10,750 c) 16.5q-5500 = 0 so q ~339 books d) changes from 339 to 6 books, a difference of 77 books 6. a) C(q)= 5q+500,000 R(q)=5.50q, P(q)=0.5q-500,000 b) $500,000 c) $5.00 per unit d) q = 1,000,000 units e) $5,500,000 Joseph F. Aieta Page 30 Printed 8/14/003

303 Appendix B: Answer Linear Activities Page C(q) = F + v q R(q) = p q P(q) = (p v) q - F 8. a) F = $3,000 b) v =$30 per student c) q = 1600 d) q = a) P(x) = 10x-800 b) sell 7 bikes c) 4 bikes 10. Let x = number of miles. a) A(5) = 3.00 B(5) =.50 Choose B b) A(15) = 4.00, B(15) = 4.50 Choose A d) both charge $3.50 for 10 miles 11. S(q)=0+0.05p, Q(q)=0.09q for 50 < q a) Speedy Printers is the only choice for 0<q< 50 and is best for over 500 pages. b) Quality Printers is best for 50 <q< a) Company A at $4.5 b) Company A at $5.75 c) Company B at $9.00 d) A(x) = x, B(x) = x, C(x) = 1.50 x e) A is best for 3 < x < 8 boxes f) B is best for 8 for more than 8 boxes g) C is best for 1 or boxes 13. a) Profit =.0.39sales 70,000 b) $179,487 c) $179,500 to nearest hundred dollars 14. a) Let x = sales in dollars, Rx) = 1.00x C(x) = 0.47x+9786, P(x) = 0.53x b) $0.47 per dollars of sales is variable cost c) $9,786 d) $33,840 e) $63,66 f) $1,614 profit g) $56,00 in sales needed to break even. 15. Need $1000 more in sales to break-even this year 16. Let x = amount invested in stocks and y = amount in MM Total return is expected to be 0.09 x y Equation and Intercepts $6,000 return contains (66,667,0) and (0, 150,000) $ 1,000 return contains (1,333,333,0) and (0, 300,000) $ 18,000 return contains (00,000, 0) and (0, 450,000) $ 4,000 return contains (66,667, 0 ) and (0, 600,000) slope of each iso-line is -.5 Joseph F. Aieta Page 303 Printed 8/14/003

304 Appendix B: Answer Linear Activities Page Let m = cases of mango Let p = cases of papaya a) Cost is 10m + 14p Equation and Intercepts 10m + 14p = 1,000 (1,00,0) and ( 0, ) b) Profit is 8m + 6p Equation and Intercepts 8m + 6p = 100 (1,50,0) and ( 0, 00) 18. a) C(q) = 400, q, R(q) = q, P(q) = q 400,000 b) q 400,000=0, q = 8 million units c) R(8,000,000) = (8,000,000) = $600, p= 18,000 (1800,0) and (0,1,58.71) 10m + 14p = 4,000 (400,0) and ( 0,1714.9) 10a + 14 =48,000 (4800,0) and ( 0, ) slope of each iso-line is -5 8m + 6p = 400 (300,0) and ( 0, 400) 8m + 6p = 3000 (375,0) and ( 0,500) 8m + 6p = 3600 (450,0) and ( 0, 600) slope of each iso-line is - 4/3 *19. a) (18-0)/(0-50) = -18/50 = 0.07 The amount of gas in the tank of the car decreases at the rate of 0.07 gallons per mile. remaining distance gallons c) gallons remaining with cruise control gallons miles traveled Joseph F. Aieta Page 304 Printed 8/14/003

305 Answers: Systems of Linear Equations and Inequalities Review 1. (1,1). (1/, /3) Appendix B: Answer Linear Activities Page No solutions, parallel lines 4. Infinite solutions, same line 5. Infinite solutions, same line Corner points (0,6) (16/7, 10/7), (8.0) Corner points (5, 1.) (5, 13.33, 10, 10) Corner points (-10, 15) (3.33, 15) (10,5) , , Corner points at (0, 0) (0, 4) (17.55, 10.91) (4,0) Corner points (0,0) ( 0, 800) (50,765) (350,475) and (350, 0) Joseph F. Aieta Page 305 Printed 8/14/003

306 Appendix B: Answer Linear Programming Activities 1 Page 306 Answers Linear Programming Activities 1 1. Nuts&Bolts1.xls a) No. 11*10+4*6 =134 which is greater than 13 so this point does not satisfy the machine C constraint. b) (0.00, 1.00) (.00, 10.80) (4.00, 9.60) (6.00, 8.40) ( ) c) No, 11*1+4*4.8 =151. which is not less than or equal to 13 so this point does not satisfy the machine C constraint. a) Yes, (6, 11.0) and (8, 9.60) b) No. The $88 profit line is above all of the constraints. Maximum profit is $84.00 c) (6.00, 11.60) and (8.00, 9.60) d) $90 at (6, 1) and (8.57, 9.43) and all points on the segment with these endpoints d) Iso-profit line that corresponds to a profit of $80 has no points in common with the feasible region e) (6, 1) 3. Burgers&fries 1.xls a) Yes, (3, 5) satisfies all constraints. b) Yes, ( 9, 5/3) satisfies all constraints c) Name two (3.00, 5.00) (5.00,3.89) (4.00,4.44) (6.00,3.33) (7.00,.78) (8.00,.) (9.00, 1.67) (10.00, 1.11) (11.00,0.56) (1.00, 0.00) d) -5/9 e) All points on the segment with endpoints (6.75, 1.5) and (9.00, 0) 4. a) No, the 10 iso- fat line is below the carbohydrate constraint. b) Yes, two such points are (6.00,.33) and (.00, 6.33) c) (, 6) and (6.75, 1.5) provide 144 grams of fat d) Name six: (, 6) (.5, 5.5) (3, 5) (3.5, 4.5) (4, 4) (4.5, 3.5) (5, 3) (5.5,.5) (6, ) (6.75, 1.5) e) The slope of the iso-value line is -1 which is the same as the slope of one of the boundary lines. f) No, the minimum cannot be less than 144 grams. Any iso-value line 18*x + 18*y = c where c is a constant would not contain any feasible points if c < a) (1.75, 6.5) b) No, the family of iso-value lines does not have the same slope as any boundary line. No two corner points correspond to the same optimal value. 6. a) Maximum value is 16.5 at (7.5,.5) b) Maximum value is at (7.5,.5) c) / 1.5 = -0.4 d) Yes e) At (1, 9) maximum Z is Joseph F. Aieta Page 306 Printed 8/14/003

307 Appendix B: Answer Linear Programming Activities 1 Page a) 8. Minimum of 6 at (6, 0) b) N7 through N10 c) We are looking for the maximum value of the objective function in cell L6 which is a) Maximum of 9 at (4, 9) 10. a) Maximum of 60 at (0, 1) b) Minimum of 45 at (0, 9) 11. a) Maximum of at (10, 0) b) Minimum of at (80, 40) 1. Corner points at (0,18) (18,6) and (30,0) Minimum value of Z=10x+6y is 108 at (0,18) 13. Corner points at (3,3) (,8) and (5.4, 4.6) Minimum value of Z= 10x+15y is 75 at (3, 3) 14. a) s = Stocks m= Money Market $15,000 $75,000 $18,000 $80,000 b) $5,000 c) $5,000 d) Maximize 0.09s m subject to non-negativity & (1) m + s > 50,000 () m + s < 100,000 (3) s < 0.5( m +s) equivalent to 5m -.75s > 0 or m 3 s > 0 s >0.10 ( m + s) equivalent to 0.10m -.90 s < 0 or m 9s < 0 e) Feasible region is a quadrilateral. Maximum is $6000 at s =$5,000 and m = $75,000 Joseph F. Aieta Page 307 Printed 8/14/003

308 Appendix B: Answer Linear Programming Activities 1 Page a) 75g Fat 00g carbo Cost is $.50 b) No, insufficient carbo c) 4 cups of French fries and 0 cups of chocolate produces only 40 grams of carbo but at least 300 grams are needed. d) Yes, provides 300 grams of fat and carbos. Cost is $40.00 e) 8 cups of French fries and 15 cups of chocolate cost $34.00 and provide 45 grams of fat and 305 grams of carbos f) Minimum cost of $3.33 occurs at 0 grams of fries and 6.67 grams of chocolate. 16. let x = lbs of House Blend let y = lbs of Deluxe Maximize Revenue =10x + 11y subject to non-negativity and java constraint :.10x +.0y < 10 mocha constraint :.90x +.80y < 7 Corner points Profit (0, 0) 0 (0, 50) 550 (64, 18) 838 maximum (80, 0) 800 a) maximum of $838 with 64 pounds of House Blend and 18 pounds of Deluxe b) at (64,18) maximum rev is $910 c) at ( 0, 50) maximum rev =$1000 d) at (64,18) maximum rev is $646 e) at (80,0) maximum rev is $ a) Minimum cost of $1,700 by shipping 8000 lbs from factory A to Distributor 1, 000 lbs from factory A to Distributor, and 5000 lbs from factory B to Distributor b) Minimum cost of $13,800 by shipping 8000 lbs from factory B to Distributor 1 and 7000 lbs from factory A to Distributor Increases cost by $1,100 from $1,700 to $13, Maximize 50N + 80R where N is dollars spent on newspaper ads and R is dollars spent on radio ads. N+ R = 1000 N >= 50 R >= 50 N>= ½ R equivalent to N R >= 0 Maximum audience exposure of 70,000 by spending $ on newspaper ads and $ on radio ads. Joseph F. Aieta Page 308 Printed 8/14/003

309 Appendix B: Answer Linear Programming Activities 1 Page 309 Answers Linear Programming Activities 1. Maximum profit of $ for 13 Model E rails and 4 Model D rails. Maximum profit is $44 for 7 super safe locks and 3 safe locks 3. a) Minimum cost is $.00 for 8 pounds of Food X and 4 pounds of Food Y b) Add constraint: Y <= ½ x equivalent to -1/x + y <= 0 or -1x + y <= 0. Optimal solution does not change. Minimum cost is $.00 for 8 pounds of Food X and 4 pounds of Food Y 4. a) Minimum cost is $7.00 for 1 lb of Cheese X and 3 lbs of Cheese Y b) add constraint: y x + 1 equivalent to x + y 1. Minimum cost is $7.5 for 1.5 lb of Cheese X and.5 lbs of Cheese Y 5. Max number of viewers is 1,1,750 when programming contains 75 minutes of comedian time and 15 minutes of commercials. 6. a) Minimum cost is $15,600 for 4 days of Mill I and 10 days of Mill F. No overproduction b) Minimum cost is $36,000 at all points on the line segment joining (4, 10) and (4,0). At points other than (4, 10) both A and AA steel are over produced. 7. Maximum of 116 at (4, 6, 8) 8. Maximum profit of $17.00 by making4 model P tires, 6. Model S tires and 8 Model E tires. 9. Min Cost of $14,750 L to A L to B Q to A Q to C truckloads Minimum cost of $8,000 using 3500 gallons of water based paint, 1000 gallons of oil base paint, and 1500 gallons of stain. 11. Yield of $ by investing 1.5 million in stock, 3 million in bond1, 3 million in bond and.5 million in the bank CD. Joseph F. Aieta Page 309 Printed 8/14/003

310 Appendix B: Answer More LP Page 310 Answers More LP KLEAN Make 6 pounds of soap A and 6 pounds of soap B for a minimum cost of $.70 MKTRES Make 360 day calls and 80 evening calls for a Minimum Cost of $1,840 HARTFORD Maximum return of $160,000 by investing $400K in Atlantic, $600K in Benson, zero $ in Carnegie, $800K in Draper, and $00K in bonds BASEBALL GLOVES Produce 500 regular gloves and 150 catcher s mitts for a profit of $7,400. There will be a slack of 175 hours in cutting/sewing POWERCO Minimum cost of $1005 Plant to area COPY EDITING One solution is to assign 800 pages to Alice and 100 to Sue for a maximum quality rating of 14,400 A A3 B1 C C CEREALS Mix 0.6 oz. of cereal A and 4.0z. of cereal B for a cost of 17.6 cents per 8 oz. package. The diet provides 0.1g more than the required 1.7 g (surplus) and also provides 0.4 g less than the allowed maximum (slack Joseph F. Aieta Page 310 Printed 8/14/003

311 Appendix B: Answer TVM Activities 1 Page 311 Answers TVM Activities 1 Give dollar amounts to the nearest cent. For daily compounding, assume an exact year of 365 days unless otherwise specified. 1. a) $1, b) $1, c) $, d) $, a) PV =FV/(1+r*t) b) r =(FV-PV)/(t*PV) c) t =(FV-PV)/(r*PV) e) $8, f) $6, $1, % % 6. 5 years months 8. $6,. 9. $ a) $77. b) $ c) $ d) $ e) $6, $6, % 13. $ a) $4,700.00, 4.6% b) $4,400.00, 9.09% c) $4,100.00, $ % % Joseph F. Aieta Page 311 Printed 8/14/003

312 Appendix B: Answer TVM Activities Page 31 Answers TVM Activities %. a) % b) % c) % d) % e) % f) % 3. a) $106, b) $106, c) No, you would lose $ % 5. $31, a) 1.36% b) 1.55% c) $90 7. $0, % % % 11. $5, a) $134, b) $134,31.64 c) $3.0 to nearest cent, $3 to nearest dollar 13. $3,173,350, a) $.7 b) $71, Bank A 4.47% effective versus 4.45% effective at Bank B % % % % 0. May not want to tie up their money longer than two years. If interest rates rise then they will lose $ on sale of T- bond. Joseph F. Aieta Page 31 Printed 8/14/003

313 Answers TVM Activities 3 Appendix B: Answer TVM Activities 3 Page a) 4 b) 6 c) 5. a) 1.50 b) c) d) e) f) g) a) 7 = $ b) log 7 3 = 1/3 c) log 64 8 = 1/ d) 8 1/3 = e) e 1 = e f) log = 3 g) y = N h) 16 1/ = 4 5. $4, About 38, % 1% 10% 9% 8% 7% 6 % 4% % years b) 1 n r = 1 a little shorter than the times with annual compounding 9. a) $ a) $1, b) $747.6 c) No. Not linear growth b) $1, c) $1, a) $11, $1, b) $11,59.00 Joseph F. Aieta Page 313 Printed 8/14/003

314 Appendix B: Answer TVM Activities 3 Page a) 4. years % b) 1.5% c) should avoid stock market if he wants a guarantee on the money 15. About 8 years by rule of About 9 years by rule of a) 5.98% compounded daily APY= % Versus6.0900% b) % 18. a) $, b) $96, $14, $60, %. cents 3. $16, The zero coupon bond, 6.16% vs. 5.90% 5. a) PV=FV*e^(-rt) 6. $5, b) r = ln(fv/pv)/t c) t = ln(fv/pv)/r 7. $5, % Years Essentially the same as daily compounding 9. $4, $3, Joseph F. Aieta Page 314 Printed 8/14/003

315 Appendix B: Answer TVM Activities 4 Page 315 Answers TVM Activities 4 1. $, $3, $, $11, $3, $1, $ $9, $13, $1, $457, $ $ a) $59,63.31 b) $9, $60, $ $9, $3, $ $15.67 Joseph F. Aieta Page 315 Printed 8/14/003

316 Answers TVM Activities 5 Appendix B: Answer TVM Activities 5 Page present value. $5.00 = 30,000*.09/1 3. unpaid balance is less in each successive month 4. the monthly payment is the same for all months 5. a) 14 b) 6% c) $10, d) $6, e) $4, f) $5, g) $5, a) 36 b) 0.75% c) $ d) $10.00 e) $ $1, a) $57.83 b) $1, a) $6.1 b) $488.6 c) $1, $1,78.10 for 0 year loan of $150,000 at 8.5%. Monthly pmt would be =$1,64.91 on 5 year loan of $175,000 at 7.5% so yes they could make the payments on this loan. 11. a) $ $46, b) $0,333 a) $194, b) $143, c) $98, a) $9, b) $9, a) $13.04 b) $ c) answers will vary 16. a) $3, b) $5, a) $573, if payments are made at the end of the year. $607,905.8, if payments are made at the beginning of the year) b) n =9.54 year 10 (end) n=8.54 year 9 (begin) 18. $ c) n =13.53, year 14 (end) n= 1.99 year 13 (begin) 19. total interest paid to first bank is $100, and total interest paid to second bank is $157, Difference is $57,16 0. a) $31,17.35 b) $3, Joseph F. Aieta Page 316 Printed 8/14/003

317 1. $1,76.6. answers will vary Appendix B: Answer TVM Activities 5 Page a) Excel table b) 58 months to pay off debt. total interest =$ a) 6.4% b) 5.17% 5. a) $8.4 b) $3, % Month BB Interest PP EB a) $6.1 b) $488.6 c) $18, d) $1, a) 7.95% b) $117, a) $4, decline 30. $3, b) $ make payments since $9, < $30, % 33. $1,13.85 vs. $1,13.86 which is about the same 34. $ % 36. a) $58,971.3 b) $34,933.7 c) month 111 (with spreadsheet) 37. a) $ % b) $1, $30, $11,76.87 Joseph F. Aieta Page 317 Printed 8/14/003

318 Answers Functions&Trendlines Activities 1 Appendix B: Answer Function & Trendlines Activities Page 318 The brief answers shown below are intended to help students compare results from their applications of Excel s Trendline Answers to several of the trend line exercises are purposely incomplete. You are expected to show how you obtain your numerical results (not simply by copying from what is given below) and to be clear and concise in your interpretations. Trends can be interpreted in a variety of ways. Not all questions related to WWW data have hard and fast answers. In certain situations, such as the AIDS pandemic, the data are restated by researchers who conclude that new methods of reporting are more accurate and useful than earlier methods. If you are asked to search the Web to obtain data that is more current than what is shown in these exercises, be sure to state your source (include the URL) and include the date of your search. 1. a) a linear fit would be a poor choice for Series 3 and Series 5. c) If we extrapolate for X = 10 we obtain 109 for Series 1, 1.13 for Series and 164 for Series 4 3. a) differences are a constant 0. b) y = 0.*x c) y= 0.x nearest hundredth. a) y = 0.0x nearest hundredth b) about $5.75 c) about $7.75 nearest quarter. d) that the linear trend continues for two years beyond the given data. 4. a) y = x x b) for a ticket price above 66, the company will start to show a loss 5. a) Scatter plot shows that quadratic is better fit than linear. b) about 04,000 cases extrapolated to 1991 (X = 8) 6. a) y = 1.89x b) defense spending is increasing by approx 13 billion dollars per year. c) trend projects much higher expenditures than actual figures 7. a) y = 0.34x (three decimal places) b) above in 1991,199, 1993, 1998, and 1999 c) about 33,85,000 d) answers will vary by state 8. a) y =.07x (nearest hundredth) b) For each additional option ordered the customer can expect to wait additional days c- d) day wait with no options. May not be meaningful. d) 43 day wait for car with 10 options 53 day wait for car with a) Decreases then increases --- does not have a constant slope. b) y = 0.181x x (3 decimal places) c) model predicts average of 49 seconds for swimmers age c) revenue is a maximum when quantity is near 100 e) equilibrium occurs near q = and price = 8.66 f) q = 636 g) demand quantity is around 1400 units h) supplier profit of about $840 at a price of $ a) for 1995 model predicts $0.38, actual was $0.38 b) for 000 model predicts $0.57, actual was $0.55 c) for 001 model predicts $0.61 which is less than actual dividend 10. a) y = 4.600x (three decimal places) slope is 4.6 b) slope for medical CPI is more than twice slope in part a) 1. a) percent increases of about 7% in 1991 and again in 001 c) linear fit predicts about 1.5 billion in 00 d) quadratic fit predicts about billion in a) 1975 estimate of about 38. million or 40.% b) 1995 estimate of about 60.9 million or 46.1% 15. a) y = x x b) rate of increase is more rapid each year c) this use of trendline is fundamentally flawed 17. a) sales below fitted line in , 1998, and 000 b) model would predict about 1.07 billion CDs in 001 c) Sales of singles peaked around c) VCR sales increase, level off, than start declining. Good linear fit to DVD sales data, model suggests about 15 million players will be sold in a) y = 4.65x (two decimal places) b) model predicts about 3.4 million students. Joseph F. Aieta Page 318 Printed 8/14/003

319 Appendix B: Answer Function & Trendlines Activities Page 319 Answers Functions&Trendlines Activities 1. a) f) Match: the quadratic function described on the left with the equation y = ax + bx + c on the right. Answers Properties Column Equation Column y=-x +4x+.5 a) Vertex at (1,4.5) and one root at (.5,0) i y=1x -4x+3 y=1x -4x+3 b) Vertex at (,-1) and one root at (3,0) ii y=-x +4x+.5 y=-1x +4x+5 c) Roots at (-1,0), (5,0) and y intercept at (0,5) iii y=x +4x y=-1x +4x-4 d) Vertex at (,0) and y intercept at (0,-4) iv y=-1x +4x y=-1x +4x e) Contains the points (0,0) and (1,3) v y=-1x +4x+5 y=x +4x f) Roots at (-,0), (0,0) and vertex at (-1,-)) vi y=-1x +4x-4 g) A relatively small change in a coefficient can have a significant impact on the graph of a function. We say that the function is highly sensitive to that coefficient. Quadratic functions are most sensitive to which one of the coefficients a, b, or c? a h) Which coefficient a, b, or c, is the most directly related to shifting the graph up or down? c i) Which coefficient is most directly related to the shifting (translation) of the graph left or right? b j) Explain how the type of curvature of a quadratic function, concave up or concave down, can be predicted from its coefficients. Give an example. If a > 0 parabola is concave up. Example y=x +4x If a < 0 parabola is concave down Example y=-1x +4x+5 k) Explain how x-intercepts, generally referred to as the roots of the function, can be determined from the coefficients a, b, and c. Give an example. b ± b 4ac x = No real roots (x- intercepts) if b 4ac < 0 a example: If x -4x+3 = 0 then 4 ± ± x = = the roots are x = 1 and 3 Joseph F. Aieta Page 319 Printed 8/14/003

320 Appendix B: Answer Function & Trendlines Activities Page 30. a) f) Match: the cubic function described on the left with the equation y = ax 3 + bx + cx + d on the right Answers Properties Column Cubics y=-x 3 +9 x +5x-15 a) Local maximum at x = local minimum at x = 3.56, inflection at x =1.5 i y=-x 3 +0 y=-x 3 +0 b) Never increases and has inflection point at (0,0)) ii y=x x +3x+ y=x x +3x+ c) Local minimum at x =0.5, local maximum at x=1, inflection at x =0.75 iii y=-1x 3 +6 x -5x+5 y=1x 3-3 x +3x-1 d) Never decreases and has inflection point at (1,0) iv y=1x 3-3 x +3x-1 y=-1x 3 +6 x -5x+5 e) Local minimum at x = 0.47, local maximum at x = 3.58, inflection at x = v y=-x 3 +9 x +5x-15 f) A cubic function has exactly one local maximum. sometimes g) A cubic function has exactly one local minimum. sometimes h) A cubic function has exactly one root. sometimes i) A cubic function increases from left to right. sometimes j) A cubic function decreases from left to right. sometimes k) A cubic function has at least one point where it is neither increasing nor decreasing. sometimes (stationary inflection point) 3. a) The graph of a cubic y = ax 3 + bx + cx + d is most sensitive to which coefficient? a b) The graph of a cubic is shifted up or down by changing the coefficient d c) The x-coordinate of the inflection point for a cubic is determined by the coefficient(s) a and b Joseph F. Aieta Page 30 Printed 8/14/003

321 Appendix B: Answer Function & Trendlines Activities Page f ( x) = -.1 x x for 0 < x < 9. Hint: Let xincr = 0.5 In Excel: = -.1*x^ *x x min x incr x max y min y max guess bounds target x new_f f(x) = -.1*x^ *x f() = f(5) = 1.03 Use Solver f(x) is a maximum at (7.14, ) to nearest hundredth Joseph F. Aieta Page 31 Printed 8/14/003

322 Appendix B: Answer Function & Trendlines Activities Page f ( x) = x x + x + 5 for -3 < x < 3 3 x min x incr x max y min y max guess bounds target point x new_f f(x) = (-1/3)*x^3-0.5*x^+x , a f(a) f(-3) =6.50 f(3) = Use Solver to find root(s), local minimum and local maximum. f(x) = 0 for x =.3876 f(x) has a local minimum at ( , 3.485) f(x) has a local maximum at ( 0.618, 5.348) Joseph F. Aieta Page 3 Printed 8/14/003

323 Appendix B: Answer Function & Trendlines Activities Page a) linear fit would be a poor choice b) the exponential fit is a better fit in early years and the quadratic fit is a better fit in latter years. c) about 11,65,000 subscribers d) about 49,845,000 subscribers e) much closer to the quadratic fit 7. a) y = x x x b) Inflection point near 1987 suggests that a cubic polynomial may be a good fit. *c) answers will vary for updated data for the early 1990s and dangers of extrapolating. 8. a) Both functions increase and level off at 0.40, exponential is concave down and logistic has an inflection point 9. a) annualized rate is the solution to (1 + r) 8.5 = so r is /8.5 1 = or 0.6 %. b) projected population of the state in 003 would be about 6,8,000 to the nearest thousand d) y = 5.37x predicts an increase of about 5 people per year e) projected population in 003 would be about 8851 f) percentage growth rate is about 30.5% annualized percentage growth rate is about 3.8% g) For annual percentage increase of 3.8% the projected population in five years would be about 916 h) For annual percentage increase of 3.8% population would double in about 1.5 years. i) Answers will vary. Find actual population in a) Annual differences are not approximately the same, they vary from 3100 to 500 b) Decay factor is EXP( ) ~ y = 1610 *0.8 x c) Answers will vary. 11. a) y = x x x projected $ amount for 1998 would be about 761 million. b) Answers will vary c) Answers will vary Joseph F. Aieta Page 33 Printed 8/14/003

324 Appendix B: Answer Function & Trendlines Activities Page a) millions y = 997/( *EXP(-.015)) years since c) Answers will vary d) About a) Sample answers: The exponential y = e x is a reasonably good fit. Since EXP(0.3049) is the growth factor is a very high which translates into 35.65% per year, The function can be expressed as y = x. In the given interval, a cubic is also a reasonably good fit and a fourth degree polynomial is even better. The logistic curve might be your choice if you expected the rate of growth to level off over time. b) Extrapolations for 199 and 1993 based upon an exponential fit are 67,54 million dollars and 91,60 million dollars, respectively. Extrapolations, based upon any fitted curve, beyond one or two years may be very far from the actual data. c) Check projections against actual sales after 199. Joseph F. Aieta Page 34 Printed 8/14/003

325 Answers Measuring Change Activities 1 short answers to selected exercises Appendix B: Answer Measuring Change Activities 1 Page a) after 5 minutes b) at about the 3 minimum point c) from minimum1.5 to minimum 3 d) from minimum 5 to 6 e) avg. speed of 3 miles in 6 minutes or 30 mph. Answers will vary 3. Since some inflation rates represent increases and others represent decreases, we will compare the absolute values of the average change over periods of two years. a) b) a) 6 degrees per hour b) 5 degrees per hour c) 3.5 degrees per hour d) 0 degrees per hour e) degrees per hour f) -3.6 degrees per hour 5.. Estimated tangent lines (eye-ball based ) a) increasing by about 7 to 9 degrees per hour b) increasing by about 4 to 6 degrees per hour c) decreasing by about 4 to 6 degrees per hour d) decreasing by about 6 to 8 degrees per hour 6. a) The slope of the tangent line to the function y = f(x) at the point P f ( x + h) f ( x) b) lim h 0 h c) m( x + h) + b ( mx + b) mx + mh + b mx b mh lim h 0 = = = m h h h a) drop of 1.4 billion dollars per year. b) decreasing at the rate of about 0.7 billion dollars per year in January At a price of $100 the demand is decreasing at the rate of about 500 units per dollar. 10. a) The resulting expression would be 0/0 which is undefined. b) Let f(x) = b where b is constant then ( b b) 0 limh 0 = = 0 h h Also the graph of f(x) is a horizontal line which has a slope of zero at all points Joseph F. Aieta Page 35 Printed 8/14/003

326 Appendix B: Answer Measuring Change Activities 1 Page f(x) = 1/x ` f(x+h) - f(x) h x f(x) h x+h f(x+h) diffquot undefined conjecture f (x) = -1/x 13. a) 0.75 b) 0 c) f(x) = LN(x) x f(x)=ln(x) h x+h f(x+h) diffquot conjecture f (x) =1/x a) Hint: find a common denominator for b) Hint: expand (x+h) 4 as 1 1 x + h x x + 4x h + 6x h + 4x h + h 15. f(a+h) -f(a) a f(a) h a+h f(a+h) h $3.10 per pound at a production level of 10 pounds. 17 slope is ( x) 5x f = f ( x) = 4x = f ( x) = 0.5x 3 x 1. 3 f ( x) = 8x + 9x. f ( x) = 1 x = 1 x 1 3. y = 1x + 4. g + x x 3 3 ( x) = x + 3x = h ( x) = 3 4 x x 6. 1 s ( x) = x 1 1 x 3 x 7. t ( x) = 1 4x, at 4a 9. 4π r ax + 3bx + cx + d 31 slope is 0 3 slope is ¼ = slope is -1/4 = x = -3/4 = x = 1 or x = 4 or a) $.50 b) minimum average cost is $.00 per unit at q = 50 Joseph F. Aieta Page 36 Printed 8/14/003

327 Answers Measuring Change Activities short answers to selected exercises Appendix B: Answer Measuring Change Activities Page endpoint minimum at (-4, -16) endpoint maximum at (4, 16) local maximum at (-, 16) local minimum at (, -16) inflection point at (-0.5, 7.5) 3. endpoint maximum at (0,0) local and absolute minimum at (1,-3). local minimum at (0,0) inflection points at (0.333, 0.034) & (1, 0.083) 4. local and absolute minimum at (1,1) 5.. local and absolute minimum at (0, 1) 6. endpoint minimum at (-,-59) endpoint maximum at (4,13) inflection point at (, 5) 7. endpoint minimum at (0, 0) and (3, 0) local and absolute maximum at (1, 16) inflection point at (, 8) 8. One example would be f(x) = x 3 +x+1 9. C ( 4) = (4) = thousand dollars C(5) C(4) = = thousand dollars. average cost of airing the first four commercials is thousand dollars 10. The profit on the sale of 1000 videocassettes is $3000 At a sales level of 1000 videocassettes, profit is declining at the rate of $3.00 per videocassette 11. a) Approximately 0.8 to 1.0 percentage points per year b) % c) The percentage of U.S. car sales that were compact cars were increasing by about 0.9 percentage point per year in a) at 64 pounds the point marginal cost is b) at about 15 pounds per day 13. a) C ( 1000) b) C (1000) = dollars per extra ton 15. f ( 0) = 4,000,000( )0 ln( ) = 77, FV (3) = 10,000( ) ln(1.0)* = $ peryear 17 x f ( x) = e 18. a) about 3.5 billion dollars per year 19. local minimum at (1, -4) local maximum at (3, 0) inflection point at (, -) 0. local minimum at (4, -16) 1. local minimum at (4, 3). inflection point at (3,7) 3. Inflection points at (0, 30) and (, 0) Local and absolute minimum at (3, 3) 4. This family of curves has a local minimum Joseph F. Aieta Page 37 Printed 8/14/003

328 Appendix B: Answer Function & Trendlines Activities 3 Page 38 Answers Measuring Change Activities 3 short answers to selected exercises 1. a) 5feet b) area function is a parabola opening down. minimum surface area of square inches when x = inches and L = 47.5 inches also, the slope function changes from positive to the left of 5 to negative on the right of 5 c) 5ft by 5 ft d) maximum area is 65 sq. ft. 3. a) 60 b) positive c) a) p = ½ =0.5 c) V(0.5) = 0.5 and V(0.1) = d) negative e) q = 90, maximum profit is a) dimensions are 4.64 feet by 56.9 feet d) optimal dimensions of W and L (in feet ) are 6. optimal dimensions are radius = 0 feet and height = 40 feet Minimum surface area is about 7540 square feet. W =SQRT((*ext*area)/(*ext +int)) and L = area/w 7. C (x) = -1500/x + 8x local minimum at 187.5^(1/3) = Maximum net of $44.91 per acre using 3.75 ounces of plant food per acre. 8. f) minimum annual cost is $, at x = 8 ft 10. a) $335,050 b) $36,00 c) $36,350 d) $109 difference 11. a) land at point D 0.89 miles from point A b) Jack's shortest time is 41. minutes 1. minimum surface area of sq cm if radius is cm and height is cm c) 47.4 minutes to sail from C to B which is 6.3 minutes longer than the best time 13. maximum volume of 59.6 cubic inches for open top box with dimensions 4/3 inches by 10/3 inches by 40/3 inches 14. a) maximum area of 7,45 sq. ft. at x = 99 ft. and y = 75 ft. 15. minimum surface area of 300 sq in. from open box with dimensions 10 inches by 10 inches by 5 inches 16. a) minimum cost of $48,0 where dimensions of unpaved corner rectangles are 44.7 feet by.36 feet b) minimum cost becomes $46, c) 1995, rate of increase was about 61,500 prisoners per year d) fastest growth in prison population was in early a) 001 increase was about 3% d) in 199 increase was about $500 dollars per year e) average credit card debt was increasing most rapidly in 1995 Joseph F. Aieta Page 38 Printed 8/14/003

329 Appendix B: Answer Function & Trendlines Activities 4 Page b) apparent inflection point near X = 4 (1996) c) 1993 d) a) Observed billion versus fitted billion b) a drop of about 0.6 billion c) debt was decreasing most rapidly near 1986 debt was increasing most rapidly near 1996 Answers Measuring Change Activities 4 short answers to selected exercises 1. a) f (x) = (x 1) b) f () ~ 1.06 x x. a) f (x) = 0.8e b) f (1/) ~ x a) f (x) = 1 8x 3 b) f () ~ a) f (x) = 8 (x 3) b) f ( 0) ~ a) 8x + 1 4x + x 3 b) f (1) ~ a) f (x) = e (x 1) b) f (-1/) ~ (x 1) = (8x 4) e (x 1) 7. a) f (x) = ( x 3x + 9) 5 b) f (4) to five decimal places 8. a) f (x) = 1 b) f ( ) = 1 x x 9. a) f ' ( x) = 5 ln 5 = 1.61(5 ) x f ''( x) = 5 (ln5) =.59(5 x ) 10. f '() = ( )( e ) = 1.6( e 0.6 ).9 b) x f '( x) = 3e f ''( x) = 3 e x c) f '( x) = 3e 0.3x 6 f ''( x) = 0.9e 0.3x 6 d) x x f '( x) = 1000(1.06) (ln1.06) ~ 58.7(1.06) x x f ''( x) = 1000(1.06) (ln1.06) ~ 3.4(1.06) Joseph F. Aieta Page 39 Printed 8/14/003

330 Appendix B: Answer Function & Trendlines Activities 4 Page a) f (x) = 1,600,000(0.01e x ) a) f (x)=3/x =1900 e x b) f (x) = 0 when x = about 197 b) c) f '( x) = x + 3 x + f '( x) = x + x 13. a) 0.15*800,000 = 10,000 b) P(t)=70,000(1- e t ) 50,000-3,000t c) loss of over $7,000 d) Run campaign for 88 days for maximum profit of $107, a) always b) never c) always d) always e) never 15. a) 53 days b) maximum profit. $88, a) about 10 years b) maximum profit approximately $5,163,500 c) maximum profit approximately $7,381, maximum profit of 8 million dollars at t ~ 6.93 years 18. b) optimal ticket price $5.00 results in net just under $33,000 c) optimal ticket price remains at $5.00 but net decreases by $1000 d) optimal ticket price remains at $5.00 but profit increases. 19. a) ()(6) + (3.5)()(3) = 34 million dollars b) GP = at a total cost of 3.1 million dollars c) GP = at a total cost of 5.75 million dollars d) GP = at a total cost of million dollars e) find first derivative of cost function using k as the cost per mile on land. 0. Hint: Start with of both sides. ln(x) x = e and then take the derivative Joseph F. Aieta Page 330 Printed 8/14/003