# (a) Let x and y be the number of pounds of seed and corn that the chicken rancher must buy. Give the inequalities that x and y must satisfy.

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1 MA 4-4 Practice Exam Justify your answers and show all relevant work. The exam paper will not be graded, put all your work in the blue book provided. Problem A chicken rancher concludes that his flock of chickens needs at least 4 pounds of protein in their winter feed, at least, pounds of total digestible nutrients (TDN) and at least 4, units (IUs) of Vitamin A. Each pound of seed provides. pounds of protein,.48 pounds of TDN, and IUs of Vitamin A. Each pound of corn supplies.65 pounds of protein,.96 pounds of TDN, and no Vitamin A. Seed costs \$ per -pound sack, while corn costs \$5 per -pound sack. (a) Let x and y be the number of pounds of seed and corn that the chicken rancher must buy. Give the inequalities that x and y must satisfy. (b) Find the objective function: Express the cost in dollars of buying x pounds of seed and y pounds of corn that the chicken rancher would like to minimize. (Don t finish the problem!!!!) Stop when you have listed the inequalities and objective function for this problem. Do not finish the linear programming problem. Solution: (a) First we will find the constraint for the protein. Each pound of seed provides. pounds of protein and each pound of corn provides.65 pounds of protein, so the function that expresses the total amount of protein in terms of x and y is.(x)+.65(y). The total protein must be at least 4 pounds, so the protein constraint is.x +.65y 4 Next we find the constraint for the TDN. Each pound of seed provides.48 pounds of TDN and each pound of corn provides.96 pounds of TDN, so the function.48(x) +.96(y) models the total amount of TDN in terms of x and y. We need at least, pounds of TDN, so our second constraint is.48x +.96y, Each pound of seed provides IUs of Vitamin A and each pound of corn provides no Vitamin A, of which we need 4, units. So we have (x) + (y) = x 4, Note this last inequality required x to hold, because it simplifies to x, But we must also ensure y because negative amounts of corn does not make sense in the context of the problem. Our family of inequalities is.x +.65y 4.48x +.96y, x 4, y

2 (b) Our objective function will be (Number of pounds of seed) ( price of seed per pound )+ (Number of pounds of corn) ( price of corn per pound ) The price of seed per pound is \$/ = \$. and the price per pound of corn is \$5/ = \$.5. So our objective function is where cost is expressed in dollars..x +.5y Problem The following picture represents a linear programming problem. Three of the boundary lines for the feasible set have the value of their slopes labeled. Find the vertex of the feasible set where the objective function 5x + 6y is maximized. Solution: We want to find the form of the family of objective function lines first: 5x + 6y = c 6y = 5x + c y = 5 6 x + c 6 y = 5 6 x + c 6 The changes in slope at the vertices A, B, C and D give us the respecive intervals ( 6 ) (,, ) ( 6,,, ), (, ) 6 6 We want to maximize the cost c, which controls the y-intercept of the line y = 5 6 x+ c 6. We start with c large and, by letting c vary downward lower the line towards the feasible set. It will touch a vertex with highest possible cost c when the slope is between the changes in slope induced by the vertices (if it were equal to one of the slopes in the picture, then it would intersect the feasible set along an entire edge). The slope 5 6

3 is in the interval ( 6, 6) which goes with the vertex B: The downward slope of the cost line is steeper than the segment between A and B, but not as steep as the segment between B and C. So the vertex B is the place on the feasible set where the objective function is maximized. Problem Let A and B be transition matrices for two Markov processes modeling frogs jumping between lily pads,, and A =.4.5., B = Let d be the initial distribution matrix. d =..5 (a) Draw transition diagrams for each process A and B. (b) Find the second distribution matrix d both under the action of Markov process A and Markov Process B. (c) Using the given initial distribution, what percent of frogs are on lily pad after running each process twice? (d) Is A regular? Is B regular? Why or why not? Solution: (a) Here are the transition diagrams: The diagram for A is first, for B second.

4 (b) Under the action of the Markov Process represented by A, the second distribution matrix d is d = A d = (.) +.5(.) +.(.5) (.) +.5(.) +.(.5) = =...(.) + (.) +.(.5)....(.) +.5(.58) +.(.).45.(.)4 +.5(.58) +.(.) =.49.(. + (.58) +.(.).4 Under the action of the Markov Process represented by B, the second distribution matrix d is d = B d = = (.) +.5(.) +.(.5) (.) +.5(.) +.(.5) = (.) + (.) +.(.5)..5 (.45) +.5(.5) +.(.5). (.45) +.5(.5) +.(.5) =.85 (.45) + (.5) +.(.5).5 =

5 (c) For each Markov process, this information is given by the second entry of the second distribution matrix d from part (b). For process A, 49.% of the frogs are on lily pad after running the process twice. For process B, 8.5% of the frogs are on lily pad after running the process twice. (d) Recall a stochastic matrix is regular if some power of the matrix has all nonzero entries. We can check to see if the matrix A is regular by computing A ; =.....(.) +.5(.4) +.(.).(.5) +.5(.5) +.().(.) +.5(.) +.(.).4(.) +.5(.4) +.(.).4(.5) +.5(.5) +.().4(.) +.5(.) +.(.).(.) + (.4) +.(.).(.5) + (.5) +.().(.) + (.) +.(.) so A = Since A has no nonzero entries, A is regular. We already know that B is not regular because B is an absorbing stochastic matrix. The identity matrix in the upper left is the identity matrix, just like on Worksheet Problem 4. So every power of B will have a in the upper left and a column of zeroes underneath; equivalently the first column of B will be given by (,, ). So, every power of B has zero entries, and B is not regular. Problem 4 Form the initial simplex tableau for the following LP problem: MInimize 5x+y subject to the constraints x + y x + y 8 x + y x y Label the columns of the tableau with the variables they represent. Indicate which variables are in Group I and Group II. Stop when you have set up the initial simplex tableau. Do not finish solving the problem. Solution: To use the simplex tableau, we must convert the objective function to a maximization problem and convert the constraints to standard form of [linear polynomial] [number]. So our new problem is Maximize 5x y subject to the constraints x y x y 8 x y x y

6 We obtain the initial tableau x y u v w M u v 8 w M 5 The Group I variables are x and y. The Group II variables are u, v, w and M (Note that I labeled the rows of the tableau with the Group II variables). Problem 5 A truck rental company in Raleigh only rents trucks on a daily basis. Rented trucks must be returned to Offices,, or. The transition diagram below represents a Markov process with regular time interval as a day and the state set as Offices,, and. The probabilities are based on the percentages of trucks returned to each location based on where they were picked up. We assume that all the truckss are rented each day. (a) Set up the stochastic matrix for this Markov process with rows and columns indexed by the state set,, and. (b) On Wednesday morning, % of the trucks are at Office, 5% of the trucks are at Office, and % of the trucks are at Office. What percentage of the trucks will be at Office on Friday morning? (c) What percentage of the trucks that are at Office Wednesday morning will be at Office Thursday morning? (d) In the long run, what can we expect the distribution of trucks to be between the offices?

7 (a) The stochastic matrix has columns indexed by the current states, and rows indexed by the next states: (b) We interpret the statistics for Wednesday as the initial distribution matrix. d =.5. To find the percentage of trucks at Office on Friday morning, we run the process twice, or in other words we find the second distribution matrix d. So, d = A d = =.5.4 The percentage of trucks at Office is given by the nd entry of d, so there are.5% of the trucks at Office on Friday morning. (c) We need to track the flow of trucks from Office to Office for one iteration of the process, and we are ignoring the flow of trucks between other states in this sub-problem, so we start with an initial distribution matrix with % of the trucks at Office : d = Thursday morning is one day later. We find the first distribution matrix: d = A d =..6. =..6. = So, 6% of the trucks that started out at Office on Wednesday morning are at Office Thursday morning. (d) To answer this we use the stable distribution of the matrix representing this Markov process: since the matrix A has no zero entries, we know it is regular and this strategy is permissible. Let X = (x, y, z) be the stable distribution. Then we can find X by solving the system of equations obtained from AX = X, x + y + z =. The condition AX = X can be written as So we have the system.6x +.y +.z = x.x +.6y +.z = y.x +.y +.z = z x +y +z =.6x +.y +.z = x.x +.6y +.z = y.x +.y +.z = z

8 We need to use GJE on this system, but we only know how to do that when the last column of our augmented matrix has numbers in it. So, let s rearrange the last three equations by moving the variable on the right of the = sign in each equation to the left hand side of the = sign, using subtraction: x +y +z =.4x +.y +.z =.x.4y +.z =.x +.y.z = Then I ll rewrite it as an augmented coefficient matrix. While I m at it, I m going to convert to fractions because I hate doing GJE on decimals for big matrices First, pivot about a : R + 5 R, R R, R 4 R Next, pivot about a : 6 4 R We ll go ahead an cancel the fourth row: 6 R 4 + R R R, 6 R + R, R R 4 4 Now we pivot about a : 6 4 R R R, 6 R 6 R

9 So the stable distribution is and in the long run, we expect 6.% of the trucks to be at Office, % of the trucks to be at office, and 4.% of the trucks to be at Office. Problem 6 Maximize 6x + 48y + z subject to the constraints x 4 y 6 4x + y + z 8 x, y, z This is a maximization problem and all the inequalities are in standard form, so we can immediately set up a simplex tableau. Note we are forced to use a simplex tableau because there are three variables in the original problem: x, y and z. The system of equations with slack variables that replaces our constraint family is x +u = 4 y +v = 6 4x +y +z +w = 8 6x 48y z +M = where x, y, z, u, v, w are all and M is as large as possible. We obtain the initial tableau x y z u v w M u 4 v 6 w 4 8 M 6 48 There are no negative numbers in the upper right box that we need to get rid of, but the solution obtained from the initial tableau is not a maximum because there are negative numbers in the left side of the bottom row. We must pivot until the lower left row has no negative numbers. To choose the pivot column, find the column with the most negative entry in the bottom row. This is the - in row. Then, to choose the pivot entry in column, compare positive ratios obtained by dividing entries in the last column by the corresponding entry in column... but in this case, there are two zeros in column so our only positive ratio comes from 8/ = 9. So we must pivot about the in column : R x y z u v w M

10 x y z u v w M u 4 R 4 + R v 6 z 9 M Our new tableau has no negative entries in the left part of the bottom row, so this tableau gives us a maximum and we re done. To read off the solution, note that our Group I variables are now x, y and w, set those equal to zero. The Group II variables index the rows now (see the tableau) and we read their values from the last column. We have x =, y =, z = 9, u = 4, v = 6 and M =. So our objective function 6x + 48y + z reaches its maximum value of at the point (,, 9). Problem The following transition diagram shows an absorbing stochastic process: (a) Write down the stochastic matrix that represents this Markov process. Since this is absorbing, the matrix should take the form [ I ] S A = R In this case, the choice of indexing the absorbing states first has been made for you. (b) Write down the sub-matrices S and R for your matrix from part (a). (c) Find the fundamental matrix (I R ). (d) Find the stable matrix [ ] I S(I R ) of the absorbing stochastic process.

11 (a) We use the same indexing as the diagram gives us to write down the absorbing stochastic matrix: (b) We have Therefore [ ] I S S = R = [ ].6, R = [. ].5..5 (c) To find (I R), we choose I to be the same dimensions as R so that the subtraction is defined: [ ] [ ] [ ] I R = = We ll use the formula for the inverse of a matrix here. The determinant D = (.8)(.5) (.)(.5) =.5 So the fundamental matrix is (I R) =.5 [ ].5.5 =..8 [.486 ] (d) We simply use the formula as we have calculated all the ingredients: [ ] [ ] [ ] S(I R) = = and [ ] I S(I R ) = Problem 8 The victims of a disease are classified into three states: cured, dead, and sick. Once cured, a person is permanently cured because they become immune. Each year 8% of the sick are cured, % remain sick, and % die. We model this as an absorbing stochastic process where the regular time interval is one year and the states are Cured, Dead, and Sick. (a) Make a transition diagram for this Markov process.

12 (b) Make the absorbing stochastic matrix [ I ] S A = R for this Markov process. Be sure to put the columns indexing the absorbing states first. Write down the sub-matrices S and R explicitly. (c) Find the fundamental matrix F = (I R). Use the fundamental matrix to estimate how long someone who is sick will stay sick. Give your answer in years rounded to two decimal places. Show your work and explain the answer in the context of the problem. Recall that the ijth entry of F is the expected number of times the process will be in nonabsorbing state i if if starts in nonabsorbing state j. (d) Find the long term-trend for the progression of the disease: in other words, what will the long-term distribution of between the absorbing states of patients who contract the disease? Hint: the formula for the stable matrix of an absorbing stochastic process is [ ] I S(I R ) (a) The transition diagram is shown below (b) Note the absorbing states are Cured and Dead, which in our diagram are the states with no strictly outbound arrows. Cured Dead Sick Cured.8 Dead. Sick.

13 so that and [ ] I S.8 A = =. R. S = [ ].8, R = [.]. (c) The fundamental matrix F = (I R) = (.) = /(.9) = /9. The rows and columns of F are indexed by the nonabsorbing states, but there is only one nonbsorbing state here, which is Sick. So, the single entry of the matrix F is telling us how long someone stays in nonabsorbing state sick once they enter non-absorbing state sick, which is /9 years, approximately. years. (d) The hint gave us the formula again, so we should use it: First [ ] [ ] 8 S(I R).8 9 = =. 9 and the stable matrix is 8 [ ] I S(I R 9 ) = 9 So in the long term, 8 9 of the patients who enter state Sick will end up in absorbing state Cured, and 9 of the patients who enter state Sick will end up in absorbing state Dead. So the long term prognosis of someone who becomes sick with the disease is that they have a 8 9 chance of surviving. 9 Problem 9 A small craft shop makes toy stuffed bears and silk flower wreaths. Each bear requires hours of preparation time, 6 hours of assembly time, and hours of finishing time. Each wreath requires hours of preparation time, 5 hours of assembly time, and hours of finishing time. There are 6 hours of preparation time, 8 hours of assembly time, and 4 hours of finishing time available each week. If each bear earns a profit of \$ and each wreath earns a profit of \$, how many of each should be crafted to maximize the profit? Solution: The table below summarizes the information in the problem: Bears Wreaths Available Prep Time labor hours labor hours 6 labor hours Assembly 6 labor hours 5 labor hours 8 labor hours Finishing labor hours labor hours 4 labor hours Profit \$ \$8 Let x be the number of bears manufactured each day, let y be the number of wreaths manufactured each day. We obtain inequalities from the table: Prep : x + y 6,

14 Assembly 6x + 5y 8, Finishing x + y 4. Also, bears and wreaths can t be represented by negative numbers, so we require x, y. Our objective function is x + 8y. In standard form the inequalities are y x + 8 y 6 5 x y x + 4 x, y Let L be y = x + 8, L be y = 6 5 x and L be y = x + 4. Then the graph of the feasible set, with the boundary lines L, L, and L color coded, is shown below. The vertex A is the origin (, ), B is the y-intercept of L so is (, 4), and E is the x-intercept of L, so is (, ). To find C, set L and L equal: Then 6 5 x = x x = 8 5 x = 6 5 () = So C has coordinates (, ). Then D is the intersection of L and L: 6 5 x = x + 8 x = 5 x = 8 Then 6 8 (8) = 6 and D has coordinates (8, 6). We check the value of each vertex in the objective function: Vertex Profit = x + 8y (, ) () + 8() = (, 4) (4) + 8() = 4 (, ) () + 8() = 6 (8, 6) (8) + 8(6) = 8 (, ) () + 8() =

15 The maximum value of \$8 occurs at the vertex (8, 6). To maximize profit each week, the craft shop should make 8 bears and 6 wreaths. Problem Form the initial simplex tableau corresponding to the following linear programming problem: Maximize x + y z subject to the constraints x + y z x y + z 8 x + y + z x, y, z Stop when you have formed the simplex tableau. Do not solve the problem. Solution: Our simplex tableau will correspond to the system of equations x +y z +u = x 6y +z +v = 8 x +y +z +w = x y +z M = where x, y, z, u, v and w are all, and M is as large as posslble. The initial tableau is x y z u v w M u v 8 w M Problem State the maximization problem corresponding to the following initial tableau, and solve it using the simplex method. Solution: x y u v M u 6 v M 4 We observe there are no negative numbers in the upper right hand box that gives values of the group II variables, we can take each row above the horizontal bar to represent an inequality in standard form: [linear polynomial] [nonnegative constraint]. Furthermore we are simply told that the objective row represents a function we d like to maximize. So the problem is: Maximize x 4y subject to 6x + y 5x + 5y 95 x, y

16 Note we are told to use the simplex method, so a graphical solution is not acceptable in this case. If you are told to use the simplex method on a two-variable problem on the exam and you use Chapter Methods, you will not get any points for that solution!!! On to the tableau: we don t need to get rid of any negative numbers in the upper right-hand box of the tableau, so we can proceed using the methods of Section 4.. First we examine the entries of the bottom row: since there is a negative number,, in column, the initial tableau does not yield a maximum and we must pivot. We choose the pivot column by choosing the column whose row entry is most negative, which is column. To choose the pivot element in column, we compare the positive ratios obtained by dividing elements of the last column by column entries in the corresponding row. Now 95/5 =, and /6 =, so we pivot about the 5 in column, row : 5 R x y u v M R 6R R + R x y u v M u x 5 M Now there are no negative entries in the bottom left box, so the current tableau is a solution to the maximization problem. Set the group I variables equal to zero, and the group II variables are written to the left of the tableau, whose values we get from the last column: x =, y =, u = 4, v =, M = 9. So, the maximum value of 9 occurs at the point (, ). Problem A major coffee supplier has warehouses in Seattle and San José. The coffee supplier receives orders from coffee retailers in Salt Lake City and Reno. The retailer in Salt Lake City needs 4 pounds of coffee, and the retailer in Reno needs 5 pounds of coffee. The Seattle warehouse has pounds available, and the warehouse in San José has 5 pounds available. The cost of shipping from Seattle to Salt Lake City is \$.5 per pound, from Seattle to Reno \$ per pound, from San José to Salt Lake City \$ 4 per pound, and from San José to Reno \$ per pound. Find the number of pounds to be shipped from each warehouse to each retailer such that the total shipping cost is minimized. Solution: Let x be the number of pounds of coffee shipped San José (SJ) to Salt Lake City (SLC), let y be the number of pounds of coffee shipped from SJ to Reno. Then, since SLC requires 4 pounds of coffee, Seattle must ship 4 x pounds to SLC, and since Reno requires 5 pounds, Seattle must ship 5 y to Reno. SJ only has 5 pounds to ship, so x+y 5. Seattle only has pounds to ship, so (4 x)+(5 y). Further, we require the amounts of coffee to be nonnegative, so we obtain the family of constraints x + y 5 (4 x) + (5 y) x, y 4 x, 5 y

17 In standard form, these become y x + 5 y x + 5 x 4 y 5 x, y The objective function that we wish to minimize is the shipping cost, which is the sum of four products, one from each shipping leg:.5(4 x) + 4x + y + (5 y) = 5 +.5x y Here is a graph of the feasible set, which has six vertices: where L is the line y = x + 5, L is the line y = x + 5, L is the line x = 4, and L4 is the line y = 5. The coordinates of the vertices are very easy to solve for so I will omit that discussion. We check the value of the objective at all six vertices: Vertex Cost = 5 +.5x y A = (, 5) 5 +.5() (5) = B = (, 5) 5 +.5() (5) = C = (5, 5) 5 +.5(5) (5) = 95 D = (4, ) 5 +.5(4) () = 55 E = (4, ) 5 +.5(4) () = 65 F = (5, ) 5 +.5(5) () = 5 So the cost is minimized at vertex B when x = and y = 5. So, to minimize the shipping cost, we ship 4 pounds of coffee from Seattle to SLC and 5 pounds from SJ to Reno. (Note: in this case, since x =, two of the shipping routes are not used!!). Problem Worksheet Solutions will be posted separately.

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