III INTEREST FORMULAS AND EQUIVALENCE. willing to pay for the use of money. The question arises why people may want to pay for use of others

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1 III 27 INTEREST FORMULAS AND EQUIVALENCE 3.0 INTEREST Interest is money earned (paid) for use of money. This definition of interest implies that people are willing to pay for the use of money. The question arises why people may want to pay for use of others money? A simple answer is that people who use our money may be able to invest it somewhere else and make more money from that investment. They just pay us a part of the money they made from the investment for use our money. This is very obvious in banking industry as banks pay interest on our deposits and in turn lend that money (our deposits) for use by others for example, for mortgage for homes or for buying automobiles, etc. The rate at which interest is paid to depositors and charged from borrowers depends on many factors such as the amount deposited or borrowed, time period, etc. A discussion about the rate of interest charged or paid is beyond the scope of this book. As stated earlier, interest is the money earned (paid) for the use of money. Alternatively, interest can be considered a charge for use of money. When money is borrowed from financial institutions, the charge for such use of money is referred to as interest. However, when owner s money is used for a business, the same charge is referred to as profit. Thus interest and profit are synonymous terms from the point of view of the use of money. 3.1 TIME VALUE OF MONEY Because money can earn interest over a period of time, money received now is preferred over the same amount of money received in future. For example, suppose we have two alternatives (i) take $10 now

2 28 or (ii) take $10 three years later. The preferred alternative (as long as interest rate is greater than zero) is take $10 now because $10 in (i) will earn interest for three more years. Note that in both the alternatives the dollar amount is same ($10) but we have more time (three years) for use of money in alternative (i) than in (ii). As money has earning power, money received (same amounts) is preferred to over money received in future. The relationship between money and time, due to earning power or interest, is referred to as the time value of money. Time also affects the purchasing power of money. In general over a period of time, due to inflation, purchasing power of money will decline as more money is needed to buy same goods and services. This will effect the choice of alternatives. This subject is dealt separately in chapter on Inflation (Chapter VII). Interest rates are generally quoted on an annual basis. For example, if interest rate is 10% per year then the amount deposited will earn 10% interest in one year. Two types of interest rates are used; simple and compound. In most real life situations, especially in dealings with financial institutions, compound interest is used more often than simple interest Simple Interest: Simple interest means that only principal, the amount under consideration, will earn interest at the specified interest rate. A formula for computation of amount of interest, I, earned is given below: I = (P)(N)(i) (3.1.1) where P = principal amount, $ N = time, years i = simple interest rate, % per year Example A person deposits $2,500 for three years in a bank. The bank pays 8% simple interest per year. Determine (i) the amount of interest earned and (ii) total money that can be

3 29 withdrawn at the end of the third year. Solution: Using Eq.(3.1.1), interest earned, I, is: I = (P)(N)(i) = (2,500)(3)(0.08) = $600 Total amount of money that can be withdrawn is the sum of the principal and interest earned in three years. Therefore, Amount of withdrawal = Principal + Interest = 2, = $3,100 Simple interest is generally quoted on an yearly basis. If money is kept for a fraction of a year, interest earned is prorated. As an example, if a person deposits $1,200 at 12% for 30 months or 2.5 years, interest earned is (1,200)(2.5)(0.12) or $ Compound Interest: In compound interest, both principal as well as interest earned earn interest. As interest earned in previous years also earns interest, total amount of interest earned is higher if interest is compounded. Suppose an individual deposits $1,500 for three years in a bank which pays 8% interest compounded yearly. Table shows amount of interest earned year by year. In first Table Amount of Interest Earned Year Amount at the start of the year Interest Amount at the end of the year Amount withdrawn at the end of the year 1 1, , , , , , , year interest earned is (1,500)(0.08) or $120. Total amount of money available at the end of the first year is principal $1,500 plus interest $120 or $1,620. As no withdrawal is made, total money in the bank is $1,620

4 30 and all of it will earn interest in the second year. Calculations for the second and third years are shown in Table The individual is able to withdraw $1, at the end of the third year. Note that if simple interest rate of 8% is used then the individual will be able to withdraw: 1, (3)(0.08) or only $1,860. Example illustrates interest calculation for amounts deposited in a bank. Similar calculations can be used to determine interest that will be paid on loans. Example A person borrows $5,000 for three years at 12% compounded yearly. The loan agreement is that $500 per year will paid back each year with all remaining amount paid back at the end of the third year. Determine the amount needed to pay off this loan at end the of the third year. Solution: Calculations for amounts due and payments made towards the loan are shown in Table Interest due at the end of the first year is (5,000)(0.12) or $600. Total amount owed is (5, ) or $5,600. Out of this, $500 is paid back at the end of the first year. Therefore, amount owed at the Table Year by Year Payment for a Loan Year Amount due at the start of the year Interest Amount due at the end of the year Amount paid back 1 5, , , , , , , start of the second year is (5, ) or $5,100. Note that amount owed or $5,100 is sum of the principal, $5,000 and the interest, $100. In compound interest, interest must be paid on both principal and interest. Therefore, interest in second year is (5,100)(0.12) or $612. Complete calculations for second and third year are shown in Table The entire amount due ($5,837.44) is paid back at the end of the 3rd year and this retires the loan. Table just represents one way of paying off a loan and different arrangements can be made to pay off a loan.

5 INTEREST FORMULAS Main purpose of developing interest formulas is to obtain equivalent cash flows. Two sets of cash flows are said to be equivalent if their value at a common point in time is same. To further illustrate the meaning of equivalence, consider following situations: I: Alternative A1: Take $10 now. Alternative A2: Take $10 three years from now. II: Alternative B1: Take $20 now. Alternative B2: Take $25 now. III: Alternative C1: Take $15 now. Alternative C2: Take $25 three years from now. Using the concept of time value of money, for I, the preferred alternative is A1 because we have same amount ($10) but three more years of 'use of money' ($10 will earn interest for three extra years) in A1 than in A2. For II, the preferred alternative is B2 because we have more money ($5 more) in B2 and equal time for 'use of money' because both the amounts are received at the same time (now). However for III, it is not clear which one is better, C1 or C2. The reason for this is that if we choose C1 we have three more years of 'use of money' but if we choose C2 we have $10 more. For determination of the preferred alternative, one must determine the amount of interest which can be earned over the three year period. Let us assume that applicable interest rate is 10% compounded yearly. This specification of interest rate allows us to calculate the amount of interest that can be earned as well as the value of $15 at the end of the 3rd year. Amount at the end of the first year = (.10) = $16.50 Amount at the end of the second year = (.10) = $18.15

6 32 Amount at the end of the third year = (.10) = $19.97 Therefore, value of $15 at the end of the third year is $ The two amounts, $15 now and $19.97 at the end of the third year are said to be equivalent as they have same value. The two alternatives for III can be written as follows: C1: Take $15 now or $19.97 three years from now. C2: Take $25 three years from now. The preferred alternative is C2 as it has more money ($25 as opposed to $19.97) and time for 'use of money' for both C1 and C2 is same as both are received at the end of the 3rd year. For more complex cash flow situations, determination of equivalent cash flows is not as easy as shown above. To facilitate computation of equivalent cash flows, which aid in selection of alternatives, interest formulas are developed and used. Before we develop interest formulas, concept of cash flow diagrams will be introduced. Cash flow diagrams facilitate determination of equivalent cash flows. A cash flow diagram is a graphical representation of a problem. Consider an example where a corporation can invest $20,000 in a machine for making a product. Life of this machine is expected to be 5 years. Net revenues (difference between revenues and expenses ) will be $3,000 per year. At the end of 5 year life, the company has no use for this machine and it can be sold for $2,500 ( salvage value is $2,500). This investment situation can be represented graphically in a cash flow diagram shown in Figure In a cash flow diagram, time periods are shown along the x-axis and the cash flows are shown by arrows pointing upwards or downwards. Cash outflows are shown by downward arrows while cash inflows are shown by upward arrows. Investment of $20,000 in the machine is a cash outflow for the corporation

7 3,000 2, , Year Figure Cash Flow Diagram for Investment in the Machine and is shown by a downward arrow in Fig Net revenues per year of $3,000 are cash inflows and so is the salvage value of the machine,$2,500. All these cash flows are shown by upward arrows. 3.3 INTEREST FORMULAS FOR SINGLE CASH FLOWS Following notations and assumptions are used in deriving interest formulas. P = present amount of money (cash flow) OR the equivalent value of one or more cash flows at time defined as present (generally t=0). F = future amount of money (cash flow) OR the equivalent value of one or more cash flows at time defined as future (generally t=n). i = interest rate, % per period. N = number of time periods, no units. It should be pointed out that all cash flows are assumed to occur only at the end of a time period, for example a year, unless specified otherwise. This is referred to as the end of period convention. If period is years, cash flows are not allowed to occur at any other time period, for example, the end of the 3rd or the 6th month. As an example, a depositor adds $500 to her account in the third year. This automatically means that deposit is made at the end of the third year unless specified as start of the 3 rd year. It is worth mentioning that end of a year is also the start of the next year. Also, in the derivation of interest formulas, the interest rate, i, is always compound interest rate.

8 Find F Given P A cash flow diagram for finding F when P is known is shown in Fig This cash flow diagram is drawn from depositor s point of view. The amount P, a deposit in a bank, is shown as a cash outflow at t=0, or at the end of year zero or start of the 1st year. The amount F is shown as cash inflow as the depositor has this amount in the account and will be able to withdraw it from the bank at t=n, or at the end of the Nth year. In this situation, the money (P) is kept in the bank for N years. The value of F, how much money is in the bank, is determined by successively finding the amount of interest earned and total amount in the bank at the end of each year using i as the interest rate. F =? 1 2 N Year P = Known Figure Cash Flow Diagram for Finding F Given P Amount at the end of the first year = principal + interest earned = P + P(i) = P(1 + i) The amount at the end of the 1 st year is the principal at the end of the 1 st year, or start of the 2nd year. Amount at the end of the 2nd year = principal + interest earned = P(1 + i) + [P(1 + i)]i = P(1 + i) 2 Similar calculations can be carried out for each year till the end of year N. It should be noted that amount at the end of 1 st year is P(1 + i) 1 and at the end of the 2 nd year it is P(1 + i) 2. By symmetry, the amount at the end of year N is as follows:

9 35 Amount at the end of Year N = P(1 + i) N The amount at the end of year N, by definition is F. Therefore, F = P(1 + i) N (3.3.1) Eq. (3.3.1) is the interest formula for finding F given P. The unknown amount, F, is determined by multiplying P, the known amount, by a multiplier or the expression (1 +i) N. Note that the interest rate, i, as well as the time period, N, should be specified. Standard notations has been developed for designation of multipliers needed to find the unknown amount for a given (known) amount. In our example, the unknown amount is F while the known amount is P. The standard notation for the multiplier needed to multiply P to find F, referred to as a factor, is (F/P, i, N). The factor, (F/P, i, N) is read as F given P. Note that the factor is based on the problem statement as we are trying to find F given P. Using standard notation for the factor, Eq. (3.3.1) is written as follows: F = P(F/P, i, N) (3.3.2) Values of the factor F given P for different values of i and N are contained in Appendix A. Example An individual deposits $2,500 in a bank for 4 years. The interest rate is 6% compounded yearly. How much money is accumulated in this account at the end of the 4th year? Solution: We first draw a cash flow diagram (Fig ) from the depositor's point of view. F=? P=2, Year Figure Cash Flow Diagram for Example (a) Interest Formula: In this problem we know P, amount at t=0, and need to find F (money in the account at a future time)

10 36 at t=4. We use the interest formula in Eq. (3.3.1). F = P(1 + i) N Substituting for P = $2,500, i = 0.06, and N=4, we get F = 2,500( ) 4 = $3, The individual has $3, in the bank account at the end of the 4th year. It should be noted that entire amount, $2,500 is kept for 4 years. (b) Factor: As the problem is to find F when P is given or known, the factor to be used is (F/P, i, N). Using this factor, we have F = P(F/P, i, N) = 2,500(F/P, 6, 4) The value of the factor (F/P, 6, 4) is obtained from Appendix A. Appendix A.9 is used because i = 6%. Then we look for the column which contains the required factor. For our example, second column is used because we need (F/P, i, N). The value of this factor is in the row with N = 4. This value is Using this value, we have F = 2,500(1.262) = 3, There are two ways to determine the value of F given P. One is to use Eq. (3.3.1) or the interest formula and other method is to use the factor (Eq ) and tabulated values in Appendix A. In general, it requires less effort if values of factors in Appendix A are used. It is worth mentioning that values of factors in Appendix A are actually obtained by using interest formulas such as Eq. (3.3.1). It is a good practice to use factors and Appendix A. Interest formulas are used only if a factor is not listed for specified

11 values of i and/or N For example, the value of factor (F/P, 6, 37) is not tabulated and therefore interest formula must be used. 37 Two cash flows F and P in Example are equivalent. Cash flow, F =$ is at t = 4, while P = $2,500 is at t=0. The value of both these cash flows, even though these are at different points in time, is same. The increase in amount at t=4 or (3, ,500) represents the time value of money in four years at the specified rate of 6%. It should be noted that these two cash flows are equivalent only at interest rate of 6% compounded yearly To Find P Given F The cash flow diagram for finding P given F is shown in Fig F=known P=? 1 2 N Year Figure Cash Flow Diagram for Finding P Given F A practical example of this problem is the determination of amount that should be deposited in a bank now, which will accumulate to $10,000 needed for a down payment for a house in 5 years if interest rate is 8% compounded yearly. Note that according to our definition, the amount to be deposited now is P and accumulated amount is F. Therefore, the problem is to find P given F. To determine P, we can solve for P in Eq. (3.3.1). F = P(1 + i) N (3.3.1) Solving for P: P = F(1 + i) -N (3.3.3) Eq. (3.3.3.) is the interest formula for finding P given F. The factor for finding P given F is (P/F,

12 38 i, N). Recall that the factor is read as P given F but written as P/F. Using this factor, P is calculated as follows: P = F(P/F, i, N) (3.3.4) Values of the factor (P/F, i, N) for different values of i and N are tabulated in Appendix A. Example A student is planning to save for a trip to Europe. The trip will cost $3,000 and will be taken 3 years from now. How much money should be deposited in a bank which pays 10% compounded yearly now so that $3,000 can be withdrawn at the end of the 3rd year? Solution: We first draw a cash flow diagram (Fig ). Note that $3,000 is needed F=$3,000 P=? Year Figure Cash Flow Diagram for Example at t=3 and is designated as F as it is a future amount. The amount needed for deposit now (t = 0) is P as it is a present amount (it occurs before F). The problem is to find P given F. Using the factor (P/F, i, N), we have P = F(P/F, i, N) (3.3.4) Substituting F = $3,000, i = 10%, N=3, we get P = 3,000(P/F, 10, 3) The value of factor (P/F, 10, 3), from Appendix A, is and P = 3,000(0.7513) = $2,253.90

13 The student needs to deposit $2, now to accumulate $3,000 in the bank at the end of the 3rd 39 year. Two interest formulas for finding equivalent cash flow to a given single cash flow have been derived. Following are the two different ways to obtain this equivalent cash flow. Find F given P Find P given F Interest formula F = P(1+i) N P = F(1+i) -N Factor F = P(F/P, i, N) P = F(P/F, i, N) We use the factors and interest formulas are used only when factors for specified values of i and/or N are not tabulated. Example How much money should be deposited now in a bank which pays 4% interest compounded yearly so that a $1,000 withdrawal can be made two years from now and a $1,500 withdrawal can be made four years from now? Solution This is a typical problem which can be solved using the concept of equivalence. We need to find an equivalent cash flow at t=0 (amount to be deposited now), which has same value as withdrawals $1,000 at t=2 and $1,500 at t=4. In other words, our deposit should allow us to make the two specified withdrawals. A cash flow diagram is shown in Fig Since the amount to be deposited is at t=0, it is designated as P in the cash flow diagram. The two withdrawals are shown by upward arrows. The first withdrawal of $1,000 will occur at the end of the 2nd year. How much money should be deposited now which will allow a withdrawal of $1,000? We need to find an equivalent cash flow (amount

14 of deposit) at t = 0 for this withdrawal of $1,000. The problem is to find P given F and factor to be used in 40 finding this P is (P/F, i, N). This value of P is designated as P 1. 1,000 1,500 P=? Year Figure Cash Flow Diagram for Example P 1 = F(P/F, i, N) = 1,000(P/F, 4, 2) = 1,000(0.925) = $925 This cash flow, $925, is at t=0 and has the same value as $1,000 at t=2. Therefore, these two cash flows are equivalent. Another way to describe this equivalence is that we need to deposit $925 now so that we can withdraw $1,000 at the end of the 2nd year. Similarly, let the equivalent cash flow for the second withdrawal of $1,500 at t=4 is P 2, then P 2 = F(P/F, i, N) = 1,500(P/F, 4, 4) = $1, The amount of money that should be deposited in the bank at t=0 to allow withdrawals of $1,000 and $1,500 is the sum of P 1 and P 2. Amount to be deposited = P 1 + P 2 = , = $2,207.20

15 41 Concept of equivalence and factors were used in determining the amount of needed deposit. This solution can be checked out by working from an initial deposit of $2, and determining if the two withdrawals can be made at specified time periods. Amount at the end of the 1st year = Principal + Interest = 2, (2,207.2)(0.04) = $2, Amount at the end of the 2nd year = 2, (2,295.45)(0.04) = $2, A withdrawal of $1,000 is made at the end of the 2nd year, therefore the remaining amount in the account is (2, ,000) or $1, Amount at the end of the 3rd year = 1, (1,387.26)(0.04) = $1, Amount at the end of the 4th year = 1, (1,442.75)(0.04) = $1, A withdrawal of $1,500 can be made at the end of the 4th year leaving no money in the account. The amount left, $0.46, is due to round off errors in factors as well as multiplication. This shows that a deposit of $2, will provide for the two specified withdrawals. The above procedure used for checking the solution is lengthy and is presented here only to show that the usage of factors or interest formulas will provide a correct solution. Actually, all interest formulas and factors are developed using the fundamental principle of time value of money used in checking this solution. In general, it is not necessary to check the solutions. 3.4 INTEREST FORMULAS FOR EQUAL PAYMENT SERIES In Section 3.3, interest formulas and factors for equivalent cash flows to single cash flows (P and F)

16 42 were developed. In real life and in many engineering applications, cash flows may be uniform over a time period. For example, 10 deposits of $1,000 each every year. The amount of each cash flow is same and after the first $1,000 all succeeding periods have a $1,000 deposit. In this section, interest formulas for such cash flows, referred to as Equal Payment Series (EPS), are developed. These cash flows are also called Uniform Series or Annuities. Cash flow diagram for a typical EPS is shown in Fig A 1 2 N Year A Figure Cash Flow Diagram for Equal Payment Series In an EPS, designated by A, an equal amount of cash flow occurs at the end of each year. For an EPS for N years, there are N equal A s with first A at t = 1 and last A at t = N (See Figure 3.4.1). Interest formulas are derived for this standard EPS. However, in general an EPS can start at any point in time To Find F Given A A cash flow diagram for finding F given A is shown in Fig We know the EPS or A and would like to find F, its value at the end of the Nth year. F=? 1 2 Year N A A Figure Cash Flow Diagram for Finding F Given A An interest formula for finding F can be derived by treating each A as a single cash flow and using the interest formulas for single cash flows (Section 3.3.1). Let us use the first A at t =1 and find its value at t = N using the interest formula for finding F given P (Eq ) or F = P(1 + i) N

17 43 We substitute P = A as this A is a single cash flow and N = N - 1 as the two single cash flows A (or P) and F are separated by (N - 1) years and not N years (See Figure 3.3.1). The resulting value of F is designated as F 1 and F 1 = A(1+i) N-1 Let F t be the future value of an A at the end of period t. Using a similar approach, treating each A in EPS as a single cash flow, F t for different time periods are as follows: F 2 = A(1 + i) N-2 F 3 = A(1 + i) N F N = A(1 + i) N - N = A Since all F t s are at t=n, these can be added together for determination of F F = F 1 + F F N or F = A(1 + i) N-1 + A(1 + i) N A(1 + i) + A (3.4.1) A closed form of Eq. (3.4.1) can be obtained by mathematical manipulation. We first multiply Eq. (3.4.1) by (1 + i). The result is shown in Eq.(3.4.2). F(1+i) = A(1 + i) N + A(1 + i) N A(1 + i) (3.4.2) Then subtract Eq. (3.4.1) from Eq. (3.4.2) F(1+i) - F = A(1+i) N - A Solving for F, we get F = A (1 + i)n 1 i (3.4.3) the factor then Eq. (3.4.3) is the interest formula for finding F given A. The factor for finding F given A is (F/A, i, N) because the problem is to find F given A. If we use

18 F = A(F/A, i, N) (3.4.4) 44 The values of the factor (F/A, i, N) for different values of i and N are tabulated in Appendix A. Example An individual is planning to deposit $500 every year for 10 years. How much money is in the account at the end of 10 th year if interest rate is 6% compounded yearly? Solution: A cash flow diagram for this problem is shown in Figure F=? 1 2 Year 10 A=500 A Figure Cash Flow Diagram for Example This series of ten deposits constitute an EPS with A = $500, first A at t =1 and last A at t =10. As we need amount of money at t = 10, F is at t =10. Note that this time is same as the time of the last deposit or the end of the 10 th year. The problem is to find F given A. Using the factor for F given A and its value from Appendix A, we get F = A(F/A, i, N) = 500(F/A, 6, 10) = 500(13.181) = $6, The account has $6,

19 To Find P Given A The purpose of this interest formula is to determine a cash flow P (at t =0) which is equivalent to A or an EPS. An example is, "how much money should be deposited now (P) such that equal annual withdrawals (A) can be made for N years? A cash flow diagram for this situation is shown in Fig A A P=? 1 2 N Figure Cash Flow Diagram for Finding P Given A Interest formula for finding P given A will be derived by first finding an equivalent F (at t = N) to A. F is found using the interest formula for finding F given A (Eq ) F = A (1 +i) i N 1 As F at t = N is known, the problem is to find P given F. P is determined using Eq. (3.4.1) as follows: P = F(1+i) -N Substituting for F from Eq.(3.4.5) we get (3.4.5) P = A (1 + i) i N 1 ( 1 + i) -N or P = A (1+i) N 1 (3.4.6) i(1+ i) N

20 46 Eq.(3.4.6) is the interest formula for finding P given A. The factor for finding P given A is (P/A, i, N). Using the factor, P is P = A(P/A, i, N) Example How much money should be deposited in a bank now which will allow withdrawals of $800 per year for 3 years (first withdrawal at t = 1)? Use an interest rate of 12% compounded yearly. Solution: It should be pointed out that withdrawals of $800 per year means that withdrawals are at the end of each year. The cash flow diagram for this problem is shown in Fig A=800 P=? Year Figure Cash Flow Diagram for Example Note that withdrawals constitute an EPS with A = $800, first A at t=1 and last A at t =3. As the problem is to find P given A, we use factor (P/A, i, N) and its value from Appendix A. P = A(P/A, i, N) = 800(P/A, 12, 3) = 800(2.402) = $1, A deposit of $1, will allow us to withdraw $ 800 per year for three years.

21 To Find A Given F to find this A. This interest formula is used to determine A which will accumulate to F at t = N. We use Eq.(3.4.3) F =A (1+i) i N 1 (3.4.3) We solve Eq. (3.4.3) for A in terms of F i A =F (1+i) N 1 (3.4.7) Eq.(3.4.7) is interest formula for finding A given F. As the problem is to find A given F, the factor to be used is (A/F, i, N). Using factor, A is A = F(A/F, i, N) Example A company is planning to set aside equal amounts of money each year for 5 years for purchasing a machine at the end of the 5th year which will cost $7,500. If interest rate is 8% compounded yearly, determine this equal amount that should be set aside each year for the purchase of this machine. Solution: The cash flow diagram for this problem is shown in Fig The problem is to find A given F (first A at t =1, last A at t = 5). Using the factor (A/F, i, N), we have A = F(A/F, i, N) = 7,500(A/F, 8, 5) = $1,278.75

22 48 F=7, Year A=? Figure Cash Flow Diagram for Example Therefore, five equal annual deposits of $1, will accumulate to $7,500 and the machine can be purchased at the end of the 5 th year To Find A Given P This interest formula is used to determine A, an EPS for N years, given P. The problem is to find A given P. We use Eq. (3.4.6) to find this A. N (1+ i) 1 P = A N i(1 + i) (3.4.6) Solving for A, we get i(1+ i) A = P (1+i) N N 1 (3.4.8) Eq.(3.4.8) is the interest formula for finding A given P. The factor for finding P given A is (A/P, i, N). Using factor A is, A = P(A/P, i, N) Example A deposit of $20,000 is made now in a bank which pays 6% interest compounded yearly. Determine the amount of equal annual withdrawal (end of year) for 7 years from this deposit. Solution: The cash flow diagram is shown in Fig The problem is to find A given P (first at t =1,last at t = 7)

23 49 A=? A Year P=20,000 Figure Cash Flow Diagram for Example Using the factor (A/P, i, N), we have A = P(A/P, i, N) = 20,000(A/P, 6, 7) = 20,000(0.1791) = $3, Withdrawals of $3, every year for 7 years can be made from a deposit of $20, SOME EXAMPLES presented. In this section, some examples for using interest formulas and concept of equivalence are Example An individual deposits $800 in a bank for 5 years. The bank pays 8% simple interest for the first 2 years and 10% compound yearly for the last 3 years. How much money is accumulated in the bank at the end of 5th year? Solution: Interest earned in the first two years is obtained using formula for simple interest (Eq.3.3.1). I = (P)(N)(i) = (800)(2)(0.08)

24 50 = $128 At the end of the 2nd year, the individual has or $928 in the account. This amount will earn interest for remaining three years at a rate of 10% compounded yearly. The problem is to find F given P with P = 928 and N=3. Using the factor for finding F given P, we have F = P(F/P, i, N) = 928(F/P, 10, 3) = 928(1.331) = $1, The individual will have $1, in this account. Example A couple has been saving money in an account in a bank for their daughter's education in a private high school. Their daughter is 11 years old now and the account has $10,000. She will enroll in a high school at the age of 14 (the end of the 14th year). Estimated expenses for four years are $3,800 per year and are needed at the start of each school year. The bank pays 6% interest compounded yearly. How much money, if any, should be added in this account now to provide for 4 years of high school education? Solution: We first draw a cash flow diagram (Fig ). In this diagram, $10,000 in the account is shown at the end of the 11th year or now. The first withdrawal of $3,800 is needed at the start of the 15th or the end of end of 14th. 3,800 3, Year 10,000 Figure Cash Flow Diagram for Example 3.5.2

25 51 There should be enough money in the account at t =11 to allow for four specified withdrawals. Using the concept of equivalence, we need to find the value of these withdrawals at t = 11. It can be seen that the four withdrawals constitute an EPS for four years (as there are four withdrawals) with A = $3,800 but the first A is at t =14 and last A is at t = 14. Recall that it a standard EPS, P is at t = 0 and F is at t = N. The first A in EPS of withdrawals is at t = 14 therefore, P is at t = 13. Also, as the last A is at t =17, F is at t =17. Using the factors or interest formulas, the value of the four withdrawals can be determined only at t =13 or 17. We determine the value of withdrawals at t = 13.This unknown value is designated as P 13 (a single cash flow at t = 13). The problem is to find P given A. Using the factor, P = A(P/A, i, N) Here, A = $3,800, i = 6% and N =4 (four withdrawals). Substitution of these value yields P 13 = 3,800(P/A, 6, 4) = 3,800(3.4651) = 13, The physical meaning of P 13 is that $13, at t =13 will allow exactly four specified withdrawals of $3,800 each. The couple needs to have enough money at t = 11 and not at t =13. Therefore, we need to find an equivalent cash flow at t =11 to the $13, at t =11. As this unknown equivalent cash flow occurs before the $13,167.38, it is designated as P 11 (a single cash flow at t =13). Note that both the known and unknown cash flows are single cash flows and the problem is to find P given F. Using the factor, P = F(P/F, i, N) Here F = 13, and i = 6% and N =2. Recall that for factors for single cash flows, N is the elapsed time between P and F. As the two time periods are 11 and 13, N = = 2. Substitution of these values yields

26 52 P = 13,167.38(P/F, 6, 2) = 13,167.38(0.8900) = $11, The couple has only $10,000 in the account and needs to add $1, now for four $3,800 withdrawals for school expenses. Example Given an EPS of 5 payments of $800 with first A at t = 3 and interest rate is 4% compounded yearly. (a) (b) Determine a single cash flow at t = 10 which is equivalent to the above given EPS. Determine an EPS (A) of 6 payments with first A at t = 2 which is equivalent to the above given EPS. Solution: Recall that two sets of cash flows are equivalent if their value at some common point is same. (a) The two cash flows are the EPS of 5 payments and a single unknown cash flow, say X at t = 10. We equate values of these two cash flows at t =10 and solve for X. Note that first A in the given EPS is at t =3 and the last A is at t = 3 +(5-1) = 7. Using the factors, we can find value of this EPS only at t = 2 (using P/A, i, N) or t =7 (using F/A, i, N). Let us find the value at t = 2. Value at t = 2; P = A(P/A, i, N) = 800(P/A, 4, 5) = 3, Value of the given EPS at t = 2 is $3, Note that this is a single cash flow. The value of this single cash flow at t = 10 is determined using the factor (F/P, i, N) as follows:

27 53 Value at t = 10; F = P(F/P, i, N) = 3,516.60(F/P, 4, 8) = $4, Value of unknown cash flow, X at t =10 is X. Equating the two values; X = $4, In this solution, t =10 was chosen as the common time. In general any time period can be chosen. Here, t =10 was selected because X is at t =10 and its value is X and no calculations are needed. (b) The two sets of cash flows are the given EPS of 5 payments and an unknown EPS(A) of six payments. Note that the two EPS start at different points in time and have different number of equal payments but end at the same time ( t = 7). Therefore, a good common point in time is t = 7. Value of known EPS of $800 at t =7 is found using the factor F given A as follows: Value at t = 7; F = A(F/A, i, N) = 800(F/A, 4, 5) = $4, Value of unknown EPS of A (six payments) at t =7 is also found using the factor F given A. Value at t = 7; F = A(F/A, 4, 6) = A(6.633) Value of A is obtained by equating the values of the two cash flows(known and unknown) at t = 7 4, = A(6.633) or A = $ INTEREST FORMULAS FOR GRADIENT SERIES In section 3.4, interest formulas for EPS were derived. In an EPS, all cash flows are equal. In this

28 54 section, two formulas for increasing cash flows are derived. One of these cash flows is called uniform gradient series. In an uniform gradient series, from a base, all succeeding cash flows increase by a constant amount (G). The other type of cash flows are known as geometric-gradient series. Here, from a base, all succeeding cash flows increase by a constant % (g). These two formulas can be easily modified for decreasing cash flows Uniform Gradient Series A cash flow diagram for a uniform gradient series (G) is shown in Fig N -1 N B (B+G) Year (B+2G) B+(N-1)G Figure Cash Flows for Uniform Gradient Series In Fig , base amount is B at t =1. Cash flows for succeeding periods increase by a constant amount, G. Therefore, following are the cash flows for different periods: Cash flow at t = 1; B Cash flow at t = 2; B + G Cash flow at t = 3; B + 2G... Cash flow at t = N; B + (N-1)G Several interest formulas for uniform gradient series can be derived. These are; Find P given G, Find F given G, and Find A given G. We derive an interest formula only for finding A given G. Other interest formulas/ factors can be obtained using interest formulas/factors for single cash flows and EPS.

29 As B, the base amount is variable, it is taken out from all the cash flows and a formula is developed N-1 N G 2G Year (N-1)G Figure Cash Flows for Uniform Gradient Series with B= 0 just using the G s, the constant increase in amount. The revised cash flow diagram, after taking B out, is shown in Fig Alternatively, we assume B = 0 at t = 1. Note that there are no cash flows at the end of year 0 and 1 and first cash flow, G, is at t = 2 or the end of the 2nd year. To find an equivalent A to an uniform gradient series cash flows with B = 0 at t =1, we note that this series of cash flows (Fig ) is a sum of a number of EPS with A = G with different number of G s. One such EPS starts at t = 2 and ends at t = N (Fig.3.6.3). This EPS has (N-1) G s. Year 1 2 N G G Figure Equal Payment Series with A=G and First G at t=2 The next EPS starts with first G at t = 3 and ends at t = N (Fig 3.6.4). This EPS has (N - 2) G s. The last such EPS has only one G at t =N. If all these series are added, we get the cash flow Year N G G Figure Equal Payment Series with A=G and First G at t=3

30 56 diagram shown in Fig To find A which is equivalent to the uniform gradient series (of G s or G), first an equivalent F for G is found and then an EPS (A) equivalent to this F is determined. To find the F, we first find an equivalent F to each component EPS (for example one in Fig ). The value of F is obtained by summing F s for each component series. F for EPS in Fig is found by using the interest formula for finding F given A. Note that A = G and as there (N - 1) G s, N in the formula is equal to (N - 1). F = G(F / A, i, N -1) = G (1+ i) 2 i N-1 1 Similarly for second EPS (Fig.3.6.4), with A = G and N = N - 2 F = G(F/ A, i, N -2) = G (1+ i) 3 i N -2 1 For the last such series, there is only one G, therefore F N = G As all F t s are at t = N, F = F 2 + F F N ( ) ( ) F=G 1+i i N-1 N-2 1 G 1+i + 1 i G A closed form of the above expression can be obtained and is as follows:

31 F = G i N (1 + i) 1 i NG i (3.6.1) 57 Using this F we find A using the interest formula for finding A given F. i A = F (1+ i) N 1 (3.6.2) Substituting For F in Eq.(3.6.2) from Eq.(3.6.1) we get A = G i N (1 + i) 1 i NG i i (1+ i) N 1 (3.6.3) The above expression can be simplified as follows: A = G 1 N i (1 + i) N 1 (3.6.4) Eq. (3.6.4) is the interest formula for finding A given G. Note that in G first G is at t = 2 and last G is at t = N. However, in the equivalent A, the first A is at t = 1 and last A is at t = N. As the problem is to find A given G, the factor for finding A given G is (A/G, i, N). Using the factor, A is A = G(A/G, i, N).

32 58 Example How much money should be deposited now so that following withdrawals can be made. (a) (b) A withdrawal of $3,000 at the end of the 1st year? Nine withdrawals, each increasing by $400, for following nine years. Interest rate is 8% compounded yearly. Solution: The cash flow diagram is shown in Fig The series of withdrawals constitute a uniform gradient series with B = 3,000, G = 400 and N= 10. The problem is find to P given G. 3,400 3,000 6, Year Figure Cash Flow Diagram for Example A given G is the only factor available, therefore we first find A given G and then find P given A. Note that for this factor, first G is at t = 2 (See Fig.3.6.2). Therefore, the cash flows (Fig ) are divided in two series; one with A 1 = B = $3,000 (Fig.3.6.6) and the other a uniform gradient series with G = $400 (Fig.3.6.7). Note that if the two series are added, we get cash flows (all 10 withdrawals) shown in Fig

33 59 3,000 3, Year Figure Equal Payment Series with A=3, , Year Figure Gradient Series with G=400 Equivalent A (designated as A 2 ) for the gradient series (Fig ) is found using the factor A given G. A = G(A/G, i, N) = 400(A/G, 8, 10) = 400(3.871) or A 2 = $1, It should be noted that N = 10 even though there are only nine non zero cash flows (400, 800,..., 3,600). The reason for this is that in the derivation of the interest formula/ factor it was assumed that the first cash flow at t = 1 is zero and this zero cash flow is counted in the number of years or N. As A 1 and A 2, are both EPS for 10 years and start at t = 1, these can be added together. A = A 1 + A 2 = 3, , = $4,548.40

34 60 Using this EPS (A), we find P using the factor P give A. P = A(P/A, 8, 10) = 4,548.40(6.710) =$30, A deposit of $30, will allow 10 withdrawals as specified in this problem. It should be obvious that no money will be left in the account after the withdrawals Geometric Gradient Series In a geometric gradient series, cash flows from a base amount B increase/decrease each year by a constant percentage (g). A cash flow diagram with base B at t =1 is shown is Fig N B Year B+Bg N-1 B(1+g) Figure Geometric Gradient Series Cash flows for each period are calculated as follows: Cash flow at t = 1; B Cash flow at t = 2; B + Bg = B(1+g) Cash flow at t = 3; B(1+g) + B(1+g)(g) = B(1+g) 2 Cash flow at t = 3; B(1+g) 2 + B(1+g) 2 (g) = B(1+g) 3... Cash flow at t = N; B(1+g) N-1 An interest formula for finding P given GG or geometric gradient series is derived. Other factors/formulas for this series can be derived using factors for the single cash flow, the EPS, and the uniform

35 61 gradient series. This formula, P given GG, is derived by treating each cash flow as a single cash flow (F) and finding P given F for each. Required value of P is the sum of all the components P s. P = B(P/F, i, 1) + B(1+g)(P/F, i, 2) + B(1+g) 2 (P/F, i, 3) B(1+g) N-1 (P/F, i, N) or P = B(1+i) -1 + B(1+g)(1+i) -2 + B(1+g) 2 (1+i) B(1+g) N-1 )(1+i) -N A closed form of the above expression, after some mathematical manipulations, can be obtained and is given below: P = B (1+ g) N (1 + g ) 1 N g (1+g ) (3.6.5) where, (1 + i) g = 1 (1 + g) Eq is the interest formula for finding P given GG. It should be noted that the term in brackets in Eq is the simply the factor (P/A, i, N) with i = g'. Values of this factor, (P/A, i, N) are already tabulated and therefore, there is no need to define another factor for P given GG. The factor (P/A, i, N) can be used to find P equivalent to a given geometric gradient series with a revised interest rate i = g'. Example A person is trying to save money for down payment for a home. The money will be needed at the end of the 5th year. First deposit of $3,000 is at the end of the 1st year. Using an interest rate of 8% compounded yearly, determine the amount of money available at the end of the 5th year for following conditions: (i) (ii) The person will increase the initial deposit each year by $300 for following 4 years. The person will decrease the initial deposit each year by $150 for following 4 years.

36 62 (iii) (iv) The person will increase the initial deposit each year by 5% for following 4 years. The person will decrease the initial deposit each year by 3% for following 4 years. Solution: (i) This is an increasing uniform gradient series with B = $3,000, G= $300 and N = 5. Note that including the initial deposit, there are total of 5 deposits and N = 5. For uniform gradient series, only factor A given G has been derived. Therefore, the given series of cash flows (deposits) is divided in two cash flows shown in Figs and Note that the sum of these two (an A and a G) is equal to given series of deposits. Year ,200 Figure Gradient Series With G=300 and N=5 Year ,000 3,000 Figure Equal Payment Series with A=3,000 and N=5 Using cash flows in Fig , A, designated as A 1, is determined using the factor A given G. A = G(A/G, i, N) or A 1 = 300(A/G, 8, 5) = 300(1.846) = This value of A 1 can be added to an EPS with A = A 2 = 3,000 shown in Fig A = A 1 + A 2

37 63 = ,000 = 3, The amount of money available at t = 5 is obtained by finding F given A as follows: F = A(F/A, i, N) = 3,553.80(F/A, 8, 5) = 3,553.8(5.867) = $20, The person has $20, in the bank for the down payment. (ii) As the initial deposit of $3,000 will be reduced by $150 each year, this series of cash flows constitutes a decreasing uniform gradient series. As the interest formula or the factor A/G, i, N) is only for increasing gradient series, the cash flows are divided in two series of cash flows shown in Figs and Note that the difference of two series A (Fig ) and G (Fig ) equals the decreasing gradient series or deposits and Fig is an increasing gradient series with G = 150 and N =5. Year ,000 3,000 Figure Equal Payment Series with A=3,000 and N= Figure Uniform Gradient Year Series with B=0, and N=5 Using the factor A given G, A is calculated as follows: A = G(A/G, 8, 5)

38 64 or A 1 = 150(1.846) = The difference between the two equal payment series, A 2 and A 1 is A = A 2 - A 1 = 3, = 2, Next, determine F given A F = A(F/A, i, N) = 2,723.1(5.867) = $15, The person will have $15, in the bank for the down payment. (iii) This is an increasing geometric gradient series with B = 3,000, g =0.05 and N = 5. For this series, we can only find P given A using the factor (P/A, i, N) or the formula in Eq. (3.6.5) with i = g. The value of g is calculated as follows: g = 1 + i g = = Using Eq.(3.6.5), we get 5 ( ) 1 ( + ) 3000 P = ( ) = 13,

39 65 Using this value of P, F can be calculated as follows: F = 13,136.07(F/P, 8, 5) = 13,136.07(1.469) = 19, The person has $19, in the bank for the home. (iv) This is a decreasing geometric gradient series with B = 3,000, g = and N = 5. First, we find g' as follows: g = 1 + i 1 1- g = = Using Eq.(3.6.5), P is calculated as follows: P = 3,000 ( ) 1 (1-0.03) ( ) = 11, Using this value of P, F is calculated as follows: F = 11,334.72(F/P, 8, 5) = 11, (1.469) = $16, The person has $16, in the bank for the down payment.

40 SUMMARY OF INTEREST FORMULAS/FACTORS Several interest formulas are derived in Sections 3.3 to 3.6. In this section, a few examples for using these formulas and factors are presented. Formulas for single cash flow include finding F given P and P given F. It should be noted that the two cash flows are separated by N years. In an EPS, the amount of each cash flow (A) is same, and there are N such cash flows (one at the end of each year). First A is at t =1 and the last A is at = N. But, in general, an EPS can start at any time period. For example, an EPS with 5 cash flows of $500 each with first A at t = 3 is shown in Fig There are 5 equal cash flows, A = 500 and N = 5. Years for EPS, from 1 to 5, are shown over the time line in Fig It should be noted that P for this EPS is at t = 2 and F is at t = 7 as shown in Fig F Year A A P Figure Equal Payment Series with First A at t=3 Just like an EPS, a uniform gradient series (G) can start at any time period. In the formula for uniform gradient series, A given G, there are N cash flows, one at the end of each year. Note that this includes B = 0 at t = 1. One such series with a base amount of $1,000 at t = 4 and increasing each year by $100 (G=100) for following three years is shown in Fig Note that total number of cash flows, including the base cash flow is 4 or N =4. Years for gradient series are shown above the time line in Fig

41 67 Year ,000 1,100 1,200 1,300 Figure Equal Payment Series with B=$1,000 & G=100 Similarly, a geometric gradient series can start at any point in time. For example, a geometric gradient series has a base of $250 at t = 2. This base amount increases each year by 5% (g = 0.05) for the following three years. Cash flows for this series are shown in Fig Here N = 4 as the base cash flow increases by 5% for following three years. Years for this geometric gradient series are shown over the time line in Fig Recall that no new factor is needed as we can use (P/A, g',n) where g' is a revised interest rate. Year Figure Geometric Gradient Series with Base of $250 and g=0.05 Example Following series of cash flows are given End of year Cash Flow($) 1,200 1, ,000 Interest rate is 10% compounded yearly. What EPS of 5 equal payments, A, with first A at t = 2 is equivalent to the above given series of cash flows?

42 Solution: Equivalent cash flows have same value at some point in time. Also, note that if two cash 68 flows are equivalent at some point in time, these are equivalent at all other points in time. For establishing equivalence, we simply find the values of cash flows at a common point in time. The choice of common point in time is arbitrary but in some cases selection of time can reduce the amount of calculations needed. For this example, t = 0 is chosen as the common point in time. Factor P given F is used to find the value of each given single cash flow as follows: P = 1,200(P/F, 10, 1) + 1,000(P/F, 10, 2) (P/F, 10, 4) = 1,200(0.9091) + 1,000(0.8264) (0.6830) = 2, (3.7.1) A cash flow diagram for the unknown EPS with 5 equal payments of A is shown in Fig As the first A is t = 2 and last A is at = 6, the value of this series can be found only at t = 1 (using P given A) or t = 6 (using F given A). We find value of this EPS at t = 1, designated as P 1 as follows: Year P A A Figure Equal Payment Series with First A at t=2 P = P 1 = A(P/A, i, N) or P 1 = A(P/A, 10, 5) = A(3.791) P 1 is the value of EPS at t = 1. But we need to find its value at t = 0. As P 1 is a single cash flow which occurs at t =1, its value at t =0 is found by using the factor P given F with F = P 1. P = P 1 (F/P, 10, 1)