TIME VALUE OF MONEY. Lecture Notes Week 4. Dr Wan Ahmad Wan Omar

TIME VALUE OF MONEY Lecture Notes Week 4 Dr Wan Ahmad Wan Omar

Lecture Notes Week 4 4. The Time Value of Money The notion on time value of money is based on the idea that money available at the present time is worth more than the same amount in the future due to its potential earning capacity. The earning capacity of money value is affected by interest rate, inflation (or deflation), and exchange rate. Interest rate will increase the value of money saved in the saving account of a bank which promises certain rate of interest after a stipulated period. In this case, there is no risk involved, as the bank promises to pay interest at the agreed rate of the principal value after lapse of the stipulated period. In the investment perspective, in the event there are alternatives for investment for the money, the interest earned is an opportunity cost for the alternative. Inflation decreases the value of money because of decreasing purchasing power, for example, the price of a meal last year was RM5.00, but increases to RM5.50 presently due to inflation. Exchange rate affects the money value if you are going oversea for a trip, say to Thailand. Assuming there is no inflation, the cost of tut-tut ride in Bangkok is Bhat10 per ride or RM1.00 at exchange rate RM1 = Bhat10. The exchange rate presently is RM1 = Bhat9, then the cost of tut-tut ride (Bhat10) has increased to RM1.10 (1.00 +10% x 1.00) because money in RM value has decreased by 10%. The same effect also occurs for buying imported goods and services. 4.1. Interest 4.1.1. Simple Interest Simple interest is charged linearly proportional to the initial amount of principal, thus total interest earned, I is 1

I = P N i (4.1) Where P = principal amount; N = number of period (e.g. years, months); i = interest rate per annum. For example, if RM10,000 was saved in a saving account of a bank that promises to pay interest 5.5% per annum, then the interest earned, I in 5 three year would be I = RM10,000 5 5.5% = RM2,750 Thus, your money in the bank at the end of year-5 is RM12,750 (RM10,000 + RM2,750), or general formula for future value, F is 4.1.2. Compound Interest F = P + P. N. i. = P(1 + N. i) (4.2) Compound interest is the recalculated interest on the cumulated principal (initial principal plus interest earned for the period) assuming that the initial principal was not withdrawn. The mechanic of compound interest can be seen in the following Table 4.1 for RM10,000 saved in the saving account of a bank that promises to pay interest at 5.5% per year on compounded basis as long as the principal is not withdrawn. Table 4-1. Compound interest calculation (A) (B) = (A)*5.5% (C) = (A) + (B) Beginning Period Remaining Year in A/C Interest Balance 1 10,000 550 10,550 2 10,550 580 11,130 3 11,130 612 11,742 4 11,742 646 12,388 5 12,388 681 13,070 Notice that the amount of money in year-5 for compound interest is RM13,070 which is higher than simple interest calculation at RM12,750. 2

4.2. The Concept of Economic Equivalence If we are indifferent with the interest rate, then the value of money now is equivalent to the value in future. In the existence of interest, the concept of economic equivalence is required to provide an economic alternative for goods or services comparable to the various interest rates available. Interest rate to the investors is an opportunity cost, but it is a cost of money in the engineering economy where we use other people s money in loan for business expansion or taking up new projects. The method to apply for economic equivalence is cash flow analysis, in which the estimation involves interest rate or required rate of returns on investment. The economic equivalence exists between the cash flows that have the same economic effect and could be traded for one another in the financial marketplace, which we assume to exist. Cash flow analysis provides economic worth of the project alternatives for economic equivalence. In practice, economic equivalence value may be determined by uniqueness of the alternatives in terms of, a) Difference in repayment plans, or b) Difference on interest rates. 4.2.1. Difference in Repayment Plans To explain the practice of economic equivalence in loan repayment plans, consider your newly start-up business is offered by a financier (banker) amounting to RM150,000 as a loan payable within 3 years period at interest rate of 10% per annum. The financier offers three plans for repayment to choose 1) Pay interest due at end of the year and principal at end of year-3; 2) Pay off the loan in three equal end-of-year instalments (principal plus interest); and 3) Pay principal and interest in one bullet payment at the end of third year. The details on calculation of repayment plans are presented in Table 4.2. 3

Table 4-2. Three plans of loan repayment for RM150,000 at interest rate of 5.0% p.a payable in 3 years (A) (B)= 10%x(A) (C)=(A)+(B) (D) (E)=(B)+(D) Amount Interest Total Owed Total Owed as for the at the end Principal year-end Year year begins year of year payment payment Plan-1: Pay interest due at the end of each year and principal at end of third year 1 150,000 15,000 165,000-15,000 2 150,000 15,000 165,000-15,000 3 150,000 15,000 165,000 150,000 165,000 450,000 45,000 150,000 195,000 Plan-2: Pay off the loan in three equal end-of-year instalments (principal plus interest) 1 150,000 15,000 165,000 45,317 60,317 2 104,683 10,468 115,151 49,849 60,317 3 54,834 5,483 60,317 54,834 60,317 309,517 30,952 150,000 180,952 Plan-3: Pay principal and interest in one payment at end of third year 1 150,000 15,000 165,000 - - 2 165,000 16,500 181,500 - - 3 181,500 18,150 199,650 150,000 199,650 496,500 49,650 150,000 199,650 The three repayment plans in Table 4.2 are economically equivalent but difference in repayment structure. Economically equivalence because in the event interest rate is fixed, you are indifferent as to which plan you intend to repay your business start-up loan to the financier. Though total interest paid in plan-3 is the highest but the ratio of total interest paid to total principal-year is equivalent at 0.10 (or 10%) for all the repayment plans. Table 4-3. Ratio of total interest paid to total RM-year Total Area under Interest Curve Ratio Plan Paid (A) RM-year (B) ( C) =(A)/(B) Plan 1 45,000 450,000 0.10 Plan 2 30,952 309,517 0.10 Plan 3 49,650 496,500 0.10 The repayment structure is different for each plan in terms of amount paid in each repayment year. The highest repayment amount in a single year is 4

in plan 3 where the loan and accumulated interest is paid in lump sum at the end of year 3. 4.2.2. Difference on Interest Rates If the interest rate is no longer fixed, say increases to 11% per annum, then the repayment plans do not economically equivalent anymore because interest payment for each plan would increase, and more so for the calculated repayment schedule (see column E in Table 4.2). Instead, each plan has to reduce the amount of principal payment, because more money needed to pay increased interest payment. Hence, there will be different in present sum for each plan at the increased interest rate. We will explain later on the concept of present sum and how to calculate it. This is to show that economic equivalence is dependent on the interest rate, meaning that changing in interest rate will remove the equivalence between plans of repayments. 4.3. Uniform Series Annuity Uniform series refers to series of receipts or disbursements with the same value over period of time, for example receipts from a uniform contract payment for 10 years at RM1.5 million a year, and equipment loan repayment disbursement amounting to RM3,500 per month. Such uniform series is called an annuity. 4.3.1. Present and Future Equivalent Values of Single Cash Flows If an amount of P Ringgit is invested at a point in time at interest rate of i% per annum (or profit growth), the future equivalent value, F by the end of N period is F = P(1 + i) N (4.3) The quantity, (1 + i) N in equation (4-3) is called the single payment compound factor. The present and future values are tabulated in the appendix C (Interest and Annuity Tables for Discrete Compounding) in your 5

textbook. For calculation using Table C, the equation (4-3) is expressed as F = P(F/P, i%, N). Example 4.1. Future equivalent of a present sum using Table C Suppose you borrow RM8,000 now, promising to repay the loan principal plus accumulated interest in 4 years at i = 10% per year. How much would you repay at the end of four years? Use Table C for calculation. Use Table C for discrete compounding, i = 10%, N = 4 on the first column (Single payment for compound amount factor, to find F given P, F/P). The value F/P is 1.4641 Solution Therefore, total repayment at the end of year 4, F = 8,000 (1.4641) = RM11,713 To find present equivalent value, P we solve equation (4-3), P = F ( 1 1 + i ) N = F(1 + i) N (4.4) The quantity (1 + i) N is called the single payment present worth factor, and equation (4-4) is expressed as P = F(P/F, i%, N) in using Table C. Example 4.2. Present equivalent of a future amount of money An investor is give an option to purchase an asset that will be worth RM25,000 in five years. If the value of asset increases at 5% per year, what is the willing amount the investor to pay now for the asset? Solution Use Table C for equation (4-4). P = RM25,000 (P/F, 5%, 6) = RM25,000 (0.7462) = RM18,655 Based on equations (4-3) and (4-4), the following three simple rules apply when conducting arithmetic calculations with cash flow: 6

Rule 1. Do not add or subtract cash flows unless they occur at the same point in time. Rule 2. To move a cash flow forward in time by one unit, multiply the magnitude of cash flow by (1+ i) to reflect time value of money. Rule 3. To move a cash flow backward in time by one unit, divide the magnitude of cash flow by (1+ i). To find the interest rate given P, F, and N, we solve equation (4-3) for i, N i = F P 1 (4.5) To find number of year, N given P, F, and i F = P(1 + i) N (1 + i) N = F P Use logarithm each side of equation, N log(1 + i) = log ( F P ) N = log(f P ) log(1 + i) (4.6) 4.3.2. Uniform Series (Annuity) in Relation to Present and Future Equivalent Values Consider a general cash flow involves a series of uniform (equal) receipts, each of amount A, flowing at the end of each period for N periods with interest at i% per period. Such uniform series is called annuity, A. A = Uniform Amount (given) 1 2 3 N - 1 End of period I = Interest rate per period P = Present equivalent N = Number of interest periods F = Future equivalent 7

The future equivalent values of each period cash flows are F = A(F/P, i%, N-1) + A(F/P, i%, N-2) +. + A(F/P, i%, 1) + A(F/P, i%,0) = A[(1 + i) N 1 + (1 + i) N 2 + + (1 + i) 1 + (1 + i) 0 ] The bracketed terms are geometric series having a common ratio (1 + i) 1. Recall that the sum of the first N terms of a geometric series is S N = a 1 ba N 1 b (b 1) where a 1 is the first term in the series, a N is the last term, and b is the common ratio. If we let b = (1 + i) 1, a 1 = (1 + 1) N 1, and a N = (1 + i) 0, then which reduces to (1 + i) N 1 1 (1 + i) F = A [ 1 1 ] (1 + i) F = A [ (1 + i)n 1 1 ] (4.7) i The value {[(1 + i)^n 1]/i} is called the uniform series compound amount factor. It is starting point to develop the remaining three uniform series interest factors. The equation (4-7) can be expressed for annuity Table C as F = A(F/A, i%,n), and the uniform series compound amount factors are given in the fourth column of the annuity table. Example 4.3. Future value of a university degree The findings from recent study revealed that a university degree provides additional worth of RM25,000 per year in salary income (A) as compared to SPM holders make. If interest rate (i) is 5% per year and you work for 30 years (N), what is the future compound amount (F) of this extra income? Solution We assume that the extra income will be invested into saving account that promises similar terms, and to be withdrawn after 30 years. 8

A = RM25,000 Graduate from University 1 2 3 29 30 End of period F = RM25,000 (F/A, 5%, 30) F = Future equivalent = RM25,000 (66.4388) = RM1,660,970 Finding P when given A, substitute F in equation (4-7) into equation (4-3) and solve for P, P(1 + i) N = A [ (1 + i)n 1 ] i Dividing both sides by (1 + i) N, we get P = A [ (1 + i)n 1 ] (4.8) i(1 + i) N The value in the bracket is called the uniform series present worth factor which is given in the 5 th column of Table C, and its general tabulated expression, P = A(P/A, i%, N). Example 4.4. Present equivalent of an annuity An investor holds a time payment purchase contract for some machine tools. Based on the contract, the investor will receive RM350 at the end of each month for a 5-year period. The first payment is due in one month. The investor offers to sell you the contract for RM12,000 cash today. If you otherwise make 1% per month on your money, would you accept or reject the investor s offer? Solution A = RM350, n = 5 x 12 = 60 month, and i = 1% per month. P = A(P/A, i%, N) = 350 (P/A, 1%, 60) = 350(44.955) 9

P = RM15,734, which is higher than offer cash price of RM12,000. Hence suggest you take up the offer. Example 4.5. Monthly rate of return In example 4.4, if you pay for the time purchase contract at RM12,000, what would be the monthly rate of return, i? Calculate by using the Table C (Annuity). P = A (P/A, i%, N) RM12,000 = RM350 (P/A, i%, 60) (P/A, i%, 60) = 12,000/350 = 34.2857 Look at the Table C (Annuity), find the values of (P/A, i%, 60) that are closest to 34.2857. Then compute the rate of return, by interpolation. Interpolation, Interest rate, i% (P/A, i%, 60) 2% 34.7609 i% 34.2857 3% 28.6756 2% i% 34.7609 34.2857 = 3% 2% 34.7609 28.6756 = 0.4752 6.0853 = 0.078 Rate of return = 2%+(0.078%) = 2.078% per month Finding A when given F taking equation (4-7) and solving for A, i A = F [ (1 + i) N 1 ] (4.9) The quantity in bracket is called the sinking fund factor which is given in column 6 of the Annuity Table C. General formula for A is A = F(A/F, i%, N) Finding A when given P taking equation (4-8) and solving for P i(1 + i)n A = P [ (1 + i) N 1 ] (4.10) 10

The quantity in bracket is called the capital recover factor which is given in column 7 of the Annuity Table C. General formula for A is A = P(A/P, i%, N) We then can use this tabulated formula on A to get the repayment of loan RM150,000 at 5% per year interest for 3 years for Plan 2 as mentioned in Example 4.1. as follows A = RM150,000 (A/P, 5%, 3) = 150,000 (0.3672) = RM55,080 per year. Finding the Number of Cash Flows in an Annuity given A, P, and i there is no specific formula for finding the number of cash flows in the annuity (N), but we can use the relationship between P and A to determine N. Example 4.6. Prepaying a loan Finding N Your company apply for loan amounting to RM350,000 to finance a new computer system for the plant. The bank offers to finance your new computer system at interest rate of 6.5% per annum for 10 years, and the yearly repayment will be RM48,685. Your company decides that they afford to pay RM65,000 per year. How many payments (in years) will the loan be paid off? Solution The loan amount is calculated by on formula: A = P(A/P, i%, N) So, A = RM350,000 (A/P, 6.5%, 10) = RM350,000 (0.1391) = RM48,685 Now, instead of paying RM48,685 per year, your company is going to pay RM65,000 per year. Common sense tells us that the loan will be paid off less than stipulated repayment period of 10 years. Use equation: P = A (P/A, i%, N) RM350,000 = RM65,000 (P/A, 6.5%, N) (P/A, 6.5%, N) = 5.3846 11

Use Table C Annuity for find N by interpolation (P/A, 6%, 7) = 5.5824 (P/A, 7%, 7) = 5.3896 Hence, (P/A, 6.5%, 7) = 5.3860. So the loan will be paid off after 7 years. Spreadsheet solution, use excel formula: NPER(rate, pmt, -pv) to get number of year, N. Finding the Interest Rate, i given A, F, and N there is also no single equation to determine interest rate, i. However, we can use the known relationship between i, A, F, and N to determine i by linear interpolation. Example 4.7. Finding the interest rate to meet an investment goal. Consider an engineering company has decided to replace existing equipment in the plant in five years with cash. Every year the company intends to put aside RM12,000 into a new equipment fund. In five years the price of new equipment for the plant is estimated at RM72,000. At what interest rate per year, must the company invest so that by the end of year- 5, they will accumulate cash as much as RM72,000 to purchase the new equipment for the plant. Solution Use general formula: F = A(F/A, i%, N) RM72,000 = RM12,000 (F/A, i%, 5) (F/A, i%, 5) = 6 Now, use Table C (Annuity) to find interest rate, i (F/A, 9%, 5) = 5.9847 (F/A, 10%, 5) = 6.1051 Linear interpolation: x% 6.0000 5.9847 = 10% 9% 6.1051 5.9847 = 0.0153 0.1204 = 0.127 x% =.127% The required interest rate: 9% + x% = 9% +.127% = 9.13% 12

Spreadsheet solution: use formula, RATE(nper, pmt, pv, fv) i = RATE (5, -12000, 0, 72000) = 9.13% 4.3.3. Summary of Interest Formulas and Relationships for Discrete Compounding Discrete compounding means that the interest is compounded at the end of each finite-length period, such as a month or a year. The formulas also assume discrete (i.e. lump sum) cash flows spaced at the end of equal time intervals on a cash-flow diagram. Discrete compound interest factors are given in Table C, where interest rate, i is assumed to remain constant during the N compounding periods. The summary of the six most common discrete compound interest factors is given in Table 4-4 below. Table 4-4. Discrete compounding-interest factors To Find Given Factor to multiply Factor Name Functional Symbol Single cash flows: F P (1 + i) N Single payment compound amount (F/P, i%, N) P F 1 (1 + i) N Single payment present worth (P/F, i%, N) Uniform series (annuities): F A (1 + i) N 1 i Uniform series compound amount (F/A, i%, N) P A (1 + i) N 1 Uniform series present worth (P/A, i%, N) i(1 + i) N A F i Sinking fund (A/F, i%, N) (1 + i) N 1 A P i(1 + i) N (1 + i) N 1 Capital recovery (A/P, i%, N) The relationships between the compound interest factors in equations, (P/F, i%, N) = (A/P, i%, N) = (A/F, i%, N) = 1 (F/P, i%, N) 1 (P/A, i%, N) 1 (F/A, i%, N) (4.11) (4.12) (4.13) (F/A, i%, N) = (P/A, i%, N)(F/P, i%, N) (4.14) 13

( P A, i%, N) = (P, i%, k) F N k=1 N (F/A, i%, N) = (F/P, i%, N k) k=1 (4.15) (4.16) (A/F, i%, N) = (A/P, i%, N) i (4.17) 4.3.4. Deferred Annuities (Uniform Series) The uniform series (annuities) that we have discussed to this point involve the ordinary annuities, whereby the first cash flow is made at the end of the first period. If the cash flow does not flow until some later period, then this type of annuity is called deferred annuity. Time present 0 1 J-1 J J+1 J+2 J+3 N-1 N End of period i % If the annuity is deferred for J periods ( J < N), the entire framed ordinary annuity has been moved away from time present by J periods, in which the first payment is made at the end of period (J + 1), assuming all periods are equal in length. The present equivalent at the end of period J of an annuity with cash flows of amount A based on equation (4-7) is A(P/A, i%, N-J). The present equivalent of the single amount at time present is P0 = A(P/A, i%, N-J) (P/F, i%, J) Example 4.8. Present equivalent of a deferred annuity Suppose that after setting up a plant for the company, the management have decided to provide major repair cost amounting RM20,000 per year on the 5 th, 6 th and 7 th year for the plant, by investing into an account bearing interest 12% per year. How much the company should invest now? 14

A = RM20,000 0 1 4 5 6 7 i = 12% p.a P0 =? P4 = F4 In this problem, the ordinary annuity of three withdrawals of RM20,000 each is involved and that the present equivalent of this annuity occurs at the 5 th year when a (P/A, i%, N-J) factor is utilised. In this problem, N = 7 and J = 4, hence P4 = A(P/A, 12%, 3) = RM20,000 (2.4018)= RM48,036 The next step is to calculate P0. With respect to P0, P4 is a future equivalent, hence it should be denoted as F4. To calculate P0 use Table C, P0 = F4(P/F, 12%, 4) = RM48,036 (0.6355) = RM30,527 is the amount that the company must invest when the plant is built. 4.3.5. Equivalence Calculations Involving Multiple Interest Formulas This section provides example involving two or more equivalence calculations to solve for unknown quantity. The convention of the end-ofyear cash-flow is still valid, again assuming constant interest rate over N time period. Example 4.9. Calculating equivalent P, F, and A values The following figure depicts an example problems with a series of year-end cash flows extending over five years. The cash flows are RM100 for the year- 1, RM200 for year-2, RM500 for year-3, and RM400 each for year-4 and year- 5. These could represent something like the expected maintenance expenditure for a certain equipment. Find, a) The present equivalent for expenditure, P0; b) The future equivalent for expenditure, F5; 15

c) The annual equivalent expenditure, A of these cash flows if the annual interest rate is 15%. Solution End of year a) 0 1 2 3 4 5 RM100 P0=RM994.49 RM200 RM500 RM400 RM400 P0 = F1 (P/F, 15%, 1) = RM100 (0.8696) = RM86.96 + F2 (P/F, 15%, 2) = RM200 (0.7561) = RM151.22 + F3 (P/F, 15%, 3) = RM500 (0.6575) = RM328.75 + A(P/A, 15%, 2) x (P/F, 15%, 3) =RM400(1.6257) (0.6575) = RM427.56 RM994.49 b) End of year 0 1 2 3 4 5 P0 RM100 RM200 RM500 RM400 RM400 F5 F5 = P0(F/P, 15%, 5) = RM994.49 (2.0114) = RM2,000.32 c) End of year 0 1 2 3 4 5 A P0=RM994.49 A A A A F5=RM2000.32 The equivalent A of the irregular cash flows can be calculated directly from either P0 or F5 as follows, A = P0(A/P, 15%, 5) = RM994.49 (0.2983) = RM296.65 16

or, A = F5(A/F, 15%, 5) = RM2000.32 (0.1483) = RM296.65 Spreadsheet Solution = NPV(E12, H15:H19) = E20*(1+E12)^E13 = PMT(E12, E13,-E20) Example 4-10. Determining an unknown annuity amount You are requested to pay the last two end-of-year (EOF) 8th and 9th loan payment of RM1,500 each. To make these payments possible, four annuity amounts will be deposited into the bank account at EOF 2, 3, 4, and 5. The bank pays interest rate at 10% per year for that account. a) Draw a cash-flow diagram for this situation. b) Determine the value of A that establishes equivalence in your cash flow diagram. c) Determine the lump sum value at the end of year 9 of the completed cash flow diagram based on your answers in (a) and (b). Solution a) Cash flow diagram. RM1,000 RM1,000 End of year 1 2 3 4 5 6 7 8 9 A =? 17

b) Because of unknown annuity, A, begins at EOY 2, hence the P- equivalent at EOY 1 of the 4 amounts is P1 = A(P/A, 10%, 4) Next, calculate the EOY 1 P-equivalent of RM1,000 at EOY 8 and RM1,000 at EOY 9 as follows, P 1 =RM1,000(P/A, 10%, 2)(P/F,10%, 6) The (P/F, 10%, 6) factor is to discount the equivalent value of the A amounts at EOY 7 to EOY 1. By equating both P-equivalents at EOY 1, we can solve unknown amount of A, P 1 = P 1 A(P/A, 10%, 4) = RM1,000(P/A,10%, 2)(P/F, 10%, 6) 3.1699A = RM1,000(1.7355)(0.5645) = RM979.69 A = RM309 Therefore, we conclude that deposits of RM309 at EOY 2, 3, 4, and 5 are equivalent to RM1,000 at EOY 8 and 9 if the interest rate is 10% per year. c) To calculate F-equivalent at EOY 9 of the RM309 annuity in year 2 to 5 and the annuity RM1,000 in year 8 and 9. -RM309(F/A, 10%, 4)(F/P, 10%, 4) + RM1,000(F/A, 10%, 2) = -RM0 4.4. Uniform Series Arithmetic In uniform series arithmetic, the cash flows are projected to increase or decrease by a uniform amount in each period, thus constituting an arithmetic sequence. The problems include, 1) Finding P when given G, 18

2) Finding A when given G, 3) Finding F when given G, 4.4.1. Finding P when given G The present equivalent, P is P = G(1) (1 + i) 2 + G(2) (1 + i) 3 + G(3) G(N 2) G(N 1) + + + (1 + i) 4 (1 + i) N 1 (1 + i) N Add dummy, G(0)/((1+i) 1 for missing cash flow at time 1, then rewrite P, N (n 1) P = G (1 + i) n n=1 P = G { 1 i [(1 + i)n 1 i(1 + i) N N (1 + i) N]} (4.18) The term in braces is called the gradient to present equivalent conversion factor, which can also be expressed as (1/i)[P/A, i%, N) N(P/F, i%, N)]. The values for this factor is tabulated in column 8 of Table C. The general form for this factor is P = G(P/G, i%, N). 4.4.2. Finding A when given G Use equation (4-18) by replacing P with A, A = P(A/P, i%, N) A = G { 1 i [(1 + i)n 1 i(1 + i) N N (1 + i) N]} (A/P, i%, N) = G i [(P/A, i%, N) N (1 + i) N)] (A/P, i%, N) = G i [1 Ni(1 + i) N (1 + i) N [(1 + i) N 1] ] = G i G [ N (1 + i) N 1 ] = G [ 1 i N (1 + i) N 1 ] 4.19) 19

The term in bracket is called the gradient to uniform series conversion factor, which is tabulated in column 9 of Table C. The general form for this factor is A = G(A/G, i%, N). 4.4.3. Finding F when given G Also using equation (4-18), we can derive equation for future equivalent, F of an arithmetic series F = P(F/P, i%, N) = G { 1 i [(1 + i)n 1 i(1 + i) N N (1 + i) N]} (A/P, i%, N) = G { 1 i [(1 + i)n 1 i = G i (F/A, i%, N) NG i N]} (4.20) The direct use of gradient conversion factors applies when there is no cash flow at the end of period 1. Example 4-11. Using the gradient conversion factors to find P and A Consider certain EOY cash flows are expected to be RM1,500 for the second year, RM3,000 for the third year, and RM4,500 for the fourth year. If the interest rate is 10% per year, calculate a) Present equivalent value at the beginning of the first year, b) Uniform annual equivalent value at the end of each of the four years. Solution RM1,500 RM3,000 RM4,500 0 1 2 3 4 End of year Given, G = RM1,500 and N = 4 a) The present equivalent, P0 = G(P/G, 10%, 4) = RM1,500(4.378) 20

P0 = RM6,567 b) The uniform annual equivalent value, A = G(A/G, 10%, 4) A = RM1,500(1.3812) = RM2,072 or alternatively as P0 is known, A = P0(A/P, 10%, 4) A = RM6,567(0.3155) = RM2,072 Example 4-12. Present equivalent of an increasing arithmetic gradient series Suppose we have the following arithmetic gradient cash flows, End of year Cash Flow (RM) 1 3,500 2 4,500 3 5,500 4 6,500 Calculate present equivalent value using gradient conversion factor table if interest rate is 10% per year. Solution P0T RM4,500 RM3,500 = + 0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 End of year RM4,500 RM6,500 P0A RM3,500 End of year P0G RM3,000 RM2,000 RM1,000 End of year P0T = P0A + P0G = A(P/A, 10%, 4) + G(P/G, 10%, 4) = RM3,500(3.1699) + RM1,000(4.378) = RM11,094.65 + RM4,378 = RM15,472.65 The annual equivalent of the original cash flows could be calculated as follows, 21

AT = A + AG = RM3,500 + RM1,000(A/G, 10%, 4) = RM3,500 + 1,381.12 = RM4,881.12 AT is equivalent to P0T because RM4,881.12(P/A, 10%, 4) = RM15,472.66, which is the same value obtained previously. 4.5. Geometric Sequences of Cash Flows Many engineering problems, particularly those relating to construction costs or maintenance costs, involve projected cash flows pattern that are changing at an average rate, f, each period. The resultant EOY cash flow pattern is referred to as a geometric gradient series. The present equivalent of geometric gradient series is P = A1(P/F, i%, 1) + A2(P/F, i%, 2) + + AN(P/F, i%, N) = A1(1 + i) -1 + A2(1 + i) -2 + + AN(1 + i) -N = A1(1 + i) -1 + A1(1+f )(1+i) -2 + + A1(1 + f ) N 1 (1 + i) -N = A 1 (1 + i) 1 [1 + x + x 2 + + x N 1 ] (4.21) Where, A N = A 1 (1 + f ) N 1 f = (A k A k 1 ) A k 1 A k = (A k 1 )(1 + f ), 2 k N x = (1 + f ) (1 + i) The expression in brackets in equation (4-21) reduces to (1 + x N )/(1 x) when x 1 or f 1. If f = i, then x = 1 and the expression in the brackets reduces to N, the number of terms in the summation. Therefore, A 1 (1 + i)(1 x N ) P = { 1 x A 1 N(1 + i) 1 f i, f = i 22

A 1 [1 (1 + i) N (1 + f ) N ] P = { i f A 1 N(1 + i) 1 f i f = i (4.22) or A 1 [1 (P/F, i%, N)(F/P, f %, N)] P = i f A 1 N(P/F, i%, 1) { f i f = i (4.23) Example 4-13. Increasing geometric gradient series equivalence. Suppose that your retirement benefits during your first year of retirement are RM50,000. Assuming that this amount is just enough to meet your cost of living for the first year. Your cost of living is expected to increase by 5% per year due to inflation. If you do not expect to receive any cost of living adjustment in your retirement period, then some of your future cost of living has to come from savings other than retirement scheme. If your savings account earns 7% interest per year, how much should you put aside in order to meet this future increases in the cost of living over 15 years? Solution Given: A 1 = RM50,000; f = 5% ; i = 7% per year, and N = 15 years RM50,000 Expected retirement pension RM50,000 Year 0 1 2 3 4 5 11 12 13 14 15 RM50,000 RM50,000(1.05) Expected cost of living Find the equivalent amount of total benefits paid over 15 years; P = RM50,000(P/A, 7%, 15) = 50,000(9.1079) = RM455,395 23

Find the equivalent amount of total cost of living with inflation, using equation (4-22); P = RM50,000 [ 1 (1 + 0.05)15 (1 + 0.07) 15 ] 0.07 0.05 Therefore, amount to be put aside, P = RM50,000(12.32504) = RM616,252 ΔP = RM616,252 RM455,395 = RM160,857. 4.6. Nominal and Effective Interest rates The notion of nominal and effective interest rates exists because the period used to describe interest rate is less than one year, for example, daily, weekly, monthly, or quarterly. In practice, the interest rate is determined by the market force which is known as market interest rate. Market interest rate is the interest rate established by the financial market, such as by Bank Negara, commercial banks and other financial institutions. This interest rate is supposed to reflect any anticipated changes in the earning power as well as purchasing power in the economy. 4.6.1. Nominal Interest Rates Financial institutions usually calculate loan contract and other financial services such as credit card by using nominal interest rates. For instance, in the credit card, the bank states that the interest payment for the credit card is to be charged at the rate of 1.5% compounded monthly on the unpaid balance. This statement means that the nominal interest rate or annual percentage rate (APR) for the credit card is 18% per year based on (1.5% per month x 12 months). Although APR is commonly used by credit card issuers and other financial institutions and is familiar to many customers, the term does not explain precisely the amount of interest that will accumulate in a year. The true 24

effect of APR will be explained in the concept of annual effective yields or annual percentage yield (APY). 4.6.2. Annual Effective Yields Annual effective yield represents the interest earned in a year. For example, we want to calculate the total annual interest payment for a credit card debt of RM1,000 by using a future worth formula, given interest rate, i = 1.5%, and N = 12, we obtain F = P(1 + i) N = RM1,000(1 + 0.015) 12 = RM1,195.62 In effect, we pay more than 18% on interest. Based on the above calculation, we pay RM195.62 to the credit card issuer at the end of the year, or at the interest rate of 19.56% (RM195.62/RM1,000) which is higher than stipulated interest rate of 18% per year. The relationship between nominal and effective interest rates is in the Figure 4-1. APR = 18% Interest period = month 18% Compounded monthly What it means 1.5% per month 12 interest periods per year APY = 19.56% Interest period = annual 19.56% Compounded annually Figure 4-1. Relationship between nominal and effective interest rates The annual effective yield, i a can be calculated as follows: 25

i a = (1 + r M ) M 1 (4 24) Where r is nominal interest rate, and M is compound periods occur during the year. It should be noted that effective interest rates or APY are always expressed on an annual basis, unless specifically stated otherwise. From the equation (4-24), it is clear that if M > 1 then i a > r Example 4-14. To determine compounding period Consider the following bank advertisement that appeared in the local newspaper: Open a Certificate of Deposit (CD) with our bank and get a guaranteed rate of return on as little as RM500. It s a smart way to manage your money for months. In this advertisement, no mention is made of specific interest compounding frequencies. Find the compounding period for each CD. Solution Given: r = 4.41% per year, i a = 4.50%. Find M Type of Certificate Interest Rate Annual % Yield Min Required to (APR) (APY) Open 1 year certificate 2.23% 2.25% RM500 2 year certificate 3.06% 3.10% RM500 3 year certificate 3.35% 3.40% RM500 4 year certificate 3.45% 3.50% RM500 5 to 10 year cert 4.41% 4.50% RM500 Metho-1: By trial and error using equation (4-24), i a = (1 + r M ) M 1 0.0450 = (1 + 0.0441 M M ) 1 (1 + 0.0441 M M ) = 1.0450 26

By using trial and error, we find M = 12. Method-2: Using Microsoft Excel function. In MS Excel cell; Effective rate = EFFECT(r, M) Nominal rate = NOMINAL(i a, M) 27