Interest and Equivalence in Engineering Economy

1. Engineering Economic Analysis Chapter 3  Interest and Equivalence http://academic.udayton.edu/ronalddeep/enm530.htm rd

2. Irrelevant Characteristics • Monetary Units Dollars Pounds Yen Marks • Effective Period • Day Month Year Century rd

3. Interest and Equivalence • Computing Cash Flows • Time Value of Money • Equivalence • Single Payment Compound Interest Formulas rd

4. Why Engineering Economy? • Should I pay off my credit card balance with borrowed money? • What are the worth of graduate studies over my career? • Are tax deductions for my home mortgage a good deal or should I accelerate my mortgage payments? • Exactly what rate of return did we make on our stock investments? • Should I buy or lease my next car, or keep the one I have now and pay off the loan? • When should I replace my present car? • Which cash flow is preferable? rd

5. Time Value of Money • The change in the amount of money over a given time period is call the the time value of money; and is the most important concept in engineering economy. • You borrow $10,000 and repay $10,700 a year later. Find the rate of interest. • i = 700/10,000 = 7% rd

6. Cash Flow P ~ Present at time 0; F ~ Future A ~ Uniform or Equal G ~ Gradient i ~ effective interest n ~ Number of pay periodsEvery problem involves at least 4 of the 5 symbols F P A I N with at least 3 of the 4 estimated or known. F A 0 1 2 3 4 5 I = 7% P rd

7. Compounding Process • Given: i = 10% n = 7 years P = $3000 F = P(1 + i)n • Find: F = 3000(1 + 0.10)7 compounding factor • = $5846.15 i = 10% • (FGP 3000 10 7) • P = F(1 + i)-n • Discounting factor • F = P(F/P, i%, n) rd

8. F/P n Start Interest End 1 P iP P(1 + i)12 P(1 + i)1 iP(1 + i)1 P(1 + i)23 P(1 + i)2 iP(1 + i)2 P(1 + i)3… ... ... ...n P(1 + i)n-1 iP(1 + i)n-1 P(1 + i)n rd

9. (F/P, i%, n) F given P; F = P(1 + i)n = 1000(1 + 0.04)5 1000(1.2167) = 1216.653  1216.70 calculator table F = 1216.65 i = 4% compounded annually n = 5 P = $1000 rd

10. (F/A, i%, n) F = A(F/A, i%, n) (F/A, i%, n) = A[(1 + i)n-1 + (1 + i)n-2 + … + (1 + i)1 + 1]Sum = [1 – (1 + i)(1 + i)n-1]/[1 – (1 + i)] = [(1 + i)n - 1]/i F 1 2 3 4 5 6 … n-2 n-1 n A rd

11. (F/A, i%, n) F given A; F = A(F/A, i%, n); F = 500(5.4163, 4%, 5) = $2708.16 (F/A, i%, n) = F = $2708.16 = (FGA 500 4 5) s5|0.04 = 5.41632 sinking fund i = 4% 0 1 2 3 4 5 A = $500 rd

12. P/A • P/A = • An|i = Sn|i * (1 + i)-n • Find the present worth of 5 yearly deposits of $1000 at 7% compounded annually. • P = A(P/A, 7%, 5) = 1000(4.100197) = $4100.20 = 1000(F/A, 7%, 5)(P/F, 7%, 5) = 1000 * 5.750749 * 0.712986 = 1000 * 4.100197 = $4100.20 rd

13. A/F & A/P rd

14. Compound Interest Factors 7% • n F/P P/F A/F A/P F/A P/A A/G P/G • 1 1.0700 0.9346 1.0000 1.0700 1.0000 0.9346 0.0000 0.0000 • 2 1.1449 0.8734 0.4831 0.5531 2.0700 1.8080 0.4831 0.8735 • 3 1.2250 0.8163 0.3111 0.3811 3.2149 2.6243 0.9549 2.5061 • 4 1.3108 0.7629 0.2252 0.2952 4.4399 3.3872 1.4155 4.7948 • 5 1.4026 0.7130 0.1739 0.2439 5.7507 4.1002 1.8650 7.6467 • 6 1.5007 0.6663 0.1398 0.2098 7.1533 4.7665 2.3032 10.9784 • 7 1.6058 0.6227 0.1156 0.1856 8.65405.3893 2.7304 14.7149 • 8 1.7182 0.5820 0.0975 0.1675 10.2598 5.9713 3.1466 18.7890 • 9 1.8385 0.5439 0.0835 0.1535 11.9780 6.5152 3.5517 23.1405 • 10 1.9672 0.5083 0.0724 0.1424 13.8165 7.0236 3.9461 27.7156 • 11 2.1049 0.4751 0.0634 0.1334 15.7836 7.4987 4.3296 32.4666 • 12 2.2522 0.4440 0.0559 0.1259 17.8885 7.9427 4.7025 37.3507 rd

15. Computing Cash Flows • You bought a machine for $30,000. You can either pay the full price now with a 3% discount, or pay $5000 now; at the end of 1 year pay $8000, then at the end of the next 4 years pay $6,000. i = 7% compound yearly.Option 1  0.97 * 30K = $29,100 • Option 2  0 1 2 3 4 5 -$5000 -$6000 -$6000 -$6000 -$6000 -$8000 Continued …  rd

16. 0 1 2 3 4 5 -$5000 -$6000 -$6000 -$6000 -$6000 Option 2 7% compounded annually -$8000 P = 6K(P/A, 7%, 4)(P/F, 7%, 1) + 8K(P/F, 7%, 1) + 5K = 6K * 3.387 * 0.9346 + 8K(0.9346) + 5K = 31,470. Option 2  $31,470 Option 1  0.97 * 30K = $29,100 => Choose Option 1. rd

17. Simple Interest • Simple interest is: P * i * n = Pin • You borrow $10,000 for 5 years at a simple interest rate of 6%. At the end of 5 years, you would repay: Principal plus simple interest F = P + Pin = 10,000 + 10,000 * 0.06 * 10 • = 10,000 + 600 • = $10,600 rd

18. Compound Interest • Compound Interest is: P(1 + i)n • You borrow $10,000 for 5 years at 6% compounded annually. At the end of 5 years, you would repay: Principal plus compound interestF = P(1 + 0.06)5 = $13,382.26versus the simple interest $10,600.00 $2,782.26 more rd

19. Equivalence • When comparing alternatives that provide the same service, equivalent basis depends on • Interest Rate • Amounts of money involved • Timing of the cash flow • Perspective (Point of View) • 12 inches = 1 foot …continued  rd

20. Equivalence • Your borrow $5,000 at 8% compounded annually. • You may return the $5,000 immediately, or pay according to Plan A or Plan B The 3 options are equivalent. • n 1 2 3 4 5 A -1400 -1320 -1240 -1160 -1080 B -400 -400 -400 -400 -5400 PWA = 1400(P/A, 8%, 5) – 80(P/G, 8%, 5) • = $5000 • PWB = 400(P/A,8%,5) + 5000(P/F,8%,5) = 1597.08 + 3402.92 = $5000. rd

21. Equivalence rd

22. Re-Payment Plans APR = 8% n owed interest total principal totalfor year owed owed payment paid ( P+I) 1 $5000 $400 $5400 $1000 $14002 4000 320 4320 1000 13203 3000 240 3240 1000 12404 2000 160 2160 1000 11605 1000 80 1080 1000 1080 $1200 $6200 (IRR '(-5000 1400 1320 1240 1160 1080))  8% rd

23. Re-Payment Plans APR = 8% owed interest total principal totaln for year owed owed payment paid 1 $5000 $400 $5400 $0 $4002 5000 400 5400 0 4003 5000 400 5400 0 4004 5000 400 5400 0 4005 5000 400 5400 5000 5400 $2000 $7000 (IRR '(-5000 400 400 400 400 5400)) 8% rd

24. Re-Payment Plans APR = 8% owed interest total principal totaln for year owed owed payment paid 1 $5000 $400 $5400 $ 852 $1252.282 4148 331 4479 921 1252.283 3227 258 3484 994 1252.284 2233 178 2411 1074 1252.285 1159 93 1252 1159 1252.28 $1260 $ 6261.40 (IRR '(-5000 1252.28 1252.28 1252.28 1252.28 1252.28))  8% rd

25. Re-Payment Plans APR = 8% owed interest total principal totaln for year owed owed payment paid 1 $5000 $400 $5400 $ 0 $ 02 5400 432 5832 0 03 5832 467 6299 0 04 6299 504 6803 0 05 6803 544 7347 5000 7347 $2347 $5000 7347 (IRR '(-5000 0 0 0 0 7347))  8% rd

26. Equivalence (IRR '(-5000 1400 1320 1240 1160 1080))  8% (IRR '(-5000 400 400 400 400 5400)) 8% (IRR '(-5000 1252.28 1252.28 1252.28 1252.28 1252.28)) 8% (IRR '(-5000 0 0 0 0 7347))  8% rd

27. Simple vs. Compound Interest • Fs = P(1 + ni) Simple interest • Fc = P(1 + i)n Compound interest • Given P = $24, i = 5%, n = 20 years vs. compounded annually. • Fs =24(1 + 20 * 0.05) = $48 • Fc = 24(1.05)20 = $63.68. • In 1626 Peter Minuit paid $24 for Manhattan Island. • In Year 2008: Fs = 24(1 + 382 * 0.05) = $482.40 • Fc = 24(1.05)381 = $2,982,108,814.51 rd

28. Effective Interest Rate Annual Percentage Rate (APR)  r; for example, 12% per year Effective interest rate  ieff APY ~ annual percent yieldieff = , where m is the number of pay periods Example: APR is 12% compounded monthly. ieff = = 12.68% effective yearly rate. All interest rates is formulas are effective interest rates commensurate with the pay periods. rd

29. Effective Interest Rate • Annual Percentage Rate (APR) is 12% • If compounded monthly, • effective monthly rate is 1% 12/12 • effective quarterly rate is 3.03% (1 + 0.03/3)3 - 1 • effective yearly rate is 12.68% (1 + 0.12/12)12 - 1 • If compounded quarterly, effective quarterly rate is 3% 12/4 • effective yearly rate is 12.55% (1 + 0.12/4)4 - 1 • ieff = rd

30. Effective Interest Rate • $1000 is invested for 5 years at 12% APR compounded monthly. Compute its future worth using effective a) annual rate b) quarterly rate c) monthly rate and d) 2-year rate.F5 years = 1000(1 + 0.126825)5 = $1816.70F20 qtrs = 1000(1 + 0.030301)20 = $1816.70F60 mths = 1000(1 + 0.01)60 = $1816.70F2.5yrs = 1000(1 + 0.2697)2.5 = $1816.70 • Effective 3 year rate is (1 + 0.36/36)36 – 1 = 43.07688%F3 yrs = 1000(1 + 0.4307688)5/3 = $1816.70 rd

31. Interest Rates You borrow $1000 and agree to repay with 12 equal monthly payments of $90.30. Find the a) effective monthly interest rate. b) nominal annual interest rate c) effective annual interest rate. a) 1000 = 90.30(P/A, i%, 12) => P/A factor = 11.0742 => imeff = 1.25% b) APR = 12 * 1.25% = 15% c) Annual effective rate = (1 + 0.15/12)12 – 1 = 16.08%. rd

32. Continuous Compounding • F = Pern where r is the APR and n is the number of years • The effective rate for continuous compounding is given by • er - 1. • Substitute er - 1 for i is in the formulas. • For example, F = P(1 + i)n = P(1 + er – 1)n = Pern rd

33. Continuous Compounding • Traffic is currently 2000 cars per year growing at a rate of 5% per year for the next 4 years. How much traffic is expected at the end of 2 years? • F = 2000 e2*0.05 = 2210.34 cars Investment is currently $2000 per year growing at a rate of 5% per year for the next 4 years. How much investment is expected at the end of 2 years? • F = 2000 e2*0.05 = $2210.34 rd

34. Problem $100 at time 0 is $110 at time pay period 1 and was $90 at time pay period –1. Find the APR for year –1 to 0 and for 0 to 1. 100 = 90(1 + i) => i = 11.11% 90 100 110 110 = 100(1 + i) => i = 10%; If $90 is invested at time –1 returns $110 at time +1, find i. 110 = 90(1 + i)2 => i = 10.55%. -1 0 1 rd

35. Problem You plan to make 2 deposits, $25K now and $30K at the end of year 6. You draw out $C each year for the first 6 years and C + 1000 each year for the next 6 years. Find C at 10% interest compounded annually. C(P/A, 10%, 12) + 1000(P/A, 10%, 6)(P/F, 10%, 6) = [25K + 30K(1.1)-6] C = $5,793.60. 1 6 12 C + 1K $C 25K 30K rd

36. Time to Double How many years until an investment doubles at 5% compounded annually? F = 2P = P(1 + 0.05)n=> 2 = 1.05n => Ln 2 = n Ln 1.05 n = Ln 2 / Ln 1.05 = 14.20669 years CheckF = 1000 (1.05)14.21 = $2000 rd

37. Computing i given P, F and n F = P(1 + i)n i = (F/P)1/n – 1 Example: What interest rate generates $3456 in 5 years by investing $1000 now? i = 3.4561/5 – 1 = 28.14886%. (IGPFN 1000 3456 5) 28.14886 rd

38. Computing n given P, F and i • F = P(1 + i)n • n = Ln (F/P) / Ln (1 + i) • How many years for $1000 deposited now to accumulate to $3456 at an APR of 28.14886%? • n = Ln 3.456 / Ln 1.2814886 • = 5 years • (NGPFI 1000 3456 28.14886) 4.999999 rd

39. Arithmetic Gradient (n – 2)G (n -1)G 2G G A 0 1 2 3 . . . n -1 n Gradient begins in Year 2 = A(P/A, i%, n) + G(P/G, i%, n) P rd

40. Given the cash flow in the diagram below, find the present worth value at time 0 with interest rate 5% per year. • PW(5%) = [1000(P/A, 5%, 4) + 100(P/G, 5%, 4)](P/F, 5%, 2) • = $3679.12 1300 1200 1100 1000 0 1 2 3 4 5 6 rd

41. Arithmetic Gradient • Find the present worth of the following cash flow at 7% compounded annually: A = 50, G = 20 • n 1 2 3 4 5 6 cf 50 70 90 110 130 150 • P = A(P/A, i%, n) + G(P/G, i%, n) • = 50(P/A, 7%, 6) + 20(P/G, 7%, 6) • = 238.326 + 219.567 = $457.89(PGG A G i n)  (PGG 50 20 7 6)  $457.89 rd

42. Geometric Gradient • A1; A2 = A1 + gA1 = A1(1 + g); • A3 = A2 +gA2 = A1(1 + g) + gA1(1 + g) = A1(1 + g)2 • An = A1(1 + g)n-1 • Pn = An(P/F, i%, n) = An(1 + i)-n • = A1(1 + g)n-1(1 + i)-n • P= A1(1 + i)-n • (P/A1, i, g, n) = rd

43. Geometric Gradient • Find the present worth of a cash flow beginning at $10K and increasing at 8% for 4 years at 6% per year interest. • (PGGG-table 10000 8 6 4) • n Cash-flow 8% PW-factor 6% PWorth • 1 10000.00 0.9434 9433.96 2 10800.00 0.8900 9611.96 3 11664.00 0.8396 9793.32 4 12597.12 0.7921 9978.10 $38,817.54 • PW = 10K[1 – (1.08)4/(1.06)4(0.06 – 0.08)] = $38,817.54 rd

44. Geometric Gradient Example: Calculate the present worth of a contract awarded at $1000 per year and increasing at a uniform rate of 10% per year for 5 years at 7% APR compounded annually. P = A1[1 – (1 + g)n(1 + i)-n]/(i – g) = 1000[1 – 1(1 + 0.1)5(1 + 0.07)-5 ]/(0.07 - 0.10) = $4942.38. (PGGG A g i n) ~ (PGGG 1000 10 7 5)  $4942.38 rd

45. Geometric Gradient • A modification costs $8K, expected to last 6 years with a $1300 salvage value. Maintenance is $1700 the first year and increasing 11% per year thereafter. The interest rate is 8%. Find the present worth. • PW = -8K – 1700[1- (1.11/1.08)6/(0.08 – 0.11)] =1300(1.06)-6 • = -$17,305.88 • (+ 8000 (PGGG 1700 11 8 6) (PGF -1300 8 6))  17305.88 rd

46. Problem 3-11 • n 0 1 2 3 4 cf -100 25 45 45 30 • Find the compound annual interest rate. • Guess and test or • (IRR '(-100 25 45 45 30)) 16.189% rd

47. Problem 3-12 • Compute the difference in borrowing $1E9 at 4.5% for 30 years versus at 5.25%. • Difference = 1E9[(F/P, 5.25%, 30) – (F/P, 4.5%, 30)] • = $896,232,956.47 rd

48. Problem 3-15 • 1903 painting valued at $600 • 1995 painting valued at $29,152,000 • Find i. • 29152000 = 600(F/P, i%, 92) = 600(1 + i)92 • 48586.7 = (1 + i)92 • 1.12445066 = 1 + i • 12.45% = i rd

49. Manhattan Island Manhattan Island was bought for $24 in 1626 from the native Americans. Compute the present day worth if they had invested the money in an 8% APR compounded yearly. F = 24 (1 + 0.08)2007-1626 = $130,215,319,909,000 Today each of the 300 millions Americans could be given $434,051.07. F is considerably more than what Manhattan Island is now worth. rd

50. Learning Curve The time to make the first unit is $1000 and the learning curve is 90%. What is the cost to make the 2nd, 4th and 8th units? T2 = 1000 * 2(log 0.90 2) = 1000 *2 -0.1520 = $900 T4 = 0.90 * $900 = $810 T8 = 0.90 * $810 = $729. rd