1. 12/19/2015rd1 Engineering Economic Analysis Chapter 3  Interest and Equivalence http://academic.udayton.edu/ronalddeep/enm530.htm

2. 12/19/2015rd2 Irrelevant Characteristics Monetary Units Dollars Pounds Yen Marks Effective Period Day Month Year Century

3. 12/19/2015rd3 Interest and Equivalence Computing Cash Flows Time Value of Money Equivalence Single Payment Compound Interest Formulas

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? 12/19/2015rd4

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 = (10.7K – 10K)/10K = 700/10000 = 7% 12/19/2015rd5

6. 12/19/2015rd6 Cash Flow Diagram P ~ Present at time 0; F ~ Future A ~ Uniform or Equal G ~ Gradient i ~ effective interest n ~ Number of pay periods F P 0 1 2 3 4 5 n A I = 7% gradient uniform

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 P = F(1 + i) -n Discounting factor F = P(F/P, i%, n) Genie command is (FGP 3000 10 7)  5846.15 12/19/2015rd7

8. 12/19/2015rd8 n Start InterestEnd 1 P iP P(1 + i) 1 2 P(1 + i) 1 iP(1 + i) 1 P(1 + i) 2 3 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 F/P

9. 12/19/2015rd9 F given P; F = P(1 + i) n = 1000(1 + 0.04) 5  1000(1.2167) = 1216.652  1216.70 calculator table F = 1216.65 i = 4% compounded annually n = 5 P = $1000(F/P 1000 4 5)  1216.65 (F/P, i%, n) formula

10. 12/19/2015rd10 F = A(F/A, i%, n) (F/A, i%, n) = A[(1 + i) n-1 + (1 + i) n-2 + … + (1 + i) 1 + 1] (Summing from right to left) = A[1 – (1 + i)(1 + i) n-1 ]/[1 – (1 + i)] = A[(1 + i) n - 1]/i F 1 2 3 4 5 6 … n-2 n-1 n A (F/A, i%, n)

11. 12/19/2015rd11 F given A; F = A(F/A, i%, n); F = 500(5.4163, 4%, 5) = $2708.16 (F/A, i%, n) = F = $2708.16 = (F/A 500 4 5) i = 4% 0 1 2 3 4 5 A = $500 (F/A, i%, n)

12. 12/19/2015rd12 P/A P/A = 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 = $4100.20 F/A P/F

13. 12/19/2015rd13 A/F & A/P

14. 12/19/2015rd14 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 8.6540 5.3893 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

15. 12/19/2015rdCW3B-15Relationships a.(F/P, i%, n) = i(F/A, i%, n) + 1 b.(P/F, i%, n) = 1 – i(P/A,i%, n) c.(A/F, i%, n) = (A/P, i%, n) – i d.(A/P i%, n) = i / [1 – (P/F, i%, n)] e.Find (F/P, 10%, 37) (F/P, 10%, 37) = (F/P, 10%, 35)(F/P, 10%, 2) = 28.1024 *1.21 = 34.0039 f. (P/A, i%, n) – (P/A, i%, n-1) = (P/F, i%, n)

16. 12/19/2015rd16 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  $31, 470 0 1 2 3 4 5 -$5000 -$8000 -$6000 -$6000 Continued … 

17. 12/19/2015rd17 Option 2 0 1 2 3 4 5 -$5000 -$6000 -$6000 -$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 7% compounded annually

18. 12/19/2015rd18 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

19. 12/19/2015rd19 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 interest F = P(1 + 0.06) 5 = $13,382.26 versus the simple interest $10,600.00 Difference$ 2,782.26

20. 12/19/2015rd20 Equivalence

21. 12/19/2015rd21 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 

22. 12/19/2015rd22 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 PW A = 1400(P/A, 8%, 5) – 80(P/G, 8%, 5) = $5000 PW B = 400(P/A,8%,5) + 5000(P/F,8%,5) = 1597.08 + 3402.92 = $5000.

23. 12/19/2015rd23 Re-Payment Plans APR = 8% n owed interest total principal total for year owed owed payment paid ( P+I) 1$5000 $400 $5400 $1000 $1400 2 4000 320 4320 1000 1320 3 3000 240 3240 1000 1240 4 2000 160 2160 1000 1160 5 1000 80 1080 1000 1080 $1200 $6200 (IRR '(-5000 1400 1320 1240 1160 1080))  8%

24. 12/19/2015rd24 Re-Payment Plans APR = 8% nowed interest total principal total for year owed owed payment paid 1$5000 $400 $5400 $0 $400 2 5000 400 5400 0 400 3 5000 400 5400 0 400 4 5000 400 5400 0 400 5 5000 400 5400 5000 5400 $2000 $7000 (IRR '(-5000 400 400 400 400 5400))  8%

25. 12/19/2015rd25 Re-Payment Plans APR = 8% nowed interest total principal total for year owed owed payment paid 1$5000 $400 $5400 $ 852 $1252.28 2 4148 331 4479 921 1252.28 3 3227 258 3484 994 1252.28 4 2233 178 2411 1074 1252.28 5 1159 93 1252 1159 1252.28 $1260 $ 6261.40 (IRR '(-5000 1252.28 1252.28 1252.28 1252.28 1252.28))  8%

26. 12/19/2015rd26 Re-Payment Plans APR = 8% nowed interest total principal total for year owed owed payment paid 1$5000 $400 $5400 $ 0 $ 0 2 5400 432 5832 0 0 3 5832 467 6299 0 0 4 6299 504 6803 0 0 5 6803 544 7347 5000 7347 $2347 $5000 7347 (IRR '(-5000 0 0 0 0 7347))  8%

27. 12/19/2015rd27 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%

28. 12/19/2015rd28 Simple vs. Compound Interest F s = P(1 + ni) Simple interest F c = P(1 + i) n Compound interest Given P = $24, i = 5%, n = 20 years vs. compounded annually. F s = 24(1 + 20 * 0.05) = $48 F c = 24(1.05) 20 = $63.68. In 1626 Peter Minuit paid $24 for Manhattan Island. In Year 2008: F s = 24(1 + 382 * 0.05) = $482.40 F c = 24(1.05) 381 = $2,982,108,814.51

29. 12/19/2015rd29 Annual Percentage Rate (APR)  r; for example, 12% per year Effective interest rate  i eff APY ~ annual percent yield i eff =, where m is the number of pay periods Example: APR is 12% compounded monthly. i eff = = 12.68% effective yearly rate. All interest rates is formulas are effective interest rates commensurate with the pay periods. Effective Interest Rate

30. 12/19/2015rd30 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 i eff =

31. 12/19/2015rd31 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) 3-year rate. F 5 years = 1000(1 + 0.126825) 5 = $1816.70 F 20 qtrs = 1000(1 + 0.030301) 20 = $1816.70 F 60 mths = 1000(1 + 0.01) 60 = $1816.70 F 2.5yrs = 1000(1 + 0.2697) 2.5 = $1816.70 Effective 3 year rate is (1 + 0.36/36) 36 – 1 = 43.07688% F 3 yrs = 1000(1 + 0.4307688) 5/3 = $1816.70

32. 12/19/2015rd32 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 => i month-eff = 1.25% b) APR = 12 * 1.25% = 15% c) Annual effective rate = (1 + 0.15/12) 12 – 1 = 16.08%.

33. Continuous Compounding F = Pe rn where r is the APR and n is the number of years (1 + 0.0001) 10000 = 2.7181 The effective rate for continuous compounding is given by e r - 1. Substitute e r - 1 for i in the formulas. For example, F = P(1 + i) n = P(1 + e r – 1) n = Pe rn 12/19/2015rd33

34. 12/19/2015rd34 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 e 2*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 e 2*0.05 = $2210.34

35. 12/19/2015rd35 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

36. 12/19/2015rd36 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 Problem 25K 30K C + 1K $C

37. 12/19/2015rd37 How many years until an investment doubles at 5% compounded annually? F = 2P = P(1 + 0.05) n => 2 = 1.05 n => Ln 2 = n Ln 1.05 n = Ln 2 / Ln 1.05 = 14.20669 years Check F = 1000 (1.05) 14.21 = $2000 (NGPFI 1 2 5)  14.2067 years Time to Double

38. 12/19/2015rd38 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 = 3.456 1/5 – 1 = 28.14886%. (IGPFN 1000 3456 5)  28.14886 Computing i given P, F and n

39. 12/19/2015rd39 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 $3465 at an APR of 28.14886%? n = Ln 3.465 / Ln 1.2814886 = 5 years (NGPFI 1000 3465 28.1488)  5

40. 12/19/2015rd40 Arithmetic Gradient 0 1 2 3... n -1 n Gradient begins in Year 2 = A(P/A, i%, n) + G(P/G, i%, n) G 2G (n -1)G P A (n – 2)G

41. 12/19/2015rd41 11.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 (P/F (PGG 1000 100 5 4) 5 2)  3679.12 0 1 2 3 4 5 6 1000 1100 1200 1300

42. 12/19/2015rd42 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 cf507090110130150 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

43. 12/19/2015rd43 Geometric Gradient A 1 ; A 2 = A 1 + gA 1 = A 1 (1 + g); A 3 = A 2 +gA 2 = A 1 (1 + g) + gA 1 (1 + g) = A 1 (1 + g) 2 A n = A 1 (1 + g) n-1 P n = A n (P/F, i%, n) = A n (1 + i) -n = A 1 (1 + g) n-1 (1 + i) -n P = A 1 (1 + i) -n (P/A 1, i, g, n) =

44. 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 12/19/2015rd44

45. 12/19/2015rd45 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 = A 1 [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

46. 12/19/2015rd46 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

47. 12/19/2015rd47 Problem 3-11 n 0 1 234 cf -100 2545 4530 Find the compound annual interest rate. Guess and test or (IRR '(-100 25 45 45 30))  16.189%

48. 12/19/2015rd48 Problem 3-12 Compute the difference in borrowing $1E9 at 4.5% for 30 years versus at 5.25%. Diff = 1E9[(F/P, 5.25%, 30) – (F/P, 4.5%, 30)] = $896,232,956.47

49. 12/19/2015rd49 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

50. 12/19/2015rd50 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) 2008-1626 = $140,632,545,502,000 Today each of the 300 millions Americans could be given $468,775.15. F is considerably more than what Manhattan Island is now worth.

51. 12/19/2015rd51 Learning Curve The time to make the first unit is $1000 and the learning curve is 90%. What is the cost to make the 2 nd, 4 th and 8 th units? T 2 = 1000 * 2 (log 0.90 2) = 1000 *2 -0.1520 = $900 T 4 = 0.90 * $900 = $810 T 8 = 0.90 * $810 = $729. (sim-lc 1000 8 90)  Unit Hours Cumulative 1 1000.00 1000.00 2 900.00 1900.00 3 846.21 2746.21 4 810.00 3556.21 5 782.99 4339.19 6 761.59 5100.78 7 743.95 5844.73 8 729.00 6573.73

52. 12/19/2015rd52 Cost of a Mortgage Interest Rate Loan Amount Payment Frequency Points (Prepaid interest of 1% of loan) Fees (application, loan origination)

53. 12/19/2015rd53 Principal Reduction and Interest Paid for each payment A = P(A/P, i%, N) PR n = A(P/F, i%, N – n + 1); Interest n = A - PR n PR 5 = 1627.45(P/F, 10%, 10 – 5 + 1) = $918.65 (P/F, i%, n) = (P/A, i%, n) – (P/A, i%, n - 1)

54. 12/19/2015rd54 Pay Num Payment Principal Interest Balance Total Interest 1 1627.45 627.45 1000.00 9372.55 1000.00 2 1627.45 690.19 937.26 8682.36 1937.26 3 1627.45 759.21 868.24 7923.15 2805.50 4 1627.45 835.13 792.32 7088.02 3597.82 5 1627.45 918.65 708.80 6169.37 4306.62 6 1627.45 1010.51 616.94 5158.86 4923.56 7 1627.45 1111.56 515.89 4047.30 5439.45 8 1627.45 1222.72 404.73 2824.58 5844.18 9 1627.45 1344.99 282.46 1479.59 6126.64 10 1627.45 1479.49 147.96 0.10 6274.60 Mortgage Payments Loan = $10,000Interest 10%/year N = 10 years A = 10K(A/P, 10%, 10) = $1627.45

55. Computing the Balance On a loan of $10,000 for 10 years at 10% per year, determine the annual payment and the balance after 6 years. 1 2 3 4 5 6 7 8 9 10 A = P(A/P, i%, N) = 10K(A/P, 10%, 10) = $1627.45 B 6 = 1627.45(P/A, 10%, 4) = $5158.81 12/19/2015rd55

56. 12/19/2015rd56 Points or No Points You finance a home for $100K at 15 year interest rate. You can either pay one point with 6.375% interest or no points with 6.75% interest compounded monthly. The 1 point implies that you get $99K but get charged as if you borrowed $100K A = 100K(A/P, 6.375/12 %, 180) = $864.25 99K = 864.25(P/A, i%, 180) looking for 114.55 P/A factor for 180 pay periods. i = 0.545 % => APR = 6.54% Thus go with paying the point as 6.54% < 6.75%.

57. 12/19/2015rd57 Monthly Payments What is the monthly payment for a 5-year car loan of $35,000 at 6% compounded monthly? Find the amount of the principal reduction of the 25 th payment. A = $35,000(A/P, 6% /12, 60) = $676.65. PR 25 = 676.65(P/F, 6%/12, 60 – 25 + 1) = $565.44.

58. 12/19/2015rd58 Mortgage Payments You take out a loan for $350,000 at 6% APR compounded monthly for 30 years. a)Your monthly payment is _________. b)The principal reduction PR 225 is ________. c) The interest paid Int 225 at this payment is _______. d)The total interest paid to date at this payment is _______. e) The balance at this payment is __________. f)The per cent of your loan paid by the 225 th is ______. Ans. a) $2098.43 b) $1064.90 c) $1033.53 d) $327,786.21 e) $205,640.17f) 41.245%

59. 12/19/2015rd59 Compound Continuously How long does it take for an investment to triple if interest rate is compounded continuously at i = 7%? F = Pe rn 3 = 1e 0.07n => n = 15.69 years You deposit $1000 a year for 5 years at 7% compounded continuously. At the end of 5 years you can withdraw _______. Substitute e r – 1 for i into the F/A factor. ans. $5779.59

60. 12/19/2015rd60 Interest Rate A credit card company charges 1.5% interest on the unpaid balance each month. Nominal annual interest rate is ________. ans. 18% Effective annual interest rate is ________. ans. 19.56%

61. 12/19/2015rd61 1 2 3 4 5 A = $500 Place an amount F 3 at year 3 which is equivalent to the cash flow below at 7%. F 3 = 500(P/A, 7%,5)(F/P,7%,3) = $2511.46. Equivalence

62. Preference Which alternative would you choose at i = 10%? i = 9%? 1) Receive $100 now. 2) $120 two years from now. 12/19/2015rd62

63. P/A Find the exact P/A factor at 9.35% compounded continuously for n = 10 pay periods. P/A = 12/19/2015rd63

64. Time Value of Money a) Joe wants to figure his annual car cost given the following cash flow at i = 7%: n 1 2 3 4 5 cf$4590135180225 (list-pgf '(0 45 90 135 180 225) 7)  $528.61 (A/P 528.61 7 5)  128.92 b)He got the 3rd and 4th year mixed. Refigure. (list-pgf '(0 45 90 180 135 225) 7)  $531.01 (AGP 531.01 7 5)  $129.51 12/19/2015rd64

65. Problem 3-19 What sum of money is equivalent to $8250 2 years later if interest is 4% compounded semiannually? 8250(1.04) -4 = $7052.13 12/19/2015rd65

66. Problem 3-23 A local bank pays 5% annual interest while an out of town bank pays 1.25% compounded quarterly. You have $3000 to deposit for 2 years. In which bank do you deposit your money? 3000(1.05) 2 = $3307.50 3000(1.0125) 8 = $3313.46 $5.96 12/19/2015rd66

67. Problem 3-24 P 1 = P 2 at interest rate i. Which cash flow is better at 2i? F 1 F2 0 1 2 0 1 2 3 P 1 = F 1 (1 + i) -2 = F 2 (1 + i) -3 = P 2 => F 2 = F 1 (1 + i) P' 1 = F 1 (1 + 2i) -2 P' 2 = F 2 (1 + 2i) -3 = F 1 (1 + i)(1 + 2i) -3 P' 2 /P' 1 = (1 + i)/(1 + 2i) P' 1 > P' 2 For example, try F 1 = $100 and F 2 = $107. 12/19/2015rd67

68. Problem 3-27 A sum of money Q will be received 6 years from now. At 5% annual interest the present worth of Q is $60. At the same interest rate, what would be the value of Q in 10 years? F = 60(1.05) 10 = $97.73 12/19/2015rd68

69. 12/19/2015rd69

70. 12/19/2015rdCW3B-70 Find A to equate the series at 10% compounded annually. 120 120 120 100 100 1 2 3 4 5 P 0 = 100(P/A,10%,2) + 120(P/A,10%,3)(P/F,10%,2) A A A A 1 2 3 4 5 P 0 = A(P/A,10%,4)(P/F,10%,1) = $420.18 => A = $145.81