1. TM 661 Engineering Economics for Managers Unit 2 Multiple/Continuous Compounding

2. Multiple Compounding Example: Consider investment of $1000 at 12% per annum Bank A compounds annually Bank B compounds semi-annually Bank C compounds monthly Find F 10 for bank A, B, and C

3. Multiple Compounding Example: Consider investment of $1000 at 12% per annum Bank A compounds annually Bank B compounds semi-annually Bank C compounds monthly Find F 10 for bank A, B, and C Bank A F 10 = 1000 (F/P, 12, 10) = $3106

4. Multiple Compounding Bank B F 10 = 1000(F/P,6,20) = $3207

5. Multiple Compounding Bank B F 10 = 1000(F/P,6,20) = $3207 Bank C F 10 = 1000(F/P,1,120) = $3300

6. Multiple Compounding Consider investing $100 @ 12% per annum Bank A compounds Annually Bank B compounds Quarterly Bank C compounds Monthly Find F 1yr for Banks A, B, C

7. Effective Interest Rate Bank A F 1yr = 100 (1+.12) 1 = 112.00 i eff = 12.00%

8. Effective Interest Rate Bank A F 1yr = 100 (1+.12) 1 = 112.00 i eff = 12.00% Bank B F 1yr = F 4 = 100 (1 +.03) 4 = 112.55i eff = 12.55%

9. Effective Interest Rate Bank A F 1yr = 100 (1+.12) 1 = 112.00 i eff = 12.00% Bank B F 1yr = F 4 = 100 (1 +.03) 4 = 112.55i eff = 12.55% Bank C F 1yr = F12 = 100 ( 1 +.01) 12 = 112.68 i eff = 12.68%

10. Effective Interest Let r= annual interest rate m= # compounding periods / year P = amount invested F 1yr = P(1+r/m) m

11. Effective Interest Let r= annual interest rate m= # compounding periods / year P = amount invested F 1yr = P(1+r/m) m Int Earned = F 1 - P

12. Effective Interest Let r= annual interest rate m= # compounding periods / year P = amount invested F 1yr = P(1+r/m) m Int Earned = F 1 - P = P(1+r/m) m - P

13. Effective Interest Let r= annual interest rate m= # compounding periods / year P = amount invested F 1yr = P(1+r/m) m Int Earned = F 1 - P = P(1+r/m) m - P = P[(1+r/m) m - 1]

14. Effective Interest Now, Interest rate = Interest Earned in Period Principal Started

15. Effective Interest Now, Interest rate = Interest Earned in Period Principal Started i eff = Interest / P

16. Effective Interest Now, Interest rate = Interest Earned in Period Principal Started i eff = Interest / P = P[(1+r/m) m - 1] / P

17. Effective Interest Now, Interest rate = Interest Earned in Period Principal Started i eff = Interest / P = P[(1+r/m) m - 1] / P = ( 1 + r/m) m - 1

18. Class Problem Find effective interest rate of 12% compounded monthly.

19. Class Problem Find effective interest rate of 12% compounded monthly. i eff = ( 1+.12/12) 12 - 1

20. Class Problem Find effective interest rate of 12% compounded monthly. i eff = ( 1+.12/12) 12 - 1 = (1 +.01) 12 -1

21. Class Problem Find effective interest rate of 12% compounded monthly. i eff = ( 1+.12/12) 12 - 1 = (1 +.01) 12 -1 =.1268 = 12.68%

22. Nominal vs. Effective Int. Consider the discrete End-of-Year cash flow tables below: PeriodCash FlowPeriodCash Flow 0- $100,000 3$30,000 1 30,000 4 30,000 2 30,000 5 30,000 Determine the Present Worth equivalent if a. the value of money is 12% compounded annually. b. the value of money is 12% compounded monthly. c. the value of money is 12% compounded continuously.

23. Solution; Annual Rate 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) = -100,000 + 30,000(P/A,.12, 5)

24. Solution; Nominal/Effective 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) = -100,000 + 30,000(P/A,.12, 5) = -100,000 + 30,000(3.6048) = $8,144

25. Solution; Compound Monthly 100,000 30,000 1 2 3 4 5 i r m eff m  11

26. Solution; Compound Monthly 100,000 30,000 1 2 3 4 5 i r m eff m  11    1 12 1 1011 12681268% 12. (.)..

27. Solution; Compound Monthly 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) = -100,000 + 30,000(P/A,.1268, 5)

28. Solution; Compound Monthly 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) = -100,000 + 30,000(P/A,.1268, 5) = -100,000 + 30,000(3.5449) = $6,346

29. Solution; Continuous Comp. 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5)

30. Solution; Continuous Comp. 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) i eff = ????

31. Continuous Compounding Consider r= 12% /yr m= 12 compounding periods n= 10 yrs Then F = P (1+r/m) mn = P (1+.12/12) 12*10 = P (1+.01) 120

32. Continuous Compounding Now suppose we use an infinite # of compounding periods (continuous). How might we find an answer to our problem of r=12% per year compounded on a continuous basis?

33. Continuous Compounding Now suppose we use an infinite # of compounding periods (continuous). How might we find an answer to our problem of r=12% per year compounded on a continuous basis? Choose n = B.A.N. F= P(1+.12/9999) 9999 (one year period) = P(1.1275) = P(1+.1275)

34. Formal Derivation In General F = P (1 + r/m) mn = P (1 + r/m) m/r *nr = P [(1 + r/m) m/r ] nr

35. Continuous Compounding Now suppose we use an infinite # of compounding periods (continuous) F= P[( 1 + r/m) m/r ] nr = P [ ( 1 + r/m) m/r ] nr lim m  m 

36. Continuous Compounding Now suppose we use an infinite # of compounding periods (continuous) F= P[( 1 + r/m) m/r ] nr = P [ ( 1 + r/m) m/r ] nr But, lim m  m  emr rm m   / )1(

37. Continuous Compounding Now suppose we use an infinite # of compounding periods (continuous) F= P[( 1 + r/m) m/r ] nr = P[ ( 1 + r/m) m/r ] nr F = Pe nr P = Fe -nr lim m  m 

38. Continuous Effective Interest Recall, F = Pe nr Interest Earned 1yr = F - P = Pe r - P

39. Continuous Effective Interest Recall, F = Pe nr Interest Earned 1yr = F - P = Pe r - P Now, Interest rate= Interest Earned Principal

40. Continuous Effective Interest Now, Interest rate= Interest Earned Principal i eff = Pe r - P P i eff = e r - 1

41. Continuous i eff Example: Suppose a bank pays interest on a CD account at 6% per annum compounded continuously. What is the effective rate?

42. Continuous i eff Example: Suppose a bank pays interest on a CD account at 6% per annum compounded continuously. What is the effective rate? Soln: i eff = e.06 - 1 =.0618 = 6.18%

43. Continuous i eff Example: Suppose a bank pays interest on a CD account at 6% per annum compounded continuously. What is the effective rate? Soln: i eff = e.06 - 1 =.0618 = 6.18% Check: Let r=6%, m=999 i eff = ( 1 + r/m) m - 1 = (1+.06/999) 999 - 1 =.0618 = 6.18%

44. Solution; Revisted 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) = -100,000 + 30,000(P/A,.1275, 5) = -100,000 + 30,000(3.5388) = $6,164

45. Solution; Continuous Comp. 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) i eff = ????

46. Solution; Continuous Comp. 100,000 30,000 1 2 3 4 5 ie e eff r    1 1 1127511275% 1..

47. Solution; Continuous Comp. 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) = -100,000 + 30,000(P/A,.1275, 5)

48. Solution; Continuous Comp. 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) = -100,000 + 30,000(P/A,.1275, 5) = -100,000 + 30,000(3.5388) = $6,164

49. Solution; Continuous Altern. 100,000 30,000 1 2 3 4 5 P = -100,000 + 30,000(P/A, i eff, 5) = $6,164    10000030000 1 1,, () e ee rn r

50. Inflation Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1) 5 = $1,610.51 But we still only have 1 ton of copper

51. Inflation Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1) 5 = $1,610.51 But we still only have 1 ton of copper $1,610 5 years from now buys the same as $1,000 now

52. Inflation Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1) 5 = $1,610.51 But we still only have 1 ton of copper $1,610 5 years from now buys the same as $1,000 now 10% inflation

53. Inflation Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1) 5 = $1,610.51 But we still only have 1 ton of copper $1,610 5 years from now buys the same as $1,000 now 10% inflation(deflation = neg. inflation)

54. Combined Interest Rate Suppose inflation equals 5% per year. Then $1 today is the same as $1.05 in 1 year Suppose we earn 10%. Then $1 invested yields $1.10 in 1 year.

55. Combined Interest Rate Suppose inflation equals 5% per year. Then $1 today is the same as $1.05 in 1 year Suppose we earn 10%. Then $1 invested yields $1.10 in 1 year. In today’s dollars $1.00 $1.10 $1.05 $1.10

56. Combined Interest Rate That is $1.10 = 1.05 (1+d) 1 1(1+.10) = 1(1+.05)(1+d)

57. Combined Interest Rate That is $1.10 = 1.05 (1+d) 1 1(1+.10) = 1(1+.05)(1+d) 1+i = (1+j)(1+d) i = d + j + dj

58. Combined Interest Rate That is $1.10 = 1.05 (1+d) 1 1(1+.10) = 1(1+.05)(1+d) 1+i = (1+j)(1+d) i = d + j + dj i = interest rate (combined) j = inflation rate d = real interest rate (after inflation rate)

59. Solving for d, the real interest earned after inflation, where i = interest rate (combined) j = inflation rate d = real interest rate (after inflation rate) Combined Interest Rate j ji d    1

60. Example Suppose we place $10,000 into a retirement account which earns 10% per year. How much will we have after 20 years?

61. Example Solution: F = 10,000(1+.1) 20 = $67,275

62. Example (cont.) How much is $67,275 20 years from now worth if the inflation rate is 3%?

63. Example (cont.) How much is $67,275 20 years from now worth if the inflation rate is 3%? Solution: F T = 67,275(P/F,3,20) = 67,275(1.03) -20 = $37,248

64. Alternate: Recall = (.1 -.03)/(1+.03) =.068 Example (cont.) j ji d    1

65. Alternate: Recall = (.1 -.03)/(1+.03) =.068 F T = 10,000(1+d) 20 = 10,000(1.068) 20 = $37,248 Example (cont.) j ji d    1

66. Alternate: Recall = (.1 -.03)/(1+.03) =.068 F T = 10,000(1+d) 20 = 10,000(1.068) 20 = $37,248 Note: This formula will not work with annuities. Example (cont.) j ji d    1

67. Retirement a. Stu wishes to deposit a certain amount of money at the end of each month into a retirement account that earns 6% per annum (1/2% per month). At the end of 30 years, he wishes to have enough money saved so that he can retire and withdraw a monthly stipend of $3,000 per month for 20 years before depleting the retirement account. Assuming there is no inflation and that he will continue to earn 6% throughout the life of the account, how much does Stu have to deposit each month? You need only set up the problem with appropriate present worth or annuity factors. You need not solve but all work must be shown.

68. Solution; Retirement a. Take everything to time period 360 FP = A(F/A, 0.5, 360) = 3,000(P/A, 0.5, 240) 0 1 2 3 4 360 1 2 3 4 240 A 3,000 FP

69. Solution; Retirement a. Take everything to time period 360 FP = A(F/A, 0.5, 360) = 3,000(P/A, 0.5, 240) A(1,004.52) = 3,000(139.58) 0 1 2 3 4 360 1 2 3 4 240 A 3,000 FP

70. Solution; Retirement a. Take everything to time period 360 FP = A(F/A, 0.5, 360) = 3,000(P/A, 0.5, 240) A(1,004.52) = 3,000(139.58) A = $416.82 0 1 2 3 4 360 1 2 3 4 240 A 3,000 FP

71. Retirement b. Suppose that the solution to the above problem results in monthly deposits of $200 with an amassed savings of $350,000 by the end of the 30th year. For this problem assume that inflation is 3% per annum. Compute the value of the retirement account in year 30 before funds are withdrawn (in today’s dollars)

72. Solution; Retirement a. 0 1 2 3 4 360 1 2 3 4 240 200 3,000 FP = 350,000 FP T = 350,000(1+j) -n

73. Solution; Retirement a. 0 1 2 3 4 360 1 2 3 4 240 200 3,000 FP = 350,000 FP T = 350,000(1+j) -n = 350,000(1+0.03) -30 = $144,195

74. Solution; Retirement a. 0 1 2 3 4 360 1 2 3 4 240 200 3,000 FP = 418,195 FP T = 418,195(1+j) -n = 418,195(1+0.03) -30 = $172,290

75. Break Time