1. TM 661 Engineering Economics for Managers Unit 1 Time Value of Money

2. Time Value-of-Money Suppose we invest $100. If the bank pays 5% interest, how much will I have after 1 year? 100 F 0 1 t

3. Time Value-of-Money Suppose we invest $100. If the bank pays 5% interest, how much will I have after 1 year? 100 F1F1 0 1 t F 1 = 100 + interest = 100 + 100 (.05) = 105

4. Time Value-of-Money Now suppose I keep that $105 in the bank for another year. Now how much do I have? 100 0 1 2 t F 2 = 105 + interest = 105 + 105 (.05) = 110.25 105 F2F2

5. Time Value-of-Money In General P 0 1 2 t F 1 = P + interest = P + iP = P(1+i) F1F1

6. Time Value-of-Money In General P 0 1 2 t F 1 = P + interest = P + iP = P(1+i) F 2 = F 1 + interest = F 1 + iF 1 = F 1 (1+i) F2F2

7. Time Value-of-Money In General P 0 1 2 t F 1 = P(1+i) F 2 = F 1 + interest = F 1 + iF 1 = F 1 (1+i) = P(1+i)(1+i) = P(1+i) 2 F2F2

8. Time Value-of-Money In General P 0 1 2 3 n F n = P(1+i) n FnFn

9. Example I would like to have $20,000 for my son to go to college in 15 years. How much should I deposit now if I can earn 10% interest? F n = P(1+i) n P 0 1 2 3 n 20,000

10. Example I would like to have $20,000 for my son to go to college in 15 years. How much should I deposit now if I can earn 10% interest? Recall, F n = P(1+i) n P 0 1 2 3 n 20,000

11. Example I would like to have $20,000 for my son to go to college in 15 years. How much should I deposit now if I can earn 10% interest? Recall, F n = P(1+i) n Then, P= F n (1+i) -n P 0 1 2 3 n 20,000

12. Example I would like to have $20,000 for my son to go to college in 15 years. How much should I deposit now if I can earn 10% interest? Then, P= F n (1+i) -n = 20,000(1+.1) -15 = 20,000(0.2394) = $2,394 P 0 1 2 3 n 20,000

13. Future Worth Given Annuity Suppose I wish to compute annual installments to save for college. 0 1234.. n A F

14. Annuities Given FutureWorth F = A(1+i) n-1 + A(1+i) n-2 + A(1+i) n-3 +...+ A(1+i) + A 0 1234.. n A F

15. Annuities Given Future Worth (1+i)F = A(1+i) n + A(1+i) n-1 + A(1+i) n-2 +... + A(1+i) F = A(1+i) n-1 + A(1+i) n-2 + A(1+i) n-3 +...+ A(1+i) + A 0 1234.. n A F

16. Annuities Given Future Worth iF = A(1+i) n + 0 + 0 +... - A (1+i)F = A(1+i) n + A(1+i) n-1 + A(1+i) n-2 +... + A(1+i) F = A(1+i) n-1 + A(1+i) n-2 + A(1+i) n-3 +...+ A(1+i) + A 0 1234.. n A F

17. Annuities Given Future Worth iF = A(1+i) n + 0 + 0 +... - A = A[(1+i) n - 1] (1+i)F = A(1+i) n + A(1+i) n-1 + A(1+i) n-2 +... + A(1+i) F = A(1+i) n-1 + A(1+i) n-2 + A(1+i) n-3 +...+ A(1+i) + A 0 1234.. n A F

18. Annuities Given Future Worth iF = A[(1+i) n - 1] 0 1234.. n A F FA i i A F A in n    ) (,,) 11(

19. Annuities Given Future Worth 0 1234.. n A F AF i i n   ()11

20. Annuities Given Future Worth 0 1234.. n A F AF i i n   ()11 A     20000 1 111 2000000315 50 15, (.) (.),(.) $629.

21. Annuities Given Present Worth PAi k k n k     1 1()     Ai k k n ()1 1, for A 1 = A 2 = A 3 =... = A 0 1234.. n A P

22. Annuities Given Future Worth 0 1234.. n A1A1 F A2A2 A3A3 AnAn A4A4 FAi k k n k     0 1 1()     Ai k k n ()1 0 1, for A 1 = A 2 = A 3 =... = A

23. Annuities Given Present Worth Suppose we wish to compute the monthly payment of a car if we borrow $15,000 at 1% per month for 36 months. 0 1234.. n A P

24. Recall: F n = A and P = F n (1 + i) -n Annuities Given Present Worth 0 1234.. n A P ()11  i i n

25. Recall: and P = F n (1 + i) -n Annuities Given Present Worth 0 1234.. n A P FA i i n n   ()11 PA i i iA i ii n n n n       () () () () 11 1 11 1

26. Annuities Given Present Worth Inverting and solving for A gives 0 1234.. n A P P A i ii n n    () () 11 1 AP ii i PAPin n n     () () (/,,) 1 11

27. A= 15,000[.01(1.01) 36 /(1.01 36 - 1)] = $498.22 Car Example We have an initial loan of $15,000 at a rate of i=1% per month for n=36 months. The monthly loan payment is then 0 1234.. 36 A P

28. Car Example; Alternative We have an initial loan of $15,000 at a rate of i=1% per month for n=36 months. The monthly loan payment is then A= 15,000(A/P,i,n) = 15,000(A/P,1,36) = 15,000(.0032) = $498.21 0 1234.. 36 A P

29. Car Example; Tables

30. Example; Home Mortgage Example: Suppose we borrow $75,000 for house at 9% for 30 years. Find monthly payment. Assume that the monthly interest rate is 9/12 = 3/4%.

31. Example; Home Mortgage

33. Example: Suppose we borrow $75,000 for house at 9% for 30 years. Find monthly payment. Assume that the monthly interest rate is 9/12 = 3/4%. Using the Formula: A = $75,000 = $603.47 Example; Home Mortgage.(.) (.) 00751 1 1 360  

34. Home Mortgage Using the Table: A= P(A/P, i, n) = 75,000(A/P, 9, 30) = 75,000(.0973) = 7,297.5 / year » $608.12 / month

35. Gradient Series Suppose we have an investment decision which is estimated to return $1,000 in the first year, $1,100 in the second, $1,200 in the third, and so on for the 7 year life of the project. P 123 0 7 1600 1000 1100 1200

36. Gradient Equivalent Flows We can replace the cash flow as the equivalent of an annuity and a constant growth P A 123 0 n AAAA + P G 123 0 n (n-1)G G 2G

37. P W = P A + P G P A = A (P/A, i, n) P G = 0(1+i) -1 + G(1+i) -2 + 2G(1+i) -3 +... Gradient Derivation P A 123 0 n AAAA + P G 123 0 n (n-1)G G 2G

38. Gradient Derivation (1+i)P G = G[ (1+i) -1 + 2(1+i) -2 + 3(1+i) -3 +4(1+i) -4 ] P G = G[ (1+i) -2 + 2(1+i) -3 + 3(1+i) -4 +4(1+i) -5 ]

39. Gradient Derivation (1+i)P G = G[ (1+i) -1 + 2(1+i) -2 + 3(1+i) -3 +4(1+i) -4 ] P G = G[ (1+i) -2 + 2(1+i) -3 + 3(1+i) -4 +4(1+i) -5 ] iP G = G[ (1+i) -1 + (1+i) -2 + (1+i) -3 +(1+i) -4 - 4(1+i) -5 ]

40. Gradient Derivation (1+i)P G = G[ (1+i) -1 + 2(1+i) -2 + 3(1+i) -3 +4(1+i) -4 ] P G = G[ (1+i) -2 + 2(1+i) -3 + 3(1+i) -4 +4(1+i) -5 ] iP G = G[ (1+i) -1 + (1+i) -2 + (1+i) -3 +(1+i) -4 - 4(1+i) -5 ] A Miracle Occurs P G = G[( 1 - (1+ni)(1+i) -n )/ i 2 ]

41. Example Estimated return of $1,000 in the first year, $1,100 in the second, $1,200 in the third, and so on for the 7 year life of the project. P G = A(P/A,10,7) +G(P/G,10,7) = 1000(4.8684) +100(12.7631) = $6,144.71 P 123 0 7 1600 1000 1100 1200

42. Determine PW of a depreciation scheme which saves $1000 in the first year and declines by $100 per year for the next 10 years if i = 10% Depreciation Example PWPW 123 0 10 1000 900 800 100

43. Example; (cont.) PW= 1000(P/A, 10, 10) - 100(P/G, 10, 10) = 1000(6.1446) - 100(22.8913) = $3,855.47 0 PAPA 123 10 1000 + PGPG 123 0 10 900 100 200

44. Gradient Alternative A = G(A/G, 10, 10) = 100(3.7255) = $372.55 P W = (1000 - 372.55)(P/A, 10, 10) = 627.45(6.1446) = $3,855.43 0 PAPA 123 10 1000 + PGPG 123 0 10 900 100 200

45. Geometric Series Suppose we start a new computer consultant business. We assume that we will start our business with a modest income but that the business will grow at a rate of 10% per year. If we assume an initial return of 1000 and a growth rate of 10% over 4 years, the cash flow diagram might appear as follows. 123 0 4 1000 1100 1210 1331

46. Geometric Series P= A 1 (1 + i) -1 + A 2 (1 + i) -2 +......+A n (1 + i) -n = A(1 + i) -1 + A(1 + j)(1 + i) -2 + A(1 + j) 2 (1 + i) -3 +..... + A(1 + j) n - 1 (1 + i) -n = A ()()11 1 1     ji tt t n P 123 0 n A A(1+j) A(1+j) 2 A(1+j) n-1

47. Geometric Series Special Case: For the special case where i = j, we have P = A = A ()()11 1 1     ii tt t n ()1 1 1     i t n

48. Geometric Series Special Case: For the special case where i = j, we have P = A = A P = ()()11 1 1     ii tt t n ()1 1 1     i t n nA i()1 

49. Geometric Series Case i j: For the case when the interest rate i and the growth rate j are not equal, we have P = A = A(P/A,i,j,n) ()()11 1 1     ji tt t n Miracle 2 Occurs  PA ji ij nn     111()()

50. Geometric Series Case i j: For the case when the interest rate i and the growth rate j are not equal, we have P = A ()()11 1 1     ji tt t n Miracle 2 Occurs  PA ji ij nn     111()()       A j j i t n t (1) 1 1 1

51. Example; Geometric Computer business has initial return of 1000 and a growth rate of 10% over 4 years, the cash flow diagram might appear as follows. If i = 10%, 123 0 4 1000 1100 1210 1331

52. Example; Geometric Computer business has initial return of 1000 and a growth rate of 10% over 4 years, the cash flow diagram might appear as follows. If i = 10%, P = 1000(P/A,i,j,n) = 1000(P/A,10,10,4) = 1000(3.6363) = $3,636 123 0 4 1000 1100 1210 1331

53. Example Example: Individual deposits of $1000 in end of year 1 and increases by 10% each year for 30 years. If the account earns 10% per year, how much will he have at end of 30 years?

54. Example (Cont). Solution: F = A(F/A, i, j, n) = 1000((F/A, 10, 10, 30) = 1000(475.8934) = $475,893 Alternative Solution: F = 1000(30)(1 +.10) 29 = $475,892

55. Summary F n = P(1 + i) n = P(F/P,i,n) P = F n (1 + i) -n = F(P/F,i,n) F n 12340n 100..

56. Summary = A(F/A,i,n) = F n (A/F,i,n) 01234.. n A F n FA i i n n   ()11 AF i i n n   ()11

57. Summary = A(P/A,i,n) = P(A/P,i,n) 0 1234.. n A P PA i ii n n    () () 11 1 AP ii i n n    () () 1 11

58. Summary P G = G = G(P/G,i,n) A= G = G (A/G, i, n) 123 0 n (n-1)G G 2G 1 i 11 n ii n   [() 111 2   ()()nii i n

59. Summary Case i j P = A = A(P/A,i,j,n) F = A = A(F/A,i,j,n) P 123 0 n A A(1+j) A(1+j) 2 A(1+j) n-1 ()()11   ij ij nn 111    ()()ji ij nn 

60. Break Time

61. Geometric Derivation 01234.. n A F n     Ai k k n ()1 0 1 F

62. Geometric Derivation 01234.. n A F n     Ai k k n ()1 0 1 F Recall the geometric series x x x k k n n       0 1 1 1 Letting x = (1+i) k

63. Geometric Derivation 01234.. n A F n     Ai k k n ()1 0 1 F Recall the geometric series x x x k k n n       0 1 1 1 Letting x = (1+i) k  Fx k  k  0 n  1

64. Geometric Derivation 01234.. n A F n    1 1 x x n  Fx k  k  0 n  1

65. Geometric Derivation 01234.. n A F n    1 1 x x n  Fx k  k  0 n  1     11 11 1() () i i n   1()i i n A

66. 1  Geometric Derivation 01234.. n A F n    1 1 x x n  Fx k  k  0 n  1     11 11 1() () i i n   1()i i n A FA i i A F A in n   () (,,) 1