1. InvestmentWorth Investment Worth

2. Given a minimum attractive rate-of-return, be able to evaluate the investment worth of a project using Net Present Worth Equivalent Annual Worth Internal Rate of Return External Rate of Return Capitalized Cost Method

3. MARR Suppose a company can earn 12% / annum in U. S. Treasury bills No way would they ever invest in a project earning < 12% Def: The Investment Worth of all projects are measured at the Minimum Attractive Rate of Return (MARR) of a company.

4. MARR is company specific utilities - MARR = 10 - 15% mutuals - MARR = 12 - 18% new venture - MARR = 20 - 30% MARR based on firms cost of capital Price Index Treasury bills

5. NPW( MARR ) > 0Good Investment

6. EUAW( MARR ) > 0Good Investment

7. NPW( MARR ) > 0Good Investment EUAW( MARR ) > 0Good Investment IRR > MARRGood Investment

8. Example: Suppose you buy and sell a piece of equipment. Purchase Price $16,000 Sell Price (5 years) $ 4,000 Annual Maintenance $ 3,000 Net Profit Contribution $ 6,000 MARR 12% Is it worth it to the company to buy the machine?

9. NPW= -16 + 3(P/A,12,5) + 4(P/F,12,5) 16,000 6,000 3,000 5 0 4,000 16,000 3,000 5 0 4,000

10. NPW= -16 + 3(P/A,12,5) + 4(P/F,12,5) = -16 +3(3.6048) + 4(.5674) 16,000 6,000 3,000 5 0 4,000 16,000 3,000 5 0 4,000

11. NPW= -16 + 3(P/A,12,5) + 4(P/F,12,5) = -16 +3(3.6048) + 4(.5674) = -2.916 = -$2,916 16,000 6,000 3,000 5 0 4,000 16,000 3,000 5 0 4,000

12. Annual Worth (AW or EUAW) AW(i) = PW(i) (A/P, i%, n) = [ A t (P/F, i%, t)](A/P, i%, n) AW(i) = Annual Worth of Investment AW(i) > 0 **OK Investment** 

13. Repeating our PW example, we have AW(12)= -16(A/P,12,5) + 3 + 4(A/F,12,5) 3,000 5 0 4,000 16,000

14. Repeating our PW example, we have AW(12)= -16(A/P,12,5) + 3 + 4(A/F,12,5) = -16(.2774) + 3 + 4(.1574) 3,000 5 0 4,000 16,000

15. Repeating our PW example, we have AW(12)= -16(A/P,12,5) + 3 + 4(A/F,12,5) = -16(.2774) + 3 + 4(.1574) = -.808 = -$808 3,000 5 0 4,000 16,000

16. AW(12) = PW(12) (A/P, 12%, 5) = -2.92 (.2774) = - $810 < 0 NO GOOD 3,000 5 0 4,000 16,000

17. Internal Rate-of-Return IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment**

18. Internal Rate-of-Return IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = A t (1 + i*) n - t 

19. Internal Rate-of-Return IRR - internal rate of return is that return for which NPW(i*) = 0 i* = IRR i* > MARR **OK Investment** Alt: FW(i*) = 0 = A t (1 + i*) n - t PW revenue (i*) = PW costs (i*) 

20. Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5) 3,000 5 0 4,000 16,000

21. Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5) 3,000 5 0 4,000 16,000

22. Example PW(i) = -16 + 3(P/A, i, 5) + 4(P/F, i, 5) i* = 5 1 / 4 % i* < MARR 3,000 5 0 4,000 16,000

25. 216% 16 yrs

26. 12316 100 216 F = P(F/P,i *,16) (F/P,i *,16) = F/P = 2.16 (1+i * ) 16 = 2.16

27. 12316 100 216 (1+i * ) 16 = 2.16 16 ln(1+i * ) = ln(2.16) =.7701

28. 12316 100 216 (1+i * ) 16 = 2.16 16 ln(1+i * ) = ln(2.16) =.7701 ln(1+i * ) =.0481

29. 12316 100 216 (1+i * ) 16 = 2.16 16 ln(1+i * ) = ln(2.16) =.7701 ln(1+i * ) =.0481 (1+i * ) = e.0481 = 1.0493

30. 12316 100 216 (1+i * ) 16 = 2.16 16 ln(1+i * ) = ln(2.16) =.7701 ln(1+i * ) =.0481 (1+i * ) = e.0481 = 1.0493 i* =.0493 = 4.93%

31. 12316 100 216 We know i = 4.93%, is that significant growth?

32. 12316 100 216 We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period.

33. 12316 100 216 We know i = 4.93%, is that significant growth? Suppose inflation = 3.5% over that same period. d ij j      1 04930350 1... d= 1.4%

34. NPW > 0 Good Investment

35. EUAW > 0 Good Investment

36. NPW > 0 Good Investment EUAW > 0 Good Investment IRR > MARR Good Investment

37. NPW > 0 Good Investment EUAW > 0 Good Investment IRR > MARR Good Investment Note: If NPW > 0 EUAW > 0 IRR > MARR

38. 1,000 4,100 5,580 2,520 n 0123 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR).

39. 1,000 4,100 5,580 2,520 n 0123 Consider the following cash flow diagram. We wish to find the Internal Rate-of-Return (IRR). PW R (i * ) = PW C (i * ) 4,100(1+i * ) -1 + 2,520(1+i * ) -3 = 1,000 + 5,580(1+i * ) -2

40. NPV vs. Interest ($5) $0 $5 $10 $15 $20 $25 0%10%20%30%40%50%60% Interest Rate Net Present Value

41. Purpose: to get around a problem of multiple roots in IRR method Notation: A t = net cash flow of investment in period t A t, A t > 0 0, else -A t, A t < 0 0, else r t = reinvestment rate (+) cash flows (MARR) i’ = rate return (-) cash flows   R t = C t =

42. Method find i = ERR such that R t (1 + r t ) n - t = C t (1 + i ’ ) n - t Evaluation If i ’ = ERR > MARR Investment is Good 

43. Example MARR = 15% R t (1 +.15) n - t = C t (1 + i ’ ) n - t 4,100(1.15) 2 + 2,520 = 1,000(1 + i ’ ) 3 + 5,580(1 + i ’ ) 1 i ’ =.1505  0 1 2 3 1,000 4,100 5,580 2,520

44. Example MARR = 15% R t (1 +.15) n - t = C t (1 + i ’ ) n - t 4,100(1.15) 2 + 2,520 = 1,000(1 + i ’ ) 3 + 5,580(1 + i ’ ) 1 i ’ =.1505 ERR > MARR  0 1 2 3 1,000 4,100 5,580 2,520

45. Example MARR = 15% R t (1 +.15) n - t = C t (1 + i ’ ) n - t 4,100(1.15) 2 + 2,520 = 1,000(1 + i ’ ) 3 + 5,580(1 + i ’ ) 1 i ’ =.1505 ERR > MARR Good Investment  0 1 2 3 1,000 4,100 5,580 2,520

46. Method 1 Let i = MARR SIR(i) = R t (1 + i) -t C t (1 + i) -t = PW (positive flows) - PW (negative flows)  

47. Relationships among MARR, IRR, and ERR If IRR < MARR, then IRR < ERR < MARR If IRR > MARR, then IRR > ERR > MARR If IRR = MARR, then IRR = ERR = MARR

48. Method #2 SIR(i) = A t (1 + i) -t C t (1 + i) -t SIR(i) = PW (all cash flows) PW (negative flows)  

49. Method #2 SIR(i) = A t (1 + i) -t C t (1 + i) -t SIR(i) = PW (all cash flows) PW (negative flows) Evaluation: Method 1: If SIR(t) > 1 Good Investment Method 2: If SIR(t) > 0 Good Investment  

50. Example SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16 = 3(3.6048) + 4(.5674) 16 =.818 < 1.0 16 0 123 4 5 3 3 33 7

51. Example SIR(t) = 3(P/A, 12%, 5) + 4(P/F, 12%, 5) 16 = 3(3.6048) + 4(.5674) 16 =.818 < 1.0 16 0 123 4 5 3 3 33 7

52. Method Find smallest value m such that where C o = initial investment m = payback period investment m t = 1 R t C o  

53. Example m= 5 years n c o 1 16 3 3 16 9 4 16 12 5 16 19 2 16 6 16 0 123 4 5 3 3 33 7 R t n 0 

54. Perpetuity (Capitalized Cost) Occasionally, donors sponsor perpetual awards or programs by a lump sum of money earning interest.Occasionally, donors sponsor perpetual awards or programs by a lump sum of money earning interest. The interest earned each period (A) equals the funds necessary to pay for the ongoing award or program.The interest earned each period (A) equals the funds necessary to pay for the ongoing award or program. The relationship is A = P( i ) This concept is also called capitalized cost (where CC = P).This concept is also called capitalized cost (where CC = P).

55. Perpetuity Example A donor has decided to establish a $10,000 per year scholarship. The first scholarship will be paid 5 years from today and will continue at the same time every year forever. The fund for the scholarship will be established in 8 equal payments every 6 months starting 6 months from now. Determine the amount of each of the equal initiating payments, if funds can earn interest at the rate of 6% per year with semi- annual compounding.

56. Perpetuity Problem Given: A = 10 000 per year, every year after Year 5 n = 8 payments @ 6 mo. intervals, starting @ 6 mo. i = 6%, cpd semi-annually Find Amount of Initiating payments (A i ):

57. Perpetuity Problem Given: A = 10 000 per year, every year after Year 5 n = 8 payments @ 6 mo. intervals, starting @ 6 mo. i = 6%, cpd semi-annually Find Amount of Initiating payments (A i ):

58. A flood control project has a construction cost of $10 million, an annual maintenance cost of $100,000. If the MARR is 8%, determine the capitalized cost necessary to provide for construction and perpetual upkeep.

59. 1 2 3 4... 100 100 10,000 P c = 10,000 + A/i = 10,000 + 100/.08 = 11,250 Capitalized Cost = $11.25 million

60. Suppose that the flood control project has major repairs of $1 million scheduled every 5 years. We now wish to re-compute the capitalized cost.

61. Compute an annuity for the 1,000 every 5 years: 1 2 3 4 5... 100 100 100 100 100 10,000 1,000

62. Compute an annuity for the 1,000 every 5 years: 1 2 3 4 5... 100 100 100 100 100 10,000 1,000 A = 100 + 1,000(A/F,8,5) = 100 + 1,000(.1705) = 270.5

63. 1 2 3 4 5... 170.5 170.5 10,000 P c = 10,000 + 270.5/.08 = 13,381 1 2 3 4 5... 100 100 100 100 100 10,000 1,000

64. How does Bill Gates change a light bulb?

65. How does Bill Gates change a light bulb? He doesn’t, he declares darkness a new industry standard!!!

66. How many Industrial Engineers does it take to change a light bulb?

67. How many Industrial Engineers does it take to change a light bulb? None, IE’s only change dark bulbs!!!!!