1. ENGM 661 Engineering Economics for Managers Risk Analysis

2. Motivation Suppose we have the following cash flow diagram. NPW = -10,000 + A(P/A, 15, 5) 1 2 3 4 5 A A A A A 10,000 MARR = 15%

3. Motivation Now suppose that the annual return A is a random variable governed by the discrete distribution: A p p p          200016 3 23 4 16,/,/,/

4. Motivation A p p p          200016 3 23 4 16,/,/,/ For A = 2,000, we have NPW = -10,000 + 2,000(P/A, 15, 5) = -3,296

5. Motivation A p p p          200016 3 23 4 16,/,/,/ For A = 3,000, we have NPW = -10,000 + 3,000(P/A, 15, 5) = 56

6. Motivation A p p p          200016 3 23 4 16,/,/,/ For A = 4,000, we have NPW = -10,000 + 4,000(P/A, 15, 5) = 3,409

7. Motivation There is a one-for-one mapping for each value of A, a random variable, to each value of NPW, also a random variable. A 2,000 3,000 4,000 p(A) 1/6 2/3 1/6 NPW -3,296 56 3,409 p( NPW ) 1/6 2/3 1/6

8. Spreadsheet

9. Risk Analysis 1 2 3 4 5 A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% Now Suppose the return in each year is a random variable governed by the some probability distribution. NPW = -10,000 + A 1 (1+i) -1 + A 2 (1+i) -2 +... + A 5 (1+i) -5

10. Risk Analysis    10000 1 5,aA t t t   1()whereai t t 1 2 3 4 5 A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% NPW = -10,000 + A 1 (1+i) -1 + A 2 (1+i) -2 +... + A 5 (1+i) -5

11. Risk Analysis Now suppose A t iid N(3,000, 250) NPW is a linear combination of normals 

12. Risk Analysis Now suppose A t iid N(3,000, 250) NPW is a linear combination of normals NPW Normal   Central Limit

13. Risk Analysis Now suppose A t iid N(3,000, 250) NPW is a linear combination of normals NPW Normal NPW N(  NPW,  NPW )   

14. Mean  NPWt t t E EaA    [][,10000 1 5 Recall: E[Z] = E[X 1 ] + E[X 2 ] E[aX+b] = aE[X] + b

15. Mean  NPWt t t E EaA    [][,10000 1 5 Recall: E[Z] = E[X 1 ] + E[X 2 ] E[aX+b] = aE[X] + b    10000 1 5,[]aEA t t t  NPW

16. Mean    10000 1 5,[]aEA t t t  NPW but, E[A t ] = 3,000    10000 1 5,[]a3,000 t t  NPW

17. Mean    10000 1 5,[]a3,000 t t  NPW = -10,000 + 3,000 a t t   1 5 (1+i) -t t   1 5 = -10,000 + 3,000(P/A, i, 5)

18. Variance  22 10000 NPW tt aA   (,) Recall:  2 (z) =  2 (x) +  2 (y)  2 (ax+b) = a 2  2

19. Variance  22 10000 NPW tt aA   (,) Recall:  2 (z) =  2 (x) +  2 (y)  2 (ax+b) = a 2  2  2 2 2 NPW t A a t  

20. Variance but,  22 250()A t   2 2 2 NPW t a   (A t )  22 2 250 NPW t a   () = (250) 2 [(1+i) -2 + (1+i) -4 +... + (1+i) -10 ]

21. Variance  22 2 250 NPW t a   () = (250) 2 [(1+i) -2 + (1+i) -4 +... + (1+i) -10 ] Note that [(1+i) -2 + (1+i) -4 +... + (1+i) -10 ] is just a 5 period annuity factor where the period is 2 years. A=(250) 2 1 2 3 4 5 6 7 8 9 10 P=  2 NPW

22. Variance  22 2 250 NPW t a   () = (250) 2 [(1+i) -2 + (1+i) -4 +... + (1+i) -10 ] A=(250) 2 1 2 3 4 5 6 7 8 9 10 P=  2 NPW = (250) 2 (P/A, i eff, 5), i eff = (1+i) 2 -1

23. Risk Analysis 1 2 3 4 5 A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15%  NPW = -10,000 + 3,000(P/A, 15, 5) = -10,000 + 3,000(3.3522) = $56 A t iid N(3,000, 250) 

24. Risk Analysis 1 2 3 4 5 A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% NPW = -10,000 + 3,000(P/A, 15, 5) = -10,000 + 3,000(3.3522) = $56 A t iid N(3,000, 250)  i eff = (1.15) 2 - 1 = 32.25%

25. Risk Analysis 1 2 3 4 5 A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% NPW = -10,000 + 3,000(P/A, 15, 5) = -10,000 + 3,000(3.3522) = $56  2 NPW = (250) 2 (P/A, 32.25, 5) = 62,500(2.3343) = 145,894 A t iid N(3,000, 250)  i eff = (1.15) 2 - 1 = 32.25%

26. Risk Analysis 1 2 3 4 5 A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% NPW = $56  2 NPW = 145,894  = 382 A t iid N(3,000, 250)  NPW  382)

27. Risk Analysis 1 2 3 4 5 A 1 A 2 A 3 A 4 A 5 10,000 MARR = 15% A t iid N(3,000, 250)  NPW  382)   N(56, 382) 

28. Risk Analysis 1 2 3 4 5 A 1 A 2 A 3 A 4 A 5 10,000 NPW  382)   N(56, 382)  PNPWP ()     0 056 382  

29. Risk Analysis 1 2 3 4 5 A 1 A 2 A 3 A 4 A 5 10,000 NPW  382)   N(56, 382)  PNPWP ()     0 056 382   = P( Z < -0.15 ) = 0.44

30. Review

31. You are given the following cash flow diagram: Class Problem  5,000,0.6 7,,0.4    If A i iid A 1 A 2 10,000 Compute the distribution for the Net Present Worth if the MARR = 15%.

32. Class Problem

33. You are given the following cash flow diagram: Class Problem If the MARR = 15%, what is the probability this investment alternative is no good? A 1 A 2 A 3 A 4 A 5 35,000  If A i iid N(10,000, 300)

34. Class Problem E[NPW] = -35,000 + 10,000(P/A, 15, 5) = -35,000 + 10,000(3.3522) = - 1,478  2 (NPW)  300 2 [(1.15)  2  (1. )  4 ...  (1. )  10 ] = 300 2 (2.3343) = 210,087 

35. Class Problem NPW N(-1,478, 458) P{NPW < 0} =  P NPW    0  (  1,478) 458       = P{Z < 3.27}  1.0

36. A Critical Thunk A 1 A 2 A 3 A 4 A 5 35,000 If A i iid  N(10,000, 300) Max A i  10,900

37. A Critical Thunk A 1 A 2 A 3 A 4 A 5 35,000 If Max A i  10,900 NPW = -35,000 + 10,900(P/A, 15, 5) = -35,000 + 10,900(3.3522) = 1,539

38. If A i iid U( 5000, 7000 ) A Twist Suppose we have the following cash flow diagram. A 1 A 2 10,000  Now how can we compute the distribution of the NPW? MARR = 15%.

39. Solution Alternatives u Assume normality

40. Solution Alternatives u Assume normality u Upper/Lower Bounds u Laplace Transforms u Transformation/Convolution u Simulation

41. Simulation A 1 A 2 10,000 Let us arbitrarily pick a value for A 1 and A 2 in the uniform range (5000, 7000). Say A 1 = 5,740 and A 2 = 6,500.

42. Simulation A 1 A 2 10,000 5,740 6,500 10,000 NPW = -10,000 + 5,740(1.15) -1 + 6,500(1.15) -2 = (93.96)

43. Simulation A 1 A 2 10,000 5,740 6,500 10,000 We now have one realization of NPW for a given realization of A 1 and A 2.

44. Simulation A 1 A 2 10,000 5,740 6,500 10,000 We now have one realization of NPW for a given realization of A 1 and A 2. Choose 2 new values for A 1, A 2.

45. A 1 = 6,820 A 2 = 6,218 A 1 A 2 10,000 6,820 6,218 10,000 NPW = -10,000 + 6,820(1.15) -1 + 6,218(1.15) -2 = 632.14

46. Summary A 1 A 2 NPW 5,7406,500(93.96) 6,8206,218632.14 Choose 2 new values.

47. A 1 = 5,273 A 2 = 6,422 A 1 A 2 10,000 5,273 6,422 10,000 NPW = -10,000 + 5,273(1.15) -1 + 6,422(1.15) -2 = (558.83)

48. Summary A 1 A 2 NPW 5,7406,500(93.96) 6,8206,218632.14 5,2736,422(558.83) Choose 2 new values.

49. Summary A 1 A 2 NPW 5,7406,500(93.96) 6,8206,218632.14 5,2736,422(558.83). 6,8555,947457.66

50. Simulation With enough realizations of A1, A2, and NPW, we can begin to get a sense of the distribution of the NPW. -1,871 0 1,380 NPW Freq.

51. Simulation What we now need is a formal method of selecting random values for A 1 and A 2 to avoid selection bias.

52. Simulation What we now need is a formal method of selecting random values for A 1 and A 2 to avoid selection bias. Recall the uniform 5,000 7,000 1/2,000 f(x)

53. Simulation The uniform has cumulative distribution given by: 5,000 7,000 1 F(x) F(x)  0,x  5,000 1,x  7, x  5, 2,     

54. Simulation Recall that 0 < F(x) < 1. Using modulo arithmetic, it is fairly easy to generate random numbers between 0 and 1 (Rand Function in Excel). Let P be a random variable uniformly from 0 to 1. P U(0,1) 

55. Simulation Recall that 0 < F(x) < 1. Using modulo arithmetic, it is fairly easy to generate random numbers between 0 and 1. Let P be a random variable uniformly from 0 to 1. P U(0,1) Algorithm: 1. Randomly generate P 2. Let P = F(x) 3. Solve for x = F -1 (p) 

56. Simulation 1. Randomly generate P U(0,1). P =.7 5,000 7,000 1 F(x)  F(x)  x  5,000 2, 5,000 < x < 7,000

57. Simulation 1. Randomly generate P U(0,1). P =.7 2. Let P = F(x). 5,000 7,000 1 F(x) F(x)  x  5,000 2,  5,000 < x < 7,000.7

58. Simulation 1. Randomly generate P U(0,1). P =.7 2. Let P = F(x). 3. x = F -1 (p). 5,000 7,000 1 F(x) F(x)  x  5,000 2,  5,000 < x < 7,000.7 6,400

59. Formal Derivation Recall, for 5,000 7,000 1 F(x) F(x)  x  5,000 2, 5,000 < x < 7,000. Then P P x    5000 7 5,,,   x5 2,,

60. Formal Derivation Solving for x = F -1 (p), 5,000 7,000 1 F(x) P x xP  50002,,

61. Formal Derivation Solving for x = F -1 (p), 5,000 7,000 1 F(x) P x xP  50002,, Note: 1. P = 0 x = 5,000 2. P = 1 x = 7,000

62. Risk Analysis (Excel)

63. Class Problem You are given the following cash flow diagram. The A i are iid shifted exponentials with location parameter a = 1,000 and scale parameter = 3,000. The cumulative is then given by 7,000 A 1 A 2 A 3 Fxe x () (,)/,   1 10003, x > 1,000

64. Class Problem You are given the first 3 random numbers U(0,1) as follows: P 1 = 0.8 P 2 = 0.3 P 3 = 0.5 You are to compute one realization for the NPW. MARR = 15%. 7,000 A 1 A 2 A 3 Fxe x () (,)/,   1 10003

65. Class Problem P  1  e  x  1000 3000       e  x  1000 3000        1  P  x  1000 3000      ln(1  P) x  1,000  3, ln(1  P)

66. Class Problem x  1,000  3, ln(1  P) A 1 = 1,000 - 3000 ln(1 -.8) = 5,828 A 2 = 1,000 - 3000 ln(1 -.3) = 2,070 A 3 = 1,000 - 3000 ln(1 -.5) = 3,079

67. Class Problem 7,000 5,828 2,070 3,079 NPW = -7,000 + 5,828(1.15) -1 + 2,070(1.15) -2 + 3,079(1.15) -3 = 1,657

68. Class Problem You are given the following cash flow diagram. The A i are iid gammas with shape parameter  = 4 and scale parameter  = 3,000. The density function is given by 7,000 A 1 A 2 A 3 fxxe x () () /         1, x > 0

69. Class Problem You are given the first 3 random numbers U(0,1) as follows: P 1 = 0.8 P 2 = 0.3 P 3 = 0.5 You are to compute one realization for the NPW. MARR = 15%. 7,000 A 1 A 2 A 3

70. Class Problem For  = integer, the cumulative distribution function is given by Set P = F(x), solve for x

71. Class Problem For general  (not integer), F(x) = not analytic

72. Class Problem For general  (not integer), F(x) = not analytic No Inverse

73. @RISK