1. ENGM 661 Engr. Economics for Managers Decisions Under Uncertainty

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 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

5. Motivation While we note that we now know the distribution of NPW, we have not made a decision. 1 2 3 4 5 A A A A A 10,000 MARR = 15% NPW p p p          -329616 23 16,/ 56/ 3409,/

6. Motivation There are lots of decision rules: 1. Maximax/Minimin 2. Maximin/Minimax 3. Weighted Max 4. Minimax Regret 5. Dominance 6. Expectation/Variance 7. Most Probable 8. Aspiration-Level 1 2 3 4 5 A A A A A 10,000 MARR = 15% NPW p p p          -329616 23 16,/ 56/ 3409,/

7. Motivation For now, let us use the expectation rule which states to use that decision which maximizes (minimizes) expected NPW (EUAW). 1 2 3 4 5 A A A A A 10,000 MARR = 15% NPW p p p          -329616 23 16,/ 56/ 3409,/

8. Motivation E[NPW] = -3,296(1/6) + 56(2/3) + 3,409(1/6) = $ 56.2 E[NPW]> 0 INVEST 1 2 3 4 5 A A A A A 10,000 MARR = 15% NPW p p p          -329616 23 16,/ 56/ 3409,/

9. Decision Tree We can model this same problem with as a decision tree. Assume our decision alternatives are to invest or not to invest. Then Invest Do not invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0

10. Decision Tree We can model this same problem with as a decision tree. Assume our decision alternatives are to invest or not to invest. Then Invest Do not invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0 Note: Sequential decisions can be modeled with Decision Trees

11. Notation A1A1 A2A2 (  11 ; p 11 ) (  12 ; p 12 ) (  21 ; p 21 ) A 11 A 12 D1D1 D2D2 (  111 ; p 111 ) (  112 ; p 112 ) (  121 ; p 121 ) (  122 ; p 122 ) V 111 V 112 V 121 V 122 V 12 V 21

12. Notation A1A1 A2A2 (  11 ; p 11 ) (  12 ; p 12 ) (  21 ; p 21 ) A 11 A 12 D1D1 D2D2 (  111 ; p 111 ) (  112 ; p 112 ) (  121 ; p 121 ) (  122 ; p 122 ) V 111 V 112 V 121 V 122 V 12 V 21 = a decision point = fork in tree where chance events can influence outcomes D i = ith decision A ik = kth alternative available given decision 1 was alternative i (  ; p) = outcome , having associated value v(  ; p), which occurs with probability p V ikj = value associated with branch ikj

13. Maximize Epected Gain A1A1 A2A2 (  11 ; p 11 ) (  12 ; p 12 ) (  21 ; p 21 ) A 11 A 12 D1D1 D2D2 (  111 ; p 111 ) (  112 ; p 112 ) (  121 ; p 121 ) (  122 ; p 122 ) V 111 V 112 V 121 V 122 V 12 V 21 1. At decision point D 2, calculate expected gains for A 11 and A 12 E[A 11 ] = p 111 V 111 + p 112 V 112 E[A 12 ] = p 121 V 121 + p 122 V 122 If E[A 11 ] > E[A 12 ] Choose A 11

14. Maximize Epected Gain A1A1 A2A2 (  11 ; p 11 ) (  12 ; p 12 ) (  21 ; p 21 ) D1D1 V 12 V 21 2. Replace D 2 with E[A 11 ] E[A 11 ]

15. Maximize Epected Gain A1A1 A2A2 (  11 ; p 11 ) (  12 ; p 12 ) (  21 ; p 21 ) D1D1 V 12 V 21 3. Calculate expected gains for alternatives A 1 and A 2 E[A 1 ] = p 11 E[A 11 ] + p 12 V 12 E[A 2 ] = p 21 V 21 = V 21 If E[A 1 ] > E[A 2 ] Choose A 1 E[A 11 ]

16. Maximize Epected Gain A1A1 A2A2 (  11 ; p 11 ) (  12 ; p 12 ) (  21 ; p 21 ) A 11 A 12 D1D1 D2D2 (  111 ; p 111 ) (  112 ; p 112 ) (  121 ; p 121 ) (  122 ; p 122 ) V 111 V 112 V 121 V 122 V 12 V 21 Summary: Choose Alternative A 1 at D 1.

17. Maximize Epected Gain A1A1 A2A2 (  11 ; p 11 ) (  12 ; p 12 ) (  21 ; p 21 ) A 11 A 12 D1D1 D2D2 (  111 ; p 111 ) (  112 ; p 112 ) (  121 ; p 121 ) (  122 ; p 122 ) V 111 V 112 V 121 V 122 V 12 V 21 Summary: Choose Alternative A 1 at D 1. Given A 1, either outcome  11 or  12 will occur.

18. Maximize Epected Gain A1A1 A2A2 (  11 ; p 11 ) (  12 ; p 12 ) (  21 ; p 21 ) A 11 A 12 D1D1 D2D2 (  111 ; p 111 ) (  112 ; p 112 ) (  121 ; p 121 ) (  122 ; p 122 ) V 111 V 112 V 121 V 122 V 12 V 21 Summary: Choose Alternative A 1 at D 1. Given A 1, either outcome  11 or  12 will occur. If  11 occurs, choose alternative A 11 at D 2.

19. Maximize Epected Gain A1A1 A2A2 (  11 ; p 11 ) (  12 ; p 12 ) (  21 ; p 21 ) A 11 A 12 D1D1 D2D2 (  111 ; p 111 ) (  112 ; p 112 ) (  121 ; p 121 ) (  122 ; p 122 ) V 111 V 112 V 121 V 122 V 12 V 21 Summary: Choose Alternative A 1 at D 1. Given A 1, either outcome  11 or  12 will occur. If  11 occurs, choose alternative A 11 at D 2. Given A 11, either outcome  111 or  112 will occur resulting in values gained of V 111 or V 112.

20. Problem Revisited 1 2 3 4 5 A A A A A 10,000 A p p p          200016 3 23 4 16,/,/,/

21. Problem Revisited 1 2 3 4 5 A A A A A 10,000 A p p p          200016 3 23 4 16,/,/,/ Invest Do not invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0

22. Problem Revisited Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0

23. Problem Revisited Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0 E[Invest] = -3,296(1/6) + 56(2/3) + 3,409(1/6) = $ 56.2 56.2

24. Problem Revisited E[Invest] = -3,296(1/6) + 56(2/3) + 3,409(1/6) = $ 56.2 E[No Invest]= 0 Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0 56.2 0

25. Problem Revisited Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0 E[Invest] > E[No Invest] Choose to Invest 56.2 0

26. K-Corp Re-engineer New Design do nothing No Competition (.8) Competition (.2) Competition (.5) No Comp. (.5) New do no No Comp. (.3) Comp. (.7) 125 -90 40 60 80 20 -50

27. K-Corp Re-engineer New Design do nothing No Competition (.8) Competition (.2) Competition (.5) No Comp. (.5) New do no No Comp. (.3) Comp. (.7) 125 -90 40 60 80 20 -50 (38) (60)

28. K-Corp Re-engineer New Design do nothing No Competition (.8) Competition (.2) Competition (.5) No Comp. (.5) New do no No Comp. (.3) Comp. (.7) 125 -90 40 60 80 20 -50 (38) (60) [60]

29. K-Corp Re-engineer New Design do nothing No Competition (.8) Competition (.2) Competition (.5) No Comp. (.5) New do no No Comp. (.3) Comp. (.7) 125 -90 40 60 80 20 -50 (38) (60) [60] (82) (50) (-50)

30. K-Corp Re-engineer New Design do nothing No Competition (.8) Competition (.2) Competition (.5) No Comp. (.5) New do no No Comp. (.3) Comp. (.7) 125 -90 40 60 80 20 -50 (38) (60) [60] (82) (50) (-50) [82]

31. K-Corp Re-engineer New Design do nothing No Competition (.8) Competition (.2) Competition (.5) No Comp. (.5) New do no No Comp. (.3) Comp. (.7) 125 -90 40 60 80 20 -50 (38) (60) [60] (82) (50) (-50) [82]

32. K-Corp Re-engineer New Design do nothing No Competition (.8) Competition (.2) Competition (.5) No Comp. (.5) New do no No Comp. (.3) Comp. (.7) 125 -90 40 60 80 20 -50 (38) (60) [60] (82) (50) (-50) [82]

33. Class Problem Perico is considering a capacity expansion project based on a 3-point sales estimate: Annual DemandProbability 200,000 0.3 150,000 0.5 75,000 0.2 Perico has 3 alternatives for expansion. The net return on each is dependent on annual sales demand.

34. Class Problem Total Return after investment (in millions) over 5 years: Annual Demand Alternative200,000150,00075,000 Build new plant $ 3.75 $ 2.50($ 2.00) Expand existing 1.50 1.00 0.75 Do nothing 0.00 0.00 0.00

35. Class Problem Each alternative also has an associated capital investment requirement: AlternativeCapital Invest Build New $1.25 mil. Expand Existing $0.75 mil. Do Nothing $0.00

36. Class Problem Set up the decision tree for Perico Inc. Use the resulting tree to determine if Perico should build a new plant, expand the existing facilities, or do nothing.

37. Class Problem

38. Decision Tree A B C Build New Expand Do Nothing 200,000 (.3) 150,000 (.5) 75,000 (.2) 200,000 (.3) 150,000 (.5) 75,000 (.2) 0 Total Revenue $ 3.75 $ 2.50 ($2.00) $ 1.50 $ 1.00 $ 0.75 $ 0.00 (1.25) (0.75) (0.0)

39. Decision Tree A B C E[R] = 1.125 E[R] = 1.250 E[R] = (0.40) E[R] = 0.450 E[R] = 0.500 E[R] = 0.150 E[R] = 0.000 Total Revenue $ 3.75 $ 2.50 ($2.00) $ 1.50 $ 1.00 $ 0.75 $ 0.00 [1.975] [1.10] [0.0] (1.25) (0.75) (0.0)

40. Decision Tree A B C E[R] = 1.125 E[R] = 1.250 E[R] = (0.40) E[R] = 0.450 E[R] = 0.500 E[R] = 0.150 E[R] = 0.000 Total Revenue $ 3.75 $ 2.50 ($2.00) $ 1.50 $ 1.00 $ 0.75 $ 0.00 [1.975] [1.10] [0.0] Net =.725 Net = 0.35 Net = 0.0

41. Decision Tree E[R] = 1.125 Net =.725 A B C E[R] = 1.250 E[R] = (0.40) E[R] = 0.450 E[R] = 0.500 E[R] = 0.150 E[R] = 0.000 Total Revenue $ 3.75 $ 2.50 ($2.00) $ 1.50 $ 1.00 $ 0.75 $ 0.00 [1.975] [1.10] [0.0] Net = 0.35 Net = 0.0

42. Value of Perfect Information Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0 56.2 0 Recall our simple cash flow example: 1 2 3 4 5 A A A A A 10,000

43. Value of Perfect Information Although we expect to make money (at 15%), it is entirely possible we will lose $3,296 if sales are low. Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0 56.2 0

44. Value of Perfect Information Although we expect to make money (at 15%), it is entirely possible we will lose $3,296 if sales are low. Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = -3,296 NPW = 0 56.2 0 Idea: If I knew exactly whether sales will be high, medium, or low, I can know precisely whether or not it is worthwhile for me to invest. [56.2]

45. Value of Perfect Information Under certainty, we would not invest if sales are low. We can then modify our decision tree as follows: Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = 0 605.5 0

46. Value of Perfect Information That is, under certainty, we will know whether to invest or not. Under these conditions, the expected NPW is now 605.5 Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = 0 605.5 0

47. Value of Perfect Information Then the expected value of perfect information is: EVPI = $605.5 - $ 56.2 = $549.3 Invest No invest High (1/6) Medium (2/3) Low (1/6) NPW = 3,409 NPW = 56 NPW = 0 605.5 0

48. Bayes’ Decision Rule E[A 1 ] > E[A 2 ] choose A 1

49. Decision Tree Develop Sell Market (0.4) No Market (.6) 400 -100 90 100 90

50. Posterior Probabilities u Suppose we can do a market survey which would cost us $50,000. Would it be worth it?

51. Posterior Probabilities u Suppose we can do a market survey which would cost us $50,000. Would it be worth it? Develop Sell Market (0.4) No Market (.6) 400 -100 90 100 90

52. Posterior Probabilities Develop Sell Market (0.4) No Market (.6) 400 -100 90 100 90 E[payoff perfect info] =.4(400) +.6(90) = 214 E[payoff w/o survey] = 100

53. Posterior Probabilities Develop Sell Market (0.4) No Market (.6) 400 -100 90 100 90 E[payoff perfect info] =.4(400) +.6(90) = 214 E[payoff w/o survey] = 100 EVPI = 214 - 100 = 114

54. Expected Value Survey u If we spend $50,000 to conduct survey, we have one of 4 possibilities u Survey says good when there is a market u Survey says bad when there is no market u Survey says good when there is no market u Survey says bad when there is a market

55. Expected Value Survey u Survey says good when there is a market u Survey says bad when there is no market u Survey says good when there is no market u Survey says bad when there is a market P(good survey | market) = 0.8 P(bad survey | no market) = 0.7

56. Expected Value Survey u Survey says good when there is a market u Survey says bad when there is no market u Survey says good when there is no market u Survey says bad when there is a market P(good survey | market) = 0.8 P(bad survey | no market) = 0.7 P(bad survey | market) = 0.2 P(good survey | no market) = 0.3

57. Expected Value Survey u Want P(market | good survey) = ? P(no market | bad survey) = ?

58. Expected Value Survey No Market Market 0.6 0.4 Survey says 0.3 market Survey says no 0.7 market Survey 0.8 says market Survey 0.2 says no market

59. Expected Value Survey No Market Market 0.6 0.4 Survey says 0.3 market Survey says no 0.7 market Survey 0.8 says market Survey 0.2 says no market Want P(Market | Survey says Good) = ?

60. Expected Value Survey Want P(Market | Survey says Good) = ? P(NM) = 0.6 P(M) = 0.4

61. Expected Value Survey Want P(Market | Survey says Good) = ? P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5

62. Expected Value Survey Want P(Market | Survey says Good) = ? P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5 P(BS) = 0.5 0.64

63. Expected Value Survey Want P(Market | Survey says Good) = ? P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5 P(BS) = 0.5 0.64 0.36

64. Expected Value Survey Want P(Market | Survey says Good) = ? P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5 P(BS) = 0.5 0.16

65. Expected Value Survey Want P(Market | Survey says Good) = ? P(NM) = 0.6 P(M) = 0.4 P(GS) = 0.5 P(BS) = 0.5 0.16 0.84

66. Decision Tree Do Survey No Survey Favorable Unfavorable develop market none market none sell market none sell

67. Decision Tree Do Survey No Survey 0.5 develop 0.64 0.36 0.16 0.84 sell 0.4 0.6 sell 350 -150 90 350 -150 90 400 -100 90

68. Bayes’ Decision Rule E[A 1 ] > E[A 3 ] > E[A 2 ] choose A 1

69. Expectation E[A 1 ] > E[A 3 ] > E[A 2 ] choose A 1

70. Expectation E[A 1 ] > E[A 2 ] > E[A 3 ] choose A 1

71. Expectation E[A 1 ] > E[A 3 ] > E[A 2 ] choose A 1

72. Expectation E[A 1 ] > E[A 3 ] > E[A 2 ] choose A 1

73. Expectation E[A 1 ] > E[A 3 ] > E[A 2 ] choose A 1

74. Expectation E[A 1 ] > E[A 3 ] > E[A 2 ] choose A 1

75. Laplace Principle If one can not assign probabilities, assume each state equally probable. Max E[P Ai ] choose A 1

76. Expectation-Variance If E[A 1 ] = E[A 2 ] = E[A 3 ] choose A j with min. variance

77. Bayes’ Decision Rule E[A 1 ] > E[A 2 ] choose A 1

78. Sensitivity Suppose probability of market (p) is somewhere between 30 and 50 percent.

79. Sensitivity Suppose probability of market (p) is somewhere between 30 and 50 percent.

80. Sensitivity Suppose probability of market (p) is somewhere between 30 and 50 percent.

81. Sensitivity

82. Sensitivity Plot 0 50 100 150 200 00.10.30.40.5 Prob. of Market Expected Value Develop Sell

83. Sensitivity Sensitivity Plot 0 50 100 150 200 00.10.30.40.5 Prob. of Market Expected Value Develop Sell

84. Aspiration-Level Aspiration: max probability that payoff > 60,000 P{P A1 > 60,000} = 0.8 P{P A2 > 60,000} = 0.3 P{P A3 > 60,000} = 0.3 Choose A 2 or A 3

85. Maximin Select A j : max j min k V(  jk ) e.g., Find the min payoff for each alternative. Find the maximum of minimums Sell Land Choose best alternative when comparing worst possible outcomes for each alternative.