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TM 661 Engineering Economics for Managers

TM 661 Engineering Economics for Managers. Risk Analysis. A A A A A. MARR = 15%. 1 2 3 4 5. 10,000. Motivation. Suppose we have the following cash flow diagram. NPW = -10,000 + A(P/A, 15, 5). 2. ,. 000. p. . 1. /. 6. .

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TM 661 Engineering Economics for Managers

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  1. TM 661Engineering Economics for Managers Risk Analysis

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

  3. 2 , 000 p  1 / 6   A  3 , 000 p  2 / 3   4 , 000 p  1 / 6  Motivation Now suppose that the annual return A is a random variable governed by the discrete distribution:

  4. 2 , 000 p  1 / 6   A  3 , 000 p  2 / 3   4 , 000 p  1 / 6  Motivation For A = 2,000, we have NPW = -10,000 + 2,000(P/A, 15, 5) = -3,296

  5. 2 , 000 p  1 / 6   A  3 , 000 p  2 / 3   4 , 000 p  1 / 6  Motivation For A = 3,000, we have NPW = -10,000 + 3,000(P/A, 15, 5) = 56

  6. 2 , 000 p  1 / 6   A  3 , 000 p  2 / 3   4 , 000 p  1 / 6  Motivation 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. A1 A2 A3 A4 A5 MARR = 15% 1 2 3 4 5 10,000 Risk Analysis Now Suppose the return in each year is a random variable governed by the some probability distribution. NPW = -10,000 + A1(1+i)-1 + A2(1+i)-2 + . . . + A5(1+i)-5

  10. A1 A2 A3 A4 A5 MARR = 15% 1 2 3 4 5 10,000 5    10 , 000  a A t t t  1 where a  ( 1  i )  t t Risk Analysis NPW = -10,000 + A1(1+i)-1 + A2(1+i)-2 + . . . + A5(1+i)-5

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

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

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

  14. 5    E [ NPW ]  E [  10 , 000  a A NPW t t t  1 Mean Recall: E[Z] = E[X1] + E[X2] E[aX+b] = aE[X] + b

  15. 5    E [ NPW ]  E [  10 , 000  a A NPW t t t  1 5     10 , 000  a E [ A ] NPW t t t  1 Mean Recall: E[Z] = E[X1] + E[X2] E[aX+b] = aE[X] + b

  16. 5     10 , 000  a E [ A ] NPW t t t  1 5     10 , 000  a [ 3,000 ] NPW t t  1 Mean but, E[At] = 3,000

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

  18.  2   2 (  10 , 000  a A ) NPW t t Variance Recall: 2(z) = 2(x) + 2(y) 2(ax+b) = a22

  19.  2   2 (  10 , 000  a A ) NPW t t 2   2  a  2 NPW A t t Variance Recall: 2(z) = 2(x) + 2(y) 2(ax+b) = a22

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

  21. 2   2  ( 250 ) 2 a A=(250)2 NPW t 1 2 3 4 5 6 7 8 9 10 P=s2NPW Variance = (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.

  22. 2   2  ( 250 ) 2 a A=(250)2 NPW t 1 2 3 4 5 6 7 8 9 10 P=s2NPW Variance = (250)2 [(1+i)-2 + (1+i)-4 + . . . + (1+i)-10] = (250)2(P/A, ieff, 5) , ieff = (1+i)2 -1

  23. A1 A2 A3 A4 A5 1 2 3 4 5   At iid N(3,000, 250) 10,000 Risk Analysis MARR = 15% mNPW = -10,000 + 3,000(P/A, 15, 5) = -10,000 + 3,000(3.3522) = $56

  24. A1 A2 A3 A4 A5 1 2 3 4 5   At iid N(3,000, 250) 10,000 Risk Analysis MARR = 15% ieff = (1.15)2 - 1 = 32.25% NPW = -10,000 + 3,000(P/A, 15, 5) = -10,000 + 3,000(3.3522) = $56

  25. A1 A2 A3 A4 A5 1 2 3 4 5   At iid N(3,000, 250) 10,000 Risk Analysis MARR = 15% ieff = (1.15)2 - 1 = 32.25% NPW = -10,000 + 3,000(P/A, 15, 5) = -10,000 + 3,000(3.3522) = $56 s2NPW = (250)2(P/A, 32.25, 5) = 62,500(2.3343) = 145,894

  26. A1 A2 A3 A4 A5 1 2 3 4 5   At iid N(3,000, 250) 10,000 Risk Analysis MARR = 15% NPW = $56 s2NPW = 145,894 s = 382 NPW  N(56, 382)

  27. A1 A2 A3 A4 A5 N(56, 382) 1 2 3 4 5   At iid N(3,000, 250) 10,000 -1,090 56 1,202 Risk Analysis MARR = 15% NPW  N(56, 382)

  28. A1 A2 A3 A4 A5 N(56, 382) 1 2 3 4 5 10,000 -1,090 56 1,202 - m - NPW 0 56 < = < NPW P ( NPW 0 ) P s 382 NPW Risk Analysis NPW  N(56, 382)

  29. A1 A2 A3 A4 A5 N(56, 382) 1 2 3 4 5 10,000 -1,090 56 1,202 - m - NPW 0 56 < = < NPW P ( NPW 0 ) P s 382 NPW Risk Analysis NPW  N(56, 382) = P( Z < -0.15 ) = 0.44

  30. Review

  31. A1 A2  5 , 000 , 0 . 6 If Ai iid    7 , 000 , 0 . 4  10,000 Class Problem You are given the following cash flow diagram: Compute the distribution for the Net Present Worth if the MARR = 15%.

  32. Class Problem

  33. A1 A2 A3 A4 A5 35,000 If Ai iid N(10,000, 300)   Class Problem You are given the following cash flow diagram: If the MARR = 15%, what is the probability this investment alternative is no good?

  34. 2 2  2  4  10  ( NPW )  300 [( 1 . 15 )  ( 1 . 15 )  . . .  ( 1 . 15 ) ] Class Problem E[NPW] = -35,000 + 10,000(P/A, 15, 5) = -35,000 + 10,000(3.3522) = - 1,478 = 3002 (2.3343) = 210,087 

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

  36. A1 A2 A3 A4 A5 35,000   If Ai iid N(10,000, 300) A Critical Thunk Max Ai 10,900

  37. A1 A2 A3 A4 A5 35,000 A Critical Thunk If Max Ai 10,900 NPW = -35,000 + 10,900(P/A, 15, 5) = -35,000 + 10,900(3.3522) = 1,539

  38. A1 A2 10,000 A Twist Suppose we have the following cash flow diagram. If Ai iid U(5000, 7000)   Now how can we compute the distribution of the NPW? MARR = 15%.

  39. Solution Alternatives • Assume normality

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

  41. A1 A2 10,000 Simulation Let us arbitrarily pick a value for A1 and A2 in the uniform range (5000, 7000). Say A1 = 5,740 and A2 = 6,500.

  42. A1 A2 5,740 6,500 10,000 10,000 Simulation NPW = -10,000 + 5,740(1.15)-1 + 6,500(1.15)-2 = (93.96)

  43. A1 A2 5,740 6,500 10,000 10,000 Simulation We now have one realization of NPW for a given realization of A1 and A2.

  44. A1 A2 5,740 6,500 10,000 10,000 Simulation We now have one realization of NPW for a given realization of A1 and A2. Choose 2 new values for A1, A2.

  45. A1 A2 6,820 6,218 10,000 10,000 A1 = 6,820 A2 = 6,218 NPW = -10,000 + 6,820(1.15)-1 + 6,218(1.15)-2 = 632.14

  46. Summary A1 A2 NPW 5,740 6,500 (93.96) 6,820 6,218 632.14 Choose 2 new values.

  47. A1 A2 5,273 6,422 10,000 10,000 A1 = 5,273 A2 = 6,422 NPW = -10,000 + 5,273(1.15)-1 + 6,422(1.15)-2 = (558.83)

  48. Summary A1 A2 NPW 5,740 6,500 (93.96) 6,820 6,218 632.14 5,273 6,422 (558.83) Choose 2 new values.

  49. Summary A1 A2 NPW 5,740 6,500 (93.96) 6,820 6,218 632.14 5,273 6,422 (558.83) . . . 6,855 5,947 457.66

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

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