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ENGM 661 Engineering Economics for Managers

ENGM 661 Engineering Economics for Managers. Risk Analysis. Learning Objectives. 1)Given a simple cash flow stream with uncertainty, be able to estimate the worth of a project using a worst case, best case, or expected value evaluation.

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ENGM 661 Engineering Economics for Managers

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  1. ENGM 661Engineering Economics for Managers Risk Analysis

  2. Learning Objectives 1)Given a simple cash flow stream with uncertainty, be able to estimate the worth of a project using a worst case, best case, or expected value evaluation. 2) Given a simple probabilistic (discrete case) cash flow stream, be able to compute the distribution of the Net Present Worth (or Equivalent Uniform Annual Worth). 3) Given an analytic cumulative distribution function and a uniform random variable, U(0,1), be able to use the inverse method to compute a cash flow At. 4) Given a set of random variables (0,1), a cash flow stream, and an inverse formula, be able compute 1 realization of the Net Present Worth (or EUAW) using simulation techniques.

  3. Solution Methodologies • Bounding • Analytic • C.L.T. (Assume Normality) • Simulation

  4. A1 A2 A3 3 , 000 p  1 / 4   Ai  4 , 000 p  1 / 2  1 2 3  5 , 000 p  1 / 4  10,000 Bounding Lower Bound

  5. A1 A2 A3 3 , 000 p  1 / 4   Ai  4 , 000 p  1 / 2  1 2 3  5 , 000 p  1 / 4  10,000 Bounding Upper Bound

  6. A1 A2 A3 3 , 000 p  1 / 4   Ai  4 , 000 p  1 / 2  1 2 3  5 , 000 p  1 / 4  10,000 Bounding Upper & Lower Bounds

  7. 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

  8. 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

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

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

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

  12. 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

  13. 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

  14. 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

  15. 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)

  16.  2   2 (  10 , 000  a A ) NPW t t Variance Recall: 2(z) = 2(x) + 2(y) 2(ax+b) = a22

  17.  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

  18. 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]

  19. 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.

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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)

  25. 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)

  26. 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)

  27. 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

  28. 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?

  29. 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 

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

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

  32. 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

  33. 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%.

  34. 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.

  35. 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)

  36. 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.

  37. 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.

  38. 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

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

  40. 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)

  41. 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.

  42. 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

  43. 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.

  44. Simulation What we now need is a formal method of selecting random values for A1 and A2 to avoid selection bias.

  45. 1/2,000 5,000 7,000 Simulation What we now need is a formal method of selecting random values for A1 and A2 to avoid selection bias. Recall the uniform f(x)

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

  47. 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) 

  48. 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) 

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

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

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