TM 661 Engineering Economics for Managers

1. TM 661Engineering Economics for Managers Risk Analysis

2. A1 A2 A3 3 , 000 p  1 / 4   Ai  4 , 000 p  1 / 2  1 2 3  5 , 000 p  1 / 4  10,000 Class Problem Suppose we have the following cash flow diagram (MARR = 15%). Determine if the project is worthwhile.

3. Solution Methodologies • Bounding • C.L.T. (Assume Normality) • Analytic • 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 3 , 000 p  1 / 4   Ai  4 , 000 p  1 / 2  1 2 3  5 , 000 p  1 / 4  10,000 Central Limit Theorem Preliminary

8. A1 A2 A3 3 , 000 p  1 / 4   Ai  4 , 000 p  1 / 2  1 2 3  5 , 000 p  1 / 4  10,000 Central Limit Theorem Distribution of NPW

9. A1 A2 A3 3 , 000 p  1 / 4   - = 867 E [ NPW ] Ai  4 , 000 p  1 / 2  1 2 3  5 , 000 p  1 / 4  10,000 Central Limit Theorem Distribution of NPW

10. A1 A2 A3 3 , 000 p  1 / 4  N(-867, 938)  - = 867 E [ NPW ] Ai  4 , 000 p  1 / 2  1 2 3  5 , 000 p  1 / 4  10,000 -3,681 -867 1,947 Central Limit Theorem Distribution of NPW

11. A1 A2 A3 N(-867, 938) 1 2 3 10,000 -3,681 -867 1,947 Central Limit Theorem Distribution of NPW

13. P{NPW >0} = .031 .063 .031 .016 .031 .016 .188

14. Analytic P{NPW > 0} = 0.188 C.L.T. P{NPW > 0} = 0.178

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

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

17. Simulation

18. Simulation

19. Simulation P{NPW > 0} = 5/20 = 0.25

20. Simulation Analytic P{NPW > 0} = 0.188 C.L.T. P{NPW > 0} = 0.178 Simulation P{NPW > 0} = 0.25

21. @Risk

22. @Risk Analytic P{NPW > 0} = 0.188 C.L.T. P{NPW > 0} = 0.178 Simulation P{NPW > 0} = 0.25 @Risk P{NPW > 0} = 0.20

23. A1 A2 A3 7,000  ( x  1 , 000 ) / 3 , 000 F ( x )  1  e Class Problem You are given the following cash flow diagram. The Ai are iid shifted exponentials with location parameter a = 1,000 and scale parameter  = 3,000. The cumulative is then given by , x > 1,000

24. A1 A2 A3 7,000  ( x  1 , 000 ) / 3 , 000 F ( x )  1  e Class Problem You are given the first 3 random numbers U(0,1) as follows: P1 = 0.8 P2 = 0.3 P3 = 0.5 You are to compute one realization for the NPW. MARR = 15%.

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

26. x  1 , 000  3 , 000 ln( 1  P ) Class Problem A1 = 1,000 - 3000 ln(1 - .8) = 5,828 A2 = 1,000 - 3000 ln(1 - .3) = 2,070 A3 = 1,000 - 3000 ln(1 - .5) = 3,079

27. 5,828 3,079 2,070 7,000 Class Problem NPW = -7,000 + 5,828(1.15)-1 + 2,070(1.15)-2 + 3,079(1.15)-3 = 1,657

28. A1 A2 A3 7,000    f ( x )  x   1 e  x /   (  ) Class Problem You are given the following cash flow diagram. The Ai are iid gammas with shape parameter  = 4 and scale parameter  = 3,000. The density function is given by , x > 0

29. A1 A2 A3 7,000 Class Problem You are given the first 3 random numbers U(0,1) as follows: P1 = 0.8 P2 = 0.3 P3 = 0.5 You are to compute one realization for the NPW. MARR = 15%.

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

31. Class Problem For general a (not integer), F(x) = not analytic

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