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Dealing with Uncertainty

Dealing with Uncertainty. Probabilistic Risk Analysis. Introduction. Statistical and probability concepts can also be used to analyze the economic consequences of some decision situations involving risk and uncertainty.

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Dealing with Uncertainty

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  1. Dealing with Uncertainty Probabilistic Risk Analysis

  2. Introduction • Statistical and probability concepts can also be used to analyze the economic consequences of some decision situations involving risk and uncertainty. • The probability that a cost, revenue, useful life, NPW, etc. will occur is usually considered to be the long-run relative frequency with which the value occurs.

  3. Random Variables • Factors having probabilistic outcomes are called random variables. • Useful information about a random variable is • expected value (average, mean), denoted by E[X] • variance, denoted by Var[X] • standard deviation, denoted by SD[X] • When uncertainty is considered, the • variability in the economic measures of merit and • the probability of loss associated with the alternative are very useful in the decision-making process.

  4. Some Important Relationships • Discrete Random Variables Probability: Pr{X=xi} = p(xi) for i = 1, ..., L where p(xi)>0 and i p(xi)=1 • Continuous Random Variables Probability:where • Expected Value: E[X] = i xi p(xi) or • Variance: Var[X] = i (xi - E[X]) 2p(xi) or

  5. Some Important Relationships (cont’d) • Variance: Var[X] = E[X2] - (E[X])2 • Standard Deviation: SD[X] = (Var[X]) 1/2 • Expected value of a sum: E[X+Y] = E[X] + E[Y] • Variance of a sum or differenceVar[X+Y] = Var[X-Y] = Var[X] + Var[Y] when X and Y are independent • Multiply by a constant: E[cX] = cE[X] and Var[cX] = c2 Var[X] • Expected Value of a function: E[h(X)] = i h(xi) p(xi) or

  6. Evaluation of Projects with Random Outcomes • We can use the expected value and variance concepts to assess the project’s worth • We might be interested in • the expected net present worth, E[NPW], or expected net annual worth, E[NAW] • the variance or standard deviation of the traditional measures, Var[NPW], Var[NAW], SD[NPW], SD[NAW] • the probability that the NPW or NAW is positive, i.e., Prob{NPW > 0} or Prob{NAW>0}

  7. Example 7 • A HVAC system has become unreliable and inefficient. Rental income is being hurt and O&M continue to increase. You decide to rebuild it. Assume MARR = 12%

  8. Example 7 (cont’d) • For year 12, NPW = -$521,000 + ($48,600+$31,000) (P/A, 12%, 12) = -$27,926 However, this useful life only has a 0.1 chance of occurring. • For year 13, NPW = -$521,000 + ($48,600+$31,000) (P/A, 12%, 13) = -$9,689 However, this useful life only has a 0.2 chance of occurring.

  9. Example 7 (cont’d) • What is E[NPW] and Var[NPW] ?

  10. Example 7 (cont’d) • E[NPW] = $9,984 • E[(NPW)2] = ($2) 577.524 x 106 • Var[NPW] = E[(NPW)2] - (E[NPW])2 = ($2)477.847 x 106 • SD[NPW] = (Var[NPW])1/2 = $21,859 • Probability{NPW > 0} = 1- (0.1+0.2) = 0.7 • The weakest indicator is SD(NPW) > 2E[NPW] !

  11. Example 8 • For the following cash flow estimates, find E[NPW], Var[NPW], and SD[NPW]. Determine Prob{ ROR < MARR}. Assume that the annual net cash flows are normally distributed and independent. Use a MARR = 15%.

  12. Example 8 (cont’d) • The investment is known. Year 0

  13. Example 8 (cont’d) • The cash flows for the years 1, 2 and 3 are not known.

  14. Example 8 (cont’d) • E[NPW] = -$7,000 + $3,500 (P/F,15%,1) + $3,000 (P/F,15%,2) + $2,800 (P/F,15%,3) = $153 • Var[NPW] = 02 + ($600)2 (P/F,15%,1)2 + ($500)2 (P/F,15%,2)2 +($400 )2 (P/F,15%,3)2 = ($2 )484,324 • SD[NPW] = $696

  15. Example 8 (cont’d) • Prob{ ROR <= MARR} = ? • Step 1: For a project having a unique ROR (simple investments are such projects), the probability that the ROR is less than the MARR is the same as the probability that the NPW is less than 0. So Prob{ ROR <= MARR} = Prob{ NPW <= 0} • Step 2: Because the NPW is normally distributed, we can normalize to a N(0,1) distribution. So Z = (NPW - E[NPW])/SD(NPW) = (0-153)/696 = -0.22 • Step 3: Using Normal Tables, we get Prob{NPW <=0} = Prob{Z <= -0.22} = 0.4129 • Therefore Prob{ ROR <= MARR} = 0.4129

  16. Decision Trees • Also called decision flow networks and decision diagrams • Powerful means of depicting and facilitating the analysis of problems involving sequential decisions and variable outcomes over time • Make it possible to break down large, complicated problems into a series of smaller problems

  17. Diagramming • Square symbol depicts a decision node • Circle symbol depicts a chance outcome node • All initial or immediate alternatives among which the decision maker wishes to choose • All uncertain outcomes and future alternatives that may directly affect the consequences • All uncertain outcomes that may provide information

  18. Diagramming Example Sales good bad Invest in new product line Decision Status Quo

  19. Example 9 • A new design is being evaluated as potential replacement for a heavily used machine. The new design involves major changes that have expected advantage, but would be $8600 more expensive. In return, annual expense savings are expected, but their extent depend on the machine’s reliability. • Use MARR = 18%. Life = 6 years. Salvage = 0.

  20. Example 9 (cont’d) A = $3,470 NPW = $3,538 25% A = $2,920 NPW = $1,614 40% 25% A = $2,310 NPW = -$520 New Design 10% A = $1,560 NPW = -$3,143 Current Design

  21. Example 9 (cont’d) • Based on a before-tax analysis (MARR = 18%, analysis period = 6 years, salvage value = 0), is the new design economically preferable to the current unit? • E[NPW] = - $8600 + 0.25 ($3470) (P/A,18%,6) + 0.4 ($2920) (P/A,18%,6) + 0.25 ($2310) (P/A,18%,6) + 0.10 ($1560)(P/A,18%,6) = $1086

  22. Example 9 (cont’d) A = $3,470 NPW = $3,538 25% $1,086 A = $2,920 NPW = $1,614 40% 25% A = $2,310 NPW = -$520 New Design 10% A = $1,560 NPW = -$3,143 Current Design $0

  23. Example 9 (cont’d) • Optimal decision based on perfect information • Expected Value of Perfect Information = $1530 - $1086 = $444

  24. Example 9 (cont’d) • Management is confident that data from an additional comprehensive test will show whether future operational performance will be favorable (excellent or good reliability) or not favorable (standard or poor reliability). The design team develop conditional probability estimates.

  25. Example 9 (cont’d) • We need to determine the joint probabilities of the design goal being met at a particular level and a certain test outcome occurring. For example, p(E, F) = p(F|E) p(E) = (0.95)(0.25) = 0.2375 p(E,NF) = p(NF|E) p(E) = (0.05)(0.25) = 0.0125

  26. Example 9 (cont’d) • The revised probabilities of each outcome are obtained from the joint probabilities and the marginal probabilities For example, when favorable p(E) = p(E,F)/p(F) = 0.2375/0.6575 = 0.3612 When not favorable p(E) = p(E,NF)/p(NF) = 0.0125/0.3425 = 0.0365

  27. Example 9 (cont’d) $1086 E: 0.3612 $3538 No test G: 0.5171 $1614 New Design S: 0.1141 -$520 P: 0.0076 -$3143 Do test Current Design Favorable E: 0.0365 $3538 G: 0.1752 $1614 New Design S: 0.5109 $-520 Unfavorable P: 0.2774 -$3143 Current Design

  28. Example 9 (cont’d) $1086 No test $2029 $2029 $2029 New Design $0 Do test Current Design Favorable -$726 New Design $0 Unfavorable Current Design $0

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