IE546Spring24_hw5_PracticeOnly_Solutions

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Industrial Engineering

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May 2, 2024

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IE546 – Spring 2024 Homework 5 Practice Only Solutions [Q1] (8.10) First, let us adopt some notation. Let NW and NL denote “Napoleon wins” and “Napoleon loses,” respectively. Also, let “P&E” denote that the Prussians and English have joined forces. The best way to handle this problem is to express Bayes’ theorem in odds form. Show that P(NW | P&E) P(NL | P&E) = P(P&E | NW) P(P&E | NL) P(NW) P(NL) . We have that P(NW) = 0.90 and P(NW | P&E) = 0.60, and so P(NL) = 0.10 and P(NL | P&E) = 0.40. Thus, we can substitute: 0.60 0.40 = P(P&E | NW) P(P&E | NL) 0.90 0.10 or P(P&E | NW) P(P&E | NL) = 1 6 . Thus, Napoleon would have had to judge the probability of the Prussian and English joining forces as six time more likely if he is to lose than if he is to win. [Q2] a. For each project, the investor appears to believe that E(profit) = 0.5 (150,000) + 0.5 (-100,000) = 75,000 - 50,000 = 25,000 b. However, since only one of the projects will succeed, he will gain $150,000 for the successful project, but lose $100,000 for each of the other two. Thus, he is guaranteed to lose $50,000 no matter what happens. c. For a set of mutually exclusive and collectively exhaustive outcomes, he appears to have assessed probabilities that add up to 1.5. d. “Knowing nothing” does not necessarily imply a probability of 0.5. In this case, “knowing nothing” is really not appropriate, because the investor does know something: Only one project will succeed. If, on top of that, he wants to invoke equally likely outcomes, then he should use P(Success) = 1/3 for each project. Note that it is also possible to work parts a and b in terms of final wealth. Assume that he starts with $300,000. For part a, expected final wealth, given that he invests in one project, would be: E(wealth) = 0.5 (450,000) + 0.5 (200,000) = 325,000. Because $325,000 is an improvement on the initial wealth, the project looks good. For part b, though, if he starts with $300,000 and invests in all three projects, he will end up with only $250,000 for the project that succeeds. As before, he is guaranteed to lose $50,000.
[Q3] (8.12 and 8.15) (a) The assessments will be based on personal judgments and will vary among students. Using the estimation of Dow Jones Industrial Average as an example, suppose the following assessments are made: P(DJIA £ 2000) = 0.05 P(DJIA > 3000) = 0.05 P(DJIA £ 2600) = 0.50 P(DJIA £ 2350) = 0.25 P(DJIA £ 2800) = 0.75 (b) This question requires students to make personal judgments. As an example, suppose the following assessments are made: P(S £ 65) = 0.05 P(S > 99) = 0.05 P(S £ 78) = 0.25 P(S £ 85) = 0.50 P(S £ 96) = 0.75 [Q4] (11.10) a. A straightforward sensitivity analysis would calculate expected values with the probabilities set at various values. The results would be a triangular grid that could be plotted in terms of p 1 = P($10,000) and p 2 = P($5000): 1.00 0.75 0.50 0.25 x P(Score £ x) 60 70 80 90 100 1.00 0.75 0.50 0.25 x P(Score £ x) 60 70 80 90 100
To use simulation to incorporate the uncertainty about the probabilities, it is possible to generate the probabilities randomly. For example, first, generate p 1 from a uniform distribution between 0 and 0.5. Second, generate p 2 from a uniform distribution between (0.5 - p 1 ) and 0.5, and finally calculate p 3 = 1 - p 1 - p 2 . The expected value of the payoffs if each event is equally likely is $5,333. For 100 iterations, the expected payoff with the uncertainty to probability include for one sample run was $5,854. b. No, it would not be possible to have each of the three probabilities chosen from a uniform distribution between zero and one, because the three probabilities would never sum to one. Some kind of building-up process, like that described in part a, must be followed. [Q5] (12.7) This decision model is saved in the Excel file “Problem 12.7.xls”. The decision tree for the decision whether to drill or not is shown in the first worksheet. The decision tree for parts a and c : . The expected value of drilling is $10 K, versus $0 for not drilling, so choose to drill. b. The influence diagram representation is shown in the second worksheet. With the arc between the uncertainty node “Strike oil” and the decision node “Drill?” the influence diagram evaluates the expected value of the decision assuming perfect information. To see the expected value without information, delete the arc. The EVPI is the difference between these two EV's, or $19,000 - $10,000 = $9,000. 0.5 0.5 p p 1 2 Drill Don‘t drill Consult clairvoyant Strike oil (0.1) Dry hole (0.9) Dry hole (0.9) Strike oil (0.1) Drill Drill Don‘t drill Don‘t drill $190 K -$10 K 0 $190 K 0 -$10 K 0 EMV = $19 K EMV = $10 K
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