IE546Spring24_hw5_PracticeOnly_Solutions
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Purdue University *
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546
Subject
Industrial Engineering
Date
May 2, 2024
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6
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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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c.
See the decision tree above or the decision tree model saved in the third worksheet. EVPI = EMV(Clairvoyant) - EMV(Drill) = $19 K - $10 K = $9 K. d.
We have: P(“good” | oil) = 0.95 P(oil) = 0.1 P(“poor” | dry) = 0.85 P(dry) = 0.9 We can find P(“good”) and P(“poor”) with the law of total probability: P(“good”) = P(“good” | oil) P(oil) + P(“good” | dry) P(dry) = 0.95 (0.1) + 0.15 (0.9) = 0.23 P(“poor”) = 1 - P(“good”) = 1 - 0.23 = 0.77. Now we can find P(oil | “good”) = P(“good” | oil) P(oil)
P(“good” | oil) P(oil) + P(“good” | dry) P(dry)
= 0.95 (0.1)
0.95 (0.1) + 0.15 (0.9)
= 0.41 P(dry | “good”) = 1 - P(oil | “good”) = 0.59 P(oil | “poor”) = P(“poor” | oil) P(oil)
P(“poor” | oil) P(oil) + P(“poor” | dry) P(dry)
= 0.05 (0.1)
0.05 (0.1) + 0.85 (0.9)
= 0.0065 Payoff
Strike oil?
Drill?
Payoff
Strike oil?
Drill?
Basic model:
Perfect information:
Payoff
Strike oil?
Drill?
Payoff
Strike oil?
Drill?
Basic model:
Perfect information:
P(dry | “poor”) = 0.9935. Now the decision tree is: The influence diagram solution: EVII = EMV(Consult Geologist) - EMV(Drill) = $16.56 K - $10 K = $6.56 K. Because EVII is less than $7000, which the geologist would charge, this is a case where the expected value of the geologist’s information is less than what it would cost. Don’t consult her. The corresponding influence diagram is shown in the fourth worksheet and the decision tree is in the fifth worksheet. Note, the values in the spreadsheet have slightly different values due to round-off error. Bayesian probability calculations are very sensitive to the significant digits carried through the calculations. Drill EMV = $10 K
Don‘t Drill
—
good“ (0.23)
—
poor“ (0.77)
Consult geologist EMV = $16.56 K
Strike oil (0.1)
Dry hole (0.9)
$190 K
-$10 K
$0 K
Drill Don‘t Drill
Strike oil (0.41)
Dry hole (0.59)
$190 K
-$10 K
$0 K
Drill Don‘t Drill
Strike oil (0.0065)
Dry hole (0.9935)
$190 K
-$10 K
$0 K
Drill?
Consult Geologist
Oil?
Payoff
Drill?
Oil?
Payoff
Without Geologist:
With Geologist:
Drill?
Consult Geologist
Oil?
Payoff
Drill?
Oil?
Payoff
Without Geologist:
With Geologist:
[Q6] (12.6) a.
Of course, different people will have different feelings on this one. Personally, I would prefer that the doctor wait to inform me until after the other tests have been performed. (This may not be possible if the further tests require additional blood samples or other interventions; I would certainly know that something was going on.) Why wait? I would worry about the outcome of the other tests. I would just as soon not know that they were even being performed. b. Suppose I know of no such defects. In this state of information, I can legitimately give my house a clean bill of health, given this state of knowledge. Now, suppose that I learn from the engineering report that the house has a defect and also that my buyer withdraws from the agreement to purchase. Now I would have to reveal the defect to any subsequent purchaser, and it would most likely result in a lower negotiated price. Under the circumstances, even though future buyers may also request an inspection that reveals the defect, I would rather not know the engineer’s report; this state of knowledge gives me a better chance at a better sales price.
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