ESI 6385 -- Decision Analysis
Mid Term Exam
Due Date: October 24 2022 @ 1:29 PM (FIRM Cutoff for Upload to Webcourses )
Instructions: This is an open book, open notes take home exam. There are 5 problems. You are
expected to do your own work, and must not discuss the problems or collaborate with other
students. Use this sheet as the cover sheet for the examination and sign the following statement:
“I certify that the work I have submitted is my own work and that I have not collaborated
with any other students on this examination.”
Name: -----------------------------------
Signed: -----------------------------------
Please type or write legibly and show all your work! If I can’t read it or see
your calculations, I can’t give you any points.
Problem 1. It is said that Napoleon assessed probabilities at the battle of waterloo in 1815. His
hopes for victory depended on keeping the English and Prussian armies separated. Believing that
they had not joined forces on the morning of the faithful battle, he indicated his belief that he had a
90% chance of defeating the English; P (Napoleon Wins = 0.90). When told later that the elements
of the Prussian force had joined the English, Napoleon revised his opinion downward on the basis
of this information, but his posterior probability was still at 60%; P(Napoleon Wins │ Prussian and
English join Forces) = 0.60.
Suppose Napoleon was using Bayes’ theorem to revise this information. To do so, he would have
to make some judgment about P(Prussian and English join Forces │ Napoleon Wins) and
P(Prussian and English join Forces │Napoleon Loses). In particular, he would have had to make
judgment about the ratio of these two probabilities. What is that ratio? (15 Points)
Solution:
Let suppose,
Event Napoleon wins is W
Event Prussian and English forces join = J
Therefore,
P (W) = 0.90
P (W|J) = 0.6
P (H) =P (W)*P (J|W) + P (Wc)*(J|Wc)
P (H) =0.9*P(J|W) + (1-0.9)*P(J|Wc )
Also as P(W|J) = P(W)*P(J|W)/P(J)
0.6 =0.9*P(J|W)/(0.9*P(J|W) + 0.1*P(J|Wc ))
(0.9*P(J|W) + 0.1*P(J|Wc )) = 0.9*P(J|W)/0.6
0.9*P(J|W) + 0.1*P(J|Wc ) = 1.5P(J|W)
0.1*P(J|Wc ) = 0.6*P(J|W)
P(J|W)/P(J|Wc ) =0.1/0.6 =1/6
Therefore, The ration of P (Prussian and English join Forces │ Napoleon Wins)
and P (Prussian and English join Forces │Napoleon Loses) =1/6
Problem 2.
a) Solve the decision tree below: (5 Points)
A1
8
5.08
0.27
7
A2
A
0.73
5.08
4.5
$8
(0.5)
$0
(0.5) $15
0.45
$4
$10
0.55
$0
B
Solution:
A2 = (0.5*0) + (0.5*15) = 7.5
Node 2 = Max (8, 7.5) = 8
EMV (A) = (0.27*8) + (0.73*4) = 5.08
EMV (B) = (0.45*10) + (0.55*0) = 4.5
Node 1 = Max (5.08, 4.5) = 5.08
b) Is one strategy stochastically dominant? Explain. (5 points)
Solution:
If one alternative dominates another, then the dominating alternative must have higher expected
value. The event A is stochastically dominant. As we can see from the decision tree, the values
of A1 = 8 and A2 = 7.5 are greater than the value of B (4.5,0). Hence it can be said that A is
stochastically dominant since A1 and A2 are dominating B. The alternative A dominates B, as
we can see that EMV (A) is greater than EMV (B)
c) Draw the influence diagram that corresponds to the decision tree. (5 points)
Solution:
Decision Tree
A1
8
5.08
0.27
7
A2
A
$8
(0.5)
(0.5) $15
0.73
5.08
4.5
$0
0.45
$4
$10
0.55
$0
B
Solution:
Problem 3. A friend can invest in a multiyear project. The cost is $15,000. Annual cash flows are
estimated to be $5000 per year for 6 years, but could vary between $2500 and $6000. Your friend
estimates that the cost of capital (interest rate) is 10%, but it could be as low as 9% and as high as
11%. The basis of the decision to invest will be whether the project has a positive net present value.
Construct a tornado diagram for this problem using an appropriate software and advise your friend
whether to invest or not. (20 points)
Solution:
Given,
The amount of initial investment is $15,000
Estimated annual revenue is $5000 per year for 6 years
The cost of the capital is 10%
Net Present Value (NPV) = -P + A (P/A,i,n)
⇒ -15,000 + 5,000 [{(1+i)n – 1}/{i(1+i)n
}]
⇒ -15,000 + 5,000 [{(1+0.10)^6 – 1}/ {0.10(1+0.10)^6
]
⇒ -15,000 + 5,000 (0.771561/0.1771561)
⇒ -15,000 + 5,000 (4.355)
⇒ -15,000 + 21,776.3035
⇒ $6,776.303
Since the value of NPV is positive, the friend should invest.
Now calculating the Net Present Values at different interest rates and at minimum & maximum
conditions,
Present Value PV can be calculated as, PV = [Cash Flow/ (1+r)^n]
Using the above formula, at r = 9% and minimum return of $2500
Year
0
PV
-15000
1
2
3
4
5
6
Net present value
4545.45
4132.23
3756.57
3415.06
3104.60
2822.36
-3785.25
Using the above formula, at r = 9% and maximum return of $6000
Year
0
1
2
3
4
5
6
Net present value
PV
-15000
5404.90
4869.73
4387.14
3952.38
3560.70
3207.84
10383.19
Using the above three NPVs, we have the following tornado diagram:
NPV at Different Interest Rate
NPV at minimum interest rate 9%
NPV at interest rate 10%
NPV at maximum interest rate 11%
-6000
-4000
-2000
0
2000
4000
6000
8000
10000
12000
Problem 4. Johnson Marketing is interested in producing and selling an innovative new food
processor. The decision they face is the typical “make or buy” decision often faced by
manufacturers. On one hand, Johnson could produce the processor itself, subcontracting
different sub-assemblies, such as the motor or the housing. Cost estimates in this case are as
follows:
Alternative: Make Food Processor
Cost Per Unit ($)
35.00
42.50
45.00
49.00
Chance (%)
25
25
37
13
The company also could have the entire machine made by a subcontractor. The
subcontractor, however, faces similar uncertainties regarding the costs and has provided
Johnson Marketing with the following schedule of costs and chances:
Alternative: Buy Food Processor
Cost per Unit ($)
Chance (%)
37.00
43.00
46.00
50.00
10
40
30
20
a) If Johnson Marketing wants to minimize its expected cost of production in this case,
should it make or buy? (10 points)
Solution:
Alternative: Make Food Processor
E(M) = (35*0.25)+(42.50*0.25)+(45*0.37)+(49*0.13)
E(M) = 8.75+10.625+16.65+6.37
E(M) = 42.395
Alternative: Buy Food Processor
E(M) = (37*0.1)+(43+0.4)+(46*0.3)+(50*0.20)
E(M) = 3.7+17.2+13.8+10
E(M) = 44.7
As we can see estimated cost of making food processor is 42.395 which is less than the estimated
cost to buy food processor is 44.7. So we can conclude that make food processor is good than
the buy food processor
b) Construct cumulative risk profiles to support your recommendation. (10 points)
Make:
Cost
35
42.5
45
49
Probability
0.25
0.25
0.37
0.13
Cumulative probability
0.25
0.5
0.87
1
Cummulativ probability
1,2
1
0,8
0,6
0,4
0,2
0
0
10
20
30
Buy
Cost
37
Probability
0.1
Cumulative probability
0.1
40
50
60
43
46
50
0.5
0.3
0.2
0.5
0.8
1
Cummulativ probability
1,2
1
0,8
0,6
0,4
0,2
0
0
10
20
30
40
50
60
c) Does one alternative deterministically dominate the other? Why or why not? (5 points)
Solution:
They are NOT alternative deterministically dominate because they do not meet the norms of the
cumulative risk profile.
d) Does one alternative stochastically dominate the other? Why or why not? (5 points)
Solution:
If one alternative dominates another, then the dominating alternative must have higher expected
value. As we can see from the graph, the buy and make graphs are crossing each other. Hence,
they are not stochastically dominant because all the values in one event are not higher than the
values in other event and the graphs are crossing each other
Problem 5. An orange grower in Florida faces a dilemma. The weather forecast is for cold
weather, and there is a 50% chance that the temperature tonight will be cold enough to freeze and
destroy his entire crop, which is worth some $50,000. He can take two possible actions to try to
alleviate his loss if the temperature drops. First, he could set burners in the orchard; this would
cost $5,000, but he could still expect to incur damage of approximately $15,000 to $20,000.
Second, he could set up sprinklers to spray the trees. If the temperature drops, the water would
freeze on the fruit and provide some insulation. This method is cheaper ($2,000), but less
effective. With the sprinklers he could expect to incur as much as $25,000 to $30,000 of the loss
with no protective action.
Compare the grower’s expected values for the three alternatives he has, considering the various
possible loss scenarios for the burner and the sprinklers. Which alternative would you suggest
the grower take? Why? (20 Points)
Solution:
There is a decision tree in the diagram. There are three nodes to it
Node 1 - Weather Normal,
Expected Value = 250000
Node 2 - Weather Cold
Pr1 = 0.5, Use of Burner
Pr2 = 0.5,
Incur damage of approximately = (15000+20000)/2= 17500
Expected Value= Pr1 *Pr2 * (50000-17500-5000)
= 6875
Node 3 - Weather Cold:
Pr1 = 0.5,
Use of Sprinklers Pr2 = 0.5,
Incur damage of approximately = (25000+30000)/2= 27500
Expected Value = Pr1 *Pr2 * (50000 - 27500 -2000)
= 5125
In case of cold weather, use of burner is far more effective than use of sprinkler as expected
value of node 2 is high.