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advertisement ```Name: Le Duc Thinh
ID: BABAIU14255
QUANTITATIVE METHODS FOR BUSINESS
-------------------------------------------------------------------------------------ASSIGNMENT: DECISION ANALYSIS
3-19: The Lubricant is an expensive oil newsletter to which many oil giants
subscribe, including Ken Brown (see Problem 3-17 for details). In the last issue, the
letter described how the demand for oil products would be extremely high.
Apparently, the American consumer will continue to use oil products even if the price
of these products doubles. Indeed, one of the articles in the Lubricant states that the
chances of a favorable market for oil products was 70%, while the chance of an
unfavorable market was only 30%. Ken would like to use these probabilities in
determining the best decision.
(a) What decision model should be used?
(b) What is the optimal decision?
(c) Ken believes that the \$300,000 figure for the Sub 100 with a favorable market is
too high. How much lower would this figure have to be for Ken to change his
decision made in part (b)?
EQUIPMENT
Sub 100
Oiler J
Texan
FAVORABLE MARKET
(\$)
300,000
250,000
75,000
UNFAVORABLE MARKET
(\$)
–200,000
–100,000
–18,000
Solution:
(a) Because this is a decision making when there are several possible states of
nature and we know the probabilities associated with each possible state; therefore,
we will use the Maximize Expected Monetary Value (EMV) model in which we
choose the alternative with the highest EMV.
(b) We have:
EMV(Sub 100) = (0.7)*(\$300,000) + (0.3)*(–\$200,000) = \$150,000
EMV(Oiler J) = (0.7)*(\$250,000) + (0.3)*(–\$100,000) = \$145,000
EMV(Texan) = (0.7)*(\$75,000) + (0.3)*(–\$18,000) = \$47,100
Depending on the Maximize Expected Monetary Value model, Kent should purchase
a Sub 100, which has the highest EMV.
(c) Ken should change his decision when Sub 100’s EMV be equal to the next best
one, Oiler J. Let X be the figure for the Sub 100 with a favorable market, we have:
EMV(Sub 100) = \$145,000 = (0.7)*X + (0.3)*(–\$200,000) = (0.7)*X – \$60,000
 X
≈
\$292,857.143 (or \$7142.857 lower)
Hence, the figure for the Sub 100 with a favorable market have to be \$7142.857
lower for Ken to change his decision.
Name: Le Duc Thinh
ID: BABAIU14255
3-20: Mickey Lawson is considering investing some money that he inherited. The
following payoff table gives the profits that would be realized during the next year for
each of three investment alternatives Mickey is considering:
(a) What decision would maximize expected profits?
(b) What is the maximum amount that should be paid for a perfect forecast of the
economy?
DECISION
ALTERNATIVE
Stock market
Bonds
CDs
Probability
STATE OF NATURE
GOOD
ECONOMY
80,000
30,000
23,000
0.5
POOR
ECONOMY
–20,000
20,000
23,000
0.5
Solution:
(a) Using the possible profits and probabilities in the table, we have the expected
profits for each decision are as follows:
EMV(Stock market) = (0.5)*(\$80,000) + (0.5)*(–\$20,000) = \$30,000
EMV(Bonds) = (0.5)*(\$30,000) + (0.5)*(\$20,000) = \$25,000
EMV(CDs) = (0.5)*(\$23,000) + (0.5)*(\$23,000) = \$23,000
Depending on the Maximize Expected Monetary Value model, Mickey Lawson
should invest in stock market, which has the highest EMV, to maximize expected
profit.
(b) Best alternative for favorable and unfavorable state of nature is invest in stock
market with a payoff of \$80,000 and invest in CDs with a payoff of \$23,000,
respectively. We have:
EV(with prefect information) = (0.5)*(\$80,000) + (0.5)*(\$23,000) = \$51,500
EVPI = EV(with prefect information) – Maximum EMV(without prefect information)
= \$51,500 – \$30,000 = \$21,500
So, the maximum Mickey Lawson should pay for a perfect forecast of the economy is
\$21,500.
3-21: Develop an opportunity loss table for the investment problem that Mickey
Lawson faces in Problem 3-20. What decision would minimize the expected
opportunity loss? What is the minimum EOL?
Solution:
Name: Le Duc Thinh
ID: BABAIU14255
Best alternative for favorable and unfavorable state of nature is invest in stock
market with a payoff of \$80,000 and invest in CDs with a payoff of \$23,000;
respectively.
We have the opportunity loss table for the investment problem that Mickey Lawson
faces as following:
DECISION
ALTERNATIVE
Stock market
Bonds
CDs
Probability
STATE OF NATURE
GOOD
POOR
ECONOMY
ECONOMY
80,000 – 80,000 = 0
23,000 – (–20,000) =
80,000 – 30,000 =
43,000
50,000
23,000 – 20,000 =
80,000 – 23,000 =
3,000
57,000
23,000 – 23,000 = 0
0.5
0.5
MAXIMUM
IN A ROW
(\$)
43,000
50,000
57,000
–
Depending on the table, we can see that investing in stock market has the lowest
opportunity loss; therefore, Mickey Lawson should invest in stock market to minimize
the expected opportunity loss.
We have:
EOL(Stock market) = (0.50)(\$0) + (0.50)(\$43,000) = \$21,500
EOL(Bonds) = (0.50)(\$50,000) + (0.50)(\$3,000) = \$26,500
EOL(CDs) = (0.50)(\$57,000) + (0.50)(\$0) = \$28,500
So, the minimum EOL will happen when Mickey invests in stock market, which
equals \$21,500.
Name: Le Duc Thinh
ID: BABAIU14255
3-24:
Today’s
Electronics
specializes
in
manufacturing
modern
electronic
components. It also builds the equipment that produces the components. Phyllis
Weinberger, who is responsible for advising the president of Today’s Electronics on
electronic manufacturing equipment, has developed the following table concerning a
proposed facility:
(a) Develop an opportunity loss table.
(b) What is the minimax regret decision?
Large facility
Medium-sized
facility
Small facility
No facility
STRONG
MARKET
PROFIT (\$)
FAIR
MARKET
POOR
MARKET
550,000
300,000
200,000
0
110,000
129,000
100,000
0
–310,000
–100,000
–32,000
0
Solution:
(a) Best alternative for strong, fair and poor market is having large facility with a
payoff of \$550,000; medium-sized facility with a payoff of \$129,000 and no facility
with a payoff of \$0, respectively. We have the opportunity loss table as following:
Large
facility
Mediumsized facility
Small
facility
No facility
STRONG
MARKET
550,000 –
550,000
=0
550,000 –
300,000
= 250,000
PROFIT (\$)
FAIR
MARKET
129,000 –
110,000
= 19,000
129,000 –
129,000
=0
550,000 –
200,000
= 350,000
129,000 –
100,000
= 29,000
550,000 – 0
= 550,000
129,000 – 0
= 129,000
POOR
MARKET
0 – (–
310,000)
= 310,000
0 – (–
100,000)
= 100,000
MAXIMUM
IN A ROW
(\$)
310,000
250,000
350,000
0 – (–32,000)
= 32,000
550,000
0–0
=0
(b) Depending on the table, we can see that having medium-sized facility has the
lowest opportunity loss; therefore, Today’s Electronics should have medium-sized
facility.
Name: Le Duc Thinh
ID: BABAIU14255
3-25: Brilliant Color is a small supplier of chemicals and equipment that are used by
some photographic stores to process 35mm film. One product that Brilliant Color
supplies is BC-6. John Kubick, president of Brilliant Color, normally stocks 11, 12, or
13 cases of BC-6 each week. For each case that John sells, he receives a profit of
\$35. Like many photographic chemicals, BC-6 has a very short shelf life, so if a case
is not sold by the end of the week, John must discard it. Since each case costs John
\$56, he loses \$56 for every case that is not sold by the end of the week. There is a
probability of 0.45 of selling 11 cases, a probability of 0.35 of selling 12 cases, and a
probability of 0.2 of selling 13 cases.
(a) Construct a decision table for this problem. Include all conditional values and
probabilities in the table.
(b) What is your recommended course of action?
(c) If John is able to develop BC-6 with an ingredient that stabilizes it so that it no
longer has to be discarded, how would this change your recommended course of
action?
Solution:
(a) We have the following decision table :
STOCK
11 cases
12 cases
13 cases
Probability
11 cases
11*\$35 = \$385
11*\$35 – 1*\$56 = \$329
11*\$35 – 2*\$56 = \$273
0.45
STOCK SOLD
12 cases
11*\$35 = \$385
12*\$35 = \$420
12*\$35 – 1*\$56 = \$364
0.35
13 cases
11*\$35 = \$385
12*\$35 = \$420
13*\$35 = \$455
0.2
(b) We have:
EMV(11 cases) = (0.45)*(\$385) + (0.35)*(\$385) + (0.2)*(\$385) = \$385
EMV(12 cases) = (0.45)*(\$329) + (0.35)*(\$420) + (0.2)*(\$420) = \$379.05
EMV(13 cases) = (0.45)*(\$273) + (0.35)*(\$364) + (0.2)*(\$455) = \$341.25
Depending on the Maximize Expected Monetary Value model, John should choose
to sell 11 cases because it has the highest EMV.
Name: Le Duc Thinh
ID: BABAIU14255
(c) If John is able to develop BC-6 with an ingredient that stabilizes it so that it no
longer has to be discarded, we have the decision table as following:
STOCK
11 cases
12 cases
13 cases
Probability
11 cases
11*\$35 = \$385
11*\$35 = \$385
11*\$35 = \$385
0.45
STOCK SOLD
12 cases
11*\$35 = \$385
12*\$35 = \$420
12*\$35 = \$420
0.35
13 cases
11*\$35 = \$385
12*\$35 = \$420
13*\$35 = \$455
0.2
We have:
EMV(11 cases) = (0.45)*(\$385) + (0.35)*(\$385) + (0.2)*(\$385) = \$385
EMV(12 cases) = (0.45)*(\$385) + (0.35)*(\$420) + (0.2)*(\$420) = \$404.25
EMV(13 cases) = (0.45)*(\$385) + (0.35)*(\$420) + (0.2)*(\$455) = \$411.25
In this situation, based on the Maximize Expected Monetary Value model, John
should choose to sell 13 cases because of their highest EMV.
```