Name: Le Duc ThinhID: 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 $300,000 figure for the Sub 100 with a favorable market is too high. How much lower would this figure have to be for Ken to change his decision made in part (b)? FAVORABLE MARKET UNFAVORABLE MARKET EQUIPMENT ($) ($) Sub 100 300,000 –200,000 Oiler J 250,000 –100,000 Texan 75,000 –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. 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? STATE OF NATURE DECISION GOOD POOR ALTERNATIVE ECONOMY ECONOMY Stock market 80.5 Solution: (a) Using the possible profits and probabilities in the table.000) = $30.500 So.000 Depending on the Maximize Expected Monetary Value model.000 Bonds 30.000 and invest in CDs with a payoff of $23. we have the expected profits for each decision are as follows: EMV(Stock market) = (0.000) = $23.000) = $25.5 0.5)*(–$20.5)*($23.000) + (0.5)*($80.500.500 – $30. What decision would minimize the expected opportunity loss? What is the minimum EOL? Solution: .000) + (0.000 –20.5)*($80. We have: EV(with prefect information) = (0.5)*($23.000. Mickey Lawson should invest in stock market.Name: Le Duc Thinh ID: BABAIU14255 3-20: Mickey Lawson is considering investing some money that he inherited.5)*($20.000 CDs 23. the maximum Mickey Lawson should pay for a perfect forecast of the economy is $21.000) + (0. respectively. 3-21: Develop an opportunity loss table for the investment problem that Mickey Lawson faces in Problem 3-20.000 = $21.500 EVPI = EV(with prefect information) – Maximum EMV(without prefect information) = $51.000 EMV(CDs) = (0. which has the highest EMV.000 20.000 EMV(Bonds) = (0. (b) Best alternative for favorable and unfavorable state of nature is invest in stock market with a payoff of $80.000 23. to maximize expected profit.000 Probability 0.5)*($30.000) = $51.000) + (0.5)*($23. 50)($0) + (0.000 50. respectively.500 EOL(Bonds) = (0.000 – 23.5 0.000) + (0.000 and invest in CDs with a payoff of $23.000 = 0 0.50)($43.000) = 43.500 So. therefore. which equals $21.000) = $21.000 CDs 80.50)($3.000 23. we can see that investing in stock market has the lowest opportunity loss. the minimum EOL will happen when Mickey invests in stock market.50)($0) = $28.000 = 57.5 Depending on the table.000 Stock market 80.000 – (–20.000 Bonds 50. Mickey Lawson should invest in stock market to minimize the expected opportunity loss.000 23.000 – 30.000 = 3.000 = 0 23.50)($57.000) = $26. We have the opportunity loss table for the investment problem that Mickey Lawson faces as following: STATE OF NATURE MAXIMUM DECISION GOOD POOR IN A ROW ALTERNATIVE ($) ECONOMY ECONOMY 80.500 EOL(CDs) = (0.000 – 20.000 – Probability 57.500.000 – 23.000 – 80.000 = 43. .000) + (0.000.50)($50. We have: EOL(Stock market) = (0.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 =0 = 100.000 = 129. .000 110.000 100.000 –100.000 –310.000 facility 310.Name: Le Duc Thinh ID: BABAIU14255 3-24: Today’s Electronics specializes in manufacturing modern electronic components.000 =0 = 19.000 – 129.000 – facility 0 – (–32. medium-sized facility with a payoff of $129.000 – Large 0 – (– 550.000) 310.000 Small 350.000 – 0 =0 = 550.000 – 129.000 Small facility 0 0 0 No facility Solution: (a) Best alternative for strong. Today’s Electronics should have medium-sized facility.000 – 0 129.000 129.000 550. who is responsible for advising the president of Today’s Electronics on electronic manufacturing equipment.000 facility 200.000 = 32. has developed the following table concerning a proposed facility: (a) Develop an opportunity loss table. therefore.000 110. 0 – (– 250.000 Medium-sized 300. It also builds the equipment that produces the components.000) = 250. (b) What is the minimax regret decision? PROFIT ($) STRONG FAIR POOR MARKET MARKET MARKET Large facility 550.000 and no facility with a payoff of $0. we can see that having medium-sized facility has the lowest opportunity loss.000 129.000 550. fair and poor market is having large facility with a payoff of $550. respectively. We have the opportunity loss table as following: PROFIT ($) MAXIMUM STRONG FAIR POOR IN A ROW MARKET MARKET MARKET ($) 550.000 100.000 300.000 sized facility 100.000 – 129.000) 200.000 No facility 0–0 550.000 = 310.000 550.000 (b) Depending on the table.000 – Medium.000 = 350.000 –32.000.000 = 29. Phyllis Weinberger. (a) Construct a decision table for this problem.35)*($364) + (0. John Kubick. John should choose to sell 11 cases because it has the highest EMV. Like many photographic chemicals. There is a probability of 0. 12. or 13 cases of BC-6 each week. president of Brilliant Color.35)*($420) + (0.45 0.45)*($329) + (0. normally stocks 11. he loses $56 for every case that is not sold by the end of the week. and a probability of 0. BC-6 has a very short shelf life. One product that Brilliant Color supplies is BC-6.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.45)*($385) + (0.2)*($420) = $379.35)*($385) + (0. Since each case costs John $56.35 of selling 12 cases. he receives a profit of $35.25 Depending on the Maximize Expected Monetary Value model. John must discard it.45 of selling 11 cases.2)*($385) = $385 EMV(12 cases) = (0. so if a case is not sold by the end of the week. For each case that John sells.45)*($273) + (0.2 (b) We have: EMV(11 cases) = (0. (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. Include all conditional values and probabilities in the table. .2)*($455) = $341. a probability of 0. how would this change your recommended course of action? Solution: (a) We have the following decision table : STOCK SOLD STOCK 11 cases 12 cases 13 cases 11 cases 11*$35 = $385 11*$35 = $385 11*$35 = $385 12 cases 11*$35 – 1*$56 = $329 12*$35 = $420 12*$35 = $420 13 cases 11*$35 – 2*$56 = $273 12*$35 – 1*$56 = $364 13*$35 = $455 Probability 0.2 of selling 13 cases.35 0.05 EMV(13 cases) = (0. 45)*($385) + (0.2)*($385) = $385 EMV(12 cases) = (0.2)*($455) = $411.45 0.45)*($385) + (0. .45)*($385) + (0.2 We have: EMV(11 cases) = (0.25 EMV(13 cases) = (0.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.2)*($420) = $404.35)*($420) + (0.35)*($385) + (0.35)*($420) + (0. John should choose to sell 13 cases because of their highest EMV. based on the Maximize Expected Monetary Value model.25 In this situation. we have the decision table as following: STOCK SOLD STOCK 11 cases 12 cases 13 cases 11 cases 11*$35 = $385 11*$35 = $385 11*$35 = $385 12 cases 11*$35 = $385 12*$35 = $420 12*$35 = $420 13 cases 11*$35 = $385 12*$35 = $420 13*$35 = $455 Probability 0.35 0.