Ex 02 Decision Analysis

March 26, 2018 | Author: Ali Akbar | Category: Applied Mathematics, Business, Mathematics


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DECISION ANALYSISQ1 The Miramar Company is going to introduce one of three new products: a widget, a hummer, or a nimnot. The market conditions, whether favorable, stable or unfavorable will determine the profit or loss the company realizes, as shown in the following payoff table: Market Conditions Product Favorable Stable Unfavorable Widget $120,000 $70,000 -$30,000 Hummer $60,000 $40,000 $20,000 Nimnot $35,000 $30,000 $30,000 Determine the best decision using the following decision criteria; a. Maximax b. Maximin c. Minimax regret d. Hurwicz (with α = 0.4) e. Equal likelihood If the probabilities of the market conditions are 0.20, 0.70 and 0.10 for favorable, stable and unfavorable conditions, respectively; a. Compute the expected value for each decision and select the best one. b. Develop the opportunity loss table and compute the expected opportunity loss for each product. c. Determine how much the firm would be willing to pay a market research firm to gain better information about future market conditions. The company is considering contracting with a market research firm to do a survey to determine future market conditions. The result of the survey will indicate positive or negative market conditions. There is a 0.60 probability of a positive report, given favorable conditions; a 0.30 probability of a positive report, given stable conditions; and a 0.10 probability of a positive report, given unfavorable conditions. There is a 0.90 probability of negative report, given unfavorable conditions; a 0.70 probability, given stable conditions; and a 0.40 probability, given favorable conditions. Using decision tree analysis and posterior probability tables; a. Determine the decision strategy the company should follow. b. Determine the expected value of the strategy. c. Determine the maximum amount the company should pay the market research firm for the survey results. d. Compute the efficiency of the survey results. Q2 The following payoff table shows the profit for a decision problem with two states of nature and two decision alternatives; State of Nature Decision Alternatives s 1 s 2 d 1 10 1 d 2 4 3 a. Use graphical sensitivity analysis to determine the range of probabilities of state of nature s 1 for which each of the decision alternatives has the largest expected value. b. Suppose p(s 1 ) = 0.2 and p(s 2 ) = 0.8. What is the best decision using the expected value approach? c. Perform sensitivity analysis on the payoffs for decision alternative d 1 . Assume the probabilities are as given in part (b) and find the range of payoffs under states of nature s 1 and s 2 that will keep the solution found in part (b) optimal. Is the solution more sensitive to the payoff under state of nature s 1 or s 2 ? 21. Hale’s TV Productions is considering producing a pilot for a comedy series in the hope of selling it to a major television network. The network may decide to reject the series, but it may also decide to purchase the rights to the series for either 1 or 2 years. At this point in time, Hale may either produce the pilot and wait for the network’s decision or transfer the rights for the pilot and series to a competitor for $100,000. Hale’s decision alternatives and profits ($1000s) are as follows: State of Nature Decisions Alternative Reject, s 1 1 Year, s 2 2 Years, s 3 Produce Pilot, d 1 –100 50 150 Sell to competitor, d 2 100 100 100 The probabilities for the states of nature are P(s 1 ) = 0.20, P(s 2 ) = 0.30, and P(s 3 ) = 0.50. For a consulting fee of $5,000, an agency will review the plans for the comedy series and indicate the overall chances of a favorable network reaction to the series. Assume that the agency review will result in a favorable (F) or an unfavorable (U) review and that the following probabilities are relevant. P(F) = 0.69 P(s 1 | F) = 0.09 P(s 1 | U) = 0.45 P(U) = 0.31 P(s 2 | F) = 0.26 P(s 2 | U) = 0.39 P(s 3 | F) = 0.65 P(s 3 | U) = 0.16 a. Construct a decision tree for this problem. b. What is the recommended decision if the agency opinion is not used? What is the expected value? c. What is the expected value of perfect information? d. What is Hale’s optimal decision strategy assuming the agency’s information is used? e. What is the expected value of the agency’s information? f. Is the agency’s information worth the $5,000 fee? What is the maximum that Hale should be willing to pay for the information? g. What is the recommended decision? [3+2+2+6+1+1+1=16] 22. Suppose that you want to invest $ 10,000 in stock market by buying shares in one of the two companies: A and B. Shares in company A, though risky, could yield a 50% return on investment during the next year. If the stock market conditions are not favourable (i.e. “bear” market), the stock may lose 20% of its value. Company B provides safe investment with 15% return in a “bull” market and only 5% in a “bear” market. All the publications you have consulted (and there is always a flood of them at the end of the year!) are predicting a 60% chance for a “bull” market and 40% for a “bear” market. a. Where should you invest your money? Now, suppose that rather than relying solely on the publications, you have decided to conduct a more personal investigation by consulting a friend who has done well in the stock market. The friend offers the general opinion of “for” or “against” investment. This opinion is further quantified in the following manner: If it is a “bull” market, there is a 90% chance the vote will be “for”. If it is a “bear” market, the chance of a “for” vote is lowered to 50%. b. How do you make use of this additional information? c. What is the expected value of this information? d. What is the efficiency of this additional information? [3+7+4=14 marks] Question 21. A machine shop owner is attempting to decide whether to purchase a new drill press, a lathe or a grinder. The return from each will be determined by whether the company succeeds in getting a government military contract. The profit or loss from each purchase and the probabilities associated with each contract outcome are shown in the following payoff table: Contract No Contract Purchase 0.40 0.60 Drill press $ 40,000 $ − 8,000 Lathe 20,000 4,000 Grinder 12,000 10,000 (a) Which machine should be purchased? The machine shop owner is considering hiring a military consultant to ascertain whether the shop will get the government contract. The consultant is a former military officer who uses various personal contacts to find out such information. By talking to other shop owners who have hired the consultant, the owner has estimated a 0.70 probability that the consultant would present a favorable report, given that the contract is awarded to the shop (P(f | c)), and a 0.80 probability that the consultant would present an unfavorable report, given that the contract is not awarded (P(u | n)). Using decision tree analysis, (b) Determine (i) the decision strategy the owner should follow, (ii) the expected value of this strategy, and (iii) the maximum fee the owner should pay the consultant. [2+17+1+2=22 marks] A1 Note: Payoff in $1000s Decision Analysis Without Probabilities (or Under Uncertainty) Product Market Conditions a. Maximax b. Maximin Favorable Stable Unfavorable Widget 120 70 -30 120 -30 Hummer 60 40 20 60 20 Nimnot 35 30 30 35 30 Product Market Conditions d. Hurwicz e. Equal Likelihood Favorable Stable Unfavorable Widget 120 70 -30 30 53.333 Hummer 60 40 20 36 40.000 Nimnot 35 30 30 32 31.667 Regret = Opportunity Loss Product Market Conditions c.Minimax Regret Favorable Stable Unfavorable Widget 0 0 60 60 Hummer 60 30 10 60 Nimnot 85 40 0 85 Decision Analysis With Probabilities (or Under Risk) Decision Tree 70 120 -30 20 40 60 30 30 35 P(f) = 0.20 EV(W) = 70 EV(H) = 42 EV(N) = 31 EV(WoMR) = 70 Without Market Research P(s) = 0.70 P(u) = 0.10 P(f) = 0.20 P(f) = 0.20 P(s) = 0.70 P(s) = 0.70 P(u) = 0.10 P(u) = 0.10 Expected Value (or Expected Value without Perfect Information) Product Market Conditions a. Expected Value Favorable p = 0.20 Stable p = 0.70 Unfavorable p = 0.10 Widget 120 70 -30 70 = EVwoPI Hummer 60 40 20 42 Nimnot 35 30 30 31 Opportunity Loss table Product Market Conditions b. Expected Opportunit y Loss Favorable p = 0.20 Stable p = 0.70 Unfavorable p = 0.10 Widget 0 0 60 6 Hummer 60 30 10 34 Nimnot 85 40 0 45 Expected Value with Perfect Information Product Market Conditions Expected Value Favorable p = 0.20 Stable p = 0.70 Unfavorable p = 0.10 Widget 120 70 -30 24 + 49 Hummer 60 40 20 Nimnot 35 30 30 3 ∑ 76 = EVwPI EVPI = EVwPI – EVwoPI = 76 – 70 = 6 (i.e. $6,000) Therefore, the amount the firm would be willing to pay a market research firm would be less than $6,000. (i.e the maximum gain from the market research is only $6,000, thus should not pay the market research firm more than $6,000). Decision Analysis With Sample Information Let P(f) = probability of favorable market condition P(s) = probability of stable market condition P(u) = probability of unfavorable market condition P(p) = probability of positive market condition P(n) = probability of negative market condition Prior probabilities (Given above in decision Analysis under risk) P(f) = 0.20 P(s) = 0.70 P(u) = 0.10 Conditional Probabilities Positive market condition P(p/f) = 0.60 P(p/s) = 0.30 P(p/u) = 0.10 Negative market condition P(n/f) = 0.40 P(n/s) = 0.70 P(n/u) = 0.90 Posterior probability table for positive market condition States of Nature Prior Probabilities Conditional Probabilities Joint Probabilities Posterior Probabilities Favorable P(f) = 0.20 P(p/f) = 0.60 0.2x0.6 = 0.12 P(f/p) = 0.353 Stable P(s) = 0.70 P(p/s) = 0.30 0.7x0.3 = 0.21 P(s/p) = 0.618 Unfavorable P(u) = 0.10 P(p/u) = 0.10 0.1x0.1 = 0.01 P(u/p) = 0.029 P(p) = 0.34 Posterior probability table for negative market condition States of Nature Prior Probabilities Conditional Probabilities Joint Probabilities Posterior Probabilities Favorable P(f) = 0.20 P(n/f) = 0.40 0.2x0.4 = 0.08 P(f/n) = 0.121 Stable P(s) = 0.70 P(n/s) = 0.70 0.7x0.7 = 0.49 P(s/n) = 0.742 Unfavorable P(u) = 0.10 P(n/u) = 0.90 0.1x0.9 = 0.09 P(u/n) = 0.137 P(n) = 0.66 70 120 -30 30 35 30 70 120 -30 40 60 20 P(s/p) = 0.618 P(u/p) = 0.029 P(f/p) = 0.353 P(s/p) = 0.618 P(u/p) = 0.029 P(u/p) = 0.029 P(s/p) = 0.618 P(f/p) = 0.353 P(f/n) = 0.121 P(s/n) = 0.742 P(u/n) = 0.137 P(u/n) = 0.137 P(s/n) = 0.742 P(f/n) = 0.121 P(f/n) = 0.121 P(s/n) = 0.742 P(u/n) = 0.137 P(n) = 0.660 P(p) = 0.340 EV(W) = 84.750 EV(H) = 46.480 EV(N) = 30.605 EV(H) = 39.680 EV(W) = 62.350 EV(N) = 31.765 EV(N) = 62.350 EV(P) = 84.750 Positive Market Condition Negative Market Condition EV(MR) = 69.966 Market Research 30 35 30 40 60 20 P(f/p) = 0.353 a. Decision Strategy: Produce the widget regardless of the report. b. EVSI = EV(with sample information) – EV(without sample information) = $69,966 - $70,000 ≈ $0 (-$34) The additional (or sample) information has no value; therefore decision is to produce the widget. c. Not to pay anything for additional survey. EVSI is almost zero. d. The efficiency of sample information: Efficiency, E = (EVSI/EVPI) x 100 % = 0/6 = 0 % A2 EV(d 1 ) = 10p + 1(1-p) = 9p + 1 EV(d 2 ) = 4p + 3(1-p) = p + 3 Plot the equations on EV of decision alternatives & Probability of states of natures a. At the point A, the EV value is equal i.e. EV(d 1 ) = EV(d 2 ); 9p + 1 = p + 3; therefore, p = 0.25 Thus d 2 is optimal for p ≤ 0.25 d 1 is optimal for p ≥ 0.25 b. Best decision is d2 c. As long as the payoff for s 1 ≥ 2, then d 2 is optimal. d 1 d 2 EV p = 0 p = 0.25 p = 1 EV 10 4 3 1 A 21. s 1 a. d 1 s 2 s 3 Favorable s 1 d 2 s 2 Agency s 3 s 1 d 1 s 2 s 3 Unfavorable s 1 d 2 s 2 s 3 s 1 d 1 s 2 s 3 No Agency s 1 d 2 s 2 s 3 b. Using node 5, EV (node 10) = 0.20 (-100) + 0.30 (50) + 0.50 (150) = 70 EV (node 11) = 100 Decision: Sell to Competitor, d 2 . Expected Value = $100 c. EVwPI = 0.20 (100) + 0.30 (100) + 0.50 (150) = $125 EVPI = $125 - $100 = $25 1 6 -100 50 150 7 100 100 100 3 2 8 -100 50 150 9 100 100 100 4 10 -100 50 150 11 100 100 100 5 d. EV (node 6) = 0.09 (-100) + 0.26 (50) + 0.65 (150) = 101.5 EV (node 7) = 100 EV (node 8) = 0.45 (-100) + 0.39 (50) + 0.16 (150) = -1.5 EV (node 9) = 100 EV (node 3) = Max (101.5, 100) = 101.5 Produce EV (node 4) = Max (-1.5, 100) = 100 Sell EV (node 2) = 0.69 (101.5 + 0.31 (100) = 101.04 If Favorable, Produce If Unfavorable, Sell EV = $101.04 e. EVSI = $101.04 – 100 = $1.04 or $ 1,040. f. No, maximum Hale should pay is $1,040. g. No agency; sell the pilot. 22. a. The decision table corresponding to the problem is the following States of Nature Decision Alternative Bull Market= s 1 Bear Market= s 2 P(s 1 )= 0.6 P(s 1 )= 0.4 d 1 = choose company A 5000 -2000 d 2 = choose company B 1500 500 Since probability information about the states of nature are available, we use the expected value approach. We first calculate the expected values of the two decisions EV(d 1 )= 5000 (0.6) + (-2000) (0.4)= 2,200 EV(d 2 )= 1500 (0.6) + (500) (0.4)= 1,100 Hence d 1 or invest in stock A is the best decision and EV= 2,200 b. In terms of probability, the given information can be represented as follows. P(I 1 / s 1 ) = 0.9, P(I 2 /s 1 ) = 0.1 P(I 1 /s 2 ) = 0.5, P(I 2 /s 2 ) = 0.5 Where I 1 = the friend says “for” investment I 2 = the friend is “Against” investment. Let us know calculate the posterior probabilities and P(I 1 ) and P(I 2 ). For I 1 State of nature Prior conditional joint Posterior Probabilities probabilities probabilities probabilities s 1 0.6 0.9 0.54 0.7297 s 2 0.4 0.5 0.20 0.2702 P(I 1 )= 0.74 Hence P(s 1 /I 1 ) = 0.73, P(s 2 /I 1 ) = 0.27 and P(I 1 )= 0.74 For I 2 State of nature Prior conditional Joint Posterior Probabilities probabilities probabilities probabilities s 1 0.6 0.1 0.06 0.231 s 2 0.4 0.5 0.20 0.769 P(I 2 )= 0.26 Hence P(s 1 /I 2 ) = 0.231, P(s 2 /I 2 ) = 0.769 and P(I 2 )= 0.26. The problem can be represented by the following decision tree. 1 2 3 4 5 6 7 I 1 I 2 d 1 s 1 s 2 d 1 The expected of the branch I 1 or opinion “for” EV(d 1 /I 1 ) node 3= 5000× P(s 1 /I 1 ) + (-200) P(s 2 /I 1 )= 5000× 0.73 + (-200) 0.27 = 3110 EV(d 2 /I 1 ) node 4= 1500× P(s 1 /I 1 ) + (500) P(s 2 /I 1 )= 1500× 0.73 + (500) 0.27 = 1230 The best decision is d 1 i.e. invest in stock A. The expected of the branch I 2 or opinion “Against” EV(d 1 /I 2 ) node 6= 5000× P(s 1 /I 2 ) + (-200) P(s 2 /I 2 )= 5000× 0.231 + (-200) 0.769 = -383 EV(d 2 /I 2 ) node 7= 1500× P(s 1 /I 2 ) + (500) P(s 2 /I 2 )= 1500× 0.231 + (500) 0.769 = 731 The best decision is d 2 i.e. invest in stock B. Decision Rule: If the friend advises to invest, then invest in stock A If the friend’s advise to not invest, then invest in stock B c. The expected value with sample information is EVWSI= P(I 1 ) 3110 + P(I 2 ) 731 = 0.74 × 3110 + 0.26 × 731= 2301.4 + 190.06 = 2491.46  Then EVSI= 2491.46- EV= 2491.46-2,200= 291.46. d. The efficiency of the information given by the friend. First we have to calculate the expected value of perfect information. We have EVWPI= 5000 (0.6) + 500 (0.4)= 3000+ 200 = 3200 Hence EVPI= EVWPI- EV= 3200- 2200= 1000 Thus Efficiency of the information = E = (EVSI / EVPI) 100%= (910/1000) 100%= 91 %. We can conclude that the information is very efficient. Answer 21. (a) (b) P(c) = probability of contract = 0.40; P(n) = probability of no contract = 0.60; P(f | c) = 0.70; P(u | c) = 0.30 P(u | n) = 0.80; P(f | n) = 0.20 Computation of posterior probabilities: If f - Favorable State of Nature P(s j ) P(f | s j ) P(f∩s j ) P(s j | f) c P(c) = 0.40 P(f | c) = 0.70 P(fc) = 0.28 P(c | f) = 0.70 n P(n) = 0.60 P(f | n) = 0.20 P(fn) = 0.12 P(n | f) = 0.30 P(f) = 0.40 If u - Unfavorable State of Nature P(s j ) P(u | s j ) P(u∩s j ) P(s j | u) c P(c) = 0.40 P(u | c) = 0.30 P(uc) = 0.12 P(c | u) = 0.20 n P(n) = 0.60 P(u | n) = 0.80 P(un) = 0.48 P(n | u) = 0.80 P(u) = 0.60 (i) Decision strategy: If report is favorable, purchase a Drill press; If report is unfavorable, purchase a Grinder. (ii) EV (strategy) = $ 16,480. (iii) EVSI = EVwSI – EvwoSI = $16,480 – $11,200 = $ 5,280
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