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Unformatted text preview: 186 12.2. 20 10 5 0.1 0.2 0.6 0.1 A B 6 A = 20 0.1 A = 10 0.2 A = 5 0.6 A = 0 0.1 20 6 A B 10 6 A B 5 6 A B 6 A B Perfect Information EMV(A) = 7.00 EMV(B) = 6.00 EMV(Info) = 8.20 EVPI = EMV(Info)  EMV(A) = 8.20  7.00 = 1.20. This decision tree model is saved in the Excel file Problem 12.2.xls. 12.3. EVPI = 0 because one would still choose A regardless of As outcome. To state it slightly differently, no matter what you found out about As outcome, your decision would still be the same: Choose A. Because the information cannot change the decision, the expected value of the information equals zero. 187 12.4. a. The following decision trees are saved in the Excel file Problem 12.4.xls. Each part is shown in a separate worksheet. 20 1010 51 0.1 0.2 0.6 0.1 0.7 0.3 A B E A B 0.7 0.3 20 51 A B 0.7 0.3 10 51 A B 0.7 0.3 51 A B 0.7 0.310 51 A = 20 (0.1) A = 10 (0.2) A = 0 (0.6) A = 10 (0.1) Perfect Information about E EMV(Info about E) = 6.24 EMV(A) = 3.00 EMV(B) = 3.20 F E EVPI(Information about E) = EMV(Info)  EMV(B) = 6.24  3.20 = 3.04 188 b. 20 1010 51 0.1 0.2 0.6 0.1 0.7 0.3 A B E B = 5 (0.7) B = 1 (0.3) Perfect Information about F EMV(Info about F) = 4.4 EMV(A) = 3.0 EMV(B) = 3.2 F F 20 1010 0.1 0.2 0.6 0.1 E B 5 20 1010 0.1 0.2 0.6 0.1 E B1 EVPI(Information about F) = EMV(Info)  EMV(B) = 4.4  3.2 = 1.2 189 c. B = 5 (0.7) 20 1010 51 0.1 0.2 0.6 0.1 0.7 0.3 A B E A = 20 (0.1) A = 10 (0.2) A = 0 (0.6) A = 10 (0.1) Perfect Information about E and F EMV(Info about E and F) = 6.42 E F F B = 1 (0.3) A B 20 5 A B 201 B = 5 (0.7) F B = 1 (0.3) A B 10 5 A B 101 B = 5 (0.7) F B = 1 (0.3) A B 5 A B1 B = 5 (0.7) F B = 1 (0.3) A B10 5 A B101 EMV(A) = 3.0 EMV(B) = 3.2 EVPI(Information about both E and F) = EMV(Info)  EMV(B) = 6.42  3.2 = 3.22. 12.5. The basic influence diagram is: Chance E Decision A or B Chance F Payoff Chance E Decision A or B Chance F Payoff EV = 3.2 Chance E Decision A or B Chance F Payoff Chance E Decision A or B Chance F Payoff EV = 3.2 190 For 12.5a, b, and c, add arrows appropriately representing the information obtained: These models are saved in separate worksheets in the Excel file Problem 12.5.xls. To find the value of the information, you need to subtract the EV of the model without information from the EV of the model with information. 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....
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This note was uploaded on 12/04/2011 for the course BUSINESS 500 taught by Professor John during the Spring '11 term at Kansas.
 Spring '11
 John

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