1. Decision Analysis EMBA 8150 Dr. Satish Nargundkar

2. Basic Terms Decision Alternatives (eg. Production quantities) States of Nature (eg. Condition of economy) Payoffs ($ outcome of a choice assuming a state of nature) Criteria (eg. Expected Value)

3. What kinds of problems? Mutually Exclusive vs. Mix of alternatives Single vs. Multiple criteria Assumptions Alternatives known States of Nature and their probabilities are known. Payoffs computable under different possible scenarios

4. Decision Environments Ignorance – Probabilities of the states of nature are unknown, hence assumed equal Risk/Uncertainty – Probabilities of states of nature are known Certainty – It is known with certainty which state of nature will occur. Trivial problem.

5. Example – Decisions under Risk S1 (Poor) S2 (Avg) S3 (Good) A1 (10 units) 300350400 A2 (20 units) -100600700 A3 (40 units) -1000-2001200 Probabilities 0.300.600.10 Assume the following payoffs in $ thousand for 3 alternatives – building 10, 20, or 40 condos. The payoffs depend on how many are sold, which depends on the economy. Three states of nature are considered - a Poor, Average, or Good economy at the time the condos are completed. P robabilities of the states of nature are known, as shown below.

6. Expected Values When probabilities are known, compute a weighed average of payoffs, called the Expected Value, for each alternative and choose the maximum value. Payoff Table S1S2S3EV A1 300350400 340 A2 -100600700 400 A3 -1000-2001200 -300 Probabilities 0.300.600.10 The best alternative under this criterion is A2, with a maximum EV of 400, which is better than the other two EVs.

7. Expected Opportunity Loss (EOL) Compute the weighted average of the opportunity losses for each alternative to yield the EOL. Opportunity Loss (Regret) Table S1S2S3EOL A1 0250800 230 A2 4000500 170 A3 13008000 870 Probabilities 0.300.600.10 The best alternative under this criterion is A2, with a minimum EOL of 170, which is better than the other two EOLs. Note that EV + EOL is constant for each alternative! Why?

8. EVUPI: EV with Perfect Information S1 (Poor) S2 (Avg) S3 (Good) A1 (10 units) 300350400 A2 (20 units) -100600700 A3 (40 units) -1000-2001200 Probabilities 0.300.600.10 If you knew everytime with certainty which state of nature was going to occur, you would choose the best alternative for each state of nature every time. Thus the EV would be the weighted average of the best value for each state. Take the best times the probability, and add them all. 300*0.3 = 90 600*0.6 = 360 1200*0.1 = 120 _____________ Sum = 570 Thus EVUPI = 570

9. EVPI: Value of Perfect Information Since EVUPI is 570, and you could have made 400 in the long run (best EV without perfect information), the value of this additional information is 570 - 400 = 170. Thus, EVPI = EVUPI – Evmax = EOLmin If someone offered you perfect information about which state of nature was going to occur, how much is that information worth to you in this decision context?

10. 0.6 | 350 | 400 | -100 | 600 | 700 | -1000 | -200 | 1200 | 300 0.1 0.3 0.6 0.1 0.6 0.3 340 400 -300 A2 400 A1 A2 A3 Decision Tree

11. Sequential Decisions Would you hire a consultant (or a psychic) to get more info about states of nature? How would additional information cause you to revise your probabilities of states of nature occurring? Draw a new tree depicting the complete problem.

12. Consultant’s Track Record S1S2S3 Favorable206070 Unfavorable804030 100 Previous Economic Forecasts Instances of actual occurrence of states of nature

13. Probabilities P(F/S1) = 0.2 P(U/S1) = 0.8 P(F/S2) = 0.6 P(U/S2) = 0.4 P(F/S3) = 0.7 P(U/S3) = 0.3 F= FavorableU=Unfavorable

14. Joint Probabilities S1S2S3Total Favorable0.060.360.070.49 Unfavorable0.24 0.030.51 Prior Probabilities 0.300.600.101.00

15. Posterior Probabilities P(S1/F) = 0.06/0.49 = 0.122 P(S2/F) = 0.36/0.49 = 0.735 P(S3/F) = 0.07/0.49 = 0.143 P(S1/U) = 0.24/0.51 = 0.47 P(S2/U) = 0.24/0.51 = 0.47 P(S3/U) = 0.03/0.51 = 0.06

16. Solution Solve the decision tree using the posterior probabilities just computed.