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©2003 Thomson/South-Western 1 Chapter 19 – Decision Making Under Uncertainty Slides prepared by Jeff Heyl, Lincoln University ©2003 South-Western/Thomson.

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Presentation on theme: "©2003 Thomson/South-Western 1 Chapter 19 – Decision Making Under Uncertainty Slides prepared by Jeff Heyl, Lincoln University ©2003 South-Western/Thomson."— Presentation transcript:

1 ©2003 Thomson/South-Western 1 Chapter 19 – Decision Making Under Uncertainty Slides prepared by Jeff Heyl, Lincoln University ©2003 South-Western/Thomson Learning™ Introduction to Business Statistics, 6e Kvanli, Pavur, Keeling

2 ©2003 Thomson/South-Western 2 Two Basic Questions  What are the possible actions (alternatives) for this problem?  What is it about the future that affects the desirability of each action?

3 ©2003 Thomson/South-Western 3 Terms  Descriptions of the future: states of nature  The state of nature items are outcomes.  The key distinction between an action and a state of nature is that the action is taken is under your control, whereas the state of nature that occurs is strictly a matter of chance.  The payoff is the result of an action (A) an a state of nature (S)

4 ©2003 Thomson/South-Western 4 Payoff Table States of Nature ActionS 1 S 2 S 3 …S n A 1  11  12  13 …  1n A 2  21  22  23 …  2n A 3  31  32  33 …  3n  A k  k1  k2  k3 …  kn Table 19.1

5 ©2003 Thomson/South-Western 5 Conservative (Minimax) Strategy  The action chosen is that action that under the worst conditions produces the lowest “loss”  The opportunity loss, L ij, is the difference between the payoff for action I and the payoff for the action that would have the largest payoff under the state of nature j

6 ©2003 Thomson/South-Western 6 Minimax Strategy  Construct an opportunity loss table by using the maximum payoff for each state of nature  Determine the maximum opportunity loss for each action  Find the minimum value of the opportunity losses found in step 2; the corresponding action is the one selected

7 ©2003 Thomson/South-Western 7 The Gambler (Maximax) Strategy The Maximax strategy is to choose that action having the largest possible payoff

8 ©2003 Thomson/South-Western 8 The Strategist (Maximizing the Expected Payoff)  This strategy assigns a probability to each state of nature  The expected payoff of each action is determined  The action chosen is that action that produces the largest expected payoff

9 ©2003 Thomson/South-Western 9 Sailtown Example Average Interest Rate Amount OrderedIncreases (S 1 )Steady (S 2 )Decreases (S 3 ) 50 (A 1 )151515 75 (A 2 )2.522.522.5 100 (A 3 )-103030 150 (A 4 )-35545 Table 19.1

10 ©2003 Thomson/South-Western 10 Sailtown Example Table 19.3 State of NatureInterest RateCorresponding Demand S 1 Increases50 S 2 Holds steady100 S 3 Decreases150

11 ©2003 Thomson/South-Western 11 Sailtown Example Revenue forLoss Due to Revenue forLoss Due to ActionBoats SoldReturned BoatsNet Payoff A 1 50 $15,0000 500= $0$15,000 A 2 50 $15,00025 500= $12,00$2.500 A 3 50 $15,00050 500= $25,000-$10,000 A 4 50 $15,000100 500= $50,000-$35,000 Table 19.4

12 ©2003 Thomson/South-Western 12 Sailtown Example ActionPayoffOpportunity Loss A 1 15L 12 = 30 - 15 = 15 A 2 22.5L 22 = 30 - 22.5 = 7.5 A 3 30L 32 = 30 - 30 = 0 A 4 5L 42 = 30 - 5 = 25 Table 19.5

13 ©2003 Thomson/South-Western 13 Sailtown Example ActionPayoffOpportunity Loss A 1 15L 12 = 45 - 15 = 30 A 2 22.5L 22 = 45 - 22.5 = 22.5 A 3 30L 32 = 45 - 30 = 15 A 4 45L 42 = 45 - 5 = 0 Table 19.6

14 ©2003 Thomson/South-Western 14 Sailtown Example Table 19.7 State of Nature ActionS 1 S 2 S 3 A 1 01530 A 2 12.57.522.5 A 3 25015 A 4 50250

15 ©2003 Thomson/South-Western 15 Sailtown Example State of NatureProbability S 1 : Interest rate increasesP(S 1 ) =.3 S 2 : Interest rate remains unchangedP(S 2 ) =.2 S 3 : Interest rate decreasesP(S 3 ) =.5 Table 19.10

16 ©2003 Thomson/South-Western 16 Sailtown Example ActionExpected Payoff A 1 : Order 50 sailboats(15)(.3) + (15)(.2) + (15)(.5) = 15 A 2 : Order 75 sailboats(2.5)(.3) + (22.5)(.2) + (22.5)(.5) = 16.5 A 3 : Order 100 sailboats(-10)(.3) + (30)(.2) + (30)(.5) = 18 A 4 : Order 150 sailboats(-35)(.3) + (5)(.2) + (45)(.5) = 13 Table 19.11

17 ©2003 Thomson/South-Western 17 Sailtown Example Table 19.16 Expected Payoff P(S 1 )P(S 2 )P(S 3 )A 1 A 2 A 3 A 4.4.2.41514.5145.4.3.31514.5141.4.1.51514.5149.5.2.31512.510-3.5.1.41512.5101.3.3.41516.5189.3.2.51516.51813

18 ©2003 Thomson/South-Western 18 Sailtown Example Expected ActionPayoff (µ i ) Risk A 1 15[(15) 2 (.3) + (15) 2 (.2) + (15) 2 (.5)] - 15 2 = 0 A 2 16.5[(2.5) 2 (.3) + (22.5) 2 (.2) + (22.5) 2 (.5)] - 16.5 2 = 84 A 3 18[(-10) 2 (.3) + (30) 2 (.2) + (30) 2 (.5)] - 18 2 = 336 A 4 13[(-35) 2 (.3) + (5) 2 (.2) + (45) 2 (.5)] - 13 2 = 1216 Table 19.17

19 ©2003 Thomson/South-Western 19 Sailtown Example Figure 18.1

20 ©2003 Thomson/South-Western 20 Utility  Utility combines the decision maker’s attitude toward the payoff and the corresponding risk of each alternative  The utility value of a particular outcome is used to measure both the attractiveness and the risk associated with this dollar amount

21 ©2003 Thomson/South-Western 21 Utility Value Step 1: Assign a utility value of 0 to the smallest payoff amount (  min ) and a value of 100 to the largest (  max ) Step 2: The utility value for any payoff under consideration is found by using:U(  ij ) = P * 100

22 ©2003 Thomson/South-Western 22 Utility Utility of 11,200 Utility 10050 0 |11,200 40,000 Profit Figure 18.2

23 ©2003 Thomson/South-Western 23 Omega Corporation Payoff 100150200300400500 Probability (P).20.40.55.75.90.97 Utility [P(100)]204055759097 Table 18.20

24 ©2003 Thomson/South-Western 24 Decision Trees and Bayes’ Rule  A decision tree graphically represents the entire decision problem, including:  The possible actions facing the decision maker  The outcomes that can occur  The relationships between the actions and outcomes

25 ©2003 Thomson/South-Western 25 Decision Trees 18 15 16.5 18 13 A1A1A1A1 A2A2A2A2 A3A3A3A3 A4A4A4A4 S1S1S1S1 S2S2S2S2 S3S3S3S3(.3)(.2) (.5) S1S1S1S1 S2S2S2S2 S3S3S3S3 (.3) (.2) (.5) S1S1S1S1 S2S2S2S2 S3S3S3S3 (.3) (.2) (.5) S1S1S1S1 S2S2S2S2 S3S3S3S3 (.3) (.2) (.5) 15 15 15 2.5 22.5 22.5 -10 30 30 -35 5 45 Actions States of nature A 1 : order 50 sailboats A 2 : order 75 sailboats A 3 : order 100 sailboats A 4 : order 150 sailboats S 1 : interest rate increases S 2 : interest rate holds steady S 3 : interest rate decreases Figure 19.7

26 ©2003 Thomson/South-Western 26 Decision Trees and Bayes’ Rule Bayes’ rule allows you to revise a probability in light of certain information that is provided P(E i |B) = = P(E i and B) P(B) ith path sum of paths

27 ©2003 Thomson/South-Western 27 Use of Bayes’ Rule A1A1A1A1 A2A2A2A2 A3A3A3A3 A4A4A4A4 S1S1S1S1 S2S2S2S2 S3S3S3S3 S1S1S1S1 S2S2S2S2 S3S3S3S3 S1S1S1S1 S2S2S2S2 S3S3S3S3 S1S1S1S1 S2S2S2S2 S3S3S3S3 15 15 15 2.5 22.5 22.5 -10 30 30 -35 5 45 Actions States of nature A 1 : order 50 sailboats A 2 : order 75 sailboats A 3 : order 100 sailboats A 4 : order 150 sailboats S 1 : interest rate increases S 2 : interest rate holds steady S 3 : interest rate decreases Figure 19.8

28 ©2003 Thomson/South-Western 28 Deriving the Posterior Probabilities Consultant predicts an increase (I 1 ).7.4.2 S1S1S1S1 S2S2S2S2 S3S3S3S3 (.3) (.2) (.5) Figure 19.9

29 ©2003 Thomson/South-Western 29 Resulting Decision Tree A1A1A1A1 A2A2A2A2 A3A3A3A3 A4A4A4A4 S1S1S1S1 S2S2S2S2 S3S3S3S3(.538)(.205) (.256) (.538) (.205) (.256) (.538) (.205) (.256) (.538) (.205) (.256) 15 15 15 2.5 22.5 22.5 -10 30 30 -35 5 45 S1S1S1S1 S2S2S2S2 S3S3S3S3 S1S1S1S1 S2S2S2S2 S3S3S3S3 S1S1S1S1 S2S2S2S2 S3S3S3S3 15 8.45 -6.28 11.72 Consultant predicts an increase (I 1 ) 18 Figure 19.10

30 ©2003 Thomson/South-Western 30 Completed Decision Tree for Sailtown Problem Figure 19.11

31 ©2003 Thomson/South-Western 31 Deriving the Posterior Probabilities S1S1S1S1 S3S3S3S3 S2S2S2S2 (.3) (.2) (.5) (.7) (.4) (.2) I1I1I1I1 I1I1I1I1 I1I1I1I1A Sum of branches= P(I 1 ) =.39 P(S 1 | I 1 ) =.21/.39=.538 P(S 2 | I 1 ) =.08/.39=.205 P(S 3 | I 1 ) =.10/.39=.256  1.0 S1S1S1S1 S3S3S3S3 S2S2S2S2 (.3) (.2) (.5) (.2) (.5) (.2) I2I2I2I2 I2I2I2I2 I2I2I2I2A Sum of branches= P(I 2 ) =.26 P(S 1 | I 2 ) =.06/.26=.231 P(S 2 | I 2 ) =.10/.26=.385 P(S 3 | I 2 ) =.10/.26=.385  1.0 S1S1S1S1 S3S3S3S3 S2S2S2S2 (.3) (.2) (.5) (.1) (.1) (.6) I3I3I3I3 I3I3I3I3 I3I3I3I3A Sum of branches= P(I 3 ) =.35 P(S 1 | I 3 ) =.03/.35=.086 P(S 2 | I 3 ) =.02/.35=.057 P(S 3 | I 3 ) =.30/.35=.857  1.0 Figure 19.12


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