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HW7_Soln - MS&E 252 Decision Analysis I Handout#23 Homework...

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MS&E 252 Handout #23 Decision Analysis I 12/7/2007 Page 1 of 23 HW#7 Solutions Homework Assignment #7- Solutions Distinctions These distinctions were prepared by the teaching team and reflect our best belief of the meanings of these terms. square4 A decision diagram shows the structure of a decision situation: the decisions, uncertainties and values involved as well as the relations between them. We represent these relations with arrows. An arrow from node A to node B shows that we wish to condition B on A. This leads to several different “types” of arrows, depending on the nodes which they connect. square4 Relevance arrows , indicating the conditioning of one uncertainty on another, show the possibility of relevance between the two uncertainties. square4 Informational arrows , indicating the conditioning of a decision on an uncertainty, show that we know the outcome of the uncertainty before we make the decision. square4 Functional arrows indicate the conditioning of a deterministic node on other nodes. The deterministic node is a function of the nodes pointing to it. square4 Influence arrows , indicating the conditioning of an uncertainty on a decision, show that the probabilities we assign to the uncertainty may change depending on the alternative we choose. Probabilistic questions 1) Solution: c Only statements I. and III. are true. If the results of a test cannot be observed, it is useless, since it is the observation of the test results which would allows us to update our beliefs about the distinction of interest and (perhaps) make a different decision. A test may be relevant but not material, but it must be relevant if it is to be material or economic. 2) Solution: b Begin by determining the posterior probabilities for the three detectors. This is illustrated in the trees that follow:
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MS&E 252 Handout #23 Decision Analysis I 12/7/2007 Page 2 of 23 HW#7 Solutions Assessed form, Rain Detector A Joint Inferential form, Rain Detector A Joint 0.85 "S" 0.79 S 0.34 0.34 0.4 S 0.43 "S" 0.15 "R" 0.21 R 0.06 0.09 0.15 "S" 0.11 S 0.09 0.06 0.6 R 0.57 "R" 0.85 "R" 0.89 R 0.51 0.51 Assessed form, Rain Detector B Joint Inferential form, Rain Detector B Joint 0.55 "S" 0.45 S 0.22 0.22 0.4 S 0.49 "S" 0.45 "R" 0.55 R 0.18 0.27 0.45 "S" 0.35 S 0.27 0.18 0.6 R 0.51 "R" 0.55 "R" 0.65 R 0.33 0.33 Assessed form, Rain Detector C Joint Inferential form, Rain Detector C Joint 0.3 "S" 0.22 S 0.12 0.12 0.4 S 0.54 "S" 0.7 "R" 0.78 R 0.28 0.42 0.7 "S" 0.61 S 0.42 0.28 0.6 R 0.46 "R" 0.3 "R" 0.39 R 0.18 0.18 Next, plug in the pre-posterior and posterior values into Kim’s party problem. We show the case of Detector C as an example. Note that the cost of the detector is not yet included in the analysis.
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MS&E 252 Handout #23 Decision Analysis I 12/7/2007 Page 3 of 23 HW#7 Solutions Kim's party problem with Detector C Value Measure U-Value 0.22 S 100 1 O 13.1517 0.22222 0.78 R 0 0 0.22 S 90 0.95043388 0.54 "S" P 47.6549 0.64464 30.7133 0.46232 0.78 R 20 0.32285562 0.22 S 40 0.56753443 >>> I 47.6549 0.64464 0.78 R 50 0.66666667 0.67234 0.61 S 50.61645 100 1 O 43.9853 0.6087 0.39 R 0 0 0.61 S 90 0.95043388 0.46 "R" >>> P 54.2557 0.70486 54.2557 0.70486 0.39 R 20 0.32285562 0.61 S 40 0.56753443 I 43.7497 0.60633 0.39 R 50 0.66666667 Finally we show a table indicating the value of the deal with free information from each detector, the CE of Kim’s original deal, the cost of each detector, and finally, the net value of information (value of information provided by the detector, minus its cost): A B C
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