Exam II Solution_2023

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1 SE 450 –Decision Analysis I Fall -202 3 Exam II • Duration: 120 minutes • Close book and notes • Calculator is allowed. “I pledge on my honor that I have not given or received any unauthorized assistance on this examination.” Your Signature: Name: _____________________________ Net ID: ___________________________ Prob. 1 2 3 4 5 TOTAL Score /15 /20 /20 /20 /25 ( Note: calculations must be accurate up to the second decimal point )

2 Problem 1 (15’): in the context of engineering decision-making, please judge true ( T ) or false ( F ) for the following statements. [ T ] A decision tree shows all possible paths that the decision maker might follow, and branches from each chance node form a set that is mutually exclusive and collectively exhaustive. [ F ] Influence diagram is a flowchart of the decision process that represents all possible future scenarios, which usually display more details of a decision problem than a decision tree. [ T ] When constructing the decision tree, entropy is used to determine how informative a particular input attribute is about the output attribute for a subset of the training data. [ F ] When we consider the value of information for a decision making problem by using multiple experts, the more interrelated and consistent the information is that they provide, the more valuable it is. [ T ] Decision makers can look to the tornado graphs to determine which variables impact the specific alternative most. Then, turn to the sensitivity diagram to understand the exact nature of the variables effect. Problem 2 (20’): A professional sports club is trying to use the ID3 algorithm to construct a decision tree for determination of the length of their outdoor training session (none, half or full) in a day based upon the weather conditions. The decision is made based on four different weather attributes, as shown in the table below, together with 14 available instances. While constructing the tree, please divide the humidity into three groups (below 0.7, 0.7 to 0.9, and above 0.9). Please help the sports club to identify the first weather attribute to be considered while constructing the decision tree for the decision making. Samples Weather Attributes Decision Outlook Temperature Humidity Windy 1 Sunny Hot 0.90 False None 2 Sunny Hot 0.87 True Half 3 Overcast Hot 0.93 True Full 4 Rain Mild 0.89 False Half 5 Rain Cool 0.80 False Half 6 Rain Cool 0.59 True None 7 Overcast Cool 0.77 True Half 8 Sunny Mild 0.91 False None 9 Sunny Cool 0.68 False Full 10 Rain Mild 0.84 False Half 11 Sunny Mild 0.72 True Full 12 Overcast Mild 0.49 True Full 13 Overcast Hot 0.74 False Half 14 Rain Mild 0.86 True None

3 ࠵?(࠵?) = − 4 14 log ! , 4 14 - − 6 14 log ! , 6 14 - − 4 14 log ! , 4 14 - = 1.5567 ࠵?(࠵?) = 5 14 3− 2 5 log ! , 2 5 - − 1 5 log ! , 1 5 - − 2 5 log ! , 2 5 -5 + 4 14 3− 2 4 log ! , 2 4 - − 2 4 log ! , 2 4 -5 + 5 14 3− 3 5 log ! , 3 5 - − 2 5 log ! , 2 5 -5 = 1.1760 ࠵?(࠵?) = 2 ∗ 4 14 3− 1 4 log ! , 1 4 - − 2 4 log ! , 2 4 - − 1 4 log ! , 1 4 -5 + 6 14 3−3 ∗ 2 6 log ! , 2 6 -5 = 1.5364 ࠵?(࠵?) = 3 14 3− 2 3 log ! , 2 3 - − 1 3 log ! , 1 3 -5 + 9 14 3− 1 9 log ! , 1 9 - − 6 9 log ! , 6 9 - − 2 9 log ! , 2 9 -5 + 2 14 3−2 ∗ 1 2 log ! , 1 2 -5 = 1.1267 ࠵?(࠵?) = 7 14 3− 3 7 log ! , 3 7 - − 2 7 log ! , 2 7 - − 2 7 log ! , 2 7 -5 + 7 14 3− 1 7 log ! , 1 7 - − 4 7 log ! , 4 7 - − 2 7 log ! , 2 7 -5 = 1.4678 Thus, Humidity provides the largest entropy again. Problem 3 (20’): For the decision tree shown below, (a) Develop the risk profile for the decision strategies A, B, C, respectively. Will any one strategy dominate the other? (b) Perform one-way sensitivity analysis and construct the tornado diagram for the decision parameters as shown below. Parameters Ranges P1 [0.01, 0.50] F1 [20, 80] F2 [120, 280] F3 [50, 150] (c) Perform two-way sensitivity analysis for P2 and F4 considering the following parameter ranges: P2 ~ (0, 1) and F4 ~ (10, 75), and identify the parameter space leading to the following decision inequalities: EMV (C) > EMV (B) > EMV (A) A C $0 $180 $100 (P3=0.3) (P4 = 0.5) (1 - P3 - P4) $36 (P2 = 0.3) (1 - P2) $0 $200 $55 B (F1) (F2) $80 (F3) (F4) C-1 C-2

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