Exam II Solution_2023

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

6 Problem 4 (20’): For the decision tree shown in problem 3, (a) Calculate the EVPI for knowing the Chance Event B. (b) Before making the decision, an expert is available to access the Chance Event C-2 . The expert can tell you whether your prospects are “high”, “moderate” or “low”. But the expert is not a perfect predictor. The table below shows the expert’s record on forecasting. For example, as shown in the table, if the true outcome for Event C-2 is 180, the conditional probability is 0.85 that the expert will say prospects are “high”. Calculate the EVII for the expert assessment on chance event C. Expert Prediction True Outcome for Event C 180 80 0 “High” 0.80 0.10 0.14 “Moderate” 0.12 0.78 0.14 “Low” 0.08 0.12 0.72 (a) Calculate the EVPI for knowing the Chance Event B. EMV (PIB) = 0.3*200+0.7*55 = 98.5. EVPI = EMV(PIB)-EMV = 98.5 – 60 = $ 38.5 (b) P(“high”) = P(“high”|$180)*P($180) + P(“high”|$80)*P($80) + P(“high”|$0)*P($0) = 0.80*0.3 + 0.10*0.5+0.14*0.2 = 0.318 P($180|”high”) = P(“high”| $180)*P($180) / P(“high”) = 0.80*0.3/0.318 = 0.7547 P($80|”high”) = P(“high”| $80)*P($80) / P(“high”) = 0.1572 P($0|”high”) = P(“high”|$0)*P($0) / P(“high”) = 0.15*0.3 /0.3510 = 0.0881 45P3 + 20*(1-P3) = 25P3 + 20 8<25P3<13 Should be (0.52, 0.48) This should be 0.8 according to the table. If this value is used it will also be considered to be correct)