Expected value for the build new plant option=? The alternative that provides We

Question

Expected value for the build new plant option=? The alternative that provides Weiss the greatest expected monetary value (EMV) is =? The value of the return under this decision is =? The expected value of perfect information for Weiss =?
Do Homework-Ronnessa Casas-Google Chrome Secure nttps//www.mathd.com/Student/PlayertHtomeworkaspxthomeworkld-502389738iquestionld... SCM 352.1001 Fall 2018 course Homework: Homework #1 Score: 0 of 1 pt Problem A.5 Save 15 of 15 (14 complete) HW Score: 85 56%, 12 83 of Question Help * Howard Weiss, Inc., is considering building a sensitive new radiation scanning device. His managers believe that there is a probability of 0.45 that the ATR Co. will come out with a competitive product. If Weiss adds an assembly line for the product and ATR Co. does not follow with a competitive product, Weiss's expected profit is $50,000: if Weiss adds an assembly line and ATR follows suit, Weiss still expects $10,000 profit. If Weiss adds a new plant addition and ATR does not produce a competitive product. Weiss expects a profit of $600,000. IF ATR does compete for this market Weiss expects a loss of $120,000. a) Expected value for the Add Assembly Line option (enter your answer es a whole number) wer box and then 4

Explanation / Answer

Expected value of add assembly = .55*50000 + .45*10000 = 32000

Expected value of new plant = .55*600000 - .45*120000 = 276000

Greatest expected EMV = new plant = 276000

The value of return = 276000 – 32000 = 244000 $

Expected value of perfect information = EVPI = EV with perfect information – EV maximum

= EV with perfect information = .55*600000 + .45*10000 = 334500

EVPI = 334500 – 276000 = 58500 $

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