Question

P(W)= 0.3 P(low/W) = 0.50 P(low/S) = 0.10 P(S)= 0.7 P(medium/W)= 0.40 P(medium/S)= 0.25 P(high/W) = 0.10 P(high/S) = 0.65 BDSC 340.001-3 d) Construct a decision tree for this problem and analyze it. e) What is McHuffter’s optimal decision? f) What is the expected value of the survey(sample) information?

Answer 1

d) Decision tree is following

Expected Values (EV) of nodes are computed as under:

EV of node E = 400*0.25+400*0.35+400*0.45 = 400

EV of node F = 100*0.25+600*0.35+600*0.45 = 500

EV of node G = -300*0.25+300*0.35+900*0.45 = 450

EV of node H = 400*0.5+400*0.4+400*0.1 = 400

EV of node J = 100*0.5+600*0.4+600*0.1 = 350

EV of node K = -300*0.5+300*0.4+900*0.1 = 60

EV of node L = 400*0.1+400*0.25+400*0.65 = 400

EV of node M = 100*0.1+600*0.25+600*0.65 = 550

EV of node K = -300*0.1+300*0.25+900*0.65 = 630

EV of node C = MAX(400, 350, 60) = 400

EV of node D = MAX(400, 550, 630) = 630

EV of node B = 0.3*400 + 0.7*630 = 561

EV of node A = MAX(400, 500, 450, 561) = 561

E) McHuffter's optimal decision is to get the survey conducted and if Survey predicts Weak demand, then undertake Small development, and if the survey predits Strong demand, then undertake Large development.

F)

Expected Value without Survey Information = MAX(400, 500, 450) = 500

Expected Value with Survey Information = 561

Expected Value of Survey Information = 561 - 500 = 61