# (Solved Homework): Question 28:The manager of the greeting card section of mazey's department store is…

Question 28:The manager of the greeting card section of mazey’s department store is consideringher order for a particular line of Christmas Cards. The cost of each box of cards is \$3;each box will be sold for \$5 during the Christmas season. After Christmas, the cards willbe sold for \$2 a box. The card section manager believes that all leftover cards can besold at that price. The estimated demand during Christmas season for the line ofChristmas cards, with associated probabilities, is shown as follows:

Demand (boxes) Probability

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25 .10

26 .15

27 .30

28 .20

29 .15

30 .10

a. Develop the payoff table for this decision situation and compute the expected value for each alternative and identify the best decision.

b. Compute the expected value of perfect information.

(Can get a step by step breakdown not an excel solution?)

a. Payoff table

 Production 25 26 27 28 29 30 Probability Demand 25 50 49 48 47 46 45 0.1 26 50 52 51 50 49 48 0.15 27 50 52 54 53 52 51 0.3 28 50 52 54 56 55 54 0.2 29 50 52 54 56 58 57 0.15 30 50 52 54 56 58 60 0.1

Alternatives

25 boxes = 50 (0.1+0.15+0.3+0.2+0.15+0.1) = 50

26 boxes = 49×0.1+52 ( 0.15+0.3+0.2+0.15+0.1) =51.7

27 boxes =48×0.1+51×0.15+54 ( 0.3+0.2+0.15+0.1) = 4.8+7.65 +40.5 =52.95

28 boxes = 47×0.1+50×0.15+53×0.3+54 ( 0.2+0.15+0.1) =4.7+7.5+15.9+24.3 =52.4

29 boxes = 46×0.1+49×0.15+52×0.3+55×0.2+58 ( 0.15+0.1) = 4.6+7.35+15.6+11+14.5 =53.05

30 boxes = 45×0.1+48×0.15+51×0.3+54×0.2+57×0.15+60×0.1 = 4.5+7.2+15.3+10.8+8.55+6 = 52.35

Hence most optimum condition is with 29 boxes, payoff being the highest.

b. EVPI = EV with perfect information – EV without perfect information ( Max. EMV)

= 0.1×50+0.15×52+0.3×54+0.2×56+0.15×58+0.1×60 – 53.05

= 5+7.8+16.2+11.2+8.7+6 – 53.05 = 54.9-53.05 =1.85

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