The manager of the greeting card section of Mazey’s department store is considering her order for a particular line of Christmas cards. The cost for each box of cards is $3; each box will be sold for $5 during the Christmas season. After Christmas, the cards will be sold for $2 a box. The card section manager believes that all leftover cards can be sold at that price. The estimated demand during the Christmas season for the line of Christmas cards, with associated probabilities, is shown as follows.Demand (boxes)Probability252627282930.10.15.30.20.15.10a. Develop the payoff table for this decision situation.b. Compute the expected value for each alternative, and identify the best decision.c. Compute the expected value of perfect information. **Problem Context:**The manager of the greeting card section at Mazey's department store is planning an order for a specific line of Christmas cards. Each box costs $3, and during the Christmas season, they will be sold for $5 a box. Post-Christmas, any leftover boxes will be sold at $2 each. Estimated demand for these Christmas cards, along with their probabilities, is as follows:**Demand and Probability Table:**- **Demand (boxes):** - 25 - 26 - 27 - 28 - 29 - 30- **Probability:** - 25 boxes - 0.10 - 26 boxes - 0.15 - 27 boxes - 0.30 - 28 boxes - 0.20 - 29 boxes - 0.15 - 30 boxes - 0.10**Tasks:**a. Develop the payoff table for this decision situation.b. Compute the expected value for each alternative and identify the best decision.c. Compute the expected value of perfect information. **Problem 28: Greeting Card Demand Analysis**The manager of the greeting card section at Mazey's department store is evaluating her order strategy for a specific line of Christmas cards. Here's the situation:- **Cost per box**: $3- **Selling price during Christmas season**: $5 per box- **Selling price after Christmas**: $2 per box- **Assumption**: All leftover cards can be sold for $2 each after the Christmas season.**Estimated Demand and Probabilities:**| Demand (boxes) | Probability ||----------------|-------------|| 25 | 0.10 || 26 | 0.15 || 27 | 0.30 || 28 | 0.20 || 29 | 0.15 || 30 | 0.10 |### Tasks:a. Develop a payoff table for this decision situation.b. Compute the expected value for each alternative and identify the best decision.c. Compute the expected value of perfect information. This analysis will help the manager optimize her ordering strategy to maximize profits.

Given data Demand Probability 25 0.10 26 0.15 27 0.30 28 0.20 29 0.15 30 0.10 To determine the expected value of perfect information by using the formulae: EVPI = EV with perfect information - EV without perfect information ( Max. EMV) 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 b) the expected value for each alternative 25 boxes = 50 (0.1+0.15+0.3+0.2+0.15+0.1) = 50 26 boxes = 49x 0.1+52 ( 0.15+0.3+0.2+0.15+0.1) =51.7 27 boxes =48x0.1+51x0.15+54 ( 0.3+0.2+0.15+0.1) = 4.8+7.65 +40.5 =52.95 28 boxes = 47x0.1+50x0.15+53x0.3+54 ( 0.2+0.15+0.1) =4.7+7.5+15.9+24.3 =52.4 29 boxes = 46x0.1+49x0.15+52x0.3+55x0.2+58 ( 0.15+0.1) = 4.6+7.35+15.6+11+14.5 =53.05 30 boxes = 45x0.1+48x0.15+51x0.3+54x0.2+57x0.15+60x0.1 = 4.5+7.2+15.3+10.8+8.55+6 = 52.35 Therefore the most optimum condition is by = 29 boxes, the payoff being the largest. c) EVPI = EV with perfect information - EV without perfect information ( Max. EMV) = 0.1x50+0.15x52+0.3x54+0.2x56+0.15x58+0.1x60 - 53.05 = 5+7.8+16.2+11.2+8.7+6 - 53.05 = 54.9-53.05 =1.85