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Problem 1. Jean Clark is the manager of the Midtown Saveway Grocery Store. She now needs to replenish her supply of strawberries. Her regular supplier can provide as many cases as she wants. However, because these strawberries already are very ripe, she will need to sell them tomorrow and then discard any that remain unsold. Jean estimates that she will be able to sell 14, 15, 16, 17, 18, or 19 cases tomorrow. She can purchase the strawberries for $8 per case and sell them for $19 per case. Jean now needs to decide how many cases to purchase Jean has checked the stores records on daily sales of strawberries. On this basis, she estimates that the demand probability distribution is Demand (cases) 14 15 16 17 18 19 Probability 0.13 0.22 0.18 0.17 0.13 0.17 (a) Develop a decision analysis formulation of this problem by identifying the decision alternatives, the states of nature, and the payoff matrix. ( How many cases of strawberries should Jean purchase if she uses the maximin payoff criterion? ) How many cases should be purchased according to the maximum likelihood criterion? (d) How many cases should be purchased according to the expected payoff decision rule? (e) How many cases should be purchased according to the minimax regret decision rule? () If Jeans utility function is given by u(t)-t,what is her optimal decision?

math   Statistics-And-Probability   
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r p(x) Demand(x) probability(p(x)) distribution: T14] 15 16 17 18 19 141 15 16 Demand,x Supply, y 14 0.13 15 0.22 16 0.18 18 19 Selling Price: cp:-19 18 0.13 Purchase price: pp:-8 11 Marginal Profit: Marginal Loss: mp-CP-pp 19 0.17Profit Table(pt) pt :ニİf (บู <x.mp . T.(mp . T-ml. . (บุ一ㄒ 154 146 138 130 122 114] | 154 165 157 149 141 133| T154 165 176 168 160 152 -submatrix (pt ,0,5,0,5)- 154 165 176 187 179 1711 154 165 176 187 198 190 154 165 176 187 198 209 1541 162.53l l166.88 167.81 165.51 160.74 Expected Profit PT p(r)-154 165 176 Profit table diagonal values: PTd:- 198 209 Loss table: ii,j [ 0 8 16 24 32 40] 11 0 8 16 24 32 22 11 0 8 16 24 33 22 11 08 16 44 33 22 11 0 8 55 44 33 22 11 0 Loss table: LT:-submatrix(lt,0,5,0,5)- Optimum stock size is 17, its expected profit is 167.81, maximum. Diagonal of the profit table are the maximum values of each demand row. so we can stock quantities correspond to diagonal values for maximum profit. Cost of lost opportunity is the difference between profit and maximum possible profit.1 Decision Tables 2 Data 3 Profit 15 0.22 154 165 157 149 141 133 16 0.18 154 165 176 168 160 152 17 0.17 154 165 176 187 179 18 0.13 154 165 176 187 198 190 19 0.17 154 165 176 187 198 209 14 4 Probability 0.13 5 14 6 15 7 16 8 17 9 18 10 19 154 146 138 130 122 114 12 13 Expected Value of Perfect Information 14 Column best-MAX(B5:B10) MAX(C5:C10) MAX(D5:D10) MAX(E5:E10) MAX(F5:F10 MAX(G5:G10) 16 17Results EMV Minimum Maximum IF ABS(SUM(SB$4:$G$4)-1) 0.01,Pro SUMPRODUCT(SB$4:$G$4,B5:G5 SUMPRODUCT(SBS4:SGS4,B6:G6 SUMPRODUCT(SB$4:$G$4,B7:G7 SUMPRODUCT($B$4:$G$4,B8:G8 SUMPRODUCT(SB$4:$G$4,B9:G9 SUMPRODUCT(ŚB$4:$G$4,B10:G10) MIN(B10:G10) MAX(B10:G10) MAX(15:110) 4 MIN(B5:G5 MIN(B6:G6) MIN(B7:G7) MIN(B8:G8 MIN(B9:G9) MAX(B5:G5 MAX(B6:G6) MAX(B7:G7 MAX(B8:G8 MAX(B9:G9 10 11 Maximum 12 MAX(J5:J10 MAX(K5:K10) SUMPRODUCT($B$4:$G$4,B14:G14) <-Expected value WITH perfect information <-Best expected value 16 114-111 <-Expected value OF perfect information 17

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