assignment 2 MGT780.docx - 3-26 The probability demand will...

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3-26 The probability demand will be 6 cases = 0.1; 7 cases = 0.3; 8 cases = 0.5; 9 cases = 0.1 The cost of every case = \$45 The price that Jason gets for each case = \$95 Cases not sold by the end of the month = \$0 Profit = \$95 - \$45 = \$50 Demand EMV 6 7 8 9 6 6x50 = 300 6x50 = 300 6x50 = 300 6x50 = 300 (0.1X300)+(0.3X300)+(0.5X300)+(0.1X300 ) = 300 7 (6x50)– (1x45) = 255 7x50 = 350 7x50 = 350 7x50 = 350 (0.1X255)+(0.3X350)+(0.5X350)+(0.1X350 ) = 340.5 8 (6x50)– (2x45) = 210 (7x50)– (1x45) = 305 8x50 = 400 8x50 = 400 (0.1X210)+(0.3X305)+(0.5X400)+(0.1X400 ) = 352.5 9 (6x50)– (3x45) = 165 (7x50 – (2x45) = 260 (8x50) – (1x45) = 355 9x50 = 450 (0.1X165)+(0.3X260)+(0.5X355)+(0.1X450 ) = 317 0.1 0.3 0.5 0.1 Megley Cheese Company should produce 8 cases each month.
3-29 LOW MEDIUM HIGH MAXIMUM IN A ROW MINIMUM IN A ROW Ardmore,OK 85 110 150 150 85 Sweetwater,TX 90 100 140 140 90 Lake Charles,LA 110 120 130 130 110 0.2 0.3 0.5 150 110 Ardmore,OK Lake Charles,LA LOW MEDIUM HIGH EMV Ardmore,OK 85x0.2 = 17 110x0.3 = 33 150x0.5 = 75 17+33+75 = 125 Sweetwater,TX 90x0.2 = 18 100x0.3 = 30 140x0.5 = 70 18+30+70 = 118 Lake Charles,LA 110x0.2 = 22 120x0.3 = 36 130x0.5 = 65 22+36+65 = 123 0.2 0.3 0.5 LOW MEDIUM HIGH MAXIMUM IN A ROW EOL Ardmore,OK 110-85 = 25 120-110 = 10 150-150 = 0 25 (0.2x25)+(0.3x10)+(0.5x0) = 8 Sweetwater,TX 110-90 = 20 120-100 = 20 150-140 = 10 20 (0.2x20)+(0.3x20)+(0.5x10) = 15 Lake Charles,LA 110-110 = 0 120-120 = 0 150-130 = 20 20 (0.2x0)+(0.3x0)+(0.5x20) = 10 0.2 0.3 0.5 20 Sweetwater,TX & Lake Charles,LA
a) Optimistic criterion: 150 (Ardmore, OK) b) Pessimistic criterion: 110 (Lake Charles,LA)