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MGT 4803 Hamptonshire case

# MGT 4803 Hamptonshire case - 1 1.

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1) a. Sheen should stock the optimal stocking quantity in this situation, which is 584 newspapers. The expected profit at this stocking quantity is \$331.44. b. Q= µ+Φ -1 (C u /(C u +C 0 ))δ Q=500+ Φ -1 (.8/(.2+.8))100 Q=500+(. .7881)(100) Q=579 This is off by 5 newspapers from the model given in the spreadsheet, which results in a \$.03 difference in profits. 2) a. With the opportunity cost of her time per hour being equal to \$10, Sheen should invest 4 hours daily into the creation of the profile section. This would raise here optimal stocking quantity to 685 newspapers and would increase her expected daily profit to \$371.33. b. Sheen’s choice of effort level, h , to be 4 hours was chosen because, in order to maximize profit, she would need an effort level that made the marginal cost of her effort equal to the marginal benefit. The marginal cost(opportunity cost) of her effort was is \$10/hr. The marginal benefit equated to (0.8 *50)/(2*√ h ). When set equal to each other, the optimal # of hours invested comes out to be 4. c. The optimal profit under this model is greater because an increase the hours invested in creating the profile section, h, is in direct relation to an increase in average daily demand. With an increase in demand/sales, and no increase in fixed or variable costs, profit will increase.

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3) a. Armentrout’s optimal stocking quantity is equal to 516 newspapers. This creates of channel profit of \$322 and makes Anna’s profit equal to \$260.20. This stocking quantity maximizes Armentrout’s profit in this situation at \$62.14.
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MGT 4803 Hamptonshire case - 1 1.

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