IEOR E4210: Assignment 5 (Solutions)
a. Using the simulation in the spreadsheet would yields Q=584 b. [pic]
a. Using solver to solve the embedded model in the Excel sheet or by trying different values for h the optimum value will be obtained as “h=4” b. Marginal Revenue = Marginal Benefit
c. Optimal profit from Problem #1 = 331
Current optimal profit = 371
The difference is due to the effect of Sheen’s effort on the demand. This relation is not surprising. Players in the different stages of a supply chain can increase demand for their product through efforts in advertisement, product development etc.
a. Armentrout’s optimal stocking quantity is 516.
b. Armentrout’s cost of overstocking ($0.80) is significantly higher than his cost of understocking ($1.00 - $0.80 = $0.20). So his optimal value would be less than mean. It is important to note how the division of profit between channel members affects the sales. Since Sheen is getting most of the profit the margin of retailer is reduced so he reduces his inventory level which leads to decline of fill rate in channel. [pic]
c. As in the integrated channel Sheen’s profit initially increases with h and later decreases. At low levels of h, the marginal benefit from additional effort is significant because demand increases significantly with each additional unit of effort. At high h the growth in demand as a function of h is slower – in other words, the marginal benefit from additional effort is lower and is less than the marginal cost. So Sheen’s profit is maximized at an intermediate value h = 2.25. The optimal h in this differentiated channel is lower than the integrated channel because Sheen’s marginal benefit is lower (a part of benefit goes to Armentrout). So she puts less effort in the paper since she has less profit per paper. d. When we lower the transfer price Armentrout’s benefit would increase...
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