1. Lens-Society, a retailer of eyewear has 5 stores in the United States. Consid
ID: 369849 • Letter: 1
Question
1.Lens-Society, a retailer of eyewear has 5 stores in the United States. Consider a new designer sunglass, “Miro”, to be released and sold in the upcoming season. Demand for Miro at each store is independent and is estimated to be Normally distributed with mean of 1000 units and standard deviation of 200 units. Traditionally, the stores have operated independently and replenishment decisions are made by the store managers. Each sunglass costs $80 and is priced to sell at $200. Lens-Society estimates that leftover inventory of Miro at the end of the season can be cleared through an end of season sale for an average price of $35.
a)Find the optimal order quantity, the average total cost and the average total profit at each store. Then, calculate the average total overall profit generated through the sale of Miro for the company.
b)Suppose Lens-Society is contemplating on selling Miro on their online store only. If so, Miro will not be available at any of the retail stores. Replenishment of Miro is then made centrally and customer orders will be filled through a central warehouse. Lens-Society estimates the demand in this case as the sum of all the demands for Miro if it were to be sold at the stores. The unit cost of the glasses is $80 as before. The Company will be selling the Miro for $195. In addition, there will be an average of $2.5 shipping cost paid by Lens-Society for the on-line orders. Finally, a wholesaler has agreed to buy all the end of season leftover inventory of Miro for the same price as part (a) ($35 per unit. No shipping or any other costs are involved here). Find the optimal order quantity and the average total Profit associated with this situation.
Explanation / Answer
1) a) Underage cost, Cu = 200-80 = 120
Overage cost, Co = 80-35 = 45
Optimal service level = Cu/(Cu+Co) = 120/(120+45) = 0.0.7273
z value = NORMSINV(0.7273) = 0.6046
Optimal order quantity at each store = mean + z * Std dev = 1000 + 0.6046*200 = 1121
Average total cost at each store = 1121*80 = $ 89,680
From standard normal table, for z = 0.6046, value of L(z) = 0.1674
Expected Lost sales (L) = *L(z) = 200*0.1674 = 33
Expected Sales, S = µ - L = 1000 - 33 = 967
Expected unsold inventory, V = 1121 - 967 = 154
Average profit at each store = 967*200 + 154*35 - 1121*80 = $ 109,110
Average total profit for the company = 109110*5 = $ 545,550
b)
Average demand = sum of 5 stores demand = 5*1000 = 5000
Std deviation of demand = 200*5 = 447
Cu = 195 - 80 - 2.5 = 112.5
Co = 80 - 35 = 45
Optimal service level = Cu / (Cu + Co) = 112.5/(112.5+45) = 0.7143
z value = NORMSINV(0.7143) = 0.5660
Optimal order quantity = 5000 + 0.566*447 = 5253
For z = 0.5660, value of L(z) = 0.1782 (from standard normal table)
Expected lost sales, (L) = *L(z) = 447*0.1782 = 80
Expected Sales, S =µ - L = 5000 - 80 = 4920
Expected unsold inventory = 5253 - 4920 = 333
Expected total profit = 4920*(195 - 2.5) + 333*35 - 5253*80 = $ 538,515
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