Teddy Bower is an outdoor clothing and accessories chain that purchases a line o

Question

Teddy Bower is an outdoor clothing and accessories chain that purchases a line of parkas at $10 each from its Asian supplier, TeddySports. Unfortunately, at the time of order placement, demand is still uncertain. Teddy Bower forecasts that its demand is normally distributed with mean of 2,100 and standard deviation of 1,200. Teddy Bower sells these parkas at $22 each. Unsold parkas have little salvage value; Teddy bower simply gives them away to a charity. a) (5 pts) What is the probability that the demand of the parka is greater than 1,800 units? b) (5 pts) What is the probability that the demand of the parka is between 1,800 and 2,500 units? c) (5 pts) How many parkas should Teddy Bower buy from TeddySports to maximize expected profit? d) (5 pts) If Teddy Bower wishes to ensure a 98.5 percent in-stock probability, how many parkas should it order? For parts e and f, assume Teddy Bower orders 3,000 parkas. e) (10 pts) Evaluate Teddy Bower’s expected profit. f) (5 pts) Evaluate Teddy Bower’s stockout probability.

Explanation / Answer

a) For 1800 units, z value = (1800 - 2100)/1200 = -0.25

Probability P(Demand <= 1800) = NORMSDIST(-0.25) = 0.401

Probability of demand greater than 1800 units = 1 - 0.401 = 59.9 %

b) For 2500 units, z value = (2500 - 2100)/1200 = 0.333

Probability P(Demand <= 2500) = NORMSDIST(0.333) = 0.630

Probability of demand greater than 1800 units and less than 2500 units = 0.630 - 0.401 = 0.229 or 22.9 %

c) Underage cost, Cu = 22 - 10 = 12

Overage cost, Co = 10

Optimal service level = Cu / (Cu+Co) = 12/(12+10) = 0.5455

z value = NORMSINV(0.5455) = 0.1143

Optimal number of parkas Teddy Bower should buy = 2100 + 1200*0.1143 = 2237 parkas

d) For 98.5% stock-in probability, z value = NORMSINV(0.985) = 2.17

Number of parkas to order = 2100 + 1200*2.17 = 4704

e) For 3000 units, z value = (3000-2100)/1200 = 0.75

For z=0.75, corresponding L(z) = 0.1312 (taken from standard normal table)

Expected lost sales (L) = *L(z) = 1200*0.1312 = 157

Expected Sales , S = µ - L = 2100 - 157 = 1943

Expected unsold inventory, V = Q - S = 3000 - 1943 = 1057

Expected profit = S*Cu - V*Co = 1943*(22-10) - 1057*10 = $ 12,746

f) Stockout probability = NORMSIST(0.75) = 0.7734 or 77.34 %

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