AspenWear, a retailer of ski wear needs to place an order for the Mirabelle, a d

Question

AspenWear, a retailer of ski wear needs to place an order for the Mirabelle, a designer ski jacket for the high-end market. The jacket retails for $600 and costs AspenWear $250 from a source in China. Due to fickle customer tastes, any surplus jackets at the end of the ski season cannot be carried over to the next season but must be disposed of. A bargain discounter has offered to buy these jackets at $150 each. Also, because of the long lead times involved in sourcing from China, there is realistically only one opportunity to place an order during each season (in November of each year so that the jackets will be ready by the following August).   From past history, AspenWear believes that demand for the Mirabelle can be represented by a normal distribution with mean 6000 and standard deviation 3600.

Compute the order quantity that will maximize AspenWear’s expected profit.

Compute the following performance measures for the above order quantity: expected sales, expected overstock, and expected profit.

Explanation / Answer

The jacket retails price(p) $600

Costs (c) $250

mean 6000

standard deviation 3600

Bargain Value (s) :$150

Cost of understocking(Cu) :(p-c)=600-250 =350

Cost of overstocking (Co)= (c-s)= 250-150 = 100

Optimal cycle service level(CSL)=Cu/(Cu + Co)

=350/(350+100)=0.78

Optimal order size (O)=NORMINV(CSL, m, s) =NORMINV(0.78,6000,3600) = 8779.90=8780

Expected profits =
(p – s)mFs((O – m)/s) – (p – s)sfs((O – m)/s)   – O(c – s)F(O, m, s) + O(p – c)[1 – F(O, m, s)]

=270000*NORMDIST(0.77,0,1,1)-1620000*NORMDIST(0.77,0,1,0)-877990*NORMDIST(8779.90,6000,3600,1)+877990*(1-NORMDIST(8779.90,6000,3600,1))

Expected overstock = (O – m)Fs((O – m)/std) + std*fs((O – m)/std)

=(8779.90-6000)*NORMDIST((8779.90-6000)/3600)+3600*NORMDIST((8779.90-6000)/3600)

(2780*NORMDIST(0.8,0,1,1)+3600*NORMDIST(0.8,0,1,0))

=3233.93

Expected understock =   (m – O)[1 – Fs((O – m)/stdev)] + stdevfs((O – m)/stdev)

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