O\'Neill sources fuse wetsuits from an Asians supplier for $175 each and sells t

Question

O'Neill sources fuse wetsuits from an Asians supplier for $175 each and sells them to customers for $480 each. Leftover wetsuits at the end of the season have no salvage value. The demand forecast is normally distributed with average 800 wetsuits and standard deviation 100 wetsuits. If O'Neill uses the newsvendor model correctly and places an order from its Asian supplier for this season, what is the expected mismatch cost?
-About $26,000? -About $18,000? -About $12,000? -About $24,000? O'Neill sources fuse wetsuits from an Asians supplier for $175 each and sells them to customers for $480 each. Leftover wetsuits at the end of the season have no salvage value. The demand forecast is normally distributed with average 800 wetsuits and standard deviation 100 wetsuits. If O'Neill uses the newsvendor model correctly and places an order from its Asian supplier for this season, what is the expected mismatch cost?
-About $26,000? -About $18,000? -About $12,000? -About $24,000?
-About $26,000? -About $18,000? -About $12,000? -About $24,000?

Explanation / Answer

Purchase cost(c) =$175

Selling cost = $480

Salvage cost (s) = 0

Marginal cost of excess Ce= c-s =175-0 =175

Marginal cost of shortage Cs = r-c = 380

Optimal service rate = Cs/(Cs+Ce) = 380/(380+175) = 0.6844

For normal distribution we can find the z value for this service level form table and MS excel too

z= 0.48

So optimal order quantity = mean + z* standard deviation= 800+0.48*100=848

For normal disitribution we can calculate the expected mismatch cost by following these formula in excel

Exected shortage/Expected lost sales = Std. Dev.*{ Normdist(z,0,1,false) -z* (1- Normdist(z,0,1,true)}

Expected excess = Order quantity – mean + Expected shortage

Expected mismatch cost = Cs*Expected shortage + Ce*Expected excess

Expected shortage = 20.40

Expected Excess = 68.40 (from excel calculations attached below)

Expected mismatch cost = 380* 20.40+ 175* 68.40 = $19724

Any doubt please comment

Service level 0.6844 z =NORM.S.INV(B2) ES =100*(NORM.DIST(0.48,0,1,FALSE)-0.48*(1-NORM.S.DIST(0.48,TRUE))) EE =848-800+B5 Cost =380*B5+175*B6

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