Question

Problem 4 PeachFarm opens a small stall next to the farm in season for people who wants to buy fresh peaches directly. On average 10 boxes are sold in a day and demand distribution follows Poisson. Since people drive to the farm to buy a fresh peach, PeachFarm doesn't want to keep a lot of peaches at stall while make sure the stockout probability being no higher than 2%. The price for each box is $25. Fresh peaches are picked up in the morning and any leftover will be given away. The cost of labor to pick up is about 3 $ per box. a) How many boxes should you prepare at the stall in the morning? b) How many boxes are given away at the end of the day on average?

Answer 1

The demand follows a Poisson distribution with mean 10 boxes per day.

Stockout probability is 2%

Price of each box =$25

Cost of pick up = $3

Salvage Value = $0

Overage Cost

Co = Cost – Salvage value = $3-$0=$3

Underage Cost

Cu= Price-Cost = $25-$3=$22

Critical Fractile = Cu//(Cu+Co)= 22/(22+3)=22/25=0.88

Now we need to look into the Poisson distribution table with mean value 10 and where the Cumulative probability just exceeds the probability of critical fractile

The cumulative probability which gives a higher value than 0.88 with mean =10 is the value = 15 (Value = 0.91)

So the value of optimal order quantity = 15 boxes

b) Probability of underage = 1-Critcal fractile = 0.22

Expected overstock

The value that exceeds the 0.22 probability is 2 with mean - 10 so expected overstock = 2 i.e. they are given away in the end of the day

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