Question

A fisherman is trying to perfect his business in order to maximize the money he can save for a new car. Daily fish sales are normally distributed, with a mean of 100 and a standard deviation of 10. He sells each fish for $0.60 and pays $0.25 for them. The fisherman has to get rid of unsold fish by paying a waste removal facility fee of $0.10 per fish. How many fish should he order each day and what % of the time will he experience a stock-out? Are there any drawbacks to the order size proposed and how could the fisherman address such issues?

Answer 1

Cost price of fish,c=$ 0.25

Selling price =p=$ 0.60

Salvage value=s=$ 0.10

Cost of understocking Cu=p-c=0.60-0.25=0.35

Cost of overstocking=Co=c-s=0.25-0.10=0.15

Optimal CSL(cycle service level) =Cu/(Cu+Co)=0.35/(0.35+0.15)=0.35/0.50=0.7

We have in normal distribution,

z=x-mean/std dev

Using excel function,we calculate z as

z=NORMSINV(CSL)=NORMSINV(0.7)

  z=0.5244

solving for x ,

0.5244=x-100/10

x=105.244

=106 (rounding to next whole number)

Number of fish to order each day to maximize profit =x=106

Percentage of the time he will experience stock out=1-optimal CSL=1-0.7=0.3 or 30 %

The order size is smaller and the chances of stockout is very high.So,the fisherman should increase his selling price and decrease the cost price,so that the optimal CSL would increase.