Question

Problem setting:

There is an ice cream store. The manager tells the salesman that

Charge $2 for each ice cream in the morning;

Charge $1.5 for each ice cream in the afternoon before 6pm;

Charge $1 for each ice

cream after 6pm.

The store opens 7am-7pm.

The demand follows a Poisson process of lambda=10/hour

The wholesale price of an icecream is $1, the salvage price is $0.5 (unsold units can be returned to the supplier for $0.5 each)

1. Compute the corresponding expected profits for the order quantity Q = 50:150.

2. How many ice creams the manager should order at the beginning of the day, i.e., find the optimal Q that gives the maximum expected profit.

Answer 1

1.) Expected Profit for the order Quantity 50.

As it is given in the question that demand is 10 units per hour and the store opens at 7 am so fifty units will be sold in 5 hours (50 Nos./10 per hour=5 Hours) i.e up to 12 pm (before afternoon) .

It is also given that sale price before after noon is $2 per unit. Whoesale price of one unit is also given which is $1 per unit .It means that profit is $1 per unit if sold before afternon.

In case 50 units are sold then expected profit would be : 50*1$= 50$

Now we will discuss case where 150 units are ordered:

It will take 15 hours to sale 150 units as it is given that demand is 10 units per hour. But the shop is open only for 12ve hours . In 12ve hours only 120 units can be sold and rest 30 units will be returned at a loss of $.50/unit.

Profit on 50 units sold before afternoon=50 units * (Sale price $ 2-Cost of $1)=$50

Profit on next 60 units sold after 12 PM before 6 pm= 60 units(Sale price $ 1.5- Cost of $ 1)=$60

Profit on Next 10 units after 6 pm before 7 pm =10 units( Sale price of $1-Cost of $1)=0

Loss on 10 Unsold units as these will not be sold in the time given for store = 10units (Salvage Price $.50- Cost of $1)= -$10

Overall profit on 150 units=$50+$60+0+(-.$10)=$100

2.)Optimal quantity order that will give Maximum Expected profit :

Store is open for 07AM to 07PM i.e for 12ve hours and demand is 10units/hour so maximim 120 units can be sold in a day. But the manager has no profit if he make sale of icecream after 6 PM as cost will be same as sale price. So only 110 units should be ordered

So maximim profit can be calculated as under:

Profit on 50 units sold before afternoon=50 units * (Sale price $ 2-Cost of $1)=$50

Profit on next 60 units sold after 12 PM before 6 pm= 60 units(Sale price $ 1.5- Cost of $ 1)=$60

Overall maximum profit he can make : $50+$60=$110

Overall maximum profit would be : $50+$60+$0