Question

A hardware store orders snow blowers during the summer for delivery in the fall. Each snow blower costs the store $400 and if sold prior to or during the winter sells for a full price of $550. The store’s manager doesn’t want to carry any unsold snow blowers in inventory from one year to the next, so he reduces the price to $350 in the spring in order to get rid of any leftovers. He will be able to sell all of these at this price. Based on past experience, the store’s manager expects that the demand for snow blowers at full price will be between 6 and 8. The store manager estimates the probabilities of selling different numbers of snow blowers at full price to be the following: probability of 6 is 0.40, the probability of 7 is 0.35 and the probability of 8 is 0.25.

a) Draw the decision tree that the hardware store manager can use to analyze this problem.

b) Using an expected value approach, how many snow blowers should the hardware store manager order to maximize its profit?

Answer 1

a)

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Expected revenue = Probability of selling at full price*full price of snow blower + Probability of selling at reduced price*reduced price of snow blower

Probability of selling at reduced price = 1 - Probability of selling at full price

Example:

Expected revenue at ordering quantity 6 = 0.4*$550 + (1-0.4)*$350 = $220 + $210 = $430

b)

From above decision tree,

we can see when ordering quantity is 6, expected revenue is $430

So per unit profit = $430-$400 = $30

Total profit = 6*$30 = $180

when ordering quantity is 7, expected revenue is $420

So per unit profit = $420-$400 = $20

Total profit = 7*$20 = $140

when ordering quantity is 8, expected revenue is $400

So per unit profit = $400-$400 = $0

Total profit = 8*$0 = $0

As, using expected value approach total profit is highest when ordering quantity is 6, hardware store manager should order 6 to maximize its profit