In August 2009 a car dealer is trying to determine how many

In August 2009, a car dealer is trying to determine how many 2010 cars to order. Each car ordered in August 2009 costs $16,000. The demand for the dealer’s 2010 models has the probability distribution shown in the dataset file (car tab). Each car sells for $21,000. If the demand for 2010 cars exceeds the number of cars ordered in August 2009, the dealer must reorder at a cost of $18,000 per car. Excess cars can be disposed of at $13,000 per car. Car tab info: Cars Demanded/probability: 25/.25, 30/.2, 35/.15, 40/.2, 45/.2

How many cars should be ordered to maximize expected profit?

Solution

25 * 0.25 + 30*0.2 + 35*0.15 + 40*0.2 + 45*0.2

expected value = 34.5

expected profit = 34.5 * 21000

expected profit = $724500