Question

Offwefly Airlines has a daily flight from Sacramento to Las Vegas with a capacity of 100 passengers. On average, 16 ticket holders cancel their reservations at the last minute, so the company intentionally overbooks the flight. Cancellations can be described by a normal distribution with a standard deviation of 2.3. Profit per passenger is $70. If a passenger arrives but cannot board due to overbooking, the company policy is to provide compensation of $152. What is the optimal probability of having one or more empty seats on the plane?

Answer 1

Co = Cost of overbooking one unit more than cancellation = $152
Cu = Cost of overbooking one unit less than cancellation = $70

Critical ratio = Cu / (Co + Cu) = 70 / (152 + 70) = 0.3153

So, F(z) value at optimality = 0.3153, or, z = normsinv(0.3153) = -0.481

So, optimum overbooking level = 16 + (-0.481)*2.3 = 14.89 or 15 seats.

The probability of having one or more empty seats = Probability of 16 or more cancellation i.e. 0.50 as the cancellation distribution mean is 16 and the distribution is normal i.e. symmetrical.

Note: if we work with the fractional optimum overbooking level, the answer will change slightly.

Then the probability will be P(Cancellation >= 15.89) = 1 - normdist(15.89, 16, 2.3, 1) = 0.518