# Binomial Distribution Problem - Airline Overbooking

Here's the problem:

A Flight has 20 seats and enough demand to sell out all flights (not enough to justify buying bigger plane). All seats sell for $200. Probability a passenger with reservation shows up = p. The probability of “No show”=1-p. The occurrence of “Show/No show” is independent among passengers. Passengers who are turned away due to “overbooking” are given$240 (the purchase price plus a 20% penalty to airline) The number of reservations made for each flight is chosen by airline (not random): n

The number of booked passengers who show up is a random variable: X

• What is the probability distribution of X?
• Write out the Revenues to the airline for a flight as a function of X
• Write out the expected Revenues to the airline
• What is the effect of increasing n on expected revenues for a given p?
• What is the effect of increasing p on expected revenues for a given n?
• When would it make no sense to overbook flights?

My thoughts: I think the answer to the first part is that it is a binomial distribution (since we have two distinct outcomes and a constant and independent probability of success). However, I am facing problem in the remaining parts. Can someone please explain it to me? Thanks so much :)

• Are all tickets non-refundable? Mar 26, 2017 at 3:36
• Yup. Sorry forgot to include that assumption Mar 26, 2017 at 4:02

# What is the probability distribution of X?

Yes, $X \sim Binom(n, p)$. That is, $$P(X=x) = \binom{n}{x}p^x(1-p)^x$$

# Write out the revenues to the airline as a function of X.

If $X \leq 20$, then it is pure profit for the airlines. If $X > 20$, then the airline loses 40 dollars for each passenger over 20. $$R(X) = 200X - \max[0, 240(X-20)]$$ Or... $$R(X) = \begin{cases} 200X & ;X \leq 20 \\ 200(20) - 40(X-20) & ;X > 20 \end{cases}$$

# Write out the expected value of R(X).

Remember that $E[R(X)] = \sum_{x=0}^n R(X)P(X=x)$. The math gets a bit tricky, but it can be done using R(X) as given above.

# What is the effect of increasing n on expected revenues for a given p?

Eventually, it is going to decrease the expected revenue for the airline. Basically if the airline overbooks too often, they will lose a lot of money. However if $p$ is small, the expected value of the revenue may be optimal for some $n > 20$.

# What is the effect of increasing p on expected revenues for a given n?

Again, it should decrease the expected revenue. If $n$ is fixed (and I'm assuming greater than 20), then increasing $p$ means the airline is more likely to have to refund somebodies money. You should be able to show this formally if you evaluate $E[R(X)]$.

# When would it make no sense to overbook flights.

If $p=1$ (or in practice, if it is close to 1), then it makes no sense. Because they will always have to refund somebodies money.

• Hi! Thank you so so much for the solution! I finally understand it now. However, I am confused with the notation you used R(X)=200X−max[0,40(X−20)]. What does the max part mean? I'm sorry I'm new to this as I am still a high school student. Mar 26, 2017 at 4:14
• Also, I was wondering if there is a method to write the revenue as a function of n? Mar 26, 2017 at 4:15
• You're correct. But even if the tickets are non-refundable, they get monetary compensation greater than the ticket price. It is thus assumed, for the bumped passengers, that the monetary compensation they receive is over and above their ticket price. I'm sorry for the confusion Mar 26, 2017 at 4:29
• Yup @FabioSomenzi! Thanks so much for helping out! Also, can you help me write the revenue as a function of n? Mar 26, 2017 at 5:02
• Agreed it's not a big deal to change. I asked the OP and that's what (s)he said. (See comments to the question.) Mar 26, 2017 at 22:27