The economics of overbooking flights have earned international renown recently because of actions of a certain airline. The reason airlines overbook flights is because the economics are attractive. The result is lower prices for non-refundable flights: people who miss flights subsidize those that make it. The strategy has cost in proportion to $E(max(D\ - \ X),0)$ where $X$ is the number of seats in an airplane and $D$ is the demand. We have found that for a normally distributed demand, the average overbookings or $E(max(D\ - \ X),0)$ is given below:

$E\left( \max\left( D - X \right),0 \right) = \int_{X}^{\infty}{\left( D - X \right)\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left( - \frac{\left( D - \mu \right)^{2}}{2\sigma^{2}} \right)}\text{dD}$

We substitute variables here: $G = D - \mu$ or $D = G + \mu$

$= \int_{X - \mu}^{\infty}{\left( \mu + G - X \right)\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left( - \frac{G^{2}}{2\sigma^{2}} \right)}\text{dG}$ $= \int_{X - \mu}^{\infty}{\left( \mu - X \right)\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left( - \frac{G^{2}}{2\sigma^{2}} \right)}dG + \int_{X - \mu}^{\infty}{G\frac{1}{\sqrt{2\pi\sigma^{2}}}\exp\left( - \frac{G^{2}}{2\sigma^{2}} \right)}\text{dG}$ $= \left( \mu - X \right)P\left( D > X \right) + \frac{\sigma}{\sqrt{2\pi}}\exp\left( - \frac{\left( X - \mu \right)^{2}}{2\sigma^{2}} \right)$

For an airline, the distribution of $D$ can be determined by the number of tickets they sell. Assuming each customer is independent of one another, the distribution of $D$ will be $N(px,\sqrt{p(1 - p)x})$ where p is the likelihood a customer will show up. You can replace $\mu$ with $\text{px}$ and $\sigma$ with $\sqrt{p(1 - p)x}$ in the above formula to find the average overbookings. The cost will be proportional by a factor $c$. Thus the optimal number of overbooking will be when the derivative of $cE(max(X\ –\ D),0)$ is equal to the ticket price, which I have neglected to calculate because it is too tedious.

Why is it so complicated? Like the COMM 341 case below, the company knows and/or controls the demand function and has production/seat constraints. So the standard critical ratio method cannot be used.

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Our COMM 341 case competition was on Sport Obermeyer which made products with uncertain demand in two batches. Over producing would result in a cost of 8% of price; and under-producing would result in a cost of 18%. Thus the most obvious solution is to produce each product to the 75th percentile $\frac{C_{u}}{C_{u}\ + \ C_{o}}$ of demand, and split that quantity into two.

However, while performing the case we realized that the uncertainty of demand occurred only prior to the first batch whereas by the time of the second batch, your estimates were significantly more accurate (See Exhibit 5). Knowing this, we tried to provide a different answer than the standard one. Assuming that by the second batch, you knew demand perfectly (which according to the scatterplot is a fairly reasonable assumption) your only risk for the first batch was to over produce (if you under-produced, you would simply make it up in the second batch). So the solution was in the first batch produce the products with the least variance and the least cost of overproduction.

At the time it was known that we needed to minimize the sum of $0.08P*E(max(X - D),0))$ across all products, where $X$ is our first batch production, $D$ is the random variable for demand and $P$ was the price. Unfortunately, at the time, we did not find a way to calculate this quantity in excel, and so instead we minimized $P(X > D)$ across all products. This is clearly wrong but closer to the truth than the standard answer. Unfortunately, we lost the case competition because of this.

It is quite a shame as at the time, I was unaware that a conditional expectation of a normal distribution could be taken. In fact, when I solved a problem such as $E\left( \left| Z \right| \right)$, instead of simply writing out the expression and integrating, I would indirectly approach it through a chi-square distribution (since $\left| Z \right|$ is quite elegantly the root of $Z^{2}$, which is the Chi-square distribution). But now knowing that you can simply integrate a normal distribution to get to a conditional expectation, we find that

$E\left( \max\left( X - D \right),0 \right) = \left( X - \mu \right)P\left( D \leq X \right) + \frac{\sigma}{\sqrt{2\pi}}\exp\left( - \frac{\left( X - \mu \right)^{2}}{2\sigma^{2}} \right)$

(the similarity to the Black Scholes model is not coincidental)

When this quantity multiplied by the price is minimized using solver, we find an answer that is not entirely different than what we showed in our presentation 4 years ago. Unfortunately, this new discovery was not able to overturn our loss.

 Price C_o C_u Cu/ Average Forecast Standard Deviation 3rd year Today (Co + Cu) First Order First Order Gail 110 8.8 19.8 0.75 1,017 194 588 686 Isis 99 7.92 17.82 0.75 1,042 323 405 507 Entice 80 6.4 14.4 0.75 1,358 248 838 974 Assault 90 7.2 16.2 0.75 2,525 340 1,863 1,978 Teri 123 9.84 22.14 0.75 1,100 381 381 430 Electra 173 13.84 31.14 0.75 2,150 404 1,400 1,378 Stepahnie 133 10.64 23.94 0.75 1,113 524 212 173 Seduced 73 5.84 13.14 0.75 4,017 556 3,079 3,181 Anita 93 7.44 16.74 0.75 3,296 1,047 1,936 1,594 Daphne 148 11.84 26.64 0.75 2,383 697 1,298 1,099 Totals 20,001 12,000 12,000