How to calculate expected loss in money when overbooking?

Question

I'm currently stuck on the following problem where we know that for an airline it is a 7% probability that a passenger will not meet up for departure. So to get a better use of the plane capacity the airline overbook the tickets. The plane for this task has 243 seats.

The thing is that the airline loses 1000\$ for each non-used seat in the plane. If a passenger does not get a spot because of the overbooking, the passenger will be compensated with 4000\$. We want to find an expression for the expected loss if we take $n$ orders. And by using the expression/graph we want to find the number of orders such that we minimize the expected loss. The numerical answer is given, which is 258 orders which results a loss of $5564\$$.

I understand that this will be a binomial distribution if we define a stochastic variable $X$ which says how many passengers that really meet ups for departure... but how do I use this fact to get the desired expression? If we do not overbook, i.e. only take 243 orders, the expected loss is to be $243 \cdot 0.07 \cdot 1000 = 17010 \$$, but if I try to do the same for $n = 258$ orders I do not get the same answer as the solution(which would be $5564\$$), why could I do that for $n = 243$, but not for $n= 258$?

Answer 1

Realize that the two scenarios correspond to different situations

scenario 1: no overbooking

scenario 2: overbooking 258 against 243 seats

For exactly one seat to go vacant in scenario 1, any one among the 243 passengers has to default on making it to the departure with probability $$ ^{243}C_1 \cdot (0.07)^1 \cdot (0.93)^{242}$$

And similarly the probability for two seats to go vacant = the probability that two people fail to arrive at the departure, which would be $$ {243}C_2 \cdot (0.07)^2 \cdot (0.93)^{241}$$ and this binomial series will go on. So you can use the expectation of the binomial random variable to arrive at the expected number of seats that will be vacant at departure which is $n \cdot p$.

But this is not the same for scenario 2. In scenario 2 for exactly one seat to remain empty exactly 16 people will have to fail to show up at the departure which has probability $$ ^{258}C_{16} \cdot (0.07)^{16} \cdot (0.93)^{242}$$ and similarly for two seats to be vacant at least 17 people of 258 would have to miss the flight with probability $= \ ^{258}C_{17} \cdot (0.07)^{17} \cdot (0.93)^{241}$, and so on. Let $k$ denote the number of people who arrive for the flight at departure, $X$ be the random variable denoting vacant seats and $Y$ be the random variable denoting number of people to be compensated because of plane being full at departure.

$$\mathbb{E}[X] = \sum_{k = 1}^{243} (243 - k) \cdot \ ^{258}C_{k} \cdot (0.07)^{258-k} \cdot (0.93)^{k}$$

$$\mathbb{E}[Y] = \sum_{k = 243}^{258} (k - 243) \cdot \ ^{258}C_{k} \cdot (0.07)^{258-k} \cdot (0.93)^{k}$$

and expected loss is $\mathcal{L} = 1000 (\mathbb{E}[X] + 4\cdot\mathbb{E}[Y])$

So you cannot directly use $n\cdot p = 258 \cdot 0.07 \cdot 1000$ to calculate the expected loss in scenario 2, both the cases have different loss functions.

Your original optimization problem is of the form $$\text{arg}\,\min\limits_{n \geq 243}\, \ \sum_{i = 1}^{243} (243-i) \cdot \ ^{n}C_{i} \cdot (0.07)^{n-i} \cdot (0.93)^{i} + \ \sum_{i = 243}^{n} (i-243) \cdot \ ^{n}C_{i} \cdot (0.07)^{n-i} \cdot (0.93)^{i}$$

Answer 2

Suppose the actual number of seats is $N$, while the allowed number of bookings is $N'\geq N$. Suppose $M$ people book the flight $(0\leq M\leq N')$, while on the day of the flight only $m$ people show up $(0\leq m\leq M)$.

Let $S$ be the money lost if a seat goes vacant. If $m< N$ then the airline loses $(N-m)S$ amount of money. If $m\geq N$ then airline loses no money if it does no more than fully reimburse those who could not obtain a seat due to overbooking.

If $p$ is the probability that a booked person shows up on the day of the flight, then the probability that $m$ people show up, given that $M$ have booked, is: \begin{align} P(m|M)=C^M_mp^m(1-p)^M,\quad 0\leq m\leq M \end{align} and zero otherwise. We assume that total number of bookings has uniform probability distribution: \begin{align} P(M)=\frac{1}{N'+1},\quad 0\leq M\leq N' \end{align} The probability that $m$ people show up irrespective of how many have booked is: \begin{align} P(m)=\sum_{M=m}^{N'}P(M)P(m|M)=\frac{1}{N'+1}\sum_{M=m}^{N'}C^M_mp^m(1-p)^M,\quad 0\leq m\leq N' \end{align} and zero otherwise. The summation begins from $M=m$ because for $M<m,P(m|M)=0$.

The expected loss is therefore: \begin{align} L & =\sum_{m=0}^{N-1}P(m)~(N-m)S \\ & =\frac{S}{N'+1}\sum_{m=0}^{N-1}(N-m)\sum_{M=m}^{N'}C^M_mp^m(1-p)^M \end{align} in which the summation on $m$ ends at $N-1$ because for $m>N-1$ the airline suffers zero loss of money.