Overbooking

Abstract

Early on, many airlines adopted the policy of not penalizing booked customers for canceling reservations at any time before departure. Some would not even penalize those that did not show up for booked flights. In essence, an airline ticket was “like money” since it could be used at full face value for a future flight or redeemed for cash at any future date. In the sixties, no-shows were becoming a problem for airlines who found that flights that were fully booked were departing with many empty seats. In response, the airlines began to overbook as a means of hedging against no-shows. If a flight had more passengers show than there were seats available, then the airlines would bump some passengers. The bumped passengers would be re-booked on a later flight. In addition, bumped passengers would be given other compensation, often a meal at the airport and a discount certificate applicable to future travel. The cost to the airline of bumping a passenger is called the denied boarding cost. The denied boarding cost would include the cost of putting a bumped passenger on another flight to her destination, the cost of any direct compensation to the bumped passenger, the cost of the meals or lodging that the airline provides to each bumped passenger, and the cost of “ill will” incurred by bumping the passenger. These costs can be different for each flight. For example, a passenger bumped from the last flight of the day will be provided with a hotel room at the airline’s expense.

Appendix

Proof of Proposition 3.1

We can write Z(N) as $Z(N) = \sum _{i=1}^N X_i$, where the X _i’s are independent Bernoulli random variables with probability q. Clearly $Z(\min (D,b+1)) - Z(\min (D,b)) = X_{b+1}\times \mathbf {1}(D \geq b+1)$, where 1(⋅) is the indicator function. Consequently, we get

$$\displaystyle \begin{aligned} \mathbb E \{ Z(\min(D,b+1)) - Z(\min(D,b)) \} = q {\,} \mathbb P \{ D \geq b+1 \}, {} \end{aligned} $$

(3.9)

Similarly, note that we always have $Z(\min (D, b+1) ) \geq Z(\min (D,b) )$. Furthermore, $Z(\min (D, b+1) )$ and $Z(\min (D,b) )$ can differ by at most 1. Thus, if $Z(\min (D,b) ) < c$, then we have $Z(\min \{ b +1, D \} ) \leq c$. On the other hand, if $Z(\min (D,b) ) \geq c$, then $Z(\min \{ b+1 , D \} ) \geq c$. In this case, we obtain

$$\displaystyle \begin{aligned} & [Z(\min(D,b+1)) - c]^+ - [Z(\min(D,b)) - c]^+ \\ {} & \qquad \qquad = \begin{cases} Z(\min(D,b+1)) - Z(\min(D,b)) & \text{if {$Z(\min(D,b) ) \geq c$}} \\ {} 0 & \text{otherwise} \end{cases} \\ {} & \qquad \qquad = \begin{cases} X_{b+1} & \text{if {$D \geq b+1$} and {$Z(\min(D,b)) \geq c$}} \\ {} 0 & \text{otherwise} \end{cases} \\ {} & \qquad \qquad = \begin{cases} X_{b+1} & \text{if {$D \geq b+1$} and {$Z(b) \geq c$}} \\ {} 0 & \text{otherwise.} \end{cases} \end{aligned} $$

Using the chain of equalities above, we get

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \mathbb E \{ [Z(\min(D,b+1)) - c]^+ - [Z(\min(D,b)) - c]^+ \} \\ &\displaystyle &\displaystyle \qquad \qquad \qquad \qquad \qquad \qquad \qquad = q {\,} \mathbb P\{ D \geq b+1 \} {\,} \mathbb P \{ Z(b) \geq c\}. {} \end{array} \end{aligned} $$

(3.10)

Using (3.9) and (3.10) in (3.1), we obtain

$$\displaystyle \begin{aligned} R(b+1) - R(b) = \mathbb P \{ D \geq b+1 \} {\,} q {\,} ( p - \theta {\,} \mathbb P \{ Z(b) \geq c\}), \end{aligned} $$

from which the formula for b ^∗ follows.

Proof of Theorem 3.2

In (3.3), we need to solve the problem

$$\displaystyle \begin{aligned} \max_{0 \leq u \leq D_j} \Big\{p_j {\,} u + V_{j-1} (y + u)\Big\}. \end{aligned} $$

We define a new decision variable z such that z = y + u. Since y is the number of reservations just before making the decisions for class j and u is the number of reservations we accept from class j, the decision variable z can be interpreted as the number of reservations after making the decisions for fare class j. After the change of variables, the problem is equivalent to

$$\displaystyle \begin{aligned} \max_{y \leq z \leq y + D_j} \Big\{ p_j {\,} z + V_{j-1} (z) \Big\} - p_j {\,} y. {} \end{aligned} $$

(3.11)

Since the last term p _j y does not affect the optimal solution, we can concentrate on the following problem

$$\displaystyle \begin{aligned} \max_{y \leq z \leq y + D_j} \Big\{ p_j {\,} z + V_{j-1} (z) \Big\}. {} \end{aligned} $$

(3.12)

Since V _j−1(⋅) is concave, the objective function of problem (3.12) above is concave. Thus, the problem above maximizes a concave function subject to the constraint that the decision variable lies in the interval [y, y + D _j].

Let $b_j^*$ be the maximizer of the concave function p _j z + V _j−1(z) over [0, ∞]. The maximizer can be computed as

$$\displaystyle \begin{aligned} b_j^* = \min\{ z \geq 0 : p_j {\,} (z+1) + V_{j-1}(z+1) \leq p_j {\,} z + V_{j-1}(z) \}. \end{aligned} $$

which yields the desired result.

We can characterize the optimal solution to the constrained problem above depending on whether $b_j^*$ is in the interval [y, y + D _j] or lies to the left or the right side of this interval. In particular, using z ^∗ to denote the solution to problem (3.12), we have

$$\displaystyle \begin{aligned} z^* = \begin{cases} y & \text{if {$b_j^* < y$}} \\ b_j^* & \text{if {$y \leq b_j^* \leq y + D_j$}} \\ y + D_j & \text{if {$b_j^* > y + D_j$}.} \end{cases} {} \end{aligned} $$

(3.13)

We show the three cases above, along with the maximizer $b_j^*$ of the function p _j z + V _j−1(z) and the interval [y, y + D _j] in Fig. 3.1. If $b_j^* < y$, then the number of reservations we have y is already larger than the optimal booking limit $b_j^*$. Thus, the only way to get as close as possible to $b_j^*$ after making the decisions for class j is not to accept any reservations from class j. In other words, we keep the number of reservations on hand at y. This situation corresponds to the first case above. If $b_j^* < y$, then it is optimal to set z ^∗ = y. If $y \leq b_j^* \leq y + D_j$, then $b_j^* - y \leq D_j$. So, we can accept $b_j^* - y$ reservations from class j to bring the number of reservations on hand to exactly $b_j^*$ after making the decisions for class j. This situation corresponds to the second case above. If $y \leq b_j^* \leq y + D_j$, then it is optimal to set $z^* = b_j^*$. Lastly, if $b_j^* > y + D_j$, then $D_j < b_j^* - y$. Thus, the only way to get as close as possible to $b_j^*$ after making the decisions for class j is to accept all of the demand from class j, in which case, the number of reservations that we have after making the decisions for class j goes up to y + D _j. This situation corresponds to the third case above. If $b_j^* > y + D_j$, then it is optimal to set z ^∗ = y + D _j. Noting the change of variables z = y + u and using (3.13), as a function of y, an optimal solution to problem (3.3) is given by the expression in the theorem.

Fig. 3.1

Optimal decision for class j

Proof of Theorem 3.3

We show the result by using induction over the classes in reverse order. Since Z(y) is a binomial random variable with parameters (y, q), we can write $Z(y) = \sum _{i=1}^y X_i$, where X ₁, X ₂, … are independent Bernoulli random variables with parameter q. In this case, we have

$$\displaystyle \begin{aligned}{}[Z(y+1) - c]^+ - [Z(y) - c]^+ & = \begin{cases} Z(y+1) - Z(y) & \text{if {$Z(y) \geq c$}} \\ 0 & \text{otherwise} \end{cases} \\ & = \begin{cases} X_{y+1} & \text{if {$Z(y) \geq c$}} \\ 0 & \text{otherwise,} \end{cases} \end{aligned} $$

which implies that $\mathbb E \{ [Z(y+1) - c]^+ - [Z(y) - c]^+ \} = q {\,} \mathbb P \{ Z(y) \geq c\}$. Since Z(y) is a binomial random variable with parameters (y, q), $\mathbb P \{ Z(y) \geq c\}$ is increasing in y. Therefore, $\mathbb E \{ [Z(y+1) - c]^+ - [Z(y) - c]^+ \}$ is increasing in y. In this case, $\mathbb E\{ [Z(y) - c]^+ \}$ is convex in y, which implies that $V_0(y) = - \theta {\,} \mathbb E\{ [Z(y) - c]^+ \}$ is concave in y, as desired. This discussion establishes the base case for the induction argument. Next, we assume that the value function V _j−1(⋅) is concave and show that V _j(⋅) is also concave.

Assume that V _j−1(⋅) is concave. By using the same change of variables used to obtain problem (3.11), we can write the dynamic program in (3.3) as

$$\displaystyle \begin{aligned} V_j(y) = \mathbb E \Bigg\{ \max_{y \leq z \leq y + D_j} [p_j {\,} z + V_{j-1} (z)] \Bigg\} - p_j {\,} y. \end{aligned} $$

We define

$$\displaystyle \begin{aligned} W_j(y,D_j) = \max_{y \leq z \leq y + D_j} [p_j {\,} z + V_{j-1} (z)], {} \end{aligned} $$

(3.14)

so that $V_j(y) = \mathbb E \{ W_j(y,D_j) \} - p_j {\,} y$. If we can show that W _j(y, D _j) is concave in y, then $\mathbb E\{ W_j(y,D_j) \}$ is concave in y as well, in which case, it follows that $V_j(y) = \mathbb E \{ W_j(y,D_j) \} - p_j {\,} y$ is concave, which is the result we are after. Thus, we proceed to showing that W _j(y, D _j) is concave in y.

By the induction assumption V _j−1(⋅) is concave. We let $b_j^*$ be the maximizer of the concave function p _j z + V _j−1(z) over the interval [0, ∞]. Since V _j−1(⋅) is concave, the discussion that we used to obtain the three cases in (3.13) still holds. In this case, letting z ^∗ be the optimal solution to problem (3.14), z ^∗ is still given by the three cases in (3.13). Noting that the optimal objective function of problem (3.14) is W _j(y, D _j), we have

$$\displaystyle \begin{aligned} W_j(y,D_j) & = \begin{cases} p_j {\,} y + V_{j-1}(y) & \text{if {$b_j^* < y$}} \\ p_j {\,} b_j^* + V_{j-1}(b_j) & \text{if {$y \leq b_j^* \leq y + D_j$}} \\ p_j {\,} (y + D_j) + V_{j-1}(y + D_j) & \text{if {$b_j^* > y + D_j$}} \end{cases} \\ & = \begin{cases} p_j {\,} y + V_{j-1}(y) & \text{if {$b_j^* < y$}} \\ p_j {\,} b_j^* + V_{j-1}(b_j^*) & \text{if {$ b_j^* - D_j \leq y \leq b_j^*$}} \\ p_j {\,} (y + D_j) + V_{j-1}(y + D_j) & \text{if {$y < b_j^* - D_j$}.} \end{cases} \end{aligned} $$

We plot the function p _j y + V _j−1(y) as a function of y on the left side of Fig. 3.2. Notice that the maximizer of this function over [0, ∞] is $b_j^*$. We plot the function W _j(y, D _j) on the right side of Fig. 3.2. Notice that the functions p _j y + V _j−1(y) and W _j(y, D _j) are identical for y in $[b_j^*,\infty ]$. For y in the interval $[b_j^* - D_j , b_j^*]$, the function W _j(y, D _j) takes the constant value $b_j^* + V_{j-1}(b_j^*)$, which is the maximum value of p _j y + V _j−1(y). Lastly, for y in the interval $[0, b_j^* - D_j]$, the function W _j(y, D _j) takes the value of p _j (y + D _j) + V _j−1(y + D _j). In other words, over the last interval, the function W _j(y, D _j) is a shifted version of the function p _j (y + D _j) + V _j−1(y + D _j). Thus, intuitively speaking, the function W _j(y, D _j) is obtained by “cutting” the function p _j y + V _j−1(y) in half at the point $y = b_j^*$, “shifting” the left portion of the function D _j units to the left, and “filling in” the middle with the constant value $b_j^* + V_{j-1}(b_j^*)$. Since $b_j^* + V_{j-1}(b_j^*)$ is the maximum value of the function p _j y + V _j−1(y), it follows that W _j(y, D _j) is concave, which is the desired result.

Fig. 3.2

Concavity of the value function for class j

Proof of Theorem 3.4

We let D _tj = 1 if there is a demand for ODF j at time period t, otherwise D _tj = 0. In this case, D _tj is a Bernoulli random variable with parameter λ _tj so that $\mathbb E \{ D_{tj} \} = \lambda _{tj}$. We let the random variable $W_{tj}^*$ be the number of accepted bookings for ODF j at time period t under the optimal policy and the random variable $X_{tj}^*$ be the number of bookings for ODF j accepted at time period t that survive until the departure time. Thus, $X_{tj}^*$ is a binomial random variable with parameters $(W_{tj}^*, Q_{tj})$. Thus, we have $\mathbb E \{ X_{tj}^* \} = Q_{tj} {\,} \mathbb E \{ W_{tj}^*\}$. Lastly, we let the random variable $Y_j^*$ be the number of denied bookings for ODF j under the optimal policy. Under the optimal policy, we have the inequalities

$$\displaystyle \begin{gathered} \sum_{t=1}^T \sum_{j \in N} a_{ij} {\,} X_{tj}^* - \sum_{j \in N} a_{ij} {\,} Y_j^* \leq c_i ~~~~ \forall {\,} i \in M \\ Y_j^* \leq \sum_{t=1}^T \sum_{j \in N} X_{jt}^* ~~~~ \forall {\,} j \in N \\ W_{tj}^* \leq D_{tj} ~~~~ \forall {\,} t=1,\ldots,T,~ j \in N. \end{gathered} $$

The first inequality states that the capacity consumption of each resource, after accounting for the denied boardings, does not exceed the available capacity of the resource. The second inequality states that the number of denied boardings for each ODF cannot exceed the accepted bookings for the ODF. The third inequality states that the number of accepted bookings for each ODF at each time period cannot exceed the demand for the ODF. Taking expectations on both sides of the inequalities above and noting that $\mathbb E \{ X_{tj}^* \} = Q_{tj} {\,} \mathbb E \{ W_{tj}^*\}$, the inequalities above imply that setting $w_{tj} = \mathbb E \{ W_{tj}^*\}$ and $z_j = \mathbb E \{ Y_j^*\}$ for all t = 1, …, T, j ∈ N provides a feasible solution to problem (3.7). The total profit from the optimal policy is $\sum _{t \in T} \sum _{j \in N} p_j {\,} W_{tj}^* - \sum _{j \in N} \theta _j {\,} Y_j^*$, in which case, taking expectations, the total expected profit from the optimal policy is $V(T,0) = \sum _{t \in T} \sum _{j \in N} p_j {\,} \mathbb E \{ W_{tj}^* \} - \sum _{j \in N} \theta _j {\,} \mathbb E\{ Y_j^*\}$. Thus, setting $w_{tj} = \mathbb E \{ W_{tj}^*\}$ and $y_j = \mathbb E \{ Y_j^*\}$ for all t = 1, …, T, j ∈ N provides a feasible solution to problem (3.7) and the objective value provided by this solution is equal to V (T, 0). In this case, it follows that the optimal objective value of problem (3.7) is at least V (T, 0), so we obtain ${\bar V}(T,0) \geq V(T,0)$.

Proof of Theorem 3.5

Since the number of requests that arrive for product j in period t is a Bernoulli random variable with success probability λ _tj, the number admitted by the PAC heuristic is a thinned Bernoulli with probability $w^*_{tj}$. From this number, a fraction Q _tj will survive, so the number of bookings for period t that survive is also thinned Bernoulli with probability $Q_{tj}w^*_{tj}$. This shows that the expected revenues associated with the PAC heuristic, aggregating over all products, is equal to $\sum _{t =1}^T \sum _{j \in N} p_{tj} Q_{tj} w^*_{tj} = {\bar V}(T,0)$, where the equality uses the fact that $\{ w_{tj}^* : t = 1,\ldots ,T,~j \in N\}$ is an optimal solution to problem (3.8).

Now, we consider the expected cost $\mathbb E[V(0, X)]$, where X is the vector of reservations on hand at the end of the horizon and V (0, x) is the optimal objective value of problem (3.6). Clearly $X_j = \sum _{t=1}^T X_{tj}$, where X _tj is Bernoulli random variable with mean Q _tj w _tj. Since the X _tj’s are independent over t, it follows that X _j has mean $\sum _{t=1}^TQ_{tj}w^*_{tj}$ and variance $\sum _{t=1}^T Q_{tj}w^*_{tj}(1- Q_{tj}w^*_{tj}) \leq \sum _{t=1}^T Q_{tj}w^*_{tj}$.

A feasible solution to program V (0, X) in (3.6) is to pay the overbooking fee θ _j for each unit of product j booking in excess of the mean, yielding the feasible solution y = {y _j : j ∈ N} with $y_j = (X_j - \mathbb E[X_j])^+$. Consequently, it follows that

$$\displaystyle \begin{aligned}\mathbb E[V(0,X)] \geq - \sum_{j \in N}\theta_j {\,} \mathbb E(X_j - \mathbb E[X_j))^+ \geq - \frac{1}{2} \sum_{j \in N} \theta_j \sqrt{\sum_{t=1}^TQ_{tj}w^*_{tj}},\end{aligned}$$

where we have used the fact that for any random variable with finite second moment $\mathbb E[(X - \mathbb E[X])^+] \leq 0.5 \sqrt {\mbox{Var}[X]}$. Thus, a lower bound on the expected revenue from the PAC heuristic is given by

$$\displaystyle \begin{aligned}V_h(T,0) = {\bar V}(T,0) + \mathbb E[V(0,X)] \geq {\bar V}(T,0) - \frac{1}{2} \sum_{j \in N} \theta_j \sqrt{\sum_{t=1}^TQ_{tj}w^*_{tj}}.\end{aligned}$$

Clearly $bw^*_{tj}$ is the solution to the linear program scaled by a factor b, so ${\bar V}^b(T,0) = b{\bar V}(T,0) \geq V^b(T,0) \geq V^b_h(T,0)$. From the bound on $\mathbb E[V(0,X)]$ we see that

$$\displaystyle \begin{aligned}V^b_h(T,0) = {\bar V}^b(T,0) - \mathbb E[V^b(0,X)] \geq {\bar V}^b(T,0) - \frac{1}{2} \sum_{j \in N} \theta_j \sqrt{b\sum_{t=1}^TQ_{tj}w^*_{tj}}.\end{aligned}$$

Dividing by ${\bar V}^b(T,0)$ and letting b →∞, we find that

$$\displaystyle \begin{aligned}\lim_{b \rightarrow \infty} \frac{V^b_h(T,0)}{V^b(T,0)} \geq \lim_{b \rightarrow \infty} \frac{V^b_h(T,0)}{{\bar V}^b(T,0)} \rightarrow 1,\end{aligned}$$

completing the proof.