Network Revenue Management with Dependent Demands

Abstract

Network revenue management models have traditionally been developed under the independent demand assumption. In the independent demand setting, customers arrive into the system with the intention to purchase a particular product. If this product is available, they purchase it. Otherwise, they leave the system. This model is reasonable when products are well differentiated so that customers do not substitute between products. The independent demand model is harder to justify when there are few differences, other than price, between fares. Indeed, a more general setting is needed when the demand for each product depends heavily on whether or not other products are available for sale. This setting gives the firms the opportunity to shape the demand for each product by adjusting the offer set made available to the customer.

Appendix

Proof of Theorem 7.1

Under the optimal policy, let ${\mathcal T}(S)$ be the set of time points over the sales horizon during which set S is offered, and for each j ∈ S, let X _j(S) denote the number of sales for ODF j during the time that the set S is offered. Note that X _j(S) is a thinned Poisson random variable with mean $\pi _j(S) \int _{{\mathcal T}(S)} \lambda _t {\,} dt$, so $\mathbb E[X_j(S)]$ is equal to $\varLambda \pi _j(S) {\,} \mathbb E[\int _{{\mathcal T}(S)} \lambda _t {\,} dt/\varLambda ] = \varLambda \pi _j(S) {\,} \tau ^*(S)$, where $\tau ^*(S) := \mathbb E[\int _{{\mathcal T}(S)} \lambda _t {\,} dt/\varLambda ]$ is the expected proportion of customers offered set S during the sales horizon by the optimal policy. By construction τ ^∗(S) ≥ 0 and ∑_S⊆N τ ^∗(S) = 1. The total expected revenue obtained by the optimal policy is given by

$$\displaystyle \begin{aligned} V(T,c) = \sum_{S \subseteq N} \sum_{j \in N} p_j {\,} \mathbb E [X_j(S) ] = \varLambda \sum_{S \subseteq N} \sum_{j \in N}p_j \pi_j(S) \tau^*(S) = \varLambda \sum_{S \subseteq N} R(S) \tau^*(S) .\end{aligned}$$

Since the optimal policy has to obey the capacity constraints, we have ∑_S⊆N∑_{j ∈ N} A _j X _j(S) ≤ c. Taking expectations, it follows that

$$\displaystyle \begin{aligned}\sum_{S \subseteq N} \sum_{j \in N} A_j {\,} \mathbb E[X_j(S)] = \varLambda {\,} \sum_{S \subseteq N} \sum_{j \in N} A_j \pi_j(S) \tau^*(S) = \varLambda {\,} \sum_{S \subseteq N} A(S) \tau^*(S) \leq c.\end{aligned}$$

It follows that {τ ^∗(S) : S ⊆ N} is a feasible solution to problem (7.5) providing the objective value V (T, c) for this problem. This implies that the optimal objective value of problem (7.5) is at least V (T, c).

Proof of Theorem 7.6

We show the result by using induction. One can check that the result holds at time period 1, which is the time period in the discrete-time dynamic programming formulation right before the departure time. Assuming that the result holds at time period t − 1, we show that the result holds at time period t as well. We let S ^∗ be the optimal subset of ODF’s to offer in state (t, x) in the exact dynamic programming formulation of the network revenue management problem. In this case, noting the dynamic program in (7.2), it follows that

$$\displaystyle \begin{aligned} V(t,x) & = \sum_{j \in N} \lambda_t {\,} \pi_j(S^*) {\,} [p_j - \varDelta_j V(t-1,x)] + V(t-1,x) \\ & = \sum_{j \in N} \lambda_t {\,} \pi_j(S^*) {\,} \Big[ \sum_{i \in M} \alpha_{ij} + V(t-1,x - A_j ) \Big] \\ & \qquad \qquad \qquad + \Big[ 1 - \sum_{j \in N} \lambda_t {\,} \pi_j(S^*) \Big] {\,} V(t-1,x) \\ & \leq \sum_{j \in N} \lambda_t {\,} \pi_j(S^*) {\,} \Big[ \sum_{i \in M} \alpha_{ij} + \sum_{i \in M} v_i^\alpha(t-1,x_i - a_{ij}) \Big] \\ & \qquad \qquad \qquad + \Big[ 1 - \sum_{j \in N} \lambda_t {\,} \pi_j(S^*) \Big] {\,} \sum_{i \in M} v_i^\alpha(t-1,x_i), \end{aligned} $$

where the second equality follows from the fact that ∑_{i ∈ M} α _ij = p _j for all j ∈ N and arranging the terms, whereas the inequality is by the induction assumption. Rearranging the terms on the right side of the chain of inequalities above, the right side of the chain of inequalities is given by

$$\displaystyle \begin{aligned} & \sum_{i \in M} \Bigg\{ \sum_{j \in N} \lambda_t {\,} \pi_j(S^*) \Big[ \alpha_{ij} + v^\alpha(t-1, x_i - a_{ij}) - v_i^\alpha(t-1,x_i) \Big] + v_i^\alpha(t-1,x_i) \Bigg\} \\ & ~\leq \sum_{i \in M} \max_{S \in {\mathcal U}_i(x_i)} \Bigg\{ \sum_{j \in N} \lambda_t {\,} \pi_j(S) \Big[ \alpha_{ij} + v^\alpha(t-1, x_i - a_{ij}) - v_i^\alpha(t-1,x_i) \Big] \\ & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad + v_i^\alpha(t-1,x_i) \Bigg\} \\ & ~ = \sum_{i \in M} v_i^\alpha(t,x_i). \end{aligned} $$

Thus, we have $V(t,x) \leq \sum _{i \in M} v_i^\alpha (t,x_i)$, completing the induction argument.

Proof of Theorem 7.2

Let τ ^∗ = {τ ^∗(S) : S ⊆ N} be an optimal solution to problem (7.5). We consider the solution ${\hat x} = \{ {\hat x}_j : j = 0,1,\ldots ,n\}$ for problem (7.6), where ${\hat x}_j$ is defined as ${\hat x}_j = \sum _{S \subseteq N} \pi _j(S) {\,} \tau ^*(S)$ for all j = 0, 1, …, n. First, we will now show that ${\hat x}$ is a feasible solution to problem (7.6) and provides an objective value of ${\bar V}(T,c)$ for this problem, which implies that ${\tilde V}(T,c) \geq {\bar V}(T,c)$.

By the definition of ${\hat x}$, $\varLambda \sum _{j \in N} a_{ij} {\,} {\hat x}_j = \varLambda \sum _{j \in N} \sum _{S \subseteq N} a_{ij} {\,} \pi _j(S) {\,} \tau ^*(S) \leq c_i$ for all i ∈ M, where the inequality uses the fact that τ ^∗ is a feasible solution to problem (7.5). Thus, the solution ${\hat x}$ satisfies the first set of constraints in problem (7.6). Similarly, using the fact that ∑_{j ∈ N} π _j(S) + π ₀(S) = 1 for all S ⊆ N and ∑_S⊆N τ ^∗(S) = 1, we obtain $\sum _{j \in N} {\hat x}_j + x_0 = \sum _{j \in N} \sum _{S \subseteq N} \pi _j(S) {\,} \tau ^*(S) + \sum _{S \subseteq N} \pi _0(S) {\,} \tau ^*(S) = \sum _{S \subseteq N} (\sum _{j \in N} \pi _j(S) + \pi _0(S)) {\,} \tau ^*(S) = 1$, which implies that the solution ${\hat x}$ satisfies the second constraint in problem (7.6). Finally, if j ∈ S, then π _j(S)∕v _j = 1∕(v ₀ + V (S)) = π ₀(S)∕v ₀, whereas if j∉S, then π _j(S)∕v _j = 0. Thus, we obtain ${\hat x}_j / v_j = \sum _{S \subseteq N} \pi _j(S) {\,} \tau ^*(S) / v_j = \sum _{S \subseteq N} \mathbf {1}(j \in S) {\,} \pi _0(S) {\,} \tau ^*(S) / v_0 \leq \sum _{S \subseteq N} \pi _0(S) {\,} \tau ^*(S) / v_0 = {\hat x}_0 / v_0$ for all j = 1, …, n, which shows that the solution ${\hat x}$ satisfies the third set of constraints in problem (7.6). The discussion so far implies that ${\hat x}$ is a feasible solution to problem (7.6). Furthermore, we have $\varLambda \sum _{j \in N} p_j {\,} {\hat x}_j = \varLambda \sum _{j \in N} \sum _{S \subseteq N} p_j {\,} \pi _j(S) {\,} \tau ^*(S) = {\bar V}(T,c)$, where the last equality uses the fact that τ ^∗ is an optimal solution to problem (7.5).

Second, let $x^* = \{ x_j^* : j = 0,1,\ldots ,n \}$ be an optimal solution to problem (7.6). We reindex the products such that $x_1^* / v_1 \geq x_2^* /v_2 \geq \ldots \geq x_n^* / v_n$. Noting the third set of constraints in problem (7.6), we also have $x_0^* / v_0 \geq x_1^* / v_1 \geq x_2^* /v_2 \geq \ldots \geq x_n^* / v_n$. Label the sets S ₀ = ∅ and S _j = {1, 2, …, j} for all j = 1, …, n. Construct a solution $\hat {\tau } = \{\hat {\tau }(S) : S \subseteq N\}$ to problem (7.5) by setting

$$\displaystyle \begin{aligned} \hat{\tau}(S_j) = \Big[ \frac{x_j^*}{v_j} - \frac{x_{j+1}^*}{v_{j+1}} \Big] {\,} V(S_j) \end{aligned} $$

for all j = 0, 1, …, n with the convention that $\hat {\tau }(S_n) = [x_n^* / v_n] {\,} V(S_n)$. Set $\hat {\tau }(S) = 0$ for all S∉{S ₀, S ₁, …, S _n}. We will show that $\hat {\tau }$ is a feasible solution to problem (7.5) and provides an objective value of ${\tilde V}(T,c)$ for this problem, in which case, we obtain ${\bar V}(T,c) \geq {\tilde V}(T,c)$.

Using the definition of $\hat {\tau }(S)$ and noting that ODF j is in the sets S _j, S _j+1, …, S _n but not in S ₀, S ₁, …, S _j−1, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \sum_{S \subseteq N} \pi_j(S) {\,} \hat{\tau}(S) = \pi_j(S_j) {\,} \hat{\tau}(S_j) + \pi_j(S_{j+1}) {\,} \hat{\tau}(S_{j+1}) + \ldots + \pi_j(S_n) {\,} \hat{\tau}(S_n) \\ &\displaystyle &\displaystyle \ \quad = v_j {\,} \Big[ \frac{x_j^*}{v_j} - \frac{x_{j+1}^*}{v_{j+1}} \Big] + v_j {\,} \Big[ \frac{x_{j+1}^*}{v_{j+1}} - \frac{x_{j+2}^*}{v_{j+2}} \Big] + \ldots + v_j {\,} \Big[ \frac{x_n^*}{v_n} \Big] = x_j^*, {} \end{array} \end{aligned} $$

(7.12)

where use the definition of π _j(S) in the second equality above. In this case, we have $\varLambda \sum _{S \subseteq N} \sum _{j \in N} a_{ij} {\,} \pi _j(S) {\,} \hat {\tau }(S) = \varLambda \sum _{j \in N} a_{ij} {\,} (\sum _{S \subseteq N} \pi _j(S) {\,} \hat {\tau }(S)) = \varLambda \sum _{j \in N} a_{ij} {\,} x_j^* \leq c_i$, where the last inequality uses the fact that x ^∗ is a feasible solution to problem (7.6). So, the solution $\hat {\tau }$ satisfies the first set of constraints in problem (7.5). On the other hand, we have

$$\displaystyle \begin{aligned} & \sum_{S \subseteq N} \hat{\tau}(S) = \hat{\tau}(S_0) + \hat{\tau}(S_1) + \ldots + \hat{\tau}(S_n) \\ {} & ~ = V(S_0) {\,} \Big[ \frac{x_0^*}{v_0} - \frac{x_1^*}{v_1} \Big] + V(S_1) {\,} \Big[ \frac{x_1^*}{v_1} - \frac{x_2^*}{v_2} \Big] + \ldots + V(S_n) {\,} \Big[ \frac{x_n^*}{v_n} \Big] \\ {} & ~ = \frac{x_0^*}{v_0} {\,} V(S_0) + \frac{x_1^*}{v_1^*} {\,} ( V(S_1) - V(S_0)) + \frac{x_2^*}{v_2^*} {\,} ( V(S_2) - V(S_1)) + \ldots \\ {} &~~~~~~~~~~~~~~~~~~~~~~~~~~ + \frac{x_n^*}{v_n} {\,} {\,} ( V(S_n) - V(S_{n-1})) ~ = x_0^* + x_1^* + x_2^* + \ldots + x_n^* = 1, \end{aligned} $$

where the third equality follows by arranging the terms, the fourth equality is by the fact that V (S _j) − V (S _j−1) = v _j, and the fifth equality uses the fact that x ^∗ satisfies the second constraint in problem (7.6). Thus, the solution $\hat {\tau }$ satisfies the second constraint in problem (7.5). So, $\hat {\tau }$ is a feasible solution to problem (7.5). Finally, we have $\varLambda {\,} \sum _{S \subseteq N} \sum _{j \in N} p_j {\,} \pi _j(S) {\,} \hat {\tau }(S) = \varLambda {\,} \sum _{j \in N} p_j {\,} x_j^* = {\tilde V}(T,c)$, where the first equality uses (7.12) and the second equality uses the fact that x ^∗ is an optimal solution to problem (7.6).

Proof of Theorem 7.3

Letting z = {z _i : i ∈ M} and β, respectively, be the dual variables associated with the first and second sets of constraints in problem (7.5) and expanding the values of R(S) and A(S) by using their definitions, the dual of problem (7.5) is

$$\displaystyle \begin{aligned}\begin{array}{r*{20}l} {\bar V}(T,c) = \min ~~~~ & \sum_{i \in M} c_i {\,} z_i + \beta {} \\ {} \mbox{s.t.} ~~~~ \varLambda & \sum_{i \in M} \sum_{j \in N} a_{ij} {\,} \pi_j(S) {\,} z_i + \beta \geq \varLambda \sum_{j \in N} p_j {\,} \pi_j(S) \qquad && \forall {\,} S \subseteq N \\ {} & z_i \geq 0,~ \beta \mbox{ is free} && \forall {\,} i \in M. \end{array}\end{aligned} $$

(7.13)

Arranging the terms, we can write the first set of constraints above succinctly as

$$\displaystyle \begin{aligned}\beta \geq \varLambda {\,} \max_{S \subseteq N}R(S, A'z)\end{aligned}$$

whereas before, R(S, A′z) =∑_{j ∈ S}(p _j −∑_{i ∈ M} a _ij z _i) π _j(S).

Because problem (7.13) is a minimization problem and the objective function coefficient of the decision variable β is positive, the decision variable β takes the value $\varLambda {\mathcal {R}}(A'z)$. Thus, problem (7.13) is equivalent to

$$\displaystyle \begin{aligned} {\bar V}(T,c) = \min_{z \in \Re_+^m} \left\{ c'z + \varLambda \max_{S \subseteq N}R(S, A'z) \right\}. {} \end{aligned} $$

(7.14)

The maximization problem above is an assortment optimization problem, where the revenue associated with ODF j is p _j −∑_{i ∈ M} a _ij z _i and the customers make their choices according to the MC choice model. Using the discussion in assortment optimization problems under the MC choice model, we know that the optimal objective value of this assortment optimization problem can be obtained by solving the linear program

$$\displaystyle \begin{aligned} & \varLambda \max_{S \subseteq N}R(S, A'z) \\ & ~= \min_{v \in \Re^n} \Bigg\{ \varLambda {\,} \sum_{j \in N} \gamma_j {\,} v_j : v_j \geq p_j - \sum_{i \in M} a_{ij} {\,} z_i ~\forall {\,} j \in N,~ v_j \geq \sum_{k=1}^n \rho_{jk} {\,} v_k ~\forall {\,} j \in N \Bigg\} , \end{aligned} $$

where the second problem above is a linear program involving the decision variables v = {v _j : j ∈ N}. Thus, problem (7.14) can be written as a two-level problem, where minimize at both levels, the decision variables at the outer level are z and the decision variables at the inner level are v. Since we minimize at both levels, we can solve this problem as a single-level problem, simultaneously minimizing over the decision variables z and v to obtain the problem

$$\displaystyle \begin{aligned}\begin{array}{r*{20}l} {\bar V}(T,c) ~=~ \min ~~~~ & \sum_{i \in M} c_i {\,} z_i + \varLambda {\,} \sum_{j \in N} \gamma_j {\,} v_j \\ \mbox{s.t.} ~~~~ & v_j \geq p_j - \sum_{i \in M} a_{ij} {\,} z_i \qquad && \forall {\,} j \in N \\ & v_j \geq \sum_{k=1}^n \rho_{jk} {\,} v_k && \forall {\,} j \in N \\ & z_i \geq 0,~ v_j \mbox{ is free} && \forall {\,} i \in M,~j \in N \end{array}\end{aligned} $$

Letting {Λ x _j : j ∈ N} and {Λ y _j : j ∈ M}, respectively, be the dual variables associated with the two sets of constraints above and writing down the dual of this problem, we immediately obtain problem (7.7). Therefore, the optimal objective value of problem (7.7) is equal to the optimal objective value of the problem above, as desired.