Network Revenue Management with Independent Demands

Abstract

In this chapter, we consider a firm that has finite capacities of several resources that can be instantly combined into different products with fixed prices. We assume that there is an independent demand stream for each of the products that arrives as a Poisson process. A requested product is purchased if available. The firm generates the revenue associated with the sale and updates the inventories of the resources consumed by the product. If the requested product is not available, then the customer leaves the system without purchasing. The objective of the firm is to decide which products to make available over a finite sales horizon to maximize the total expected revenue from fixed initial inventories that cannot be replenished during the sales horizon.

Appendix

Proof of Proposition 2.5

Consider an ODF j ∈ F, and notice that $\beta ^*_j \geq p_j - \sum _{i \in M} a_{ij} {\,} z_i^* > 0$, where we use the fact that (z ^∗, β ^∗) is feasible to problem (2.8). Therefore, the second set of constraints for ODF j in problem (2.6) has a strictly positive dual variable, implying that its slack must be zero and we obtain $y_j^* = \varLambda _j$. The last equality implies that j ∈ F′. So, F ⊆ F′. Similarly, consider an ODF j ∈ R. Thus, we have $\sum _{i \in M} a_{ij} {\,} z_i^* + \beta _j^* - p_j > 0$, where we use the fact that $\sum _{i \in M} a_{ij} {\,} z_i^* > p_j$ and $\beta ^*_j \geq 0$. Therefore, the first set of constraints for ODF j in problem (2.8) has a strictly positive slack, which implies that the dual variable associated with this constraint must be zero at the optimal solution and we obtain $y_j^* = 0$. The last equality implies that j ∈ R′. So, R ⊆ R′. Since F ∪ P ∪ R = F′∪ P′∪ R′ = N, it also follows that P′⊆ P.

Proof of Theorem 2.7

Consider a variant of the PAC heuristic based on always admitting requests for product j ∈ N with probability $y^*_j/\varLambda _j$. Notice that these probabilities are independent of the scaling parameter b, since the solution to the primal and dual linear programs are insensitive to the scaling parameter. Notice also that this variant of the PAC heuristic ignores inventory considerations, and it may end up overbooking the resources. Because the demand for product j is Poisson with parameter bΛ _j and we admit demands with probability y ^∗∕Λ _j, sales for product j form a thinned Poisson process with mean $by^*_j$ for each j ∈ N. Consequently, the expected revenue from this variant of the PAC heuristic is $b \sum _{j \in N}p_jy^*_j = \bar {V}^b(T,c) = b\bar {V}(T,c)$. From this expected revenue, we need to deduct the cost for overbooking capacity. Let S _j denote the random sales for product j, from our earlier discussion this is a Poisson random variable with mean $by^*_j$ and variance $by^*_j$. Suppose that we are charged an overbooking cost θ _i for each unit of capacity of resource i that we consume in excess of capacity. Then the overbooking costs are equal to

$$\displaystyle \begin{aligned}\sum_{i \in M} \theta_i [\sum_{j \in N} a_{ij}S_j - bc_i]^+.\end{aligned}$$

Consequently, the expected revenue of this variant of the PAC heuristic, net of overbooking costs is of the form

$$\displaystyle \begin{aligned}\underline{\varPi}^b(T,c) = b \bar{V}(T,c) - \sum_{i \in M} \theta_i {\,} \mathbb E[\sum_{j \in N} a_{ij}S_j - bc_i]^+.\end{aligned}$$

We now claim that if we select

$$\displaystyle \begin{aligned}\theta_i \geq \max\{p_j: a_{ij} = 1,j \in N\},\end{aligned}$$

then

$$\displaystyle \begin{aligned} \underline{\varPi}^b(T,c) \leq \varPi^b(T,c) \leq V^b(T,c) \leq \bar{V}^b(T,c), \end{aligned} $$

(2.21)

where ${\bar V}^b(T,c)$ is the optimal objective value of the deterministic linear program with a scaling factor of b. The first inequality above follows from a sample path argument. Notice that as long as the capacities are not violated, both the PAC heuristic and the alternative policy make the same decisions. If the alternative policy sells a ticket for an ODF and violates the capacity, then it incurs a penalty that is larger than the revenue from the sold ODF, losing revenue from the sale. Due to this decision, the alternative policy may also consume capacities of other available resources. Thus, the alternative policy not only loses money from the sale, but it is also left with even less capacity than the PAC heuristic. So, the revenue net of overbooking costs is always smaller than the revenue generated by the PAC heuristic. The second inequality follows because the PAC is a heuristic and its performance is bounded above by the expected revenue of the optimal policy. The last inequality follows because the optimal objective value of the deterministic linear program is an upper bound on the optimal total expected revenue.

Dividing the string of inequalities in (2.21) by $\bar {V}^b(T,c)$ results in the following string of inequalities:

$$\displaystyle \begin{aligned} 1 - \frac{ \sum_{i \in M} \theta_i {\,} \mathbb E[\sum_{j \in N} a_{ij}S_j - bc_i]^+ }{\bar{V}^b(T,c)} = \frac{\underline{\varPi}^b(T,c)}{\bar{V}^b(T,c)} \leq \frac{\varPi^b(T,c)} {\bar{V}^b(T,c)} \leq 1. \end{aligned} $$

Consequently, if we can show that

$$\displaystyle \begin{aligned}\lim_{b \rightarrow \infty} \frac{ \sum_{i \in M} \theta_i {\,} \mathbb E[\sum_{j \in N} a_{ij}S_j - bc_i]^+ }{\bar{V}^b(T,c)} = 0,\end{aligned}$$

then it would follow that

$$\displaystyle \begin{aligned}1 = \lim_{b \rightarrow \infty} \frac{\varPi^b(T,c)}{\bar{V}^b(T,c)} \leq \lim_{b \rightarrow \infty} \frac{\varPi^b(T,c)}{V^b(T,c)} \leq 1.\end{aligned}$$

For each i, consider the random variable $Z_i = \sum _{j =1}^na_{ij}S_j$ corresponding to the aggregate demand for resource i under the variant of the PAC heuristic that accepts request for product j with probability $y^*_j/ \varLambda _j$ regardless of capacity. Note that $\mathbb E[Z_i] = b \sum _{j \in N}a_{ij}y^*_j \leq bc_i$, where we use the fact that $\{y_j^* : j \in N\}$ is a feasible solution to problem (2.6). Also, the variance of Z _i satisfies $\mbox{Var}[Z_i] = b\sum _{j \in N}a_{ij}y^*_j$, where we have used the fact that $a_{ij}^2 = a_{ij}$ and that both the mean and the variance of S _j are equal to $by^*_j$. We now use the bound on partial expectations

$$\displaystyle \begin{aligned}\mathbb E[(Z-z)^+] \leq 0.5(\sqrt{\sigma^2+(z-\mu)^2}-(z-\mu)) \leq \frac{1}{2} \sigma + \frac{1}{2} (|z-\mu|- (z-\mu)),\end{aligned}$$

that holds for all random variables Z with mean μ and variance σ ², and arbitrary constant z; see Gallego (1992). Notice that the last term vanishes when z ≥ μ. Applying the bound to the random variable Z _i and to the constant $z_i = bc_i \geq \mathbb E [Z_i]$, we obtain

$$\displaystyle \begin{aligned}\mathbb E[ \sum_{j =1}^na_{ij}S_j - c_i]^+ \leq \frac{1}{2} \sqrt{b\sum_{j \in N}a_{ij}y^*_j}.\end{aligned}$$

Multiplying by the last expression by θ _i adding over i and dividing by $b \sum _{j =1}^n p_jy^*_j$, we see that

$$\displaystyle \begin{aligned}\frac{\sum_{i \in M} \theta_i {\,} \mathbb E[\sum_{j \in N} a_{ij}D_j - bc_i]^+ }{\bar{V}^b(T,c)} \leq \frac{\frac{1}{2} \sum_{i \in M} \theta_i \sqrt{\sum_{j \in N}a_{ij}y^*_j}}{ \sqrt{b} \sum_{j \in N} p_j {\,} y^*_j}.\end{aligned}$$

Notice that the ratio goes to zero at rate $1/\sqrt {b}$ as b →∞.

Proof of Theorem 2.8

The proof of the first inequality is essentially identical to that of Theorem 2.2 and we omit it. To see the second inequality, let ${\{ y_{tj}^* : t=1,\ldots ,T,~ j \in N\}}$ and ${\{ x_{ti}^* : t = 1,\ldots ,T,~ i \in M\}}$ be an optimal solution to problem (2.10). For each i ∈ M, adding the first two sets of constraints overall t = 2, …, T yields $\sum _{i \in N} a_{ij} {\,} \sum _{t=2}^T y_{tj}^* + x_{1i}^* = c_i$. On the other hand, for each i ∈ M, adding the third set of constraints for t = 1 overall j ∈ N yields $\sum _{i \in N} a_{ij} {\,} y_{1j}^* \leq \sum _{j \in N} \lambda _{1j} {\,} x_{1i}^* \leq x_{1i}^*$. Combining the inequalities, we obtain $\sum _{i \in N} a_{ij} {\,} \sum _{t=1}^T y_{tj}^* \leq c_i$ for all i = 1, …, n, implying that the solution $\{ \sum _{t=1}^T y_{tj}^* : j \in N\}$ satisfies the first set of constraints in problem (2.6). Furthermore, adding the fourth set of constraints in problem (2.10) overall t = 1, …, T, we obtain $\sum _{t=1}^T y_{tj}^* \leq \sum _{t=1}^T \lambda _{tj}$ for all j ∈ N, so that the solution $\{ \sum _{t=1}^T y_{tj}^* : j \in N\}$ satisfies the second set of constraints in problem (2.6) as well. Also, we have $\sum _{i \in N} p_j \sum _{t=1}^T {\,} y_{tj}^* = {\tilde V}(T,c)$ by the definition of $\{ y_{tj}^* : t=1,\ldots ,T,~ j \in N\}$. Therefore, $\{ \sum _{t=1}^T y_{tj}^* : j \in N \}$ is a feasible solution to problem (2.6) and it provides an objective value of ${\tilde V}(T,c)$ for this problem, which imply that the optimal objective value of problem (2.6) can only be larger than ${\tilde V}(T,c)$, yielding ${\bar V}(T,c) \geq {\tilde V}(T,c)$.

Proof of Lemma 2.15

The function [p _j −∑_{i ∈ M} α _τij]⁺ is convex in α. Noting (2.17), it is enough to show that $v_i^\alpha (t,x_i)$ is a convex function of α. The dynamic program in (2.16) characterizes $v_i^\alpha (t,x_i)$. Thus, by the discussion in Sect. 2.8, the value functions $\{ v_i^\alpha (t,\cdot ) : t =1,\ldots ,T\}$ can be obtained by solving the linear program

$$\displaystyle \begin{aligned} \min ~~~ & \nu_i(t , x_i) \\ \mbox{s.t.} ~~~ & \nu_i(\tau,x_i) \geq \sum_{j \in N} \lambda_{\tau j} \Big\{ \alpha_{\tau ij} {\,} w_{ij} + \nu_i(\tau -1, x_i- {\,} w_{ij} {\,} a_{ij} ) \Big\} \\ & \forall {\,} \tau=1,\ldots,T,~x_i \in C_i,~ w_i \in {\mathcal U}_i(x_i), \end{aligned} $$

where the decision variables are {ν _i(τ, x _i) : τ = 1, …, T, x _i ∈ C _i}. The optimal objective value of the problem above provides $v_i^\alpha (t,x_i)$. The set of Lagrange multipliers α appear only on the right side of the constraints above. Thus, the optimal objective value of the problem above is convex in α by linear programming duality and the desired result follows.