Assortment Optimization

Abstract

A fundamental question in revenue management involves deciding which fares to offer in response to a request from an origin to a destination. The solution depends on available capacity, the time of departure, and how consumers make choices. This problem needs to be solved in real time as travel requests arrive. The assortment optimization problem is crucial in other RM applications such as hotels and car rentals, and is becoming more important in retailing and e-commerce. The fundamental tradeoff in assortment optimization is that broad assortments result in demand cannibalization and spoilage, while narrow assortments result in disappointed consumers that may walk away without purchasing. The formulation can be interpreted broadly to include more strategic decisions such as the location of stores within a city. The profitability of an assortment can be best captured through a choice model that provides sale probabilities as a function of the set of products contained in the assortment. In this chapter, we formulate and solve assortment optimization problems for many of the choice models presented in the previous chapter.

Appendix

Proof of Theorem 5.1

To see that a nested-by-revenue assortment solves problem (5.2), notice that (5.2) implies ${\mathcal {R}}^* \geq \sum _{j \in S} p_j {\,} v_j / (v_0 + V(S))$ for all sets S ⊆ N with equality holding at optimal assortments. The last inequality is equivalent to $v_0 {\,} {\mathcal {R}}^* \geq \sum _{j \in S} (p_j -{\mathcal {R}}^*) {\,} v_j$. Therefore, $v_0 {\,} {\mathcal {R}}^* \geq \sum _{j \in S} (p_j -{\mathcal {R}}^*) {\,} v_j$ for all sets S ⊆ N with equality holding at optimal assortments. The last statement implies that an optimal assortment can be recovered by solving the problem

$$\displaystyle \begin{aligned} \max_{S \subseteq N} \Bigg\{ \sum_{j \in S} (p_j -{\mathcal{R}}^*) {\,} v_j \Bigg\}. \end{aligned} $$

There is a simple solution to the problem above. It is optimal to offer each product whose revenue exceeds ${\mathcal {R}}^*$. Therefore, the optimal solution is $S^* = \{ j \in N : p_j \geq {\mathcal {R}}^*\}$. Notice that the products are indexed such that p ₁ ≥ p ₂ ≥… ≥ p _n, so S ^∗ given in the last expression has be to one of the sets E ₀, E ₁, …, E _n.

Proof of Theorem 5.7

The fact that H(E _i) ≥ H(E _i−1) follows directly from (5.12). Clearly H(E ₁) = λ ₁ p ₁ > 0, so $\tilde {E}_1 = E_1 = \{1\}$ and $R(\tilde {E}_1) = H(E_1)$. Assume that the result holds for i − 1, so $H(E_{i-1}) = R(\tilde {E}_{i-1}) \geq R(E)$ for all E ⊆ E _i−1. We will show that $H(E_i) = R(\tilde {E}_i) \geq R(E)$ for all E ⊆ E _i. From the algorithm,

$$\displaystyle \begin{aligned} H(E_{i}) & = H(E_{i-1}) +\lambda_{i}(p_{i}-H(E_{i-1}))^+ = R(\tilde{E}_{i-1}) + \lambda_i (p_i - R(\tilde{E}_{i-1}))^+. \end{aligned} $$

If p _i ≤ H(E _i−1), then we have $H(E_i) = H(E_{i-1}) = R(\tilde {E}_{i-1}) = R(\tilde {E}_i)$ on account of $\tilde {E}_i = \tilde {S}_{i-1}$. On the other hand, if p _i > H(E _i−1), then we have $\tilde {E}_i = \tilde {E}_{i-1} \cup \{i\}$, and

$$\displaystyle \begin{aligned} H(E_i) = (1 - \lambda_i) R(\tilde{E}_{i-1}) + \lambda_i p_i = R(\tilde{E}_i), \end{aligned} $$

where the last equality follows from Eq. (5.11). To complete the proof, we need to show that $R(\tilde {E}_i) \geq R(E)$ for all $E \subseteq \tilde E_i$. If i∉E, then we have $R(E) \leq R(\tilde {E}_{i-1}) \leq R(\tilde {E}_i)$. On the other hand, if i ∈ E, then we can write E = T ∪{i} for some T ⊆ E _i−1. In this case, we get

$$\displaystyle \begin{aligned} R(E) & = R(T \cup \{i\}) = (1-\lambda_i) {\,} R(T) + \lambda_i {\,} p_i \\ & \leq (1 -\lambda_i) {\,} R(\tilde{E}_{i-1}) + \lambda_i {\,} p_i = (1 - \lambda_i) {\,} H(E_{i-1}) + \lambda_i {\,} p_i \\ & \leq H(E_{i-1}) + \lambda_i {\,} (p_i - H(E_{i-1})^+ \\ & = H(E_i) = R(\tilde{E}_i), \end{aligned} $$

as desired.

Proof of Theorem 5.4

For notational brevity, we let $V_i^* = V_i(x_i^*)$, $\varPi _i^* = R_i(x_i^*)$, ${\hat V}_i = V_i({\hat x}_i)$ and ${\hat \varPi }_i = R({\hat x}_i)$, where $x_i^*$ is an optimal solution to problem (5.7) and ${\hat x}_i$ is an optimal solution to problem (5.9). It is enough to show that

$$\displaystyle \begin{aligned} {\hat V}_i^{\gamma_i} {\,} ({\hat \varPi}_i - z) \geq (V_i^*)^{\gamma_i} {\,} (\varPi_i^* - z). \end{aligned} $$

First, assume that $z \geq \varPi _i^*$. By the definition of $u_i^*$, we have $u_i^* = z$. Also, the definition of ${\hat x}_i$ implies that ${\hat V}_i{\,} ({\hat \varPi }_i - u_i^*) \geq 0$ because not offering any of the products provides an objective value of zero for problem (5.9) so that the optimal objective value of this problem should at least be zero. Using the definition of $u_i^*$ and the fact that $z \geq \varPi _i^*$, the last inequality yields ${\hat V}_i^{\gamma _i} {\,} ({\hat \varPi }_i - z) = {\hat V}_i^{\gamma _i} {\,} ({\hat \varPi }_i - u_i^*) \geq 0$. On the other hand, since $z \geq \varPi _i^*$, we have $0 \geq (V_i^*)^{\gamma _i} {\,} (\varPi _i^* - z)$. Thus, the last two inequalities yield ${\hat V}_i^{\gamma _i} {\,} ({\hat \varPi }_i - z) \geq (V_i^*)^{\gamma _i} {\,} (\varPi _i^* - z)$, as desired.

Second, assume that $\varPi _i^* > z$. By the definition of ${\hat x}_i$, we have ${\hat V}_i {\,} ({\hat \varPi }_i - u_i^*) \geq V_i^* {\,} (\varPi _i^* - u_i^*)$. Using the definition of $u_i^*$ and the fact that $\varPi _i^* > z$, we have $u_i^* = \gamma _i {\,} z + (1 - \gamma _i) {\,} \varPi _i^*$. Using the last equality, the last inequality can equivalently be written as ${\hat V}_i {\,} ({\hat \varPi }_i - \gamma _i {\,} z - (1 - \gamma _i) {\,} \varPi _i^* ) \geq \gamma _i {\,} V_i^* {\,} (\varPi _i^* - z)$. Multiplying both sides of this inequality by ${\hat V}_i^{\gamma _i - 1}$ yields ${\hat V}_i^{\gamma _i} {\,} ({\hat \varPi }_i - \gamma _i {\,} z - (1 - \gamma _i) {\,} \varPi _i^* ) \geq \gamma _i {\,} V_i^* {\,} {\hat V}_i^{\gamma _i-1} {\,} (\varPi _i^* - z)$. Arranging the terms, we write this inequality as

$$\displaystyle \begin{aligned} {\hat V}_i^{\gamma_i} ({\hat \varPi}_i - z) \geq [\gamma_i {\,} V_i^* {\,} {\hat V}_i^{\gamma_i-1} + (1 - \gamma_i) {\,} {\hat V}_i^{\gamma_i}] {\,} (\varPi_i^* - z). \end{aligned} $$

Since γ _i ∈ [0, 1], $u^{\gamma _i}$ is a concave function of u. Using the subgradient inequality for this concave function, we have ${\hat V}_i^{\gamma _i} + \gamma _i {\,} {\hat V}_i^{\gamma _i -1} {\,} (V_i^* - {\hat V}_i) \geq (V_i^*)^{\gamma _i}$, which can also be written as $(1 - \gamma _i) {\,} {\hat V}_i^{\gamma _i} + \gamma _i {\,} V_i^* {\,} {\hat V}_i^{\gamma _i-1} \geq (V_i^*)^{\gamma _i}$. Using this inequality in the inequality displayed above, it follows that

$$\displaystyle \begin{aligned} {\hat V}_i^{\gamma_i} ({\hat \varPi}_i - z) \geq [\gamma_i {\,} V_i^* {\,} {\hat V}_i^{\gamma_i-1} + (1 - \gamma_i) {\,} {\hat V}_i^{\gamma_i}] {\,} (\varPi_i^* - z) \geq (V_i^*)^{\gamma_i} {\,} (\varPi_i^* - z), \end{aligned} $$

which is the desired result.

Proof of Theorem 5.9

The proof has two steps. First, we give an equivalent formulation of problem (5.16). In particular, using ${\mathcal {R}}^*$ to denote the optimal objective value of problem (5.16), we claim that problem (5.16) is equivalent to the linear program

$$\displaystyle \begin{aligned}\begin{array}{r*{20}l} \max ~~~ & \sum_{j \in N} (p_j - {\mathcal{R}}^*) {\,} \frac{v_j}{v_0} {\,} x_j {} \\ \mbox{s.t.} ~~~ & \sum_{j \in N} a_{ij} {\,} x_j \leq b_i \qquad && \forall {\,} i \in L \\ & 0 \leq x_j \leq 1 && \forall {\,} j \in N \end{array}\end{aligned} $$

(5.27)

in the sense that an optimal solution to problem (5.16) can be recovered by using an optimal solution to problem (5.27) and the two problems share the same optimal objective value. To see the claim, we define the feasible set ${\mathcal F} = \{ x \in \{0,1\}^n: \sum _{j \in N} a_{ij} {\,} x_j \leq b_i ~ \forall {\,} i \in L\}$ to capture the feasible set of problem (5.16). Since ${\mathcal {R}}^*$ is the optimal objective value of problem (5.16), we have ${\mathcal {R}}^* \geq \sum _{j \in N} p_j {\,} v_j {\,} x_j / (v_0 +\sum _{j \in N} v_j {\,} x_j)$ for all $x \in {\mathcal F}$ and the inequality holds as equality at the optimal solution to problem (5.16). Writing the last inequality equivalently as ${\mathcal {R}}^* \geq \sum _{j \in N} (p_j - {\mathcal {R}}^*) {\,} v_j {\,} x_j / v_0$, it follows that we have ${\mathcal {R}}^* \geq \sum _{j \in N} (p_j - {\mathcal {R}}^*) {\,} v_j {\,} x_j / v_0$ for all $x \in {\mathcal F}$ and the last equality holds as equality at the optimal solution. Therefore, we can obtain an optimal solution to problem (5.16) by solving

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \max_{x \in \{0,1\}^n} \Bigg\{ \sum_{j \in N} (p_j - {\mathcal{R}}^*) {\,} v_j {\,} x_j / v_0 : x \in {\mathcal F} \Bigg\} \\ &\displaystyle &\displaystyle \qquad \qquad \qquad = \max_{x \in \{0,1\}^n} \Bigg\{ \sum_{j \in N} (p_j - {\mathcal{R}}^*) {\,} v_j {\,} x_j / v_0 : \sum_{j \in N} a_{ij} {\,} x_j \leq b_i~ \forall {\,} i \in L \Bigg\} \end{array} \end{aligned} $$

and the optimal objective value of these problems would be ${\mathcal {R}}^*$. The objective function and constraints in the last problem are linear. Since the constraint matrix is TU, we can relax the binary constraints without loss of optimality. If we relax the binary constraints in the last problem, then we obtain problem (5.27), which establishes the claim. Thus, it is enough to show that problems (5.17) and (5.27) are equivalent to each other.

Let $y^* = \{ y_j^* : j \in N\}$ and $y_0^*$ be an optimal solution to problem (5.17) with the corresponding optimal objective value ζ ^∗. Let x ^∗ be an optimal solution to problem (5.27). By the discussion above, the optimal objective value of problem (5.27) is ${\mathcal {R}}^*$. In the second part of the proof, we show that ${\mathcal {R}}^* = \zeta ^*$. We construct a solution $({\hat y},{\hat y}_0)$ to problem (5.17) as follows. For all j ∈ N, ${\hat y}_j = v_j {\,} x_j^* / (v_0 + \sum _{i \in N} v_i {\,} x_i^*)$ and ${\hat y}_0 = 1 - \sum _{j \in N} {\hat y}_j = v_0 / (v_0 + \sum _{j \in N} v_j {\,} x_j^*)$. By definition, $({\hat y},{\hat y}_0)$ satisfies the first constraint in problem (5.17). Furthermore, we have

$$\displaystyle \begin{aligned} \sum_{j \in N} a_{ij} {\,} \frac{{\hat y}_j}{v_j} = \frac{\sum_{j \in N} a_{ij} {\,} x_j^*}{v_0 + \sum_{j \in N} v_j {\,} x_j^*} \leq \frac{b_i}{v_0 + \sum_{j \in N} v_j {\,} x_j^*} = \frac{b_0}{v_0} {\,} {\hat y}_0 \end{aligned} $$

for all i ∈ L, where the equalities use the definition of ${\hat y}_j$ and ${\hat y}_0$, whereas the inequality uses the fact that x ^∗ is a feasible solution to problem (5.27). Thus, $({\hat y},{\hat y}_0)$ satisfies the second set of constraints in problem (5.17). Also, we have ${\hat y}_j / v_j = x_j^* / (v_0 + \sum _{j \in N} v_j {\,} x_j^*) \leq 1 / (v_0 + \sum _{j \in N} v_j {\,} x_j^*) = {\hat y}_0 / v_0$ for all j ∈ N, indicating that $({\hat y},{\hat y}_0)$ satisfies the third set of constraints in problem (5.17). Therefore, $({\hat y},{\hat y}_0)$ is a feasible solution to problem (5.17). In this case, the objective value provided by this feasible solution for problem (5.17) can at most be ζ ^∗, so that we obtain

$$\displaystyle \begin{aligned} \zeta^* \geq \sum_{j \in N} p_j {\,} {\hat y}_j = \frac{ \sum_{j \in N} p_j {\,} v_j {\,} x_j^*}{ v_0 + \sum_{j \in N} v_j {\,} x_j^*} = {\mathcal{R}}^*, \end{aligned} $$

where the first equality is by the definition of ${\hat y}_j$ and the second equality uses the fact that x ^∗ is an optimal solution to problem (5.27), whose optimal objective value is ${\mathcal {R}}^*$. So, we have $\zeta ^* \geq {\mathcal {R}}^*$. To get a contradiction, assume that $\zeta ^* > {\mathcal {R}}^*$. We construct the solution ${\hat x}$ to problem (5.27) as ${\hat x}_j = (y_j^* / v_j) / (y_0^* / v_0)$ for all j ∈ N. Noting the third set of constraints in problem (5.17), we have $0 \leq {\hat x}_j \leq 1$. Furthermore, we have

$$\displaystyle \begin{aligned} \sum_{j \in N} a_{ij} {\,} {\hat x}_j = \sum_{j \in N} a_{ij} {\,} \frac{y_j^* / v_j}{y_0^* / v_0} \leq b_i \end{aligned} $$

for all i ∈ L, where the inequality uses the fact that y ^∗ is a feasible solution to problem (5.17). Thus, ${\hat x}$ is a feasible solution to problem (5.27). In this case, the objective value provided by ${\hat x}$ for problem (5.27) can at most be ${\mathcal {R}}^*$, yielding

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle {\mathcal{R}}^* \geq \sum_{j \in N} (p_j - {\mathcal{R}}^*) \frac{v_j}{v_0} {\,} {\hat x}_j = \sum_{j \in N} (p_j - {\mathcal{R}}^*) \frac{y_j^*}{y_0^*} \\ &\displaystyle &\displaystyle \qquad \qquad \qquad = \frac{1}{y_0^*} \sum_{j \in N} p_j {\,} y_j^* - \frac{1}{y_0^*} {\mathcal{R}}^* ( 1 - y_0^*) > \frac{\zeta^*}{y_0^*} - \frac{\zeta^*}{y_0^*} ( 1 - y_0^*) = \zeta^*, \end{array} \end{aligned} $$

where the second inequality uses the fact that $\sum _{j \in N} p_j {\,} y_j^* = \zeta ^*$ and the assumption that $\zeta ^* > {\mathcal {R}}^*$. The chain of inequalities contradict the assumption that $\zeta ^* > {\mathcal {R}}^*$. Thus, we must have $\zeta ^* = {\mathcal {R}}^*$ and the solutions $({\hat y},{\hat y}_0)$ and ${\hat x}$ must be optimal for problems (5.17) and (5.27), respectively.

Proof of Theorem 5.11

Let z ¹ and z ² be two distinct vectors and let α ∈ [0, 1]. Then

$$\displaystyle \begin{aligned} \begin{array}{rcl} {\mathcal{R}}(\alpha A'z^1 + (1- \alpha)A'z^2) &\displaystyle = &\displaystyle \max_{S \subseteq N}R(S,A'(\alpha z^1 + (1- \alpha)z^2))\\ &\displaystyle = &\displaystyle \max_{S \subseteq N}[\alpha R(S,A'z^1) + (1- \alpha)R(S, A'z^2)]\\ &\displaystyle \leq &\displaystyle \alpha \max_{S \subseteq N} R(S,A'z^1) + (1-\alpha) \max_{S \subseteq N} R(S, A'z^2)\\ &\displaystyle = &\displaystyle \alpha {\mathcal{R}}(A'z^1) + (1-\alpha) {\mathcal{R}}(A'z^2). \end{array} \end{aligned} $$

Now suppose that assortment S is optimal for each of {z ^k : k = 1, …, K}. Then ${\mathcal {R}}(A'z^k) = R(S,A'z^k)$ for k = 1, …, K. Let α be a vector in the K-dimensional simplex, so α ≥ 0 and e′α = 1. Let $z = \sum _{k=1}^K \alpha _k z^k$. So,

$$\displaystyle \begin{aligned} \begin{array}{rcl} R(S, A'z) &\displaystyle \leq &\displaystyle {\mathcal{R}}(A'z) \\ &\displaystyle \leq &\displaystyle \sum_{k = 1}^K \alpha_k{\mathcal{R}}(A'z^k)\\ &\displaystyle = &\displaystyle \sum_{k=1}^K \alpha_k R(S,A'z^k)\\ &\displaystyle = &\displaystyle R(S, A'z). \end{array} \end{aligned} $$

The first inequality follows from the definition of ${\mathcal {R}}$. The second from the convexity of ${\mathcal {R}}$. The first equality is from the definition of S, and the last equality is from the definition of z. Thus, all inequalities above are equalities, which implies that S is optimal for $z = \sum _{k =1}^K \alpha ^k z^k$ for any α in the K-dimensional simplex.

Proof of Theorem 5.12

We will first show that for any $\rho \in \Re ^m_+$, the problem that computes Q(ρ) has a solution where t(S) > 0 only for sets $S \in {\mathcal {E}}$. To see this, suppose that $\rho \in \Re ^m_+$ and there is a solution to Q(ρ) =∑_S⊆N R(S) t(S) with t(S) > 0 for some $S \notin {\mathcal {E}}$. Consider the problem that computes Q(A(S)). Then, we have R(S) < Q(A(S)) =∑_U⊆N R(U)t(U) with ∑_U⊆N A(U)t(U) ≤ A(S). We can then substitute S by this convex combination and strictly improve the solution without violating the capacity constraint contradicting the assumed optimality.

Suppose the problem of maximizing R(S, A′z) has a solution, say S ^∗ such that R(S ^∗, A′z) > R(E, A′z) for all $E \in {\mathcal {E}}$. Then for all $E \in {\mathcal {E}}$, we have

$$\displaystyle \begin{aligned}R(E, A'z) < R(S^*,A'z) = R(S^*) - z'A(S^*) < Q(A(S^*)) - z'A(S^*),\end{aligned}$$

where the first inequality follows from the assumption that all efficient sets are suboptimal, and the second inequality follows from the fact that S ^∗ is not efficient. But then there exist a convex combination of efficient sets such that $R(S^*) < Q(A(S^*)) = \sum _{E \in {\mathcal {E}}} R(E) t(E)$ with $A(S^*) \geq \sum _{E \in {\mathcal {E}}}A(E) t(E)$. This implies that

$$\displaystyle \begin{aligned}R(S^*) - z'A(S^*) < \sum_{E \in {\mathcal{E}}}t(E) (R(E) - z'A(E)).\end{aligned}$$

For this last inequality to be true, there must be an $E \in {\mathcal {E}}$ such that

$$\displaystyle \begin{aligned}{\mathcal{R}}(A'z) = R(S^*, A'z) = R(S^*) - z'A(S^*) < R(E) - z'A(E) = R(E, A'z)\end{aligned}$$

contradicting the optimality of S ^∗.