Assortment Optimisation Under a General Discrete Choice Model: A Tight Analysis of Revenue-Ordered Assortments

Abstract

The assortment problem in revenue management is the problem of deciding which subset of products to offer to consumers in order to maximise revenue. A simple and natural strategy is to select the best assortment out of all those that are constructed by fixing a threshold revenue $\pi$ and then choosing all products with revenue at least $\pi$. This is known as the revenue-ordered assortments strategy. In this paper we study the approximation guarantees provided by revenue-ordered assortments when customers are rational in the following sense: the probability of selecting a specific product from the set being offered cannot increase if the set is enlarged. This rationality assumption, known as regularity, is satisfied by almost all discrete choice models considered in the revenue management and choice theory literature, and in particular by random utility models. The bounds we obtain are tight and improve on recent results in that direction, such as for the Mixed Multinomial Logit model by Rusmevichientong et al. (Prod Oper Manag 23(11):2023–2039, 2014). An appealing feature of our analysis is its simplicity, as it relies only on the regularity condition. We also draw a connection between assortment optimisation and two pricing problems called unit demand envy-free pricing and Stackelberg minimum spanning tree: these problems can be restated as assortment problems under discrete choice models satisfying the regularity condition, and moreover revenue-ordered assortments correspond then to the well-studied uniform pricing heuristic. When specialised to that setting, the general bounds we establish for revenue-ordered assortments match and unify the best known results on uniform pricing.

Appendices

Appendix A

See Table 1 for a (non-exhaustive) list of discrete choice models that satisfy the regularity axiom.

Table 1 List of discrete choice models that satisfy the regularity axiom (non-exhaustive)

Appendix B

In this section we provide the proofs missing from the main text.

Proof of Lemma 4.2

Recall from Sect. 4.1 that every discrete choice model based on random utility can also be described by means of a probability distribution over all rankings of the elements in $\mathcal {C} \cup \{0\}$ (see for instance Block and Marschak [11]). Let $\mathcal {S}_{N+1}$ denote the set of all such rankings, and let $\alpha _i$ be the probability of choosing the ranking $\prec _i \in \mathcal {S}_{N+1}$. Then the system of choice probabilities for our RUM can be written as

$$\begin{aligned} P(x,S)=\sum _{\prec _i \in \mathcal {S}_{N+1}}\alpha _i \mathbb {1} \{ x \prec _i y \quad \forall y \in (S \setminus \{x\}) \cup \{0\}\} \end{aligned}$$

for all $S \subseteq \mathcal {C}$ and $x \in S$.

Consider now, for every ranking $\prec _i \in \mathcal {S}_{N+1}$ , the function $D_i(S):2^{\mathcal {C}}\rightarrow \{0,1\}$ defined as

$$\begin{aligned} D_i(S)= \sum _{x \in S} \mathbb {1} \{ x \prec _i y \quad \forall y \in (S \setminus \{x\}) \cup \{0\}\}. \end{aligned}$$

Observe that $D_i(S)=0$ if $\prec _i$ puts element 0 before all elements in S, and $D_i(S)=1$ otherwise. Let us show that $D_i(S)$ is submodular. That is, for every $S \subseteq S' \subseteq \mathcal {C}$ and $x \in \mathcal {C}$,

$$\begin{aligned} D_i(S'\cup \{x\}) - D_i(S') \leqslant D_i(S\cup \{x\}) - D_i(S). \end{aligned}$$

(7)

First, observe that $D_i$ is monotone (i.e. $D_i(S) \leqslant D_i(S')$ if $S \subseteq S'$). Second, observe that if the LHS in (7) is equal to zero, the inequality directly follows. Consider then the case in which the LHS is equal to 1. This implies that $D_i(S')=0$, and thus $D_i(S)=0$, and that $D_i(S'\cup \{x\})=1$. Hence, $x \prec _i y$ for all $y \in S' \cup \{0\}$, which means that $x \prec _i y$ for all $y \in S \cup \{0\}$. Therefore, $D_i(S\cup \{x\})=1$, and the inequality (7) follows.

To finish the proof of the lemma, observe that

$$\begin{aligned} f(S)= & {} \sum _{x \in S} \mathcal {P}(x,S) \\= & {} \sum _{x \in S} \sum _{\prec _i \in \mathcal {S}_{N+1}}\alpha _i \mathbb {1} \{ x \prec _i y \quad \forall y \in (S \setminus \{x\}) \cup \{0\}\} \\= & {} \sum _{\prec _i \in \mathcal {S}_{N+1}}\alpha _i \sum _{x \in S} \mathbb {1} \{ x \prec _i y \quad \forall y \in (S \setminus \{x\}) \cup \{0\}\} \\= & {} \sum _{\prec _i \in \mathcal {S}_{N+1}}\alpha _i D_i(S). \end{aligned}$$

Since the sum of submodular functions is also submodular, the proof is now complete. $\square$

Proof of Lemma 4.5

Let $p: [n] \rightarrow \mathbb {R}_{>0}$ denote an optimal price assignment and suppose there is an item $x\in [n]$ such that $p(x)=q \notin \{v_1,\dots ,v_m\}$. If no consumer buys product x at price q then the seller would obtain the same revenue if the price for product x were changed to $v_m$, as is easily checked. If, on the other hand, there is a consumer that buys product x at price q, we do the following modification. Let w denote the lowest consumer valuation that is greater or equal to q, that is, $w = \min \{ v_j : j \in [m] \text { and } v_j \geqslant q \}$. Suppose now that the seller modifies the price assignment p by setting $p(x) :=w$. First, observe that every consumer buying a product different from x is unaffected by this modification since the price of product x has not decreased. Second, every consumer that was buying product x before either still buys it under the new price assignment, or buys another product y from her set with price $q \leqslant p(y) \leqslant w$. Hence, the seller’s revenue does not decrease. Repeating this argument at most n times, we eventually obtain a price assignment that satisfies this lemma. $\square$

Proof of Corollary 4.8

To be precise, this corollary follows from the proof of Theorem 4.6, as we now explain. Consider the assortment problem instance associated to a given $UDP_{min}$ instance described in that proof. As in Theorem 3.3, let $S^{*} \subseteq \mathcal {C}$ denote an optimal solution to this instance and let

$$\begin{aligned} N_i:=\sum _{\begin{array}{c} x\in S^{*}, \\ r(x) \geqslant r_{i} \end{array}} \mathcal {P}(x, S^{*}) \end{aligned}$$

for each $i \in [k]$. For convenience, let us scall these quantities by a factor m: For each $i \in [k]$, let

$$\begin{aligned} \hat{N}_i:=N_i \cdot m. \end{aligned}$$

Then $\hat{N}_1$ is exactly the number of consumers of the $UDP_{min}$ instance that buy some item in the optimal solution. Thus,

$$\begin{aligned} \hat{N}_1 \leqslant m. \end{aligned}$$

(8)

Moreover, letting $\ell \in [k]$ be maximum such that $\hat{N}_{\ell } > 0$, we have that $\hat{N}_{\ell }$ is the number of consumers that buy the most expensive item that was sold in the optimal solution. Hence,

$$\begin{aligned} \hat{N}_{\ell } \geqslant 1. \end{aligned}$$

(9)

By Theorem 3.3, the revenue-ordered assortments strategy approximates the optimum to within a factor of

$$\begin{aligned} \frac{1}{1 + \ln (N_{1} / N_{\ell })} = \frac{1}{1 + \ln (\hat{N}_{1} / \hat{N}_{\ell })} \leqslant \frac{1}{1 + \ln m}, \end{aligned}$$

where the last inequality follows from (8) and (9). The claim then follows from Theorem 4.6. $\square$

Proof of Lemma 4.14

Let $G :=\mathrm {greedy}_{M}(F, L)$ and $G':=\mathrm {greedy}_{M}(F', L)$. To see the first property, suppose for a contradiction that $|G'| < |G|$. Enumerate the elements of $F'$ as $e_1, \dots , e_k$ following the ordering L. Let $e_i \in G - G'$ be such that $G' \cup \{e_i\}$ is independent. In particular, $(G' \cap \{e_1, \dots , e_{i-1}\}) \cup \{e_i\}$ is also independent. This implies that when greedy considered the i-th element of $F'$, $e_i$, it added $e_i$ in the set being built, a contradiction.

Let us now prove the second property. Enumerate the elements of F this time as $e_1, \dots , e_k$ according to L. Arguing by contradiction, suppose that $e_i \in G' - G$ for some i. We may assume that index i is minimum with this property. Thus, $G' \cap \{e_1, \dots , e_{i-1}\} \subseteq G$.

Let $I:=G \cap \{e_1, \dots , e_{i-1}\}$. Thus $I \cup \{e_i\}$ is not independent, by definition of the greedy algorithm. Let K be an independent set satisfying $I \subseteq K \subseteq I \cup G'$ and inclusion-wise maximal with this property. Let us point out that $e_i \notin K$, since $I \subseteq K$.

Case 1 $|K| < |G'|$. Since $G'$ is independent there exists $e\in G'-K$ such that $K\cup \{e\}$ is independent, which contradicts the maximality of K.

Case 2 $|K| > |G'|$. Since K is independent there exists $e\in K- G' = I - G'$ such that $G'\cup \{e\}$ is independent. However, when building $G'$ the greedy algorithm did not pick element e because the subset $J \subseteq G'$ of elements it already picked when considering e was such that $J \cup \{e\}$ is not independent. Since $J \cup \{e\} \subseteq G'\cup \{e\}$, this contradicts the fact that $G'\cup \{e\}$ is independent.

Case 3 $|K| = |G'|$. Here, we use that $|K| > |G' - \{e_i\}|$, which implies that there exists $e\in K - (G' - \{e_i\})$ such that $(G' - \{e_i\}) \cup \{e \}$ is independent. Since $e_i \notin K$, we have $K - (G' - \{e_i\}) = K - G' = I - G'$, and it follows that $e=e_j$ for some $j <i$, meaning that $e_j$ was considered before $e_i$ by the greedy algorithm when building set $G'$. Yet the algorithm did not pick $e_j$ when it considered $e_j$, because the subset $J \subseteq G'$ of elements it already picked at that time could not be extended into an independent set by adding $e_j$. However, $J \cup \{e\} \subseteq (G' - \{e_i\}) \cup \{e_{j} \}$, contradicting the independence of $(G' - \{e_i\}) \cup \{e_{j} \}$.

Since in each of the three cases we derived a contradiction, we deduced that the second property holds, as claimed. $\square$

Proof of Lemma 4.15

This is a consequence of the following more general property of the greedy algorithm: Suppose that E is partitioned into $\ell$ blocks $E_1, \dots , E_{\ell }$, and that L and $L'$ are two linear orderings of the elements in E that agree on this block partition, in the sense that $e <_{L} f$ and $e <_{L'} f$ for all $i\in \{1, \dots , k-1\}$ and $e\in E_i, f\in E_{i+1}$. Then

$$\begin{aligned} |\mathrm {greedy}_{M}(E, L) \cap E_i| = |\mathrm {greedy}_{M}(E, L') \cap E_i| \quad \forall i \in \{1, \dots , k\}. \end{aligned}$$

(10)

To see that it implies the lemma, it suffices to take the partition of E obtained by partitioning R according to the costs of the red elements and B according to the prices of the blue elements, and ordering the resulting blocks in non-decreasing order of costs/prices, giving priority to blue over red in case of ties.

Thus it remains to prove (10). Arguing by contradiction, suppose that the property does not hold and let i be the smallest index such that $|\mathrm {greedy}_{M}(E, L) \cap E_i| \ne |\mathrm {greedy}_{M}(E, L') \cap E_i|$. Say without loss of generality $|\mathrm {greedy}_{M}(E, L) \cap E_i| < |\mathrm {greedy}_{M}(E, L') \cap E_i|$. By our choice of i, $|\mathrm {greedy}_{M}(E, L) \cap (E_1 \cup \cdots \cup E_{i-1})| = |\mathrm {greedy}_{M}(E, L') \cap (E_1 \cup \cdots \cup E_{i-1})|$. Since $|\mathrm {greedy}_{M}(E, L) \cap (E_1 \cup \cdots \cup E_{i})| < |\mathrm {greedy}_{M}(E, L') \cap (E_1 \cup \cdots \cup E_{i})|$, by the last of the matroid axioms there exists $e \in (\mathrm {greedy}_{M}(E, L') - \mathrm {greedy}_{M}(E, L)) \cap (E_1 \cup \cdots \cup E_{i})$ such that $(\mathrm {greedy}_{M}(E, L) \cap (E_1 \cup \cdots \cup E_{i})) \cup \{e\}$ is independent. Hence, the greedy algorithm w.r.t. ordering L could have picked e when considering the elements in $E_i$ but did not, a contradiction. $\square$

Proof of Theorem 4.16

Define the set $\mathcal {C}$ of products for the assortment problem as follows:

$$\begin{aligned} \mathcal {C} :=B \times \{c_1,\dots ,c_k\}. \end{aligned}$$

Thus $\mathcal {C}$ consists of all pairs of a blue element and a cost $c_i$ taken by some red element. The revenue function $r: \mathcal {C} \rightarrow \mathbb {R}_{>0}$ for the assortment problem is defined by setting

$$\begin{aligned} r( (e, q) ) :=|B| \cdot q \end{aligned}$$

for all $(e, q) \in \mathcal {C}$.

Next we define the system of choice probabilities $\mathcal {P}$. To do so, it is convenient to first define an auxiliary matroid $M'$ with ground set $R \cup \mathcal {C}$, where $X \subseteq R \cup \mathcal {C}$ is independent if only if the following two conditions are satisfied:

• (1) $|\{(e, q): q\in \{c_{1}, \dots , c_{k}\} \}| \leqslant 1$ for each element $e \in B$, and

• (2) $(R\cap X) \cup \{e \in B: (e, q) \in \mathcal {X} \text { for some } q\in \{c_{1}, \dots , c_{k}\}\}$ is independent in M.

We leave it to the reader to check that $M'$ is indeed a matroid. In order to give some intuition about $M'$, we remark that in the case of the Stackelberg Minimum Spanning Tree problem, where M is the graphical matroid of the input graph, $M'$ is simply the graphical matroid of the graph obtained by replacing each edge e with k parallel copies of e.

For simplicity, let us say that the cost of element $x \in R \cup \mathcal {C}$ is c(x) if $x\in R$, and q if $x=(e, q) \in \mathcal {C}$. Let L be a linear ordering of the elements in $R \cup \mathcal {C}$ that is consistent with their associated costs (elements appear in non-decreasing order of cost), where priority is given to elements in $\mathcal {C}$ over those in R (that is, if $(e, q) \in \mathcal {C}$, $f\in R$, and $c(f) = q$, then $(e, q) <_{L} f$).

We define $\mathcal {P}$ as follows. For each $S\subseteq \mathcal {C}$ and $(e,q)\in \mathcal {C}$, let

$$\begin{aligned} \mathcal {P}((e,q),S) :=\left\{ \begin{array}{ll} \displaystyle \frac{1}{|B|} &{} \quad \text {if }(e, q)\in \mathrm {greedy}_{M'}(R\cup S, L) \\ 0 &{} \quad \text {otherwise}\\ \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \mathcal {P}(0,S) :=1 - \sum _{(e,q) \in S} \mathcal {P}((e,q),S). \end{aligned}$$

Let us prove that $\mathcal {P}$ is a regular discrete choice model. Clearly $\mathcal {P}(y, S) \geqslant 0$ for every $y\in \mathcal {C} \cup \{0\}$ and $S\subseteq \mathcal {C}$, thus axiom (i) is satisfied. If $(e, q)\in \mathcal {C}$ and $S\subseteq \mathcal {C} \setminus \{(e,q)\}$ then $\mathcal {P}((e, q), S) = 0$ since $(e, q) \notin \mathrm {greedy}_{M'}(S ,L)$ trivially. Hence, axiom (ii) is satisfied. Also, for each $S\subseteq \mathcal {C}$ we have $\sum _{(e,q) \in S}\mathcal {P}((e,q), S) \leqslant |\mathrm {greedy}_{M'}(R\cup S, L)| / |B| \leqslant 1$, since $|\mathrm {greedy}_{M'}(R\cup S, L)| \leqslant |B|$. This implies that axiom (iii) holds. Therefore, it only remains to check axiom (iv), the regularity condition. Let thus $S\subseteq S'\subseteq \mathcal {C}$, and let $y\in S \cup \{0\}$. We want to show that $\mathcal {P}(y, S) \geqslant \mathcal {P}(y, S')$.

First suppose that $y=(e, q) \in S$. If $\mathcal {P}((e,q), S') = 0$, then trivially $\mathcal {P}((e,q), S) \geqslant \mathcal {P}((e,q), S')$. If $\mathcal {P}((e,q), S') = 1/|B|$, then $(e,q) \in \mathrm {greedy}_{M'}(R\cup S', L)$. By Lemma 4.14, we also have $(e,q) \in \mathrm {greedy}_{M'}(R\cup S, L)$, and thus $\mathcal {P}((e,q), S) = 1/|B|$. Hence, $\mathcal {P}((e,q), S) \geqslant \mathcal {P}((e,q), S')$ holds in both cases.

Now assume that $y=0$ is the no-choice option. We know from Lemma 4.14 that $(R \cup S) \cap \mathrm {greedy}_{M'}(R \cup S', L) \subseteq \mathrm {greedy}_{M'}(R \cup S, L)$. Since $|\mathrm {greedy}_{M'}(R \cup S', L)| \geqslant |\mathrm {greedy}_{M'}(R \cup S, L)|$ by the same lemma, and since $(R \cup S') - (R \cup S) = S' - S \subseteq \mathcal {C}$, we deduce that

$$\begin{aligned} |\mathrm {greedy}_{M'}(R \cup S', L) \cap \mathcal {C}| \geqslant |\mathrm {greedy}_{M'}(R \cup S, L) \cap \mathcal {C}|. \end{aligned}$$

This implies that $\sum _{(e,q)\in S'}\mathcal {P}((e, q), S') \geqslant \sum _{(e,q)\in S}\mathcal {P}((e, q), S)$, and we conclude that

$$\begin{aligned} \mathcal {P}(0, S) = 1 - \sum _{(e,q)\in S}\mathcal {P}((e, q), S) \geqslant 1 - \sum _{(e,q)\in S'}\mathcal {P}((e, q), S') = \mathcal {P}(0, S'). \end{aligned}$$

Therefore, axiom (iv) is satisfied.

Next we prove that the maximum revenue achievable on each instance is the same. Given a set $S\subseteq \mathcal {C}$ we define a corresponding price assignment $p_S$ by setting

$$\begin{aligned} p_S(e) :=\left\{ \begin{array}{ll} \min \{q: (e, q) \in S\} &{} \quad \text {if }\exists q\mathrm{s.t. }(e, q) \in S \\ + \infty &{} \quad \text {otherwise}\\ \end{array} \right. \end{aligned}$$

for each element $e \in B$. The revenue resulting from choosing assortment S can be expressed as follows:

$$\begin{aligned} \sum _{y \in S}\mathcal {P}(y,S) \cdot r(y) = |B| \sum _{(e,q) \in S}\mathcal {P}((e,q),S) \cdot q = \sum _{(e, q)\in \mathrm {greedy}_{M'}(R\cup S, L)} q. \end{aligned}$$

In the Stackelberg Matroid instance with price assignment $p_S$, the customer buys element $e\in B$ if and only if the greedy algorithm chooses element e when considering matroid M with some linear ordering $L^*$ of its elements $R\cup B$ that is consistent with their costs/prices, where priority is given to elements in B in case of ties. That is, letting $c'(e) :=c(e)$ if $e\in R$ and $c'(e) :=p_S(e)$ if $e\in B$, the ordering $L^*$ used by the customer satisfies:

$$\begin{aligned} \text {if }c'(e)< c_S(f)\text { then }e <_{L^*} f\text { for every }e,f \in R \cup B,\text { and} \end{aligned}$$

(11a)

$$\begin{aligned} \text {if }c'(e) = c_S(f)\text { then }e <_{L^*} f\text { for every }e \in B\text { and }f\in R. \end{aligned}$$

(11b)

There might be more than one linear ordering satisfying the above properties; however thanks to Lemma 4.15 we know that all such orderings yield the same revenue. Since we are not interested in the particular elements bought by the customer but only by the resulting revenue, for the purpose of the following analysis we may assume that $L^{*}$ ‘breaks ties’ in the same that L does, that is,

$$\begin{aligned}&{\forall e,f\in B: e<_{L^*} f \Leftrightarrow (e, p_S(e))<_L (f, p_S(f));}\\&{\forall e,f\in R: e<_{L^*} f \Leftrightarrow e <_L f.} \end{aligned}$$

Element $e\in B$ is bought by the customer if and only if $e\in \mathrm {greedy}_{M}(R\cup B, L^*)$, which by our assumption on $L^{*}$ is the same condition as $(e, p_S(e))\in \mathrm {greedy}_{M'}(R\cup S, L)$. Furthermore, if $(e, q)\in \mathrm {greedy}_{M'}(R\cup S, L)$ for some q then this q is unique and by definition of the price functin $p_S$ we have $p_S(e)=q$. Hence, the revenue resulting from price function $p_S$ equals

$$\begin{aligned} \sum _{e\in \mathrm {greedy}_{M}(R\cup B, L^*)} p_S(e) = \sum _{(e, q)\in \mathrm {greedy}_{M'}(R\cup S, L)} q, \end{aligned}$$

and is thus equal to the revenue given by assortment S. This shows that the maximum revenue achievable on the Stackelberg Matroid instance is at least that achievable on the assortment problem we defined. We now prove the converse statement, and hence that the two quantities are the same.

Let p be some price function for the elements in B and let $L^*$ be a linear ordering of the elements in $R \cup B$ that a customer could use when running the greedy algorithm on M under this price assignment. As recalled above, $L^*$ is thus any linear ordering satisfying (11a) and (11b), where in this case $c'(\cdot )$ is modified by setting $c'(e) :=p(e)$ if $e\in B$.

We may assume that all prices assigned by p to elements in B are in the set $\{c_1, \dots , c_k\} \cup \{+\infty \}$. Indeed, if $c_{i-1}< p(e) < c_i$ for some $e\in B$ and $i\in \{1, \dots , k\}$ (where we let $c_0:=0$) then we can increase p(e) to $c_i$ without changing $L^*$ being a valid ordering for these prices, which can only improve the resulting revenue. Similarly, if $p(e) > c_k$ then we may as well set $p(e) :=+\infty$ since element e will never be bought by the customer, as follows from the existence of a base of M in R. Since our goal is to bound from above the revenue resulting from price assignment p by the optimal revenue achievable on the assortment problem, we may thus iteratively modify p as described until it has the desired form.

Relying on the fact that p takes values in $\{c_1, \dots , c_k\} \cup \{+\infty \}$, we define the following corresponding assortment $S \subseteq \mathcal {C}$:

$$\begin{aligned} S :=\{(e, p(e)): e\in \mathrm {greedy}_{M}(R\cup B, L^*) \}. \end{aligned}$$

The ordering $L^*$ of the elements in $R \cup B$ induces in a natural way an ordering L of the elements in $R\cup S$. The revenue given by assortment S is then

$$\begin{aligned} \sum _{y \in S}\mathcal {P}(y,S) \cdot r(y)= & {} |B| \sum _{(e,q) \in S}\mathcal {P}((e,q),S) \cdot q =\sum _{(e,q)\in \mathrm {greedy}_{M'}(R\cup S, L)} q \\= & {} \sum _{e\in \mathrm {greedy}_{M}(R\cup B, L^*)} p(e) \end{aligned}$$

and is thus equal to the revenue resulting from price function p. This shows that the optimal revenue achievable on the assortment problem is least that achievable on the Stackelberg Matroid instance, as desired. $\square$

Next we provide a proof of Theorem 5.1. In order to do so, we need to introduce some definitions and a lemma.

Given $\delta \in \mathbb {R}$ such that $r_k + \delta \geqslant 0$, we define $\mathcal {L}^*(\delta )$ to be the set of indices $\ell \in [k]$ with the property that, if we add $\delta$ to the revenue of all items, then choosing all items of revenue at least $r_{\ell } + \delta$ gives an optimal revenue-ordered assortment (w.r.t. the modified revenue function) for the usual assortment problem. (Note that whenever the revenue of an item becomes negative, the item will never be chosen by the revenue ordered assortment strategy.) That is, $\mathcal {L}^*(\delta )$ is the set of indices $\ell \in [k]$ such that

$$\begin{aligned} \sum _{x=1}^{j(\ell )}\mathcal {P}(x,\{1,\dots ,j(\ell )\})(r(x) + \delta ) = \max _{\ell ' \in [k]}\sum _{x=1}^{j(\ell ')}\mathcal {P}(x,\{1,\dots ,j(\ell ')\})(r(x) + \delta ). \end{aligned}$$

Lemma 1

Let $\delta _1, \delta _2 \in \mathbb {R}$ with $\delta _1 + r_k \geqslant 0$ and $\delta _1 \leqslant \delta _2$. Then, $\min \mathcal {L}^*(\delta _2) \leqslant \min \mathcal {L}^*(\delta _1)$.

Proof

Let $\ell _1:= \min \mathcal {L}^*(\delta _1)$, $\ell _2 := \min \mathcal {L}^*(\delta _2)$ and $\Delta := \delta _2 - \delta _1 \geqslant 0$. For the purpose of contradiction, suppose that $\ell _2 > \ell _1$. This means that $j(\ell _2) < j(\ell _1)$ since products in $\mathcal {C}$ are enumerated in decreasing order of revenue.

We will show that if we add $\delta _2$ to the revenue of all items (w.r.t. revenue function r), the expected revenue resulting from the assortment $\{1,\dots ,j(\ell _2)\}$ is at most that of the assortment $\{1,\dots , j(\ell _1)\}$. This implies $\ell _2 \leqslant \ell _1$, a contradiction.

We have

$$\begin{aligned}&\sum _{x=1}^{j(\ell _2)} \mathcal {P}(x, \{1,\dots ,j(\ell _2)\}) (r(x)+ \delta _2) \\&\quad = \sum _{x=1}^{j(\ell _2)} \mathcal {P}(x, \{1,\dots ,j(\ell _2)\}) (r(x) + \delta _1) + \Delta \sum _{x=1}^{j(\ell _2)} \mathcal {P}(x, \{1,\dots ,j(\ell _2)\}) \\&\quad \leqslant \sum _{x=1}^{j(\ell _1)} \mathcal {P}(x, \{1,\dots ,j(\ell _1)\}) (r(x) + \delta _1) + \Delta \sum _{x=1}^{j(\ell _2)} \mathcal {P}(x, \{1,\dots ,j(\ell _2)\}) \\&\quad \leqslant \sum _{x=1}^{j(\ell _1)} \mathcal {P}(x, \{1,\dots ,j(\ell _1)\}) (r(x) + \delta _1) + \Delta \sum _{x=1}^{j(\ell _1)} \mathcal {P}(x, \{1,\dots ,j(\ell _1)\}) \\&\quad = \sum _{x=1}^{j(\ell _1)} \mathcal {P}(x, \{1,\dots ,j(\ell _1)\}) (r(x)+ \delta _2) \end{aligned}$$

The first inequality holds because $\{1,\dots ,j(\ell _1)\}$, by definition, yields the highest expected revenue among all revenue-ordered assortments when $\delta _1$ is added to the revenue of each item. The second inequality follows from Lemma 2.1 and the assumption that $\ell _2 > \ell _1$. This concludes the proof. $\square$

Proof of Theorem 5.1

We begin by proving that $\ell ^*_t(q) \leqslant \ell ^*_{t}(q-1)$ if $q \geqslant 2$. For $t' \geqslant 0$ and $q' \geqslant 1$ let

$$\begin{aligned} \Delta \mathcal {J}_{t'}(q') := \mathcal {J}_{t'}(q') - \mathcal {J}_{t'}(q'-1), \end{aligned}$$

that is, $\Delta \mathcal {J}_{t'}(q')$ is the marginal value of the capacity when there are $t'$ time periods remaining. (Let us point out that $\Delta \mathcal {J}_{t'}(q') = \mathcal {J}_{t'}(q') = \mathcal {J}_{t'}(q'-1)=0$ if $t'=0$.)

We can express $\mathcal {J}_t(q)$ as follows:

$$\begin{aligned} \mathcal {J}_t(q)= & {} \max _{\ell \in [k]} \left\{ \sum _{x=1}^{j(\ell )} \mathcal {P}(x,\{1,\dots ,j(\ell )\})(r(x) + \mathcal {J}_{t-1}(q-1))\right. \nonumber \\&\left. + \mathcal {P}(0,\{1,\dots ,j(\ell )\}) \mathcal {J}_{t-1}(q) \right\} \nonumber \\= & {} \max _{\ell \in [k]} \left\{ \sum _{x=1}^{j(\ell )} \mathcal {P}(x,\{1,\dots ,j(\ell )\}) (r(x) - \Delta \mathcal {J}_{t-1}(q)) \} + \mathcal {J}_{t-1}(q) \right\} \end{aligned}$$

(12)

Observe that

$$\begin{aligned} \ell ^*_t(q-1) = \min \mathcal {L}^*(- \Delta \mathcal {J}_{t-1}(q-1)). \end{aligned}$$

(13)

In other words, $\ell ^*_t(q-1)$ is the largest revenue ordered assortment which is optimal when $\Delta \mathcal {J}_{t-1}(q-1)$ is subtracted from the revenue of each item. Since the most revenue one could obtain from an extra unit of capacity is the revenue of the most expensive product (i.e. $r_k$), we have that $r_k - \Delta \mathcal {J}_{t-1}(q-1) \geqslant 0$ as needed.

Suppose that we subtract $\Delta \mathcal {J}_{t-1}(q)$ to the original revenue of each item. By Equation (12) we know that

$$\begin{aligned} \ell ^*_t(q) = \min \mathcal {L}^*(-\Delta \mathcal {J}_{t-1}(q)) \end{aligned}$$

(14)

Again, since the most revenue one could obtain from an extra unit of capacity is the revenue of the most expensive product (i.e. $r_k$), we have that $r_k - \Delta \mathcal {J}_{t-1}(q) \geqslant 0$ as desired.

Talluri and Van Ryzin [44, Lemma 4] proved that $\Delta \mathcal {J}_{t-1}(q-1) \geqslant \Delta \mathcal {J}_{t-1}(q)$ always holds, regardless of the discrete choice model under consideration (that is, the three axioms (i), (ii) and (iii) are enough for this property to hold). Therefore, by Lemma 1, we have that

$$\begin{aligned} \min \mathcal {L}^*(-\Delta \mathcal {J}_{t-1}(q-1)) \geqslant \min \mathcal {L}^*(-\Delta \mathcal {J}_{t-1}(q)) \end{aligned}$$

(15)

Combining (14), (15) and (13) we have that

$$\begin{aligned} \ell ^*_t(q)&= \min \mathcal {L}^*(-\Delta \mathcal {J}_{t-1}(q)) \leqslant \min \mathcal {L}^*(-\Delta \mathcal {J}_{t-1}(q-1)) = \ell ^*_t(q-1) \end{aligned}$$

as desired.

We now proceed to prove that $\ell ^*_t(q)$ is non-increasing in t, which carries on in a similar way.

Suppose that we subtract $\Delta \mathcal {J}_{t-2}(q)$ to the original revenue of each item. By equation (12) we know that

$$\begin{aligned} \ell ^*_{t-1}(q) = \min \mathcal {L}^*(- \Delta \mathcal {J}_{t-2}(q)) \end{aligned}$$

(16)

Again, since the most revenue one could obtain from an extra unit of capacity is the revenue of the most expensive product (i.e. $r_k$), we have that $r_k - \Delta \mathcal {J}_{t-2}(q) \geqslant 0$ as desired.

Talluri and Van Ryzin [44, Lemma 5] proved that $\Delta \mathcal {J}_{t-1}(q) \geqslant \Delta \mathcal {J}_{t-2}(q)$ always holds, regardless of the discrete choice model under consideration. Therefore, by Lemma 1, we have that

$$\begin{aligned} \min \mathcal {L}^*(-\Delta \mathcal {J}_{t-2}(q)) \leqslant \min \mathcal {L}^*(-\Delta \mathcal {J}_{t-1}(q)) \end{aligned}$$

(17)

Combining (16), (17) and (14) we have that

$$\begin{aligned} \ell ^*_{t-1}(q) = \min \mathcal {L}^*(- \Delta \mathcal {J}_{t-2}(q)) \leqslant \min \mathcal {L}^*(-\Delta \mathcal {J}_{t-1}(q)) = \ell ^*_t(q) \end{aligned}$$

as desired. $\square$