Retailers’ product location problem with consumer search

Abstract

With few exceptions, today’s retailers sell products across multiple categories. One strategic consideration of such retailers is product location, which determines how easy or difficult different categories are for customers to access. For example, grocery or department stores determine which products will be located closer to the entrance of the store versus at the back of it, while online retailers decide which products to feature on the homepage, and which will require scrolling or keyword search to get to. In this paper, we study how a retailer should optimally locate products within a store, when the locations chosen affect consumer search costs. We show that the retailer has an incentive to prioritize products with lower utility, contrasting with prior work. The intuition for our result is that the consumer may be willing to search less preferred products only at the lower cost, while the more preferred products will be searched even at higher search costs. This strategy benefits the retailer by increasing the number of products the consumer searches and thus, the ones she may buy. Our finding is robust to several extensions: (i) a retailer determining not only product locations, but also prices, (ii) independent (e.g. categories), as well as substitute products, and (iii) a focal retailer that faces competition. From a managerial perspective, we show that allocating products in the store without taking into account how this affects consumer search costs, might mean consumers overlook products they would otherwise purchase.

Appendix:

Before proving Proposition 3 that Algorithm 1 maximizes the number of products searched by the consumer, we first demonstrate Lemma 1 below, which requires the following notation. Consider locations in increasing order of their search cost. Let $C_{l}(\{L_{j}, j \in \overline {S}_{l}\})$ be the maximum number of products that can be searched starting with location l, given the set of products not yet allocated $\overline {S}_{l}$ and their search cutoffs L_j.

Lemma 1

For any l,$C_{l}(\{L_{j}, j \in \overline {S}_{l}\}) \geq C_{l}(L_{j}, j \in \overline {S}_{l}\setminus \{a\} \cup \{b\})$ifL_a ≥ L_b.

In other words, swapping a yet-to-be-allocated product for a different product with weakly smaller search cutoff cannot increase the total number of products searched.

Proof Proof of Lemma 1

Any allocation of products that maximizes the number of products searched from the set $\overline {S}_{l} \setminus \{a\} \cup \{b\}$ can also be used to generate an equal number of products searched from $\overline {S}_{l}$, if it does not place product b in any location. If the search maximizing product allocation does place product b in a location, it can be swapped for product a, because L_a ≥ L_b. □

Using the result in Lemma 1, we can now prove Proposition 3.

Proof Proof of Theorem 3

Let S and Q be the sets of products resulting from Algorithm 1 and the optimal solution, respectively. If S = Q, then S is optimal and we have proven the claim. Suppose instead that S≠Q. If Q ⊂ S, then Q cannot be optimal, because S results in more searches. The case S ⊂ Q is also not possible. Any $j \in Q \cap \overline {S}$ must have L_j ≥ L_k, ∀k ∈ S, because Algorithm 1 always selects products with the lowest L_j. So any such j could always be appended to S at the end. Because Algorithm 1 stopped, it means there are no such products. Finally, if |S| = |Q|, even though S≠Q (given that Algorithm 1 does not produce a unique allocation), then the claim is also proven.

Then there must be at least one element in S that is not in Q and vice versa. Consider locations in increasing order of their search cost. Let l be the first location in which $\overrightarrow {j}^{S}$ differs from $\overrightarrow {j}^{Q}$, i.e. ${j_{l}^{S}} \neq {j_{l}^{Q}}$. It must be that $L_{{j_{l}^{S}}} \leq L_{{j_{l}^{Q}}}$, because Algorithm 1 chose to allocate ${j_{l}^{S}}$. Then for location l, the set of products not allocated as per Algorithm 1, $\overline {S}_{l}$ is equal to the set of products not allocated as per the optimal algorithm, $\overline {Q}_{l}$ except that the product ${j_{l}^{S}}$ was swapped for ${j_{l}^{Q}}$, that is $\overline {S}_{l}=\overline {Q}_{l} \setminus \{{j_{l}^{S}}\} \cup \{{j_{l}^{Q}}\}$. Because $L_{{j_{l}^{S}}} \leq L_{{j_{l}^{Q}}}$, we can apply Lemma 1 to show that $C_{l}(\{L_{j}, j \in \overline {S}_{l}\}) \geq C_{l}(\{L_{j}, j \in \overline {Q}_{l}\})$. Therefore, by allocating ${j_{l}^{S}}$ and discarding ${j_{l}^{Q}}$ in the optimal solution, we obtain a solution $\overrightarrow {j}^{Q^{\prime }}$ that results in $N_{Q^{\prime }}\geq N_{Q}$ products being searched, and which differs from $\overrightarrow {j}^{S}$ by one less product. By repeatedly applying this transform, $\overrightarrow {j}^{Q}$ can be transformed to $\overrightarrow {j}^{S}$ with no decrease in total number of products searched. This shows that S is optimal. □

Proof Proof of Theorem 4

⁷ Suppose Algorithm 2 produces a set of allocated products S, while the optimal algorithm produces a set Q. If S = Q, then S is optimal and we have proven the claim. It is also clear that neither Q ⊂ S (contradicts Q being optimal), nor S ⊂ Q (could append any $j \in Q \cap \overline {S}$ to S at the end) are possible. Then there must be at least one element in S that is not in Q and vice versa.

We first show that there exist feasible product allocations $\overrightarrow {j}^{S}$ and $\overrightarrow {j}^{Q}$, such that all products that are included in both S and Q are placed in the same location. Let j be placed in location l in $\overrightarrow {j}^{S}$ and in location l^′ in $\overrightarrow {j}^{Q}$. If l < l^′, then replace j in l^′ in $\overrightarrow {j}^{S}$. Now, j is placed in the same location in the two allocations. If l > l^′, the same transformation can be applied to $\overrightarrow {j}^{Q}$. Therefore, we can transform any two feasible product allocations into allocations for which all common products are placed in the same location.

Denote by ϕ_j the expected payoff of a product j. Let a be the highest expected payoff product that is included in S and not in Q, that is a ∈ S, a ∉ Q. Then, it must be that ϕ_a ≥ ϕ_b, ∀b ∈ Q, b ∉ S. Otherwise, if ϕ_b > ϕ_a, Algorithm 2 would consider product b before product a, and include it in S.

Consider now the location at which product a is placed in $\overrightarrow {j}^{S}$. Let c be the product placed in the same location in $\overrightarrow {j}^{Q}$. Because all common elements are placed in the same location under both algorithms, we know that a ∈ S, a ∉ Q and c ∈ Q, c ∉ S. Swapping c with a in Q cannot decrease the total payoff from Q because ϕ_a ≥ ϕ_c, and it cannot increase the total payoff from Q because Q is optimal. Thus, swapping c with a, gives a feasible product allocation that differs from the set S in one less product than did Q. Through repetition, the set Q can be transformed to S with no decrease in payoff. Therefore, the set S was optimal. □

Proof Proof of Theorem 5

⁸ If $\overrightarrow {j}^{S}$ has allocated products with lower expected net utility to lower search cost locations, then the claim has been proven. Suppose instead that this does not hold. Then ∃j, k ∈ S, such that γ_j < γ_k, but k was located in a lower cost location than j. Without loss of generality, suppose k is located at l and j at l + h, where h > 0. Because γ_j < γ_k, then it follows that L_j ≤ L_k. Product j is located at l + h, which means that L_j ≥ l + h. Because L_j ≤ L_k, then it must also hold that L_k ≥ l + h. Thus, we can swap j and k in S. The resulting allocation is still feasible, because the consumer would be willing to search both products in their new locations. Also, this cannot increase the retailer’s expected payoff, because S is optimal. Thus, if $\overrightarrow {j}^{S}$ has not allocated products with lower expected net utility to lower search cost locations, it can be arranged in such a way, while preserving optimality. □