Managing Variety on the Retail Shelf: Using Household Scanner Panel Data to Rationalize Assortments

Abstract

We propose a model for the rationalization of retail assortment and stocking decisions for retail category management. We assume that consumers are heterogenous in their intrinsic preferences for items and are willing to substitute less preferred items to a limited extent if their preferred items are not available. We propose that the appropriate objective function for a far-sighted retailer includes not only short-term profits but also a penalty for disutility incurred by consumers who do not find their preferred items in the available assortment. The retailer problem is formulated as a constrained integer programming problem. We demonstrate an empirical application of our proposed model using household scanner panel data for eight items in the canned tuna category. Our results indicate that the inclusion of the penalty for disutility in the retailer’s objective function is informative in terms of choosing an assortment to carry. We find that customer disutility can be significantly reduced at the cost of a small reduction in short term profits. We also find that the optimal assortment behaves non-monotonically as the weight on customer disutility in the retailer’s objective function is increased.

Keywords

Appendix

Proof of Proposition 4.1.

Without loss of generality, we will illustrate this for the general case rather than the special case of fixed depth of search d.

First consider the constraint (10. 6a^′′ ). The constraint for j = 1 will be

$$ 1-\left({x}_{s2}+{x}_{s3}+\dots +{x}_{sK}+{x}_{s0}\right)\ge {y}_1 $$

However, from (10.6c) we know that

$$ {x}_{s1}+{x}_{s2}+\dots +{x}_{jK}+{x}_{s0}\le 1 $$

Actually, given a “no purchase” option, the above is an equality; i.e.,

$$ {x}_{s1}+{x}_{s2}+\dots +{x}_{jK}+{x}_{s0}=1 $$

Using this we rewrite $1-\left({x}_{s2}+{x}_{s3}+\dots +{x}_{sK}+{x}_{s0}\right)\ge {y}_1$ as simply x ₁ ≥ y ₁. Similarly, we can write (10. 6a″) for j = k as

$$ {x}_{s1}+{x}_{s2}+\dots +{x}_{sk}\ge {y}_k. $$

Using this, for any arbitrary customer segment s that prefers K products in the ordinal order (without loss of generality) the constraint sets (10. 6a′) and (10. 6a″) are

Table 7

Consider normalized constraint (1′) and (1″):

$$ {x}_{s1}+\left(\frac{V_{s2}}{V_{s1}}\right){x}_{s2}+\cdots \left(\frac{V_{sk}}{V_{s1}}\right){x}_{sk}\kern2em \mathrm{and}\kern2em {x}_{s1}\ge {y}_1. $$

Since (1^′′) and (1^′) are identical in x _s1 dimension and (1^′) has k − 1 extra variables (degrees of freedom), constraint (1^′′) is tighter than constraint (1^′). Using similar arguments one can show that constraints (2^′′) to ((k − 1)^′′) will be tighter than (2^′) to ((k − 1)^′). Constraint (k ^′′) may be identical to (k ^′). The argument can be repeated for other segments. Thus problem (P1) with (10. 6a^′′ ) is a tighter formulation than (P1) with (10. 6a^′ ).

To see that relaxation of x _sj still leads to an integer solution, first consider (1^′′). If y ₁ = 0, then x _s1 = 0 using (10.6c). If y ₁ = 1, then x _s1 = 1. Now consider (2^′′). Suppose y ₁ = 0. If y ₂ = 0 then x _s2 = 0; otherwise (y ₂ = 1), x _s2 = 1. However, if y ₁ = 1, then (10.6b) ensures that x _s2 = 0. Following this argument, we can show that x _sj is integer. ■