Here, we consider the problem of allocating limited inventory across demand sites, given the availability of a lower-quality substitute that can be used when inventory runs out. We find that it can be optimal to send the premium item to only a subset of locations.
Problem and general solution strategy
In our first post on allocation, we argued that to optimize net sales across a system, we should distribute inventory across sites so that each has the same probability of stockout. When inventory is scarce relative to demand, this optimal strategy leaves every site (equally) underserved. Such cases motivate asking when it might be better to send a premium product to only a subset of sites — serving those well — while the remaining sites receive a more abundant, lower-quality substitute.
To make progress, suppose we have already equilibrated some of the sites to a common stock out probability,
\begin{equation} \label{phi} \tag{1} \phi = 1 - P(d_i \leq Q_i). \end{equation}
Here, \(d_i\) and \(Q_i\) are the demand and units initially allocated to site \(i\), respectively, and \(P(d_i \leq Q_i)\) is the probability that demand at site \(i\) is not more than its allocated inventory. Recall that the reason these should be equilibrated across sites is that \(1 - P(d_i \leq Q_i)\) is the increase in expected sales at site \(i\) when it is allocated one additional unit – see our prior post for derivation. Equilibrating the partials at right in (\ref{phi}) to the common \(\phi\) value then prevents arbitrage.
Now consider adding one more site to receive the the premium product. Suppose doing so would require \(Q_* \equiv Q_* (\phi)\) units to equilibrate. To fulfill this demand, those units will have to be pulled from the other sites. The sales lost from these other locations due to the reallocation will then be
\begin{equation}\label{cost} \tag{2} \text{Cost to serve} = Q_* \times \phi. \end{equation}
Here, we again use the fact that \(\phi\) is the first partial of sales to units at each equilibrated site, and we assume a large system size so that \(\phi\) does not change appreciably as we reallocate the \(Q_*\) units.
The net benefit of serving the additional site is the expected sales from the site, minus the cost (\ref{cost}). We should proceed if this beats assigning the substitute:
\begin{equation}\label{equilibrium} \tag{3} E(sales \vert Q_*) - Q_* \times \phi \geq E(sales \vert \text{substitute}). \end{equation}
All terms here are specific to the candidate site.
Setting (\ref{equilibrium}) to equality and solving for \(\phi\) gives a "critical \(\phi\)" value for each site. If the system-wide \(\phi\) exceeds this, the site should receive the substitute. Otherwise, it should receive the premium product.
This leads to a simple prioritization rule: sort sites by their critical \(\phi\) values (largest to smallest), and allocate in that order. At step \(K\), we assign the premium product to the top \(K\) sites and evaluate \(\phi\). If this is no larger than the next site’s critical \(\phi\), we then add that site and re-equilibrate. Otherwise, we stop and the remaining sites receive the substitute.
Example: equivalent sites
In a simple limit, all sites have equivalent demand. In this case, each site will have the same critical \(\phi\) value. Rather than sorting, we simply need to ask how many sites to give the premium item to. Suppose there are \(Q_T\) items total, \(N\) total sites, and \(K\) sites assigned the premium product. In this case, (\ref{equilibrium}) maps to
\begin{equation}\label{equilibrium_simple} \tag{4} E \left (sales \vert \frac{Q_T}{K} \right ) - \frac{Q_T}{K} \times \phi \geq E(sales \vert \text{substitute}). \end{equation}
If we set \(K \to N\) and the left side of (\ref{equilibrium_simple}) is still larger than the right, all sites will share the premium product. However, if this is not the case, a split solution is optimal.
The resulting lift in sales can vary widely depending on parameters. For example, it can be arbitrarily large when substitute quality is acceptable, demand is high, and premium inventory is scarce.