# Optimal Sequence in Which to Service Orders

27 Jul

What is the optimal order in which to service orders assuming a fixed budget?

Let’s assume that we have to service orders o_1,…,…o_n, with the n orders iterated by i. Let’s also assume that for each service order, we know how the costs change over time. For simplicity, let’s assume that time is discrete and portioned in units of days. If we service order o_i at time t, we expect the cost to be c_it. Each service order also has an expiration time, j, after which the order cannot be serviced. The cost at expiration time, j, is the cost of failure and denoted by c_ij.

The optimal sequence of servicing orders is determined by expected losses—service the order first where the expected loss is the greatest. This leaves us with the question of how to estimate expected loss at time t. To come up with an expectation, we need to sum over some probability distribution. For o_it, we need the probability, p_it, that we would service o_i at t+1 till j. And then, we need to multiply p_it with c_ij. So framed, the expected loss for order i at time t =
c_it – \Sigma_{t+1}_{j} p_it * c_it

However, determining p_it is not straightforward. New items are added to the queue at t+1. On the flip side, we also get to re-prioritize at t+1. The question is if we will get to the item o_i at t+1? (It means p_it is 0 or 1.) For that, we need to forecast the kinds of items in the queue tomorrow. One simplification is to assume that items in the queue today are the same that will be in the queue tomorrow. Then, it reduces to estimating the cost of punting each item again tomorrow, sorting based on the costs at t+1, and checking whether we will get to clear the item. (We can forgo the simplification by forecasting our queue tomorrow, and each day after that till j for each item, and calculating the costs.)

If the data are available, we can tack on clearing time per order and get a better answer to whether we will clear o_it at time t or not.