Economic Lot-Sizing Problem with Remanufacturing Option: Complexity and Algorithms

Abstract

In a single item dynamic lot-sizing problem, we are given a time horizon and demand for a single item in every time period. The problem seeks a solution that determines how much to produce and carry at each time period, so that we will incur the least amount of production and inventory cost. When the remanufacturing option is included, the input comprises of number of returned products at each time period that can be potentially remanufactured to satisfy the demands, where remanufacturing and inventory costs are applicable. For this problem, we first show that it cannot have a fully polynomial time approximation scheme (FPTAS). We then provide a pseudo-polynomial algorithm to solve the problem and show how this algorithm can be adapted to solve it in polynomial time, when we make certain realistic assumptions on the cost structure. We finally give a computational study for the capacitated version of the problem and provide some valid inequalities and computational results that indicate that they significantly improve the lower bound for a certain class of instances.

Appendices

A Proof of Lemma 1

Proof

Suppose in $(\mathbf {p}^*,\mathbf {q}^*)$, we produce something not from this set $\{0, \tilde{R}_t\}$. Hence, some intermediate return stock of $0< a < \tilde{R}_t$ is carried, which also means that we are remanufacturing at time t in the optimal solution. Since $t<t^*$, there exist a time period after t in the optimal solution where we remanufacture. Let the $\tilde{t}$ be the first time period after t, when we remanufacture in the optimal solution. We are also carrying a non-zero return inventory until this time period. If we remanufacture at least a in time $\tilde{t}$, then we could have remanufactured this a in time t and carried a units of manufactured inventory until time $\tilde{t}$ with no additional cost, since return inventory cost is higher than manufactured inventory cost. If we produced less than a in time $\tilde{t}$, say $\tilde{a}$, then we could have produced $\tilde{a}$ in time t and produced nothing in time $\tilde{t}$ and continue with our argument. If $\tilde{t} = t^*$ and $\tilde{a} < a$, then we would have new optimal solution with t being the last time period of remanufacturing.    $\square$

B Proof of Lemma 3

Proof

We prove this lemma again through induction. For $t=1$, the lemma’s claim is that the inventory level of serviceable goods after time $t=1$ will belong to the set $\{0, \bigcup _{i=2}^T \{\sum _{k=2}^iD_k\}, \bigcup _{i=2}^{\ell ^*}\bigcup _{j=2}^i\{(\sum _{k=2}^{i}D_k - \sum _{k=2}^{j}R_k)^+\}, (R_1-D_1)^+ \}$. In order to show this, we do a case analysis:

• Case 1 Manufacturing takes place at $t=1$ : In this case, either remanufacturing does not take place or it does take place and we have $R_1 < D_1$, otherwise we could manufacture everything we manufactured in time period 1 in time period 2 and get a cheaper solution by saving on the inventory cost. Let us take k, where $1<k\le \ell ^*$, as the first time period after time $t=1$ when we manufacture again. Now the demand for all the intermediate periods $\sum _{i=2}^{k-1}D_i$ needs to come from either the manufactured goods in period 1 or the return products from periods 1 to $k-1$. From Lemma 2, we know that the only possible return inventory levels between the time periods 2 to $k-1$ are $\{0, \bigcup _{i=2}^{k-1} \{\mathcal {R}_{i} - R_1\}\}$ and we know from lemma 1 that at any of these time periods, we either remanufacture all available inventory or none of them. The remaining demand then needs to be manufactured at time period 1. This has to be true for all values of $k=2\dots \ell ^*$. So we get the claim.

• Case 2 Manufacturing does not take place at $t=1$ : This would mean that $R_1 \ge D_1$, otherwise, we would not have a feasible solution and the possible inventory levels in this case is $(R_1-D_1)^+$.

In order to see that the lemma is true, we invoke the induction hypothesis and do a similar case analysis as above.    $\square$