A note on lot scheduling on a single machine to minimize maximum weighted tardiness

Abstract

We study a single machine lot scheduling problem. Customers’ orders may be processed simultaneously in the same lot. The order sizes are assumed to be order-dependent, and the lots have identical size and identical processing time. Orders may be split, i.e., their processing may be performed on two consecutive lots. We assume order-dependent due-dates and weights, and the goal is to minimize the maximum weighted tardiness among all orders. A polynomial time solution algorithm is introduced.

Appendix A

Strong NP-hardness of the case of no order splitting.

Theorem

Problem $1\left|Lot, No-split\right|{T}_{max}^{W}$ is NP-hard in the strong sense even when all the orders share a common due-date and all orders have identical weights.

Proof

We prove by reduction from the strongly NP-hard problem, 3-Partition.

3-partition: Given $3k$ positive integers, ${a}_{j}, j=1,\dots ,3k$, and a positive integer $B$ such that $\frac{B}{4}<{a}_{j}<\frac{B}{2}, j=1,\dots 3k$ and ${\sum }_{j=1}^{3k}{a}_{j}=kB$, does there exist disjoint sets ${A}_{1}, {A}_{2},\dots ,{A}_{k}$ such that ${\sum }_{j\in {A}_{i}}{a}_{j}=B, i=1,\dots ,k$?

Based on 3-Partition, we create the following instance of P2:

There are $n=3k$ orders, with processing times ${p}_{j}={a}_{j}, j=1,\dots ,3k$. The size of each lot is $B$ and the processing time of each lot is 1. There is a common due-date for all the orders: ${d}_{j}=kB, j=1,\dots ,3k$. We show that 3-Partition has a YES answer if and only if the above instance has maximal weighted tardiness of zero.

Assume first that 3-Partition has a Yes answer. Then, there are $k$ disjoint sets ${A}_{1}, {A}_{2},\dots ,{A}_{k}$, such that ${\sum }_{j\in {A}_{i}}{a}_{j}=B, i=1,\dots ,k$. Schedule each set in a single lot, hence, the orders are processed in exactly $k$ lots, and the last lot is completed at time $kB$, implying that there are no tardy orders i.e., ${T}_{max}^{W}=0$.

Assume now that 3-Partition does not have a Yes answer. Then, there must be at least one lot which is not fully used, i.e., the total size of the orders assigned to this lot is strictly smaller than $B$. It follows that the solution must contain at least $k+1$ lots, and therefor ${T}_{max}^{W}>0$. $\hfill \square$