A note: maximizing the weighted number of Just-in-Time jobs for a given job sequence

Abstract

We study a single machine scheduling problem to maximize the weighted number of Just-in-Time jobs, i.e., jobs which are completed exactly at their due-dates. We focus on the case that the job sequence is given. A pseudo-polynomial solution algorithm has been published for this problem, and we introduce here an efficient polynomial time dynamic programming algorithm. The new algorithm is tested numerically, and the results are compared to those obtained by the old algorithm. While the running times required by both algorithms for solving large instances are extremely small, it appears that the new algorithm performs better in two aspects: (1) The resulting running time is independent of the actual density of the due-dates, (2) The required memory is significantly reduced. An extension to a two-machine flowshop is provided, and this case is shown to remain polynomially solvable.

Appendix: Algorithm DP2 introduced by Lann and Mosheiov (1996)

Jobs are numbered according to the given sequence.

State variables:

$j$: index of the current job,

$t$: the completion time of the job $j-1$.

The return function:

$f\left(t,j\right):$ minimum cost of scheduling jobs $j,...,n$, where the starting time of job $j$ does not precede $t$. Thus, the recursion is given by:

$$ f\left( {t,j} \right) = \left\{ {\begin{array}{*{20}c} {\min \left\{ {w_{j}^{E} + f\left( {t + p_{j} , j + 1} \right), f\left( {d_{j} ,j + 1} \right)} \right\}} & {{\text{if}}\quad t + p_{j} \le d_{j} } \\ {w_{j}^{T} + f\left( {t + p_{j} , j + 1} \right)} & {{\text{otherwise}}} \\ \end{array} } \right.. $$

[Note that the first line reflects the case that the current job can be scheduled on time, so it is either scheduled as early as possible, i.e., at time $t$, or scheduled to be completed at the due-date. The second line reflects the case that the job must be tardy, and then it is scheduled as early as possible.]

The boundary conditions are: $f(t, n + 1) = 0$ for all values of $t$.

The optimal cost is given by $f(0, 1)$.

Running time: Let ${d}_{\mathrm{max}}=\mathrm{max}\{{d}_{j}, j=1,\dots ,n\}$, and recall that $P={\sum }_{j=1}^{n}{p}_{j}.$ Thus, an upper bound on the value of $t$ is $\mathrm{max}\{{d}_{\mathrm{max}}, P\}$. $j$ is bounded by $n$. The computation of $f(t,j)$ requires a constant time. It follows that the running time of DP2 is $O(n \mathrm{max}\{{d}_{\mathrm{max}}, P\}$.