Maximizing the number of jobs scheduled at their baseline starting times in case of machine failures

Abstract

We investigate the problem of keeping the maximum number of starting times of a baseline schedule if some machines happen to be out of order when the baseline schedule is to be implemented. If the machines are identical, we show that the problem is polynomially solved when no deadline is imposed on the reactive schedule and is strongly NP-hard otherwise. If the number of unrelated machines is fixed and if no deadline is imposed on the reactive schedule, a polynomial algorithm, based on a state graph, has been developed. We conclude with an open complexity question and some further research directions for this class of problems.

Correctness proof of ALG(X, m, f)

Proof

We denote by L the list $(1,\ldots ,n)$.

Let H be the subset returned by ALG(X, m, f), and let $H^*$ be an optimal x-compatible subset. Let i be the first job of L that belongs to H but not to $H^*$. From the definition of i, every job $k\in \{1,\ldots ,i-1\}$ either belongs to H and $H^*$, or does not belong either to H or $H^*$. So $H^*$ contains a job $j>i$ that does not belong to H since otherwise the number of jobs of $S^*$ would be less than the number of jobs of H. Let us consider such a job j ($j\in H^*$, $j\not \in H$) with a minimum $x_j$ (see Fig. 10) and consider the set $\bar{H}=H^*\setminus \{j\}\cup \{i\}$.

Fig. 10

The exchange process $\bar{H}=H^*\setminus \{j\}\cup \{i\}$

From the definition of j, every job $k>i$ such that $x_k<x_j$ satisfies either ($k\not \in H^*$, $k\not \in H$) or ($k\in H^*$, $k\in H$) or ($k\not \in H^*$, $k\in H$). Since $x_j+p_j\ge x_i+p_i$, after time $x_j$, the number of active jobs in ${\bar{H}}$ is at most equal to the number of active jobs in $H^*$. Before time $x_j$, the active jobs of ${\bar{H}}$ are also the active jobs of H. So, ${\bar{H}}$ is also an optimal solution. Further, we can repeat that exchange process until there is no such job i and conclude that H is an optimal x-compatible subset. $\square$