Rescheduling due to machine disruption to minimize the total weighted completion time

Abstract

We investigate a single machine rescheduling problem that arises from an unexpected machine unavailability, after the given set of jobs has already been scheduled to minimize the total weighted completion time. Such a disruption is represented as an unavailable time interval and is revealed to the production planner before any job is processed; the production planner wishes to reschedule the jobs to minimize the alteration to the originally planned schedule, which is measured as the maximum time deviation between the original and the new schedules for all the jobs. The objective function in this rescheduling problem is to minimize the sum of the total weighted completion time and the weighted maximum time deviation, under the constraint that the maximum time deviation is bounded above by a given value. That is, the maximum time deviation is taken both as a constraint and as part of the objective function. We present a pseudo-polynomial time exact algorithm and a fully polynomial time approximation scheme.

Appendices

A Proof of Lemma 2.1

Proof

By contradiction, assume $(J_i, J_j)$ is the first pair of jobs for which $J_i$ precedes $J_j$ in $\pi ^*$, i.e., $p_i/w_i \le p_j/w_j$, but $J_j$ immediately precedes $J_i$ in the earlier schedule of $\sigma ^*$. Let $\sigma '$ denote the new schedule obtained from $\sigma ^*$ by swapping $J_j$ and $J_i$. If $C_j(\sigma ') \ge C_j(\pi ^*)$, then $C_i(\sigma ') \ge C_i(\pi ^*)$ too, and thus $\Delta _j(\sigma ', \pi ^*) \le \Delta _i(\sigma ', \pi ^*) < \Delta _i(\sigma ^*, \pi ^*)$ due to $p_j > 0$; if $C_j(\sigma ') < C_j(\pi ^*)$ and $C_i(\sigma ') \le C_i(\pi ^*)$, then $\Delta _i(\sigma ', \pi ^*) \le \Delta _j(\sigma ', \pi ^*) < \Delta _j(\sigma ^*, \pi ^*)$ due to $p_i > 0$; lastly if $C_j(\sigma ') < C_j(\pi ^*)$ and $C_i(\sigma ') > C_i(\pi ^*)$, then $\Delta _i(\sigma ', \pi ^*) < \Delta _i(\sigma ^*, \pi ^*)$ and $\Delta _j(\sigma ', \pi ^*) < \Delta _j(\sigma ^*, \pi ^*)$. That is, $\sigma '$ is also a feasible reschedule.

Furthermore, the weighted completion times contributed by $J_i$ and $J_j$ in $\sigma '$ is no more than those in $\sigma ^*$, implying the optimality of $\sigma '$. It follows that, if necessary, after a sequence of job swappings, we will obtain an optimal reschedule in which the jobs in the earlier schedule are in the same order as they appear in $\pi ^*$. This proves the part (a).

For the second part (b) of the lemma, similarly by contradiction we assume $(J_i, J_j)$ is the first pair of jobs for which $J_i$ precedes $J_j$ in $\pi ^*$, i.e., $p_i/w_i \le p_j/w_j$, but $J_j$ immediately precedes $J_i$ in the later schedule of $\sigma ^*$. Let $\sigma '$ denote the new schedule obtained from $\sigma ^*$ by swapping $J_j$ and $J_i$. If $C_j(\sigma ') \ge C_j(\pi ^*)$, then $C_i(\sigma ') \ge C_i(\pi ^*)$ too, and thus $\Delta _j(\sigma ', \pi ^*) \le \Delta _i(\sigma ', \pi ^*) < \Delta _i(\sigma ^*, \pi ^*)$ due to $p_j > 0$; if $C_j(\sigma ') < C_j(\pi ^*)$ and $C_i(\sigma ') \le C_i(\pi ^*)$, then $\Delta _i(\sigma ', \pi ^*) \le \Delta _j(\sigma ', \pi ^*) < \Delta _j(\sigma ^*, \pi ^*)$ due to $p_i > 0$; lastly if $C_j(\sigma ') < C_j(\pi ^*)$ and $C_i(\sigma ') > C_i(\pi ^*)$, then $\Delta _i(\sigma ', \pi ^*) < \Delta _i(\sigma ^*, \pi ^*)$ and $\Delta _j(\sigma ', \pi ^*) < \Delta _j(\sigma ^*, \pi ^*)$. That is, $\sigma '$ is also a feasible reschedule.

Furthermore, the weighted completion times contributed by $J_i$ and $J_j$ in $\sigma '$ is no more than those in $\sigma ^*$, implying the optimality of $\sigma '$. If follows that, if necessary, after a sequence of job swappings we will obtain an optimal reschedule in which the jobs in the earlier schedule are in the same order as they appear in $\pi ^*$. $\square$

B Proof of Lemma 2.2

Proof

Item (a) is a direct consequence of Lemma 2.1, since the jobs in the earlier schedule are in the same order as they appear in $\pi ^*$, which is the WSPT order. That is, for the job $J_j$ in the earlier schedule of $\sigma ^*$, if all the jobs $J_i$, for $i = 1, 2, \ldots , j-1$, are also in the earlier schedule, then $C_j(\sigma ^*) = C_j(\pi ^*)$; otherwise, $C_j(\sigma ^*) < C_j(\pi ^*)$.

For item (b), assume the machine idles before processing the jobs $J_{j_1}$ and $J_{j_2}$, with $J_{j_1}$ preceding $J_{j_2}$ in the earlier schedule. We conclude from Lemma 2.1 and item (a) that if $\Delta _{j_2} < \Delta _{j_1} \le \Delta _{\max }$ (or $\Delta _{j_1} < \Delta _{j_2} \le \Delta _{\max }$, respectively), then moving the starting time of the job $J_{j_2}$ ($J_{j_1}$, respectively) one unit ahead will maintain the maximum time deviation and decrease the total weighted completion time, which contradicts the optimality of $\sigma ^*$. It follows that we must have $\Delta _{j_1} = \Delta _{j_2}$; in this case, if there is any job $J_j$ with $j_1< j < j_2$ in the later schedule, it can be moved to the earlier schedule to decrease the total weighted completion time, which again contradicts the optimality of $\sigma ^*$. Therefore, there is no job $J_j$ with $j_1< j < j_2$ in the later schedule, which together with $\Delta _{j_1} = \Delta _{j_2}$ imply that the machine does not idle before processing the job $J_{j_2}$.

Item (c) is clearly seen for the same reason used in the last paragraph, that firstly their time deviations have to be the same, and secondly if this time deviation is less than $\Delta _{\max }$, one can then move their starting time one unit ahead to decrease the total weighted completion time while maintaining the maximum time deviation, thus contradicting the optimality of $\sigma ^*$.

Item (d) is implied by Item (c), given that all the job processing times are positive.

Let the job in the earlier schedule right after the idle time period be $J_j$. clearly, there is a job $J_a$ with $S_a(\pi ^*) < S_j(\sigma ^*)$ in the later schedule of $\sigma ^*$, due to the machine idling. The time deviation for $J_a$ is $\Delta _a > T_2 - S_a(\pi ^*)$. If $S_j(\pi ^*) < T_2$, then $\Delta _{\max } = \Delta _j< T_2 - S_j(\sigma ^*)< T_2 - S_a(\pi ^*) < \Delta _a$, a contradiction. This proves item (e) that $S_j(\pi ^*) \ge T_2$.

Item (f) is clearly seen from Lemma 2.1 and the optimality of $\sigma ^*$, that if the machine idles then it can start processing the jobs after the idling period earlier to decrease the total weighted completion time, while maintaining or even decreasing the maximum time deviation.

From item (f) and the second part of Lemma 2.1, we see that the deviation times of the jobs in the later schedule are non-increasing. Therefore, item (g) is proved. $\square$