Rescheduling with new orders and general maximum allowable time disruptions

Abstract

We study the rescheduling with new orders on a single machine under the general maximum allowable time disruptions. Under the restriction of general maximum allowable time disruptions, each original job has an upper bound for its time disruption (regarded as the maximum allowable time disruption of the job), or equivalently, in every feasible schedule, the difference of the completion time of each original job compared to that in the pre-schedule does not exceed its maximum allowable time disruption. We also consider a stronger restriction which additionally requires that, in a feasible schedule, the starting time of each original job is not allowed to be scheduled smaller than that in the pre-schedule. Scheduling objectives to be minimized are the maximum lateness and the total completion time, respectively, and the pre-schedules of original jobs are given by EDD-schedule and SPT-schedule, respectively. Then we have four problems for consideration. For the two problems for minimizing the maximum lateness, we present strong NP-hardness proof, provide a simple 2-approximation polynomial-time algorithm, and show that, unless $\text {P}= \text {NP}$, the two problems cannot have an approximation polynomial-time algorithm with a performance ratio less than 2. For the two problems for minimizing the total completion time, we present strong NP-hardness proof, provide a simple heuristic algorithm, and show that, unless $\text {P}= \text {NP}$, the two problems cannot have an approximation polynomial-time algorithm with a performance ratio less than 4/3. Moreover, by relaxing the maximum allowable time disruptions of the original jobs, we present a super-optimal dual-approximation polynomial-time algorithm. As a consequence, if the maximum allowable time disruption of each original job is at least its processing time, then the two problems for minimizing the total completion time are solvable in polynomial time. Finally, we show that, under the agreeability assumption (i.e., the nondecreasing order of the maximum allowable time disruptions of the original jobs coincides with their scheduling order in the pre-schedule), the four problems in consideration are solvable in polynomial time.

Appendix: \(\sum C_j\)-minimization under (R1) and (R\(^*\)1)

The due-index constraint in scheduling was first introduced in Zhao and Yuan (2007). Under the due-index constraint, each job $J_j$ is assigned an index $k_j\in \{1, 2, \ldots , n\}$, called the positional due index of $J_j$. A schedule is feasible under the due-index constraint if $\sigma ^{-1}(j) \le k_j$ for all job $J_j\in {\mathcal {J}}$, that is, each job $J_j$ must be scheduled in a position with index in $\{1, 2, \ldots , k_j\}$. Then $1|\sigma ^{-1}(j) \le k_j| \sum C_j$ is used to denote the scheduling problem for minimizing the total completion time under the due-index constraint.

Gao and Yuan (2015) presented an $O(n^2)$-time algorithm for the bi-criteria scheduling problem $1|\sigma ^{-1}(j) \le k_j| \sum C_j: f_{\max }\le Q$, where $f_{\max }$ is an arbitrary nondecreasing max-form scheduling objective function. By setting $Q= +\infty$, Gao and Yuan’s algorithm also solves problem $1|\sigma ^{-1}(j) \le k_j| \sum C_j$ in $O(n^2)$ time. We next show that the time complexity can be reduced to $O(n\log n)$.

A modification of Gao and Yuan’s algorithm for problem $1|\sigma ^{-1}(j) \le k_j| \sum C_j$ on the job set ${\mathcal {J}}=\{J_1, J_2, \ldots , J_n\}$, which we call Index-SPT, can be stated as follows.

Index-SPT Beginning from the nth position (i.e., the last position), schedule the jobs backwardly. In every iteration, we keep a list ${\mathcal {L}}$ to store the available jobs (subject to the due-index constraint) in the SPT order and break ties by the increasing order of their indices. Then the last job in ${\mathcal {L}}$ (which must be a available job with the maximum processing time) is scheduled as the last job.

Optimality of algorithm Index-SPT is guaranteed by the discussion in Gao and Yuan (2015). Moreover, the $O(n\log n)$-time complexity of Index-SPT can be proved by a same discussion as that for algorithm Dual-Relaxing-SPT in Sect. 2.2. Then Index-SPT solves problem $1|\sigma ^{-1}(j) \le k_j| \sum C_j$ in $O(n\log n)$-time.

Now we consider the two rescheduling problems $1|\text {(R1)}|\sum C_j$ and $1|(\hbox {R}^*1)|\sum C_j$. We define the positional due index $k'_j$ of each job $J_j\in {\mathcal {J}}= {\mathcal {J}}_O\cup {\mathcal {J}}_N$ by the following way:

$$\begin{aligned} k'_j =\left\{ \begin{array}{ll} j+k_j, &{}\text {if }J_j\in {\mathcal {J}}_O,\\ n,&{}\text {if }J_j\in {\mathcal {J}}_N.\end{array}\right. \end{aligned}$$

(7)

We will show that the schedule, denoted by $\sigma$, obtained by algorithm Index-SPT for problem $1|\sigma ^{-1}(j) \le k'_j| \sum C_j$ on the job set ${\mathcal {J}}= {\mathcal {J}}_O\cup {\mathcal {J}}_N$ is optimal for both problems $1|\text {(R1)}|\sum C_j$ and $1|(\hbox {R}^*1)|\sum C_j$. As a consequence, the two problems are simultaneously solved by algorithm Index-SPT for problem $1|\sigma ^{-1}(j) \le k'_j| \sum C_j$ in $O(n\log n)$ time.

To this end, note that (R1) is a relaxation of $(\hbox {R}^*1)$ and, from (7), $(\hbox {R}^*1)$ is equivalent to the restriction “$\sigma ^{-1}(j) \le k'_j$ for all $J_j\in {\mathcal {J}}$ and $\sigma ^{-1}(j) \ge j$ for all $J_j\in {\mathcal {J}}_O$”. Then we only need to show that, for the scheduled $\sigma$ obtained by algorithm Index-SPT for problem $1|\sigma ^{-1}(j) \le k'_j| \sum C_j$, we have

$$\begin{aligned} \sigma ^{-1}(j) \ge j \text { for all } J_j\in {\mathcal {J}}_O. \end{aligned}$$

(8)

The following proof is similar to but simpler than the proof of (6).

Suppose to the contrary that the statement in (8) is violated. Let $j\in \{1, 2, \ldots , n_O\}$ be the maximum index of original jobs such that $\sigma ^{-1}(j) < j$. Then some original job $J_i\in {\mathcal {J}}_O$ with $i<j$ is scheduled after $J_j$ in $\sigma$. This means that $p_i\le p_j$. Let $J_x$ be the job scheduled directly after $J_j$ in $\sigma$. Since $k'_j=j+k_j\ge j \ge \sigma ^{-1}(x)$, at position $\sigma ^{-1}(x)$, both $J_j$ and $J_x$ are available. Then the scheduling strategy of Index-SPT at position $\sigma ^{-1}(x)$ implies that $x>j$ and $p_x\ge p_j$.

If $J_x\in {\mathcal {J}}_N$, then $x>i$ and $J_x \prec _{\sigma } J_i$. From the scheduling strategy of Index-SPT at position $\sigma ^{-1}(i)$, we have $p_x < p_i$ and so $p_x< p_j$, contradicting the fact that $p_x\ge p_j$.

If $J_x\in {\mathcal {J}}_O$, since $x>j$, we have $\sigma ^{-1}(x)=\sigma ^{-1}(j) +1 \le j <x$. This contradicts our assumption that j is the maximum index of the original jobs such that $\sigma ^{-1}(j) < j$. This completes the discussion.