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Journal of Scheduling

, Volume 16, Issue 4, pp 439–441

Erratum to: Minimizing total tardiness on parallel machines with preemptions

• Svetlana A. Kravchenko
• Frank Werner
Erratum

Abstract

In this note, we correct a mistake which was made in the paper “Minimizing Total Tardiness on Parallel Machines with Preemptions” (see J Schedul 15:193–200, 2012).

Keywords

Parallel machines Total tardiness  Preemptions  NP-hardness

Introduction

In Sect. 3 of paper (Kravchenko and Werner 2012), a mistake in the NP-hardness proof for the problem $$P\mid \text{ pmtn}\mid \sum T_j$$ was made. A counterexample was given in Prot et al. (2012). In Sect. 2 of this note, we present a corrected proof. In Sect. 3, we give some further remarks.

Corrected proof

In this section, we show that problem $$P\mid \text{ pmtn}\mid \sum T_j$$ is NP-hard, i.e., we consider the problem of scheduling $$n$$ jobs on $$m$$ machines $$\{1,\ldots ,m\}$$ with preemptions to minimize the total tardiness value $$\sum _{j=1}^n T_j$$. Thus, for each job $$j$$, we know in advance its processing time $$p_j$$ and its due date $$d_j.$$

The proof is obtained by a reduction from PARTITION: Given a set of positive integers $$a_1,\ldots , a_k, b$$ with $$\sum _{i=1}^k a_i = 2b.$$ Does there exist a subset $$I\subseteq \{ 1, \ldots , k\}$$ such that $$\sum _{i\in I} a_i =b$$?

PARTITION is ordinary NP-hard (Karp 1972), and it can be shown that it remains so under the assumption that $$3k<b/2$$. To see this, it is sufficient to replace $$a_1,\ldots , a_k, b$$ by $$6k a_1,\ldots , 6k a_k, 6k b.$$

Given any instance of PARTITION, we define a corresponding instance of the problem $$P\mid \text{ pmtn}\mid \sum T_j$$ as follows. Let $$n=3k-1$$ and $$m=k.$$ We set $$L=10b^3$$. There are three classes of jobs:
• the class of $$a$$-jobs comprises $$k$$ jobs with processing times $$a_i$$ and due dates $$d_i=L$$ for $$i=1,\ldots , k;$$

• the class of $$ba$$-jobs comprises $$k$$ jobs with processing times $$9b^3+a_i b^2 - a_i$$ and due dates $$d_i=L-a_i$$ for $$i=1,\ldots , k;$$

• the class of $$b$$-jobs comprises $$k-1$$ jobs with processing times $$b^3$$ and due dates $$b^3.$$

We claim that PARTITION has a solution if and only if there exists a schedule with $$\sum T_j\le b^3+b.$$
First, we show that, if PARTITION has a solution, then for the corresponding instance of the problem $$P\mid \text{ pmtn}\mid \sum T_j$$ a schedule exists with $$\sum T_j\le b^3+b.$$ Without loss of generality, suppose that $$\{ a_1,\ldots , a_z\}$$ is a solution for PARTITION, i.e., we suppose that $$\sum _{i=1}^z a_i = b$$ holds. Then:
1. 1.

we schedule each $$ba$$-job $$i$$, $$i=1,\ldots , k,$$ with processing time $$9b^3+a_ib^2-a_i$$ for $$9b^3-a_i$$ time units in the interval $$[b^3,L-a_i]$$ on machine $$i$$, and each $$ba$$-job $$i, i=1,\ldots , z,$$ for $$a_ib^2$$ time units in the interval $$[L,L+a_ib^2]$$ on machine $$i$$;

2. 2.

we schedule each $$a$$-job $$i$$, $$i=1,\ldots , k,$$ with processing time $$a_i$$ for $$a_i$$ time units in the interval $$[L-a_i,L]$$ on machine $$i$$;

3. 3.

we schedule each $$b$$-job $$i, i=1,\ldots , k-1,$$ with processing time $$b^3$$ for $$b^3$$ time units in the interval $$[0,b^3]$$ on machine $$i$$; and

4. 4.
we schedule the remaining parts of the $$ba$$-jobs $$i, i=z+1,\ldots , k,$$ in the interval $$[0,b^3]$$ on machine $$k,$$ see Fig. 1.

Note that after steps 1–3, all $$a$$-jobs, all $$b$$-jobs, and all $$ba$$-jobs for $$i=1,\ldots , z$$ are completely scheduled whereas for $$i=z+1,\ldots , n$$, each $$ba$$-job is scheduled for $$9b^3-a_i$$ time units. Thus, before step 4, the only unscheduled jobs are the $$ba$$-jobs for $$a_ib^2$$ time units, $$i=z+1,\ldots , n$$.

Since, before step 4, machine $$k$$ is idle for the time interval $$[0,b^3]$$ and $$\sum _{i=z+1}^n a_ib^2 = b^3$$ holds, step 4 completes the construction of a feasible schedule.

Besides, for the constructed schedule we obtain
\begin{aligned}&\sum _{b{\text{-jobs}}} T_j= 0, \quad \sum _{a{\text{-jobs}}} T_j= 0, \qquad \mathop {\sum _{ba{\text{-jobs}}}}\limits _{j\in \{1,\ldots ,z\}} T_j= b^3+b,\\&\quad \mathop {\sum _{ba{\text{-jobs}}}}\limits _{j\in \{z+1,\ldots ,n\}} T_j= 0. \end{aligned}
Thus, for the constructed schedule $$\sum T_j= b^3+b$$ holds.

Next, we show that, if there exists a schedule for which $$\sum T_j\le b^3+b$$ holds, then PARTITION has a solution. Consider a schedule, say $$s$$, with $$\sum T_j\le b^3+b$$.

The $$b$$-jobs have to be scheduled within the time interval $$[0,2b^3+b],$$ since otherwise inequality $$\sum T_j> b^3+b$$ holds. Thus, after $$L$$ only $$a$$-jobs and $$ba$$-jobs can be scheduled.

All $$a$$-jobs have to be scheduled before their due dates, i.e., before the time $$L=10b^3$$. To see this, consider a $$ba$$-job, say $$f$$, with the maximal completion time $$C_f$$ among all $$ba$$-jobs. There are two cases:

Case 1. If there is an $$a$$-job scheduled (partially or completely) within the interval $$[C_f,+\infty [,$$ one can switch this part of the $$a$$-job with the $$ba$$-job $$f$$ scheduled within the interval $$[2b^3+b,9b^3].$$ The value $$\sum T_j$$ does not increase.

Case 2. However, if there are no $$a$$-jobs scheduled within the interval $$[C_f,+\infty [,$$ take any $$a$$-job processed within the interval $$[L,C_f[$$ for some time, say for $$\Delta$$ time units. Denote this part of the $$a$$-job by $$a_\Delta .$$ Now one can take the part of the $$ba$$-job $$f$$ processed within the interval $$[2b^3+b,9b^3]$$ for $$\Delta$$ time units. Denote this part of the job by $$f_\Delta .$$ Finally, one can replace $$f_\Delta$$ by $$a_\Delta ,$$ and schedule $$f_\Delta$$ in the interval $$[C_f,C_f+\Delta ]$$. The value of $$\sum T_j$$ does not increase.

As a result, we obtain a new schedule, say $$s^{\prime }$$, where all $$a$$-jobs are processed before $$L$$ and the completion time of the $$ba$$-job $$f$$ is $$C_f+X,$$ where $$X$$ is the total amount of the processing of $$a$$-jobs, scheduled after $$L$$ in $$s$$. Since the completion times are known for all jobs, the schedule $$s^{\prime }$$ can be constructed without applying the described transformation, but using a network flow model (Horn 1974) with a vertex for each job and for each completion time.

Thus, after the time $$L=10b^3$$, only $$ba$$-jobs can be scheduled. Denote this set of jobs by $$A,$$ i.e., $$A=\{ i \mid i \text{ is} \text{ a} ba{\text{-job} \text{ and} } C_i\ge L\}$$. Denote by $$l_i$$ the total length of a $$ba$$-job $$i,$$ scheduled after $$L$$.

Moreover, since the sum of all processing times is $$10b^3k+b^3,$$ we have $$\sum _{i\in A} l_i\ge b^3$$ and therefore, all $$b$$-jobs have to be scheduled not in the interval $$[0, 2b^3+b]$$ but in the interval $$[0, b^3+b].$$

Thus, within the interval $$[b^3+b,10b^3]$$, only $$ba$$-jobs and $$a$$-jobs can be scheduled. Since the total length of the $$a$$-jobs is $$2b,$$ it follows that each $$ba$$-job has to be processed in the interval $$[b^3+b,10b^3]$$ for at least $$9b^3-3b$$ time units and thus, inequality $$l_i\le p_i-9b^3+3b$$ holds, and since $$p_i=9b^3+a_ib^2-a_i,$$ we obtain the inequality $$l_i\le a_ib^2+3b-a_i,$$ i.e., $$l_i\le a_ib^2+3b$$ holds.

Now, by assumption, we have $$b^3+b\ge \sum _{i\in A} T_i$$. Since $$\sum _{i\in A} T_i\ge \sum _{i\in A} a_i + \sum _{i\in A} l_i$$ and $$\sum _{i\in A} l_i\ge b^3,$$ we obtain
\begin{aligned} b^3+b\ge \sum _{i\in A} T_i\ge \sum _{i\in A} a_i + \sum _{i\in A} l_i\ge \sum _{i\in A} a_i+b^3 \end{aligned}
and therefore, $$\sum _{i\in A} a_i \le b$$ holds.
Moreover, since inequalities $$\sum _{i\in A} l_i \ge b^3$$ and $$l_i\le a_i b^2 +3b$$ hold, we get
\begin{aligned} b^3\le \sum _{i\in A} l_i \le \sum _{i\in A} (a_ib^2+3b). \end{aligned}
Now, we have $$\sum _{i\in A} (a_ib^2+3b) = (\sum _{i\in A} a_i)b^2+|A|3b.$$ Since $$|A|\le k,$$ then
\begin{aligned} b^3 \le \left(\sum _{i\in A} a_i \right)b^2+3kb. \end{aligned}
Thus, one can see that, if $$3k<b/2,$$ then inequality $$\sum _{i\in A} a_i \ge b$$ holds.

Hence, we obtain $$\sum _{i=1}^z a_i = b,$$ i.e., the set of $$ba$$-jobs completed after time point $$L$$ defines the solution for PARTITION.

Remarks

In Kravchenko and Werner (2012), the polynomial transformation from PARTITION to the problem $$P\mid r_j, p_j=p, \text{ pmtn}\mid \sum T_j$$ substantially used the decision version of the problem $$P\mid \text{ pmtn}\mid \sum T_j.$$ In fact, the decision version of the problem $$P\mid \text{ pmtn}\mid \sum T_j$$ was polynomially reduced to the decision version of the problem $$P\mid r_j, p_j=p, \text{ pmtn}\mid \sum T_j.$$ In the corrected proof, we changed the decision version of the problem $$P\mid \text{ pmtn}\mid \sum T_j.$$ However, it is easy to see that the new decision version of the problem $$P\mid \text{ pmtn}\mid \sum T_j$$ can be polynomially transformed to the problem $$P\mid r_j, p_j=p, \text{ pmtn}\mid \sum T_j$$ using the same scheme given in Kravchenko and Werner (2012).

Thus, both problems $$P\mid \text{ pmtn}\mid \sum T_j$$ and $$P\mid r_j, p_j=p, \text{ pmtn}\mid \sum T_j$$ are NP-hard.

References

1. Horn, W. A. (1974). Some simple scheduling algorithms. Naval Research Logistics Quarterly, 21, 177–185.
2. Karp, R. M. (1972). Reducibility among combinatorial problems. In R. E. Miller & J. W. Thatcher (Eds.), Complexity of computer computations (pp. 85–103). New York: Plenum Press.
3. Kravchenko, S. A., & Werner, F. (2012). Minimizing total tardiness on parallel machines with preemptions. Journal of Scheduling, 15, 193–200.
4. Prot, D., Bellenguez-Morineau, O., & Lahlou, C. (2012). A note on the paper “Minimizing total tardiness on parallel machines with preemptions” by Kravchenko and Werner (2012). Journal of Scheduling doi:.

Copyright information

© Springer Science+Business Media New York 2013

Authors and Affiliations

1. 1.United Institute of Informatics ProblemsMinskBelarus
2. 2.Fakultät für MathematikOtto-von-Guericke-UniversitätMagdeburgGermany