Open Shop Scheduling with Synchronization
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Abstract
In this paper, we study open shop scheduling problems with synchronization. This model has the same features as the classical open shop model, where each of the n jobs has to be processed by each of the m machines in an arbitrary order. Unlike the classical model, jobs are processed in synchronous cycles, which means that the m operations of the same cycle start at the same time. Within one cycle, machines which process operations with smaller processing times have to wait until the longest operation of the cycle is finished before the next cycle can start. Thus, the length of a cycle is equal to the maximum processing time of its operations. In this paper, we continue the line of research started by Weiß et al. (Discrete Appl Math 211:183–203, 2016). We establish new structural results for the twomachine problem with the makespan objective and use them to formulate an easier solution algorithm. Other versions of the problem, with the total completion time objective and those which involve due dates or deadlines, turn out to be NPhard in the strong sense, even for \(m=2\) machines. We also show that relaxed models, in which cycles are allowed to contain less than m jobs, have the same complexity status.
Keywords
Open shop Synchronization Complexity1 Introduction
Scheduling problems with synchronization arise in applications where job processing includes several stages, performed by different processing machines, and all movements of jobs between machines have to be done simultaneously. This may be caused by special requirements of job transfers, as it happens, for example, if jobs are installed on a circular production unit which rotates to move jobs simultaneously to machines of the next stage (see Soylu et al. 2007; Huang 2008; Waldherr and Knust 2014). Alternatively, there may be health and safety regulations requiring that no machine is in operation while jobs are being removed from or moved to a machine. Similar synchronization takes place in the context of switchbased communication systems, where senders transmit messages to receivers in a synchronous manner, as this eliminates possible clashes for receivers (see Gopal and Wong 1985; Rendl 1985; Kesselman and Kogan 2007).
Synchronization arises naturally in assembly line systems where each assembly operation may start only after all preceding operations are completed, see Doerr et al. (2000), Chiang et al. (2012), and Urban and Chiang (2016), and the survey by Boysen et al. (2008). In the context of shop scheduling models, synchronization aspects were initially studied for flow shops (Soylu et al. 2007; Huang 2008; Waldherr and Knust 2015), and later for open shops (Weiß et al. 2016). In the latter paper the makespan problem is addressed, with the main focus on the underlying assignment model. In the current paper we continue that line of research, elaborating further the study of the makespan minimization problem and addressing other variants of the model with different objective functions.
Formally, the open shop model with synchronization is defined as follows. As in the classical open shop, n jobs \(J_{1},J_{2},\ldots ,J_{n}\) have to be processed by m machines \(M_{1},M_{2},\ldots ,M_{m}, n\ge m\). Each job \( J_{j}, 1\le j\le n\), consists of m operations \(O_{ij}\) for \(1\le i\le m\), where \(O_{ij}\) has to be processed on machine \(M_{i}\) without preemption for \(p_{ij}\) time units. The synchronization requirement implies that job processing is organized in synchronous cycles, with operations of the same cycle starting at the same time. Within one cycle, machines which process operations of smaller processing times have to wait until the longest operation of the cycle is finished before the next cycle can start. Thus, the length of a cycle is equal to the maximum processing time of its operations. Similar to the classical open shop model, we assume that unlimited buffer exists between the machines, i.e., jobs which are finished on one machine can wait for an arbitrary number of cycles to be scheduled on the next machine.
The goal is to assign the nm operations to the m machines in n cycles such that a given objective function f is optimized. Function f depends on the completion times \(C_{j}\) of the jobs \(J_{j}\), where \(C_{j}\) is the completion time of the last cycle in which an operation of job \(J_{j}\) is scheduled. Following the earlier research by Huang (2008) and Waldherr and Knust (2015), we denote synchronous movement of the jobs by “synmv” in the \(\beta \)field of the traditional threefield notation. We write Osynmvf for the general synchronous open shop problem with objective function f and Omsynmvf if the number m of machines is fixed (i.e., not part of the input). The most common objective function is to minimize the makespan \(C_{\max }\), defined as the completion time of the last cycle of a schedule. If deadlines \(D_{j}\) are given for the jobs \(J_{j}\), the task is to find a feasible schedule with all jobs meeting their deadlines, \(C_{j}\le D_{j}\) for \(1\le j\le n\). We use the notation \(Osynmv,C_{j}\le D_{j}\) for the feasibility problem with deadlines. In problem \(Osynmv\sum C_{j}\) the sum of all completion times has to be minimized.
Usually, we assume that every cycle contains exactly m operations, one on each machine. In that case, together with the previously stated assumption \(n\ge m\), exactly n cycles are needed to process all jobs. However, sometimes it is beneficial to relax the requirement for exactly m operations per cycle. Then a feasible schedule may contain incomplete cycles, with less than m operations. We denote such a relaxed model by including “rel” in the \(\beta \)field. Similar to the observation of Kouvelis and Karabati (1999) that introducing idle times in a synchronous flow shop may be beneficial, we will show that a schedule for the relaxed problem Osynmv, relf consisting of more than n cycles may outperform a schedule for the nonrelaxed problem Osynmvf with n cycles.
The synchronous open shop model is closely related to long known optimization problems in the area of communication networks: the underlying model is the maxweight edge coloring problem (MEC), restricted to complete bipartite graphs, see Weiß et al. (2016) for the link between the models, and Mestre and Raman (2013) for the most recent survey on MEC and other versions of maxcoloring problems. As stated in Weiß et al. (2016), the complexity results from Rendl (1985) and Demange et al. (2002), formulated for MEC, imply that problems \(OsynmvC_{\max }\) and \(Osynmv,relC_{\max }\) are strongly NPhard if both n and m are part of the input. Moreover, using the results from Demange et al. (2002), Escoffier et al. (2006), Kesselman and Kogan (2007), de Werra et al. (2009), and Mestre and Raman (2013), formulated for MEC on cubic bipartite graphs, we conclude that these two open shop problems remain strongly NPhard even if each job is processed by at most three machines and if there are only three different values for nonzero processing times.
On the other hand, if the number of machines m is fixed, then problem \(OmsynmvC_{\max }\) can be solved in polynomial time as mdimensional assignment problem with a nearly Monge weight array of size \(n\times \cdots \times n\), as discussed in Weiß et al. (2016) and in Sect. 2 of the current paper. The relaxed version \(Omsynmv,relC_{\max }\) admits the same assignment model, but with a larger mdimensional weight array extended by adding dummy jobs. As observed in Weiß et al. (2016), the number of dummy jobs can be bounded by \((m1)n\). Both problems, \(OmsynmvC_{\max }\) and \( Omsynmv,relC_{\max }\), are solvable in \({{\mathcal {O}}}(n)\) time, after operations are sorted in nonincreasing (or nondecreasing) order of processing times on all machines. However, this algorithm becomes impractical for larger instances, as the constant term of the linear growth rate exceeds \((m!)^{m}\).
The remainder of this paper is organized as follows. In Sect. 2, we consider problem \(O2synmvC_{\max }\) and establish a new structural property of an optimal solution. Based on it we formulate a new (much easier) \({{\mathcal {O}}}(n)\)time solution algorithm, assuming jobs are presorted on each machine. Then we address in more detail problem \( Osynmv,relC_{\max }\) and provide a tight bound on the maximum number of cycles needed to get an optimal solution. In Sects. 3 and 4 we show that problems \(O2synmv, C_j \le D_j\) and \( O2synmv\sum C_j\) are strongly NPhard. Finally, conclusions are presented in Sect. 5.
2 Minimizing the makespan
In this section, we consider synchronous open shop problems with the makespan objective. Recall that problem \(OmsynmvC_{\max }\) with a fixed number of machines m can be solved in \({{\mathcal {O}}}(n)\) time (after presorting) by the algorithm from Weiß et al. (2016). In Sect. 2.1 we elaborate further results for the twomachine problem \(O2synmvC_{\max }\), providing a new structural property of an optimal schedule, which results in an easier solution algorithm. In Sect. 2.2 we study the relaxed problem \(Osynmv,relC_{\max }\) and determine a tight bound on the maximum number of cycles in an optimal solution.
2.1 Problem \(O2synmvC_{\max }\)
Note that the problem \({{\mathrm {AP}}}_{\infty }\) in its more general form is the subject of our related paper, Weiß et al. (2016). In the current paper we focus on the formulation \({{\mathrm {AP}}}_{{\mathcal {F}}}\), which is equivalent to \({{\mathrm {AP}}}_{\infty }\) for costs \(c_{ij}\) of form (2). Formulation \({{\mathrm {AP}}}_{{\mathcal {F}}}\) allows us to produce stronger results, see Theorems 1–2 in the next section. The main advantage of formulation \({{\mathrm {AP}}}_{{{\mathcal {F}}}}\) is the possibility to use finite wvalues for all pairs of indices, including \(w_{ij}\)’s defined for forbidden pairs \((i,j) \in {{\mathcal {F}}}\).
Example 1
The entries in bold font in \({{\mathcal {W}}}\) and \(\mathcal {C}\) correspond to the optimal solution illustrated in Fig. 1. Here, \(x_{12}=1\) for the pair of jobs \(J_{1},J_{2}\) assigned to the same cycle, and \(x_{23}=x_{34}=x_{41}=1\) for the other cycles. The makespan is \(7+6+3+3=19\).
The corridor property is proved in Weiß et al. (2016) in its generalized form for the case of the mdimensional assignment problem with a nearly Monge array (this is an array where \(\infty \)entries are allowed and the Monge property has to be satisfied by all finite entries). Thus, this property also holds for the mmachine synchronous open shop problem. It appears that for the case of \(m=2\) the structure of an optimal solution can be characterized in a more precise way, which makes it possible to develop an easier solution algorithm.
In the following, we present an alternative characterization of optimal solutions for \(m=2\) and develop an efficient algorithm for constructing an optimal solution. Note that the arguments in Weiß et al. (2016) are presented with respect to problem \(\mathrm {AP} _{\infty }\); in this paper our arguments are based on the formulation \(\mathrm {AP }_{{{\mathcal {F}}}}\) and on its relaxation \(\mathrm {AP}_{\mathcal {F=}\emptyset }\) , with the condition “\(x_{ij}=0\ \)for \((i,j)\in {{\mathcal {F}}}\) ” dropped.
Theorem 1
(“Small Block Property”): There exists an optimal solution to problem \(\mathrm {AP}_{\mathbf {\mathcal {F}}}\) in blockdiagonal structure, containing only blocks of type (6).
This theorem is proved in Appendix 1. The small block property leads to an efficient \({{\mathcal {O}}}(n)\)time dynamic programing algorithm to find an optimal solution. Here we use formulation \(\mathrm {AP}_{\infty }\) rather than \(\mathrm {AP}_{{{\mathcal {F}}}}\), as infinite costs can be easily handled by recursive formulae. The algorithm enumerates optimal partial solutions, extending them repeatedly by adding blocks of size 1, 2, or 3.

extending \(S_{i}\) by adding a block of size 1 with \(x_{i+1,i+1}=1\); the cost of the assignment increases by \(w_{i+1,i+1}\);

extending \(S_{i1}\) by adding a block of size 2 with \(x_{i,i+1}=x_{i+1,i}=1\); the cost of the assignment increases by \(w_{i,i+1}+w_{i+1,i}\);
 extending \(S_{i2}\) by adding a block of size 3 with the smallest cost:Let \(w(S_{i})\) denote the cost of \(S_{i}\). Then
 (i)
\(x_{i1,i+1}=x_{i,i1}=x_{i+1,i}=1\) with the cost \(w_{i1,i+1}+w_{i,i1}+w_{i+1,i}\), or
 (ii)
\(x_{i1,i}=x_{i,i+1}=x_{i+1,i1}\) with the cost \(w_{i1,i}+w_{i,i+1}+w_{i+1,i1}\).
 (i)
Theorem 2
Problem \(O2synmvC_{\max }\) can be solved in \({{\mathcal {O}}}(n)\) time.
 (i)
there is no more than one \(\infty \)entry in every row and every column of the cost matrix \(\mathcal {C}\), and
 (ii)
matrix \(\mathcal {C}\) can be transformed into a Monge matrix by modifying only the \(\infty \)entries, keeping other entries unchanged.
Finally, we observe that while the assignment matrices arising from the multimachine case are completable in the same way as for the twomachine case (see Weiß et al. 2016), it remains open whether this can be used to obtain an improved result for more than two machines as well. The technical difficulties of that case are beyond the scope of this paper.
2.2 Problem \(Osynmv,relC_{\max }\)
In this section, we consider the relaxed problem \(Osynmv,relC_{\max }\) where more than n cycles are allowed, with unallocated (idle) machines in some cycles. This problem can be transformed to a variant of problem \(OsynmvC_{\max }\) by introducing dummy jobs, used to model idle intervals on the machines. Dummy jobs have zerolength operations on all machines, and it is allowed to assign several operations of a dummy job to the same cycle. Thus, in a feasible schedule with dummy jobs, all cycles are complete, but some of the m operations in a cycle may belong to dummy jobs.
Similar to the observation of Kouvelis and Karabati (1999) that introducing idle times in a synchronous flow shop may be beneficial, we show that a schedule for the relaxed open shop problem \(Osynmv,relC_{\max }\) consisting of more than n cycles may outperform a schedule for the nonrelaxed problem \(OsynmvC_{\max }\) with n cycles.
Example 2
Clearly, for algorithmic purposes it is desirable to have the number of added dummy jobs as small as possible. As discussed in Waldherr et al. (2015), for the synchronous flow shop problem \(Fsynmv, relC_{\max }\), instances exist where for an optimal solution \((n1)(m2)\) dummy jobs are needed. In the following we show that for the open shop problem \(Osynmv, relC_{\max }\) at most \(m1\) dummy jobs are needed to obtain an optimal solution.
Theorem 3
There exists an optimal solution to problem \(Osynmv,relC_{\max }\) with at most \(m1\) dummy jobs, so that the number of cycles is at most \(n+m1\).
Proof
Let S be an optimal schedule with \(\xi \) dummy jobs, \(\xi \ge m\). We construct another schedule \(\widetilde{S}\) with \(C_{\max }(\widetilde{S} )\le C_{\max }(S)\) and \(\xi 1\) dummy jobs. Notice that it is allowed to assign several operations of the same dummy job to any cycle.
Case 1 If there exists a cycle \(I^{\prime }\) which consists solely of dummy operations of the same job \(J_{d}\in \{J_{n+1},J_{n+2},\ldots ,J_{n+\xi }\}\), then that dummy job can be eliminated and \(\widetilde{S}\) is found.
Case 2 If there exists a cycle \(I^{\prime }\) which consists solely of dummy operations, some of which belong to different dummy jobs, then we can achieve Case 1 by selecting a dummy job \(J_{d}\) arbitrarily and swapping its operations from outside \(I^{\prime }\) with the dummy operations in \(I^{\prime }\). The resulting schedule is feasible and has the same makespan.
Case 3 Suppose no cycle in S consists purely of dummy operations. Let \(I^{\prime }\) be the shortest cycle and let \(\nu \) be the number of actual operations in \(I^{\prime }, 1\le \nu \le m\). We demonstrate that each actual operation processed in \(I^{\prime }\) can be swapped with a dummy operation from another cycle.
Consider an actual operation \(O_{ij}\) in cycle \(I^{\prime }\) with machine \(M_{i}\) processing job \(J_{j}\). Select another cycle \(I^{\prime \prime }\) (its existence is demonstrated below) such that it does not involve an operation of \(J_{j}\) and has a dummy operation on \(M_{i}\). Swap operations on \(M_{i}\) in \(I^{\prime }\) and \(I^{\prime \prime }\), reducing the number of actual operations in \(I^{\prime }\) by 1. Clearly, after the swap both cycles are feasible, because introducing a dummy operation into \(I^{\prime }\) cannot cause a conflict, and because no operation of \(J_j\) was processed in \(I^{\prime \prime }\) before the swap. After the swap, both cycles \(I^{\prime }\) and \(I^{\prime \prime }\) have either the same length as before or cycle \(I^{\prime }\) becomes shorter. Performing the described swaps for each actual operation \(O_{ij}\) in cycle \(I^{\prime }\), we arrive at Case 1 or 2.

there are at least \(\xi \) cycles with a dummy operation on \(M_{i}(\xi \ge m)\) and those cycles are different from \(I^{\prime }\);

there are exactly \(m1\) cycles with \(J_{j}\) processed on a machine that differs from \(M_{i}\), and those cycles are different from \(I^{\prime }\). \(\square \)
We continue by demonstrating that the bound \(m1\) is tight.
Example 3
Consider an instance of problem \(Osynmv,relC_{\max }\) with m machines, \(n=m+1\) jobs and processing times \(p_{ij}=2m\) for \(i=1,\ldots ,m, j=1,\ldots ,n1,\) and \(p_{in}=1\) for \(i=1,\ldots ,m\).
An optimal schedule consists of m complete cycles of length 2m each, containing operations of the jobs \(\left\{ J_{1},J_{2},\ldots ,J_{m}\right\} \) only, and m incomplete cycles with the single actual job \(J_{m+1}\) grouped with \(m1\) dummy jobs, see Fig. 3. The optimal makespan is \(C_{\max }^{\mathrm {opt}}=2m^{2}+m\). In any schedule with less than \(m1\) dummy jobs, at least one operation of job \(J_{m+1}\) is grouped with another operation of an actual job, the length of such a cycle being 2m. Thus, a schedule with less that \(m1\) dummy jobs consists of at least \(m+1\) cycles of length 2m, so that the makespan is at least \(2m(m+1)>C_{\max }^{\mathrm {opt}}\).
3 Scheduling with deadlines
In this section, we consider problem \(Osynmv,C_{j}\le D_{j}\), where each job \(J_{j}, 1\le j\le n\), has a given deadline \(D_{j}\) by which it has to be completed. We prove that finding a feasible schedule with all jobs meeting their deadlines is NPcomplete in the strong sense even if there are only two machines and each job has only one nonzero processing time. Furthermore, we show that problem \(O2synmv,C_{j}\le D_{j},D_{j}\in \{D^{\prime },D^{\prime \prime }\}\), where the set of all deadlines is limited to two values, is at least NPcomplete in the ordinary sense. The proofs presented below are based on the ideas of Brucker et al. (1998) who established the complexity status of the parallel batching problem with deadlines.
Consider the 3PARTITION problem (3PART) known to be strongly NPcomplete, cf. Garey and Johnson (1979). Given a set \(Q=\{1,\ldots ,3q\}\), a bound E and natural numbers \(e_{i}\) for every \(i\in Q\), satisfying \(\sum _{i\in Q}e_{i}=qE\) and \(\frac{E}{4}<e_{i}<\frac{E}{2}\), can Q be partitioned into q subsets \(Q_{k}, 1\le k\le q\), such that \(\sum _{i\in Q_{k}}e_{i}=E\)?
Lemma 1
If there exists a solution \(Q_{1},Q_{2},\ldots ,Q_{q}\) to an instance of 3PART, then there exists a feasible schedule for the instance I(q) of the twomachine synchronous open shop problem with q deadlines.
Proof
We construct a schedule \(S^{*}\) consisting of q components \(\varGamma _{1},\varGamma _{2},\ldots ,\varGamma _{q}\), each of which consists of 3q cycles, not counting zerolength cycles. In component \(\varGamma _{k}, 1\le k\le q\), machine A processes 3q component\(_{k}\) Ajobs, one job of each type \(l, l=1,2,\ldots ,3q\). Machine B processes 3 long Bjobs and \(3\left( q1\right) \) short Bjobs, also one job of each type \(l, l=1,2,\ldots ,3q\).
Within one component, every cycle combines an Ajob and a Bjob of the same type \(l, 1\le l\le 3q\). The ordering of cycles in each component is immaterial, but component \(\varGamma _{k}\) precedes component \(\varGamma _{k+1}, 1\le k\le q1\). If \(Q_{k}=\left\{ l_{1},l_{2},l_{3}\right\} \) is one of the sets of the solution to 3PART, then the three long Bjobs \(J_{q+1,l_{1}}, J_{q+1,l_{2}},J_{q+1,l_{3}}\) are assigned to cycle \(\varGamma _{k}\).
Finally, there are \(3q^{2}\) cycles of length zero. We assume that each zerolength operation is scheduled immediately after the nonzero operation of the same job.
The resulting schedule \(S^{*}\) is shown in Fig. 4.
It is easy to verify that if \(Q_{1},Q_{2},\ldots ,Q_{q}\) define a solution to the instance of 3PART, then the constructed schedule \(S^{*}\) is feasible with all jobs meeting their deadlines. \(\square \)
We now prove the reverse statement. The proof is structured into a series of properties where the last one is the main result of the lemma.
Lemma 2
 (1)
each cycle of nonzero length contains an Ajob of type l and a Bjob of the same type \(l, l=1,2,\ldots ,3q\); without loss of generality we can assume that each zerolength operation is scheduled in the cycle immediately after the nonzerolength operation of the same job;
 (2)
no component\(_{j}\) Ajob is scheduled on machine A before any component\(_{i}\) Ajob, with \(1\le i\le j1\); hence S is splittable into components \(\varGamma _{1},\varGamma _{2},\ldots ,\varGamma _{q}\) in accordance with Ajobs;
 (3)
each component \(\varGamma _{j}, 1\le j\le q\), defines a set \(Q_{j}\) of indices that correspond to long Bjobs scheduled in \(\varGamma _{j}\); the resulting sets \(Q_{1},Q_{2},\ldots ,Q_{q}\) define a solution to the instance of 3PART.
Proof
Note that the above especially shows that zerolength operations are only paired in cycles with other zerolength operations. Therefore, we can assume without loss of generality that zerolength operations are scheduled immediately after the nonzerolength operations of the same job. Indeed, if this is not the case, we can change the order of cycles, and possibly the assignment of zerolength operations to the zerolength cycles in order to achieve the assumed structure, without changing the feasibility of the schedule.
The second property implies that on machine A all component\(_{1}\) Ajobs are scheduled first, followed by all component\(_{2}\) Ajobs, etc. Thus, the sequence of jobs on machine A defines a splitting of the schedule S into components \(\varGamma _{1},\varGamma _{2},\ldots ,\varGamma _{q}\).
(3) Given a schedule S satisfying the first two properties, we first define sets \(Q_{1},Q_{2},\ldots ,Q_{q}\) and then show that they provide a solution to 3PART.
Schedule S consists of components \(\varGamma _{j}, 1\le j\le q\). In each component \(\varGamma _{j}\) machine A processes all component\(_{j}\) Ajobs \(J_{j,l} (1\le l\le 3q)\), each of which is paired with a Bjob of the same type l. Recall that a Bjob \(J_{q+1,l}\) of type l is long, with processing time \(lW+qe_{l}\). All other Bjobs \(J_{j,l}, q+2\le j\le 2q\), of type l are short, with processing time lW. Considering the long Bjobs of component \(\varGamma _{j}\), define a set \(Q_{j}\) of the associated indices, i.e., \(l\in Q_{j}\) if and only if the long Bjob \(J_{q+1,l}\) is scheduled in component \(\varGamma _{j}\). Denote the sum of the associated numbers in \(Q_{j}\) by \(e(Q_{j}):=\sum _{l\in Q_{j}}e_{l}\).
Lemmas 1 and 2 together imply the following result.
Theorem 4
Problem \(O2synmv,C_{j}\le D_{j}\) is NPcomplete in the strong sense, even if each job has only one nonzero operation.
Similar arguments can be used to formulate a reduction from the PARTITION problem (PART) to the twomachine synchronous open shop problem, instead of the reduction from 3PART. Notice that in the presented reduction from 3PART all Bjobs have the same deadline, while Ajobs have q different deadlines, one for each component \(\varGamma _{j}\) defining a set \(Q_{j}\) . In the reduction from PART we only require two different deadlines \( D,D^{\prime }\), one for each of the two sets corresponding to the solution to PART. Similar to the reduction from 3PART, we define component\(_{1}\) Ajobs with deadline D and component\(_{2}\) Ajobs with deadline \(D^{\prime }\) which define a splitting of the schedule into two components \(\varGamma _{1},\varGamma _{2}\). For each of the natural numbers of PART we define one long Bjob and one short Bjob and show that the distribution of the long jobs within the two components of the open shop schedule corresponds to a solution of PART. Omitting the details of the reduction, we state the following result.
Theorem 5
Problem \(O2synmv,C_{j}\le D_{j},D_{j}\in \{ D^{\prime },D^{\prime \prime }\} \) with only two different deadlines is at least ordinary NPcomplete, even if each job has only one nonzero operation.
At the end of this section we note that the complexity of the relaxed versions of the problems, which allow incomplete cycles modeled via dummy jobs, remains the same as stated in Theorems 4 and 5. Indeed, Property 1 of Lemma 2 stating that each nonzero operation of some job is paired with a nonzero operation of another job, still holds for the version with dummy jobs. Therefore, in the presence of dummy jobs a schedule meeting the deadlines has the same component structure as in Lemmas 1 and 2, so that the same reduction from 3PART (PART) works for proving that \(O2synmv,rel,C_{j}\le D_{j}\) is strongly NPcomplete and \(O2synmv,rel,C_{j}\le D_{j},D_{j}\in \left\{ D^{\prime },D^{\prime \prime }\right\} \) is at least ordinary NPcomplete.
4 Minimizing the total completion time
In this section, we prove that the synchronous open shop problem with the total completion time objective is strongly NPhard even in the case of \(m=2\) machines. The proof uses some ideas by Röck (1984) who proved NPhardness of problem \(F2nowait\sum C_{j}\). Note that the latter problem is equivalent to the synchronous flow shop problem \(F2synmv\sum C_{j}\).
For our problem \(O2synmv\sum C_{j}\) we construct a reduction from the auxiliary problem AUX, which can be treated as a modification of the HAMILTONIAN PATH problem known to be NPhard in the strong sense (Garey and Johnson 1979).

first add to \(G^{\prime }\) a universal vertex 0, i.e., a vertex connected by an edge with every other vertex; denote the resulting graph by \(G=(V,E)\);

then replace each edge of graph G by two directed arcs in opposite directions; denote the resulting directed graph by \(\overrightarrow{G}=(V, \overrightarrow{E})\).
Processing times of the jobs in instance SO
We call each operation with a processing time of L a “long operation” and each operation with a processing time of less than L a “short operation.” Further, we refer to a job as a long job if at least one of its operations is long and as a short job if both of its operations are short.
Theorem 6
Problem \(O2synmv\sum C_{j}\) is strongly NPhard.
Proof
Consider an instance AUX and the corresponding scheduling instance SO. We prove that an instance of problem AUX has a solution, if and only if the instance SO has a solution with \(\sum C_{j}\le \varTheta \).

In Part 1, machine \(M_{1}\) processes \(2\sigma +1\) vertexjobs and \(2\sigma \) arcjobs in the order that corresponds to traversing \(\epsilon \). Machine \(M_{2}\) starts with processing the forcing job \(F_{0}\) in cycle 1 and then proceeds in cycles \(2, 3, \ldots , 4\sigma +1\) with the same sequence of vertexjobs and arcjobs as they appear in cycles \(1, 2, \ldots , 4\sigma \) on machine \(M_{1}\). Notice that in Part 1 all vertex and arcjobs are fully processed on both machines except for job \(\hbox {Ve}_{0}^{d(v)}\) which is processed only on \(M_{1}\) in the last cycle \(4\sigma +1\).

In Part 2, machine \(M_{1}\) processes the forcing jobs \(F_{0}, F_{1}, \ldots , F_{2n^{9}}\) in the order of their numbering. Machine \(M_{2}\) processes in the first cycle of Part 2 (cycle \(4\sigma +2\) ) the vertexjob \(\hbox {Ve}_{0}^{d(v)}\) which is left from Part 1. Then in the remaining cycles \(4\sigma +3, \ldots , 4\sigma +2+2n^{9}\), every job \(F_{i} (i=1,\ldots ,2n^{9})\) on \(M_{1}\) is paired with job \(F_{i+1}\) on \(M_{2}\) if i is odd, and with job \(F_{i1}\), otherwise.
In Fig. 6 we present an example of the described schedule based on graph \(\overrightarrow{G}\) of Fig. 5. Notice that there are \(n=5\) vertices in \(\overrightarrow{G}\), and parameter \(\sigma \) equals 8. Traversing the Eulerian tour \(\epsilon =(0,1,0,2,0,3,0,4,2,4,3,2\), Open image in new window ) incurs the sequence of vertexjobs and arcjobs \((\hbox {Ve}_{0}^{0}, \hbox {Ar}_{01}, \hbox {Ve}_{1}^{1}, \hbox {Ar}_{10}, \hbox {Ve}_{0}^{1}, \hbox {Ar}_{02}, \ldots , \hbox {Ve}_{1}^{2}, \hbox {Ar}_{12}, \hbox {Ve}_{2}^{4}, \hbox {Ar}_{23}, \hbox {Ve}_{3}^{3}, \hbox {Ar}_{34}, \hbox {Ve}_{3}^{4}, \hbox {Ar}_{40}, \hbox {Ve}_{0}^{4})\). There are \(2\sigma +1=17\) vertexjobs, \(2\sigma =16\) arcjobs, and \(2n^{9}+1=2 \times 5^{9}+1\) jobs \(F_i\), so that all jobs are allocated in \(34+2\times 5^{9}\) cycles. The schedule is represented as a sequence of cycles, where the operations on machines \(M_{1}\) and \(M_{2}\) are enframed and the lengths of the corresponding operations are shown above or below. Operations of equal length in one cycle are shown as two boxes of the same length; the sizes of the boxes of different cycles are not to scale.
We demonstrate that the constructed schedule satisfies \(\sum C_{j}=\varTheta \). Observe that most cycles have equal workload on both machines, except for the n cycles that correspond to the vertexjobs of the Hamiltonian path; in each such cycle the operation on \(M_{1}\) is one unit longer than the operation on \(M_{2}\).
First consider the short jobs. The initial vertexjob \(\hbox {Ve}_{0}^{0}\) that corresponds to the origin \(v_{0}=0\) of \(\epsilon =\left( v_{0},v_{1},\ldots ,v_{2\sigma }\right) \) completes at time \(\xi \). Each subsequent vertexjob that corresponds to \(v_{i}, 1\le i\le 2\sigma n\), where we exclude the last n vertices of the Hamiltonian path, completes at time \((2i+1)\xi +iK\). Consider the next \(n1\) vertexjobs \(v_{i}\) with \(2\sigma n+1\le i\le 2\sigma 1\) (excluding the very last vertexjob \(\hbox {Ve}_{0}^{d(0)}\) as it is a long job); every such job \(v_{i}\) completes at time \((2i+1)\xi +iK+(n+i2\sigma )\).
The remaining short jobs correspond to arcjobs. The completion time of the ith arcjob \(\text {Ar}_{v_{i1}v_{i}}\) is \(2i\xi +iK2v_{i}\) for \(1\le i\le 2\sigma n\) and \(2i\xi +iK2v_{i}+(n+i2\sigma )\) for \(2\sigma n+1\le i\le 2\sigma \).
 1.
In each cycle in S, both operations are either short or long.
 2.
All long operations are scheduled in the last \(2n^{9}+1\) cycles. This defines the splitting of schedule S into Parts 1 and 2, with cycles \(1, 2, \ldots , 4\sigma +1\) and \(4\sigma +2, \ldots , 4\sigma +2+2n^{9}\).
 3.
The sum of completion times of all long jobs is at least \(\varTheta _{2}\).
 4.
In S, machine \(M_{1}\) operates without idle times.
 5.In Part 1 of S, job \(\hbox {Ve}_{0}^{0}\) is processed in the first two cycles which are of the form , where represents a short operation. While the order of these two cycles is immaterial, without loss of generality we assume that precedes ; otherwise the cycles can be swapped without changing the value of \(\sum C_{j}\).
 6.
The two operations of each vertexjob and the two operations of each arcjob are processed in two consecutive cycles, first on \(M_{1}\) and then on \(M_{2}\).
 7.
In Part 1 of S, machine \(M_{1}\) alternates between processing arcjobs and vertexjobs. Moreover, an operation of a vertexjob corresponding to v is followed by an operation of an arcjob corresponding to an arc leaving v. Similarly, an operation of an arcjob for arc (v, w) is followed by an operation of a vertexjob for vertex w. By Property 6, the same is true for machine \(M_{2}\) in Part 1 and in the first cycle that follows it.
 8.
The first arcjob that appears in S corresponds to an arc leaving 0. Among the vertexjobs, the last one is \(\text {Ve}_{0}^{d(0)}\).
Properties 1–2 allow the splitting of S into two parts. Part 2 plays an auxiliary role. Part 1 is closely linked to problem AUX.
The sequence of arc and vertexjobs in Part 1 of S defines an Eulerian tour in \(\overrightarrow{G}\). Indeed, all arcjobs appear in S and by Property 7 the order of the arc and vertexjobs in S defines an Eulerian trail in \(\overrightarrow{G}\). Since for every vertex v, its indegree equals its outdegree, an Eulerian trail must be an Eulerian tour. Denote it by \(\epsilon =\left( v_{0},v_{1},\ldots ,v_{2\sigma }\right) \). Due to Property 8 and by the assumption of Property 5 the Eulerian tour \(\epsilon \) starts and ends at \(v_0=0\).
In Fig. 7 we present the structure of Part 1 of schedule S, where \(\hbox {Ve}_{v}^{*}\) represents one of the vertexjobs \(\hbox {Ve}_{v}^{1}\), \(\hbox {Ve}_{v}^{2}\),..., \(\hbox {Ve}_{v}^{d(v)}\), with processing time \(\xi +K2v+1\) or \(\xi +K2v\) on machine \(M_{1}\), depending on whether the upper index is d(v) or a smaller number. Part 2 is as in the proof of “ \(\Rightarrow \)”.
Summary of the results
Problem  Complexity  References 

\(OsynmvC_{\max }\)  str. NPh.  Weiß et al. (2016) 
\(OmsynmvC_{\max }\)  \({{\mathcal {O}}}(n)^{*}\)  Weiß et al. (2016) 
\(O2synmvC_{\max }\)  \({{\mathcal {O}}}(n)\)  Section 2 
\(O2synmv,C_{j}\le D_{j}\)  str. NPc.  Section 3 
\(O2synmv,C_{j}\le D_{j}, D_{j}\in \{D^{\prime },D^{\prime \prime }\}\)  NPc.  Section 3 
\(O2synmv\sum C_{j}\)  str. NPh.  Section 4 
By the main assumption of the part “\(\Leftarrow \)”, AUX does not have a solution where the last n vertices form a Hamiltonian path. Therefore, the last n vertices of any Eulerian tour \(\epsilon =\left( v_{0},v_{1},\ldots ,v_{2\sigma }\right) \) have at least two occurrences of the same vertex v and therefore in the associated schedule, among the last n vertexjobs there are at least two vertexjobs \(\hbox {Ve}_{v}^{i}, \hbox {Ve}_{v}^{j}\) associated with v. Thus, it is impossible to have \(n1\) jobs from \(\vartheta \) allocated to the last \(n1\) oddnumbered cycles and to achieve the required threshold value \(\varTheta _{1}\). \(\square \)
At the end of this section we observe that the proof of Properties 1–8 can be adjusted to handle the case with dummy jobs. Indeed, in an optimal solution of the instance, even if we allow dummy jobs, dummy operations are not allowed to be paired with actual operations of nonzero length (see Property 4). We conclude therefore that the complexity status of the relaxed problem is the same as that for the standard one.
Theorem 7
Problem \(O2synmv, rel \sum C_j\) is strongly NPhard.
5 Conclusions
In this paper we studied synchronous open shop scheduling problems. The results are summarized in Table 2. Note that the polynomial time results in lines 2 and 3 do not include presorting of all jobs.
All results from Table 2 also hold for the relaxed versions of the scheduling problems, in which cycles may consist of less than m jobs.
For problem \(O2synmvC_{\max }\) we proved a new structural property, namely the small block property. Using it, we formulated a much easier solution algorithm than previously known. Unfortunately, we were unable to prove an improved structural property for any fixed \(m>2\). In Table 2 we quote a previously known algorithm, which is based on the corridor property. Our result for two machines gives hope that this general result for fixed m may also be improved and highlights possible approaches for such an improvement.
The NPcompleteness results of Sect. 3 imply that if instead of hard deadlines \(D_{j}\) soft due dates \(d_{j}\) are given (which are desirable to be met, but can be violated), then the corresponding problems Osynmvf with the traditional regular due daterelated objectives f such as the maximum lateness \(L_{\max }=\max _{1\le j\le n}\{C_{j}d_{j}\}\), the number of late jobs \(\sum _{j=1}^{n}U_{j}\), or the total tardiness \(\sum _{j=1}^{n}T_{j}\) are NPhard, even if there are only two values of the due dates, \(d_{j}\in \{d,d^{\prime }\}\). The corresponding problems become strongly NPhard in the case of arbitrary due dates \(d_{j}\).
Finally, due to the symmetry known for problems with due dates \(d_{j}\) and those with release dates \(r_{j}\), we conclude that problem \( O2synmv,r_{j}C_{\max }\) is also strongly NPhard and remains at least ordinary NPhard if there are only two different values of release dates for the jobs.
In Sect. 4 we show that \(O2synmv\sum C_{j}\) and its relaxed version are strongly NPhard. Thus, due to the reducibility between scheduling problems with different objectives, the open shop problem with synchronization is NPhard for any traditional scheduling objective function, except for \(C_{\max }\).
Overall the synchronized version of the open shop problem appears to be no harder than the classical version, with two additional positive results for it: 1) \(OmsynmvC_{\max }\) is polynomially solvable for any fixed m while \(OmC_{\max }\) is NPhard for \(m\ge 3\) (Gonzalez and Sahni 1976); 2) \( Osynmv,n=n'C_{\max }\) is polynomially solvable for any fixed number of jobs \(n'\) (due to the symmetry of jobs and machines), while \(On= n^{\prime }C_{\max }\) is NPhard for \(n^{\prime } \ge 3\). Moreover, in a solution to \(OsynmvC_{\max }\) with \(n\le m\) all jobs have the same completion time, so that an optimal schedule for \(C_{\max }\) is also optimal for any other nondecreasing objective f. It follows that we can solve problem \(Osynmv, n=n'f\), with a fixed number of jobs \(n'\), for any such objective f.
Finally, comparing the open shop and flow shop models with synchronization, we also observe that the open shop problem is no harder, with a positive result for \(OmsynmvC_{\max }\) with an arbitrary fixed number of machines m , while its flow shop counterpart \(FmsynmvC_{\max }\) is NPhard for \( m\ge 3\), see Waldherr and Knust (2015).
Notes
Acknowledgments
The work of S. Knust and S. Waldherr was supported by the Deutsche Forschungsgemeinschaft, KN 512/71. The work of N.V. Shakhlevich was supported by the EPSRC grant EP/K041274/1. We are very grateful for the comments of two anonymous reviewers who helped us to improve the presentation of the paper.
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