Resource-constrained flowshop scheduling with separate resource recycling operations

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Resource-constrained ﬂowshop scheduling with separate resource

recycling operations

T.C.E. Cheng

a,1

, B.M.T. Lin

b,

, H.L. Huang

b a

Department of Logistics and Maritime Studies, The Hong Kong Polytechnic University, Kowloon, Hong Kong

b_{Institute of Information Management, Department of Information and Finance Management, National Chiao Tung University, Hsinchu 300, Taiwan}

a r t i c l e

i n f o

Available online 7 August 2010 Keywords: Relocation problem Resource constraint Flowshop scheduling Recycling operations Complexity

a b s t r a c t

This paper considers the relocation problem arising from public re-development projects cast as a two-machine flowshop scheduling problem. In such a project, some buildings need to be torn down and re-constructed. The two processes of tearing down and re-constructing each building are often viewed as a single operation. However, under certain circumstances, the re-construction process, i.e., the resource recycling process, can be viewed as a separate operation. In this paper we regard these two processes as separate on the assumption that they are handled by different working crews. We formulate the problem as a resource-constrained two-machine flowshop scheduling problem with the objective of finding a feasible re-development sequence that minimizes the makespan. We provide problem formulations, discuss the complexity results, and present polynomial algorithms for various special cases of the problem.

1. Introduction

Scheduling is a decision-making process concerning the allocation of limited resources to activities over time so as to optimize one or more objective functions. The most commonly considered resources are machines that are deployed to perform (process) the activities (tasks or jobs). When resource-constrained scheduling is considered, extra resources such as money, personnel, energy, etc., are required for the machines to process their assigned tasks. Resource constraints have been considered in scheduling problems[1–3]. A sequence of the tasks is called feasible if all constraints, especially the resource constraints, are satisﬁed and all operations are completed. In this paper we consider a variant of the relocation problem that can be converted into a two-machine ﬂowshop scheduling problem with generalized resource constraints.

The relocation problem was ﬁrst proposed and formulated by Kaplan and his colleagues [4–7] for a public house re-development project in Boston [8]. It was a successful operations research application in the public sector. In the basic setting of the relocation problem, there is a set of buildings to be re-constructed. Each building is speciﬁed by two capacity-related parameters: the number of current tenants and the number of tenants that can be accommodated after re-development. When a

building is under re-development, the evacuated tenants must be temporarily housed. The tenants are not required to reside at the same site after re-development. Given an initial budget of temporary housing, the authority has to determine a re-develop-ment sequence of the buildings such that all tenants can be successfully housed during the re-development process.

In the scheduling literature resources can be classiﬁed by type and category. For the relocation problem, all jobs need a single type of resource that is discrete and non-renewable. The original capacity of a building speciﬁes the amount of the resource required to start a job (i.e., temporary housing to accommodate the evacuated tenants) and the new capacity after re-develop-ment is the amount of the resource returned by the job upon its completion. The resource constraints involved in the relocation problem differ from the commonly adopted resource constraints in that the amount of the resource a job returns upon its completion is not necessarily equal to the amount of the resource the job acquired for its processing.

The basic relocation problem is concerned with the existence of a feasible re-development sequence. The optimization counter-part of the problem is to determine the minimum initial budget that guarantees the existence of a feasible sequence. Kaplan and Amir[6]showed that the optimization problem is mathematically equivalent to the two-machine ﬂowshop scheduling problem to minimize the makespan[9]. The ﬁnancial planning problem[10]

can be regarded as a special case of the relocation problem. More applications and discussion could be found in Amir and Kaplan[5]

and Cheng and Lin[11]. Kononov and Lin[12]investigated the relocation problem on identical parallel machines. They settled Contents lists available atScienceDirect

journal homepage:www.elsevier.com/locate/caor

Computers & Operations Research

Corresponding author. Tel.: + 886 3513 1472; fax: +886 3572 3792. E-mail addresses: LGTcheng@polyu.edu.hk (T.C.E. Cheng), bmtlin@mail.nctu.edu.tw, bmtlin@iim.nctu.edu.tw (B.M.T. Lin).

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the computational complexity of the problem and proposed approximation algorithms with analyses of their performance ratios. Lin and Liu[13]incorporated generalized due dates into the relocation problem to maximize the total rewards. They proposed a branch-and-bound algorithm equipped with two dominance rules and two lower bounds. Kononov and Lin[14]

proved the strong NP-hardness of the problem to minimize the total weighted completion time and introduced an equivalence property for the unit-weighted case and the unit processing time case. Sevastyanov et al.[15]considered the relocation problem to minimize the makespan subject to release dates. In addition to the complexity analyses of several problem settings, they proposed a multi-parametric dynamic programming algorithm to deal with the case with a ﬁxed number of distinct release dates. The running time of the dynamic program is pseudo-polynomial, which is tight because the case with two distinct due dates is NP-hard.

The first study on the relocation problem with the resource recycling issue incorporated is due to Lin and Huang [16]. In previous studies of resource-constrained scheduling, once a job is completed, the resource it returns is immediately available, i.e., resource recycling is assumed to be included in the job processing time. It is very similar to the situations where we assume that the setup time and removal time of a job are embedded in the processing time of the job. Under certain circumstances, the resource recycling process of a job can be viewed as a separate operation that can be performed concur-rently with the normal processing of the other jobs. For the relocation problem, the re-development of a building consists of two stages, namely demolishing and re-construction of the building. For the basic relocation problem, the two stages are considered as an aggregate job. In this paper we treat the resource recycling process as a separate operation that has to be processed by a dedicated machine. Although the model considered in this paper stems from the relocation problem, it is applicable to situations where there are separate resource recycling operations. Given a certain amount of the resource, our problem is to determine a feasible schedule that minimizes the makespan. We denote the studied problem by F2jrpjCmax using the three-field notation for scheduling problems[17]. The first field indicates that the machine configuration is a two-machine flowshop, where the first machine is for processing the jobs (demolishing the buildings) and the second machine is dedicated to the resource recycling operations (erecting new buildings). The second field highlights that the job characteristics are due to the context of the relocation problem. Lin and Huang[16]first proposed this problem setting and presented NP-hardness proofs, a branch-and-bound algorithm, and an approximation algorithm. This paper continues this line of study by focusing on the theoretical nature of the problem. One unique nature of the problem is due to the existence of two flowshops that interact in a kind of orthogonal manner. We will elaborate on this nature in the next section.

This paper is organized as follows: We present the problem formulation and introduce several preliminary observations in Section 2. While most previous studies assumed permutation schedules, we discuss the issue of non-permutation schedules in Section 3. We examine in Section 4 the complexity status of several special cases and present a complexity hierarchy of the problem. Finally, we conclude the paper and suggest some topics for future research in Section 5.

2. Problem formulation and preliminary properties

In this section we ﬁrst introduce the notation that will be used throughout the paper. We then present a formal problem

statement and discussions of several preliminary properties. An integer programming model follows.

Notation:

J ¼ fJ1,J2, . . . ,Jng the set of jobs to be processed;

pi processing time of job Jion machine M1;

qi processing time of job Jion machine M2;

a

i resource requirement of job Ji;

b

i amount of the resource returned by job Ji;

d

i¼

b

i

a

i contribution of job Ji;

s

¼ ð

s

1,

s

2, . . . ,

s

nÞ a particular sequence of the jobs (assumed for the case of permutation schedules); S a particular schedule that assigns a starting

time to each of the 2n operations;

V initial resource level;

vt resource level at time t Z0;

Zð

s

Þ,ZðSÞ makespan of feasible schedule

s

and S, respectively;

sm,i(S) starting time of job Jion machine Mmin

schedule S, m ¼1,2;

Cm,i(S) completion time of job Jion machine Mmin

schedule S, m ¼1,2.

We formally state the problem as follows: From time zero onwards, a set of jobs J ¼ fJ1,J2, . . . ,Jngis available to be processed in a two-stage ﬂowshop consisting of machines M1and M2. Initially,

the common resource pool contains V units of a single type of resource. Job JiAJ can start processing only if machine M₁is not occupied and the resource level in the pool is not less than

a

i. Once job Ji starts processing, it immediately consumes

a

i units of the resource from the resource pool and takes piunits of time on M1.

After the machine-one operation is completed, qiunits of time is

required to complete its resource recycling operation on M2. Upon

the completion of job Jion machine M2, it returns

b

i units of the resource to the resource pool. No preemption on either machine is permitted. The goal is to minimize the makespan, i.e., to ﬁnd a feasible schedule that completes all jobs in the shortest time. 2.1. Preliminary properties

We ﬁrst present some known results on the basic relocation problem to facilitate discussion. In the basic relocation problem, the jobs are to be processed on a single machine and each job Ji is

associated with two parameters

a

i and

b

i. The minimum initial resource level guaranteeing the existence of a feasible sequence for the jobs of J is called the minimum resource requirement of set J . It is clear that no temporal parameters are involved. In fact, such a feasibility issue is linked to the classical two-machine flowshop scheduling problem to minimize the makespan. Given job set J in the basic relocation problem, we define job set ^J ¼ f^J1,^J2, . . . ,^Jngfor the classical two-machine flowshop scheduling problem by letting

a

iand

b

iof job Jias the processing times of job ^Jion the stage-one and stage-two machines, respectively. The following lemma establishes the equivalence between the relocation problem and the two-machine ﬂowshop scheduling problem.

Lemma 1 (Kaplan and Amir[6]). The minimum resource require-ment of job set J for the basic relocation problem is equivalent to the minimum total idle time on machine M2 for the two-machine

ﬂowshop scheduling of job set ^J .

Therefore we can determine the minimum resource requirement of a given job set in O(n log n) time using Johnson’s algorithm. Cheng and Lin [11] further elaborated on the relation between the two-machine ﬂowshop scheduling problem and the relocation problem, and highlighted the potential use of the notion of the

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relocation problem in the study of two-machine flowshop schedul-ing. The relocation problem provides a more intuitive interpretation of Johnson’s algorithm for makespan minimization. To perform a set of projects, it is beneficial to undertake the projects that make profits ð

d

i¼

b

i

a

iZ0Þ ﬁrst and the remaining projects ð

d

_i¼

b

i

a

io0Þ follow. The projects that make proﬁts are arranged in non-decreasing order of their investment requirement ð

a

iÞ. On the other hand, the remaining projects follow in non-increasing order of their returns ð

b

iÞ.

In this section we consider permutation schedules, i.e., the processing sequences on both machines are identical. In a later section we will consider the non-permutation case. To illustrate the problem deﬁnition, we consider the following instance of three jobs with an initial resource level V ¼ 5.

Job pi qi

a

i

b

i

J1 1 1 5 2

J2 2 3 2 5

J3 3 2 3 1

Fig. 1shows schedule S1, corresponding to sequence (1,2,3), and

schedule S2, corresponding to sequence (2,3,1). Given the

objective of makespan minimization, we apply Johnson’s algorithm by considering piand qiand obtain sequence (1,2,3).

As can be seen from the Gantt charts, S1is inferior to S2. In fact,

the makespan of S1is the worst.

A unique feature of F2jrpjCmax stems from the interaction between two two-machine flowshops. The feasibility issue in the basic relocation problem involves only resource constraints and is equivalent to the idle time issue in the context of two-machine flowshop scheduling. The second flowshop emerges from the separation of the resource consumption (or job processing) operation from the resource recycling operation of a job and the serial-processing requirement of the coupled operations. We may consider the F2jrpjCmaxproblem by reversing the roles of the two flowshops. Assume that the processing times are f

a

igand f

b

igand the resource-related parameters are {pi} and {qi}. The project is

required to be completed by a deadline D ¼ V þPni ¼ 1

b

i. We want to determine the minimum initial resource level required to ensure the existence of a schedule that is feasible with respect to the resource constraints and the deadline constraint. The orthogonal type of interaction of the two ﬂowshops makes the problem hard to solve but may spur theoretical interest in the structural properties of the problem.

Now we introduce the time symmetry property of F2jrpjCmax. Given an instance I and a schedule S, we construct a mirror

instance Iu as follows: Let pui¼qi, qui¼pi,

a

_iu¼

b

i,

b

iu¼

a

i, and Vu ¼ V þPJiAJ

d

i. Denote the job set by J u ¼ fJ1u,J2u, . . . ,Jnug. We look

at the Gantt chart of schedule S from the right hand side and generate a mirror schedule Su of instance Iu by setting

s1,iðSuÞ ¼ ZðSÞC2,iðSÞ and C2,iðSuÞ ¼ ZðSÞs1,iðSÞ: ð1Þ Assume that preemptions are not allowed in schedule S, i.e., if job Jiustarts at s_1,iðSuÞ (respectively, s_2,iðSuÞ), then the processing of its ﬁrst (respectively, second) operation occupies the time interval ½s1,iðSuÞ,s1,iðSuÞ þpiuÞ(respectively, ½s2,iðSuÞ,s2,iðSuÞ þ qiuÞ). Later, we will also address some results for the case without this assumption. Without preemptions, (1) implies the following equalities:

C1,iðSuÞ ¼ ZðSÞs2,iðSÞ and s2,iðSuÞ ¼ ZðSÞC1,iðSÞ: ð2Þ Inspecting the Gannt charts, we know, by (1) and (2), that instance Iu and schedule Su are mirror to instance I and schedule S, and vice versa. Since schedule Su also satisﬁes the two-machine ﬂowshop scheduling constraints

s2,iðSuÞC1,iðSuÞ ¼ s2,iðSÞC1,iðSÞ Z 0, ð3Þ it has the same makespan as schedule S, i.e., ZðSÞ ¼ ZðSuÞ. The following lemma conﬁrms the feasibility of schedule Su. The proof is adapted from Kononov and Lin[12].

Lemma 2. Schedule Su is feasible with respect to the initial resource level Vu.

Proof. We show that the resource requirement is fulﬁlled at any time point t A ½0,ZðSuÞ. First of all, we examine the time point t ¼ ZðSuÞ, at which all jobs are completed. The resource level at ZðSuÞ is given by vZðSuÞðSuÞ ¼ Vu þ X 1r i r n ð

b

u_i

a

iuÞ ¼Vu X 1r i r n ð

b

i

a

iÞ ¼V Z0:

If toZðSuÞ, then Pfijs1,iðSuÞo tg

a

ui is the total amount of resource

consumed and P_fijC

2,iðSuÞr tg

b

iu is the total amount of resource

returned. Therefore, we have vtðSuÞ ¼ Vu

X fijs1,iðSuÞo tg

a

_iuþ X fijC2,iðSuÞr tg

b

_iu ¼ V þ X 1r i r n ð

b

i

a

iÞ ! X

fijs1,iðSuÞo tg

b

iþ X fijC2,iðSuÞr tg

a

i

¼V þ X

fijs1,iðSuÞ Z tg

b

i X fijC2,iðSuÞ 4 tg

a

i

¼V þ X

fijs1,iðSuÞ Z tg

ð

b

i

a

iÞ þ

X fijs1,iðSuÞr t r C2,iðSuÞg

a

i:

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By letting

t

¼ZðSÞt and (1), we have vtðSuÞ ¼ V þ X fijC2,iðSÞrtg ð

b

i

a

iÞ X fijs1,iðSÞrtr C2,iðSÞg

a

i: ð4Þ

Let

t

0o

t

be the latest time point where some job completes or ﬁnishes in schedule S. Eq. (4) leads to

vtðSuÞ ¼ V þ X fijC2,iðSÞrt0g

ð

b

i

a

iÞ

X fijs1,iðSÞrt0o C2,iðSÞg

a

i¼vt0ðSÞ Z0: ð5Þ

The last inequality follows from the fact that schedule S is feasible with respect to V. The proof is complete. &

With the fact that ZðSÞ ¼ ZðSuÞ and Lemma 2, solution algorithms and complexity results for instance I can be applied to instance Iu, and vice versa, because the transformation between the two instances takes only linear time.

2.2. Integer linear programming formulation

In the following we provide an integer linear program formulation of the F2jrpjCmax problem using positional variables. We use the binary variable uki to indicate if job Jiis scheduled in

position k or not. The auxiliary variable Cm,k denotes the

completion time of the k-th job on machine Mmfor m ¼1 or 2.

We use the binary variables y‘,kand bi,‘,kto handle the resource availability relation between any two jobs. If y‘,k¼1 for ‘ok, then the ‘th job completes on machine M2before the k-th job starts

on machine M1. If bi,‘,k¼1, then job Jiis scheduled in position ‘

and it is completed on M2 earlier than the k-th job starts

processing on M1. The variables y‘,kand bi,‘,kare related. We use these variables to help identify whether the resource returned by the ‘th job could be used by the later k-th job. If the resource is sufﬁcient for processing the later k-th job, variable bi,‘,kdoes not have to be 1. Similarly, it is possible to have y‘,k¼0 even if ‘th job is completed before k-th job starts.

F2RP-Perm. Minimize C2,n ð6Þ subject to C1,1¼ Xn i ¼ 1 piui1, ð7Þ C1,kC1,k1 Xn i ¼ 1 piuikZ0, 1okrn, ð8Þ C2,kC2,k1 Xn i ¼ 1 qiuikZ0, 1okrn, ð9Þ C2,kC1,k Xn i ¼ 1 qiuikZ0, 1okrn, ð10Þ Xn i ¼ 1 ui k¼1, 1rkrn, ð11Þ Xn k ¼ 1 ui k¼1, 1rirn, ð12Þ VX k r ¼ 1 Xn i ¼ 1

a

iuirþ Xk1 r ¼ 1 Xn i ¼ 1

b

ibi,r,kZ0, 1rkrn, ð13Þ ui

‘þy‘,k2bi,‘,kZ0, 1rirn,1r‘okrn, ð14Þ

C1,kC2,‘ Xn i ¼ 1 piuikþ ð1y‘,kÞM Z0, 1r‘okrn, ð15Þ ui kAf0,1g, 1ri,krn, ð16Þ y‘,kAf0,1g, 1r‘okrn, ð17Þ

bi,‘,kAf0,1g, 1rirn,1r‘ okrn, ð18Þ

Cm,kZ0, 1rkrn,m ¼ 1,2: ð19Þ

The objective function (6) is to minimize the makespan, i.e., the completion time of the job scheduled last. Constraints (7)–(10) confine all jobs to the following restrictions of flowshop scheduling: any job cannot start until its preceding job is completed and the machine-two operation of any job cannot be processed unless its machine-one operation is finished. Constraints (11) and (12) specify the one-to-one correspondence between jobs and positions, i.e., any position can accommodate at most one job and any job can occupy only one position. Constraints (13) require that the resource requirement of the k-th job cannot be greater than the current resource level, which is the initial resource level plus the resource returned by the jobs already completed before it can start. Constraints (14) ensure that if job Jiis scheduled in position ‘ and completes before the k-th

job starts, then bi,‘,k¼1. Constraints (15) check whether the ‘th job completes before the k-th job starts. The remaining constraints specify the ranges of the values of the variables.

3. Non-permutation scheduling

One of the main constraints characterizing a flowshop scheduling problem is whether all machines have the same processing sequence of the jobs. A schedule is called permutational if the job sequences on all machines in the shop are the same. For the classical two-machine flowshop problem F2JCmax, it suffices to consider only permutation schedules[9]. However, the optimal schedule for the m-machine flowshop problem, FmjjCmax, is not necessarily permuta-tional, although the sequences on the first two machines or the last two machines are the same. For the F2jrpjCmaxproblem studied by Lin and Huang [16], permutation sequences were assumed. The issue of whether there exists at least one permutation optimal schedule for the F2jrpjCmax problem has not been investigated before. In this section we discuss the properties of permutation and non-permutation schedules.

Lemma 3. The optimal schedule for F2jrpjCmax is not necessarily permutational.

Proof. Consider the following instance of four jobs with an initial resource level V ¼ 6. pi qi

a

i

b

i J1 0 4 3 10 J2 4 5 3 10 J3 1 1 9 10 J4 5 0 11 10

Fig. 2 shows the optimal permutation and non-permutation

schedules. It is easy to verify that the optimal solution is obtained by the non-permutation schedule for the given instance. This example also shows that the optimal sequence may not

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necessarily be permutational even if all jobs return the same amount of the resource. &

The instance given above to illustrate Lemma 3 for non-permutation optimal schedules contains four jobs. It can be veriﬁed that instances with two or three jobs always permit permutational optimal schedules. Therefore, the four-job instance given above constitutes the smallest instance for which optimal schedules are not permutational.

Therefore, if we do not require schedules to be permutational, an optimal schedule may have different job sequences on the two machines. In the IP formulation (F2RP-Perm) in Section 2.2, we only considered permutation schedules. Below, we propose another formulation that considers non-permutation schedules. Since the job sequences may not necessarily be the same on both machines, we have to re-deﬁne the positional variables. If job Jiis scheduled in position k on machine Mm, then uim,k ¼1; 0 otherwise. F2RP-Non-Perm. Minimize C2,n ð20Þ subject to C1,1¼ Xn i ¼ 1 piui1,1, ð21Þ C1,kC1,k1 Xn i ¼ 1 piui1,kZ0, 1okrn, ð22Þ C2,kC2,k1 Xn i ¼ 1 qiui2,kZ0, 1okrn, ð23Þ

C2,kC1,jqiþ ð1ui2,kÞM þð1ui1,jÞM Z 0, 1ri,j,krn, ð24Þ Xn i ¼ 1 ui m,k¼1, 1rkrn,m ¼ 1,2, ð25Þ Xn k ¼ 1 ui m,k¼1, 1rirn,m ¼ 1,2, ð26Þ VX k r ¼ 1 Xn i ¼ 1

a

iui1,rþ Xk1 ‘ ¼1 Xn i ¼ 1

b

ibi,‘,kZ0, 1rkrn, ð27Þ ui

2,‘þy‘,k2bi,‘,kZ0, 1rirn,1r‘okrn, ð28Þ

C1,kC2,‘ Xn i ¼ 1 piui1,kþ ð1y‘,kÞM Z 0, 1r‘okrn, ð29Þ ui m,kAf0,1g, 1ri,krn,m ¼ 1,2, ð30Þ y‘,kAf0,1g, 1r‘okrn, ð31Þ Cm,kZ0, 1rkrn,m ¼ 1,2: ð32Þ

Most constraints in this formulation are similar to those for the permutational case (F2RP-Perm). Constraints (21)–(24) deﬁne the processing characteristics of the two-machine ﬂowshop. The difference between constraints (21)–(24) and (7)–(10) lies in the necessity for new variables, ui

m,k, which are required

because the two operations of the same job might be scheduled in different positions on the two machines. We use the new variables to determine the completion times of a job on different machines. Constraints (24) verify whether job Ji’s machine-two

operation starts after its machine-one operation by checking all pairs of positions on different machines.

We have shown that permutation schedules do not always provide optimal solutions in the general situation. Even when all jobs return the same amount of the resource, as shown in the instance of Lemma 3, the optimal schedule is not necessarily permutational. For some special cases, however, permutation schedules do provide the optimal solutions. The following lemma gives a special case that admits permutation optimal schedules. Lemma 4. If machine M1is dominant, i.e., minfpig Zmaxfqig, then it sufﬁces to consider permutation schedules only.

Proof. Consider an optimal non-permutation schedule S. Assume that the jobs are re-numbered by their processing order on machine M1. Let Ji be the job with the smallest index whose

machine-two operation does not start immediately after the completion of its machine-one operation, C1,i. Since the recycling

process does not demand any resource, it could be started once its machine-one operation is completed and machine M2is available.

By assumption, machine M2is idle while job Ji + 1is processed on

machine M1. We shift the recycling operation of job Jibackward to

start at the completion of its machine-one operation. Clearly, the new schedule remains feasible. Moreover, since machine M1 is

dominant, C2,irC1,i þ 1 must hold, implying that C2,i + 1 remains

unchanged. Repeating the same argument, we will obtain a permutation optimal schedule. The proof is complete. &

By Lemmas 2 and 4, we can say that if machine M2 is

dominant, there exists an optimal schedule that is permutational. Lemma 4a. If machine M2is dominant, i.e., maxfpigrminfqig, then it sufﬁces to consider permutation schedules only.

4. Complexity results

Lin and Huang[16]showed that F2jrpjCmaxwithout preemp-tion is NP-hard in the strong sense even if only one parameter is ﬁxed. The complexity status of the preemptive case ðF2jrp,pmtnjCmaxÞand some special cases, such as more than one parameter are assumed to be ﬁxed, has not been settled. In this section we provide the complexity results for F2jrp,pmtnjCmaxand various special cases of the F2jrpjCmax problem.

4.1. Hard cases

First we present a strong NP-hardness proof for the preemptive case, F2jrp,pmtnjCmax, by a reduction from the 3-Partition problem, which is known to be NP-complete in the strong sense[18].

J2 J3 J₂ J3 J4 J1

Permutation schedule with Cmax = 15.

J2 J1 J3 J3 J2 J4 Non-permutationschedulewith Cmax = 11.

Fig. 2. Gantt charts of permutation and non-permutation schedules. (a) Permuta-tion schedule with Cmax¼15. (b) Non-permutation schedule with Cmax¼11.

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3-Partition. Instance: A positive integer B, a set of 3m elements fx1,x2, . . . ,x3mg such that B=4oxioB=2 for each index i A N ¼ f1,2, . . . ,3mg andP3mi ¼ 1xi¼mB.

Question: Can index set N be partitioned into m disjoint sets N1,N2,y,Nmsuch that for each 1rkrm,Pi A Nkxi¼B?

Theorem 1. The decision version of the F2jrp,pmtnjCmaxproblem is NP-complete in the strong sense even if all the jobs have the same resource requirement.

Proof. The decision version of F2jrp,pmtnjCmax is F2jrp,pmtn,CmaxrCj, where C is the threshold parameter impos-ing an upper bound on the schedule length. To establish the membership of this decision problem in NP, we have to show that the number of preemptions is bounded by a polynomial in the length of the problem input and that if preemptions occur at non-integer time points, the encoding of the length of each fractional part of an operation in an optimal schedule is also bounded by a polynomial in the length of the problem input. The two issues have been addressed by[19]for much more general scheduling problems with resource constraints and for a wide variety of objective functions, including all the classical ones. Given a schedule of the studied problem, the makespan can be calculated in polynomial time. Hence this decision problem belongs to NP.

Given an instance of 3-Partition, we construct an instance of F2jrp,pmtn,CmaxrCj with an initial resource level V¼2B as follows:

3m ordinary jobs Ji,1rir3m: pi¼xi,qi¼0,

a

i¼B;

b

i¼B þ xi. m enforcer jobs Ji, 3m þ 1rir4m: pi¼0,qi¼B,

a

i¼B,

b

i¼0. The threshold is C ¼ mB. We claim that a partition exists for the 3-Partition problem if and only if there exists a feasible schedule whose makespan is not greater than C .

Let subsets N1,N2,y, Nm constitute a partition of set N in

3-Partition. We ﬁrst schedule an enforcer job, say J3m + 1, followed

by the three jobs, say J1, J2, J3, deﬁned by the elements of set N1.

On machine M2, the processing of enforcer job J3m + 1is preempted

when job J1completes on machine M1. The preemption prevents

job J2from being blocked due to a shortage of resource. Similarly,

job J2 preempts the processing of job J3m + 1on machine M2 to

facilitate the processing of job J3. When the four jobs are ﬁnished,

the resource level is again 2B. Repeating the procedure for dispatching an enforcer job J3m + kfollowed by the three ordinary

jobs deﬁned by the subset Nk, 1rkrm, we can develop a feasible

schedule with a makespan of mB.

To show the if-part of the claim, we assume that there is a feasible schedule S with the makespan equal to mB. While the total machine load of either machine is mB, the assumption implies that no idle time is allowed on either machine. Since all enforcers are identical, we assume that they are arranged in increasing order of their job indices. To avoid any idle time on machine M2, enforcer job J3m + 1

must be scheduled ﬁrst. We consider the set of ordinary jobs scheduled to start on machine M1in the time interval ½0,C2,3m þ 1ðSÞÞ. Let N1 be the set of elements deﬁning these ordinary jobs. If

P

i A N1xi4B, then there must be some ordinary job completed on

machine M1later than B, which is the completion time of the ﬁrst

enforcer job J3m + 1. That is, idle time occurs on machine M2. On the

other hand, ifPi A N1xioB, then the resource level at the completion

time of these ordinary jobs is B þP_{i A N}

1xio2B. Then the second

enforcer job J3m + 2 starts its processing and reduces the resource

level to ðB þPi A N1xiÞB, which is less than B. At this moment, none

of the unscheduled jobs, either ordinary or enforcer, can be successfully started. ThereforePi A N1xioB cannot be true. Equality

P

i A N1xi¼B must hold. Following the same line of reasoning, we can

come up with sets N2, y, NmwithPi A Nkxi¼B,2rkrm, satisﬁed.

A partition is identiﬁed for the 3-Partition problem and the proof is complete. &

It can be seen that for the instance of the F2jrp,pmtn,CmaxrCj problem constructed for a given instance of the 3-Partition problem, there is no optimal schedule without preemption. Thus the Preemption Redundancy Property[19]does not hold for this scheduling problem. Otherwise, if it did, then the NP-completeness of this decision problem would directly follow from its Preemption Redundancy Property and from the NP-completeness of its non-preemptive counterpart /F2jrp,CmaxrCjS proved earlier by Lin and Huang [16]. Another note for the theorem is that all jobs in the instance constructed in the proof have the same resource requirement

a

i¼B. By Lemma 2, we have:

Corollary 1. The decision version of F2jrp,pmtnjCmaxis NP-complete in the strong sense even if all jobs return the same amount of resource.

In the following we discuss some special cases where preemption is not allowed. Lin and Huang [16] showed four special cases of F2jrpjCmaxto be NP-hard in the strong sense, i.e., (1) pi¼p for all i, (2) qi¼q for all i, (3)

a

i¼

a

for all i, and (4)

b

i¼

b

for all i. That is, the problem remains hard to solve if one of the four parameters is assumed to be ﬁxed. In the following we show that several further restricted cases remain NP-hard.

The following theorem addresses the case where all jobs are almost identical except for the processing times of the stage-two operations. In the following discussion we only outline how the instances of our scheduling problems are constructed from 3-Partition and omit the detailed analysis.

Theorem 2. F2jrp,pi¼p,

a

i¼

b

i¼1jCmax is strongly NP-hard. Proof. An instance of F2jrp,pi¼p,

a

i¼

b

i¼1,C j with 4m jobs, initial resource level V¼4, and threshold C ¼ ð4m þ 1ÞB is given as follows:

3m ordinary jobs Ji, 1rir3m: pi¼B,qi¼xi,

a

i¼1,

b

i¼1; m enforcer jobs Ji, 3m þ 1rir4m: pi¼B,qi¼3B,

a

i¼1,

b

i¼1. A schedule with a makespan of C corresponds to a partition as required. &

The next result readily follows from the time symmetry property.

Corollary 2. F2jrp,qi¼q,

a

i¼

b

i¼1jCmax is strongly NP-hard. The next theorem concerns the special case where each job requires one unit of time for its stage-one operation and there are only two distinct stage-two processing times among all jobs. This restricted case remains strongly NP-hard.

Theorem 3. F2jrp,pi¼1,qiAfq₁,q₂gjC_maxis strongly NP-hard. Proof. An instance of F2jrp,pi¼1,qiAfq₁,q₂g,C_maxrCj with an initial resource level V ¼ 6B is given as follows:

3m ordinary jobs Ji, 1rir3m: pi¼1,qi¼0,

a

i¼B þ xi,

b

i¼4xi; m enforcer jobs Ji, 3m þ 1rir4m: pi¼1,qi¼3,

a

i¼2B,

b

i¼2B. A desired partition of 3-Partition exists if and only if there is a feasible schedule whose makespan is not greater than C ¼ 4m. & 4.2. Solvable cases

In this section we study some polynomially solvable cases by developing exact solution algorithms. In Theorem 2 and Corollary 2,

(7)

we showed that the problem remains strongly NP-hard even if all jobs are common in the three parameter values ðpi,

a

i,

b

iÞor ðqi,

a

i,

b

iÞ. It is interesting to consider other special cases where all jobs have common values of ðpi,qi,

a

iÞor ðpi,qi,

b

iÞ. In the following we show that the two special cases can be solved in polynomial time. Lemma 5. There exists an optimal schedule for F2jrp,pi¼p,qi¼q,

a

i¼

a

jCmax in which the jobs are scheduled in non-increasing order of

b

i.

Proof. By Lemma 4, there exists a permutation optimal schedule. Let

s

¼ ð

s

1, . . . ,

s

nÞ be an optimal sequence in which

b

si þ 14

b

si.

Then, evidently, swapping the two jobs in the sequence retains the feasibility of the permutation schedule with the same length (and thus optimal). Repeating the swapping argument, if necessary, we can obtain a schedule with the sequence as required. &

Due to the time symmetry property, we have a similar result for the case F2jrp,pi¼p,qi¼q,

b

i¼

b

jCmax.

Lemma 6. The F2jrp,pi¼p,qi¼q,

b

i¼

b

jCmaxproblem can be solved in O(n log n) time by arranging the jobs in non-decreasing order of

a

i. By Lemma 5, we can determine an optimal schedule for F2jrp,pi¼p,qi¼q,

a

i¼

a

jCmax by sorting the values of

b

i in non-decreasing order. Similarly, by Lemma 6, we can determine an optimal schedule for F2jrp,pi¼p,qi¼q,

b

i¼

b

jCmaxby sorting the values of

a

iin non-increasing order. The following theorem thus follows.

Theorem 4. For any instance of problem F2jrp,pi¼p,qi¼q,

a

i¼

a

jCmax or problem F2jrp,pi¼p,qi¼q,

b

i¼

b

jCmax with n jobs, there exists an optimal permutation schedule that could be found in O(n log n) time. 5. Conclusion

In this study we considered the relocation problem with separate recycling operations. We examined several properties concerning whether the optimal schedule is permutational or not. We presented integer linear programming models for both the permutational and non-permutational settings. We settled the complexity status of several special cases of the problem, although some cases remain unresolved. Summarizing the results presented in this paper, we depict the complexity status of all studied cases of the problem in

Fig. 3. It is seen that the complexity of the case where the processing

times are ﬁxed on both machines remains unknown. Furthermore, it is hard to determine the complexity of the most simpliﬁed unit execution time case F2jrp,pi¼qi¼1jCmax. For further studies, theoretical analysis of approximation algorithms for F2jrpjCmaxwill be an interesting topic.

Acknowledgements

The authors are grateful to the reviewers for their professional comments, which have improved the presentation of the paper. This research was supported in part by The Hong Kong Polytechnic University under Grant number G-U471. Lin was also supported by the National Science Council of Taiwan under the Grant NSC96-2416-H-009-012-MY2.

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