Scheduling of a two-stage differentiation flowshop to minimize weighted sum of machine completion times

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Computers & Operations Research

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Scheduling of a two-stage differentiation flowshop to minimize weighted sum of

machine completion times

T.C.E. Cheng

a

_{, B.M.T. Lin}

b,∗

_{, Y. Tian}

b

a_{Department of Logistics and Maritimes Studies, The Hong Kong Polytechnic University, Hung Hom, 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 A B S T R A C T

Available online 20 February 2009

Keywords:

Flowshop

Machine completion time Approximation algorithm Performance ratio

This paper considers the problem of scheduling a two-stage flowshop that consists of a common crit-ical machine in stage one and two independent dedicated machines in stage two. All the jobs require processing first on the common critical machine. Each job after completing its critical operation in stage one will proceed to the dedicated machine of its type for further processing in stage two. The objective is to minimize the weighted sum of stage-two machine completion times. We show that the problem is strongly NP-hard, and develop an O(n3_{) polynomial time algorithm to solve the special case where} the sequences of both types of jobs are given. We also design an approximation algorithm with a tight performance ratio of 4

3 for the general case.

1. Introduction

Flowshop scheduling is one of the most widely studied topics in the scheduling literature[6,23]. This broad topic contains many different settings and special cases, reflecting a wide range of appli-cations. In this paper, we consider a special kind of the two-stage flowshop, where all the products require processing first on a com-mon critical machine in the first stage, after which each product proceeds to a dedicated machine for further processing in the sec-ond stage. We call this model the two-stage differentiation flowshop and present the msachine configuration of such a flowshop model in Fig. 1. Many manufacturing environments that produce multiple final products are extensions of this basic model. In particular, this model can be applied to describe the production setting for delayed prod-uct differentiation, which is one of the approaches taken to achieve mass customization[4,26]. Kyparisis and Koulamas[15]remarked that “applications of the proposed flowshop model are encountered in manufacturing settings, where all jobs must first go through the same main process, and then they require a finishing operation spe-cial to the job”. For a specific application, consider a production line for furniture manufacturing (e.g., chairs). The main bodies of the chairs are manufactured in the first stage. Several different head-supports are then assembled in the second stage. Thus, each type of chair products proceeds to a different dedicated machine in the

∗ Corresponding author. Tel.: +886 3 5131472; fax: +886 3 5729915.

E-mail address:[email protected](B.M.T. Lin).

second stage. Clearly, the second stage may be an assembly opera-tion as described above, a painting operaopera-tion (painting the product in one of several colors), or a packaging operation (wrapping up the product in one of several packages), etc. Another application is pot-tery production. The main glazing process is performed in the first stage. Several heating processes to produce different figures or sur-face effects are applied in the second stage. In other words, each type of pottery products proceeds to a corresponding dedicated machine for the required baking process after being glazed. The second stage may consist of re-glazing, various thermal treatments, or packaging. The machine setting studied in this paper is similar to the hybrid flowshop, where multiple parallel machines are available at each stage and jobs can be processed by any of the machines. The reader is referred to Linn and Zhang[17]for a review of scheduling in hy-brid flowshops. In the differentiation model considered in this study, stage two consists of two machines, and jobs are routed to only the dedicated machines of their types. When there is only one type of jobs, the differentiation model reduces to the classical two-machine flowshop proposed by Johnson[13]. The model studied in this paper probably first appeared in Herrmann and Lee [9], in which three objectives, namely the makespan, the number of tardy jobs and the maximum tardiness, were investigated and the corresponding problems were shown to be strongly NP-hard. For the problem to minimize the makespan with two fixed sequences for the two types of jobs, they transformed it into the problem to minimize the max-imum lateness such that Jackson's earliest due date (EDD) rule[11] can solve the makespan problem in O(n log n) time. Given an arbi-trary number of stage-two machines in a job shop, Drobouchevitch and Strusevich[5]designed a heuristic algorithm to minimize the makespan with a performance ratio of 3

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Common Critical Machine

Dedicated Machine 2 Dedicated Machine 1

Fig. 1. Machine configuration.

[15]investigated the model with m types of jobs that need to be processed on m dedicated machines in the second stage, subject to a technical constraint, called the block assumption, whereby jobs of the same type must be processed contiguously on the critical ma-chine in the first stage. The authors proposed an O(m(n log n_{+log m))} algorithm to minimize the makespan. Subsequently, Moshelov and Yovel[20]made several comments on Kyparisis and Koulamas's al-gorithm and improved the time complexity to O(n log n) under the assumption that m

n. Cheng and Kovalyov[2]incorporated setup times on the common machine whenever the processing of jobs on it is switched from one type to the other. They proposed a dynamic programming algorithm that is polynomial in the number of jobs for makespan minimization. A model that exhibits a reverse production flow has m dedicated machines for m types of jobs installed in stage one and contains a critical machine that is common for all the jobs in stage two. With regard to makespan minimization, the two sym-metric models are equivalent. Specifically, the reverse model with

m_{=2 types of jobs was independently studied by Lin}[16], Neumytov and Sevastyanov[21]and Oguz et al.[22].

The goal of makespan minimization, as an internal management metric, is to maximize machine utilization. Another objective is to minimize the weighted total job completion time (wiCi), which, as a metric of WIP inventory or customer's waiting time, is based upon job completion times. The objective to minimize the weighted sum of machine completion times studied, in this paper, is non-classical. It aims at attaining better machine utilization where dif-ferent machines may have difdif-ferent operating costs. This objective borrows the concept of the weighted total job completion time, but it generalizes the goal of makespan minimization that focuses on in-ternal management. Bagga[1]first studied the situation where the machine (rental) cost depends on the duration between a machine starts processing its jobs and when it finishes processing all the jobs. Ho and Gupta[10] investigated permutation flowshop scheduling with dominant machines to minimize the total machine completion time. Fondrevelle et al.[7]investigated this objective in a multi-stage flowshop incorporating the constraint of minimum and maxi-mum time lags between successive operations. They showed that the classical two-stage (F2) problem can be solved using Johnson's rule and that the problem with three or more stages, i.e., Fm, m

3, is strongly NP-hard. For the parallel-machine model (Pm), we can min-imize the weighted sum of machine completion times in polynomial time because the problem can be formulated as the classical assign-ment problem, which is polynomially solvable. The objective of to-tal machine completion time is different from the objective of toto-tal machine load, which was introduced by Mosheiov[19]to measure the total time in which the machines are engaged in the production process. The total machine load objective has the potential to deal with the objective of the total completion time as demonstrated in Jeng and Lin[12]. To the best of our knowledge, the three-machine two-stage flowshop scheduling model with the objective of

mini-mizing the weighted sum of machine completion times, denoted by

F(1, 2)_WMT, under study has not been addressed in the scheduling

literature. Note that only stage-two machine completion times are involved in the objective function because the completion time of the stage-one machine is fixed once the input instance is given.

The rest of the paper is organized as follows. In Section 2 we present the notation that is used throughout the paper, and give a numerical example to illustrate the problem definition. Section 3 presents the result that the problem under study is strongly NP-hard. We dedicate Section 4 to the development of a polynomial time algorithm for the special case with a fixed sequence for each of the two types of jobs. We develop an approximation heuristic and analyze its performance ratio in Section 5. We present some concluding remarks and suggest some topics for future research in Section 6.

2. Problem definition and notation

In this section, we give a formal definition of the F(1,2)_WMT problem and define the notation that will be used. We then give an example of two schedules for illustration.

The problem setting of F(1,2)_WMT is formally defined as follows. There is a set of n jobs

J

={J1, J2, . . . , Jn} available from time zero to be

processed on a three-machine two-stage flowshop, where M0is the

stage-one machine and M1, M2are two different dedicated machines

in the second stage. The jobs belong to two different types: type 1,

J

1= {J1, J2, . . . , Jn1} and type 2

J

2= {Jn1+1, Jn1+2, . . . , Jn1+n2} with

n1+n2=n. Each job in

J

1consists of two operations, of which the first

is performed on the common critical machine M0, and the second

is performed on the dedicated machine M1, as in the classical

two-machine flowshop. Similarly, the jobs of

J

2 are processed first on

the common critical machine M0and then on the dedicated machine

M2. Each machine can process at most one operation at any time,

and no preemption is allowed. The goal is to find a schedule that minimizes the weighted sum of machine completion times. Applying the job-interchange argument to jobs of the same type, we can show that it suffices to consider only permutation schedules, i.e., jobs of the same type have the same processing sequence on the critical machine and on their dedicated machine.

Notation:

J

= {J1, . . . , Jn} =

J

1∪

J

2: the set of jobs to be processed;

J

1= {J1, . . . , Jn1}: the set of type-1 jobs;

J

2= {Jn1+1, . . . , Jn1+n2}: the set of type-2 jobs, where n1+

n2= n;

M0: the stage-one common critical

ma-chine;

M1: the stage-two dedicated machine

for jobs of

J

1;

M2: the stage-two dedicated machine

for jobs of

J

2;

pki: the processing time on machine Mk,

k= 0, 1, 2, of job Ji∈

J

;

wk: the weight of machine Mk, k= 1, 2;

= (

1,

2, . . . ,

n): a particular schedule,

i ∈

J

, 1

i

n;

C_ki(

): the completion time of job Ji∈

J

on

machine Mk, k=0, 1, 2, under sched-ule

;

C(k)₍

_): _{the completion time of machine}

Mk, k= 0, 1, 2, under schedule

;

Z(

)= w1C(1)(

)+ w2C(2)(

): the weighted sum of machine

com-pletion times under schedule

;

Z∗: the optimal weighted sum of

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Fig. 2. Two example schedules with different objective values.

In the above notation,

may be omitted if no ambiguity would arise, and Cki will replace Cki(

). To avoid possible confusion con-cerning missing operations in the flowshop, we assume that all the processing times pki are strictly positive. Note that processing times p1i (resp. p2i) are only defined for i

n1 (resp. i

>

n1). The

machine weights w_kare positive, too. Furthermore, if it is not nec-essary to specify the starting times of the jobs, then a sequence, instead of a schedule, of the jobs on machine M0will be referred to.

Fig. 2shows four jobs, two of each type, to be scheduled. Consider schedules

1= (J3, J1, J2, J4) and

2= (J1, J3, J4, J2). Jobs of each type

are sequenced in the Johnson's order in both schedules. The two schedules correspond to different interleaved Johnson's sequences and result in different weighted sums of machine completion times.

3. Complexity results

In the following, we show that the F(1,2)_WMT problem is NP-hard in the strong sense. We first introduce an optimality property of the studied problem. A job sequence on machine M0can be

di-vided into blocks, each of which contains jobs of the same type. The following property resolves the sequencing issue within each block.

Lemma 1. There is an optimal schedule in which the jobs of the same block on machine M0are sequenced by Johnson's rule.

Proof. The validity can be established by applying the

job-interchange argument to two consecutive jobs of the same type not following Johnson's order.

The following NP-hardness proof is based on a reduction from 3-PARTITION, which is known to be strongly NP-hard[8].

3-PARTITION: Given a negative integer B and a set of 3m non-negative integers A_{= {x}1, x2, . . . , x3m} with B/4

<

xi

<

B/2 for each xi and_x

i∈Axi= mB, is there a partition A1, A2, . . . , Amof set A such that for each subset Aj,xi∈Ajxi= B?

+B M2 M₀ M₁ I₀ A₁ J₁ J₁ J₂ J₂ A₂ A₂ A_m A_m J_m-1 J_m J_m +2B +BM +B +BM +2B +2BM +2B +mB (M+1)+2 B J_3m+1<A₁> +mBM +mB

Fig. 3. Configuration of the optimal schedule in Theorem 1.

Theorem 1. The F(1,2)_WMT problem is NP-hard in the strong sense even if the stage-two machines are equally weighted.

Proof. The verification process can be easily done in polynomial

time. Therefore, the decision version of the F(1,2)_WMT problem is clearly in NP. From a given instance of 3-PARTITION, we construct the following instance of F(1,2)_WMT consisting of 4m+ 1 jobs:

n1= 3m + 1 type-1 jobs:

p0i= xi, p1i= xi(M+ 1), i = 1, ... , 3m, where M

>

mB,

p0,3m+1=

, p1,3m+1= 2B where 0

<

1.

n2= m type-2 jobs:

p0i= BM, p2i= B(M + 1), i = 3m + 2, ... , 4m,

p0,4m+1= BM, p2,4m+1= 2B.

The two dedicated machines have the same weights. Without loss of generality, we assume that m

>

2. Given a 3-PARTITION in-stance and the constructed inin-stance, it can be shown that the an-swer to 3-PARTITIONis affirmative if and only if there is a schedule for the F(1,2)_WMTproblem whose objective value is no greater than 2mB(M+ 1) + 4B + 2

.

If there is a partition A1, A2, . . . , Amof set A, we construct a schedule

by arranging the jobs on machine M0as

(J3m+1,A1, J3m+2,A2, J3m+3, . . . , J4m,Am, J4m+1),

where_{Ai denotes the type-1 jobs corresponding to the elements of}

Ai. It is easy to verify that the weighted sum of machine completion times is exactly 2mB(M+ 1) + 4B + 2

. The configuration is shown in Fig. 3.

We now assume that there is a schedule

with Z(

)

2mB(M+ 1)+ 4B + 2

. First, notice that on machine M1 the total processing

length of all the jobs is mB(M_{+ 1) + 2B and that on machine M}2the

total processing length of all the jobs is (m− 1)B(M + 1) + 2B. With these observations, we readily have that the sum of idle times on machine M1and machine M2cannot exceed B(M+ 1) + 2

. Because

all the type-2 jobs have the same operation on machine M0and job

J4m+1has the shortest processing time on M2, we assume that the

type-2 jobs are arranged in increasing order of their indices. The validity of this assumption comes from Lemma 1.

Assume that type-2 job J_3m+2 is scheduled first on M0. Then,

an idle time of BM will be incurred on both dedicated machines, resulting in 2BM total idle time, a contradiction. Therefore, some type-1 jobs must be scheduled before job J_3m+2 on M0. LetA1 ∪

{J3m+1} denote the set of type-1 jobs that precede job J3m+2on M0.

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is inevitable for the first job J_3m+2 on M2. By Lemma 1, we assume

that type-1 job J3m+1is scheduled first because p0,3m+1=

. The idle

time on M1is at least

. Consider the following two cases:

Case 1:_x

i∈A1xi

>

B

The starting time of job J3m+2 on machine M2 is

+xi∈A1xi+

BM

_{+ (B + 1) + BM}

>

B(M_{+ 1) +}

. That is, the idle time on M2is

greater than B(M+ 1) +

. Combining the idle time

on M1, the total

idle time on both dedicated machines is greater than B(M+ 1) + 2

, a contradiction.

Case 2:_x

i∈A1xi

<

B

On machine M1, the completion time of the last job ofA1 is

+ 2B + (M + 1)xi∈A1xi. Because the completion time of job J3m+2 on machine M0 is

+xi∈A1xi+ MB, the first type-1 job following A1 on machine M1has an idle time of at least

⎛ ⎝

+ xi∈A1 xi+ MB ⎞ ⎠ − ⎛ ⎝

+ 2B + (M + 1) xi∈A1 xi ⎞ ⎠ = M ⎛ ⎝B − xi∈A1 xi ⎞ ⎠ − 2B

M− 2B

>

mB− 2B

>

B.

Therefore, the total idle time on M1is greater than

+ B. On the

other hand, the total idle time on M2is at least

+ BM. Hence the

sum of the idle times on the two stage-two machines is greater than

B(M+ 1) + 2

, a contradiction.

From the analysis of the above two cases, we know that the equality_x

i∈A1xi= B must hold. Let the elements corresponding to the type-1 jobs included inA1 form a subset A1. When job J3m+2is

finished, the completion times of the three machines are

_{+ BM + B,}

+BM+3B and

+2BM+2B, respectively. We can continue the same analysis to obtain sets A2, A3, . . . , Amand complete the proof.

Before closing this section, we note that the instance constructed in the proof exhibits three side conditions: (1) The dedicated ma-chines are equally weighted. (2) Agreeable condition: for any jobs

Ji and Jj of the same type, if p0i

<

p0j, then p1i

p1j (type 1) or

p2i

p2j(type 2). (3) Equal-processing-time condition: all the

type-2 jobs have the same processing time on M0. In other words, the

F(1,2)_WMT problem remains computationally intractable even

if the input instance satisfies these three assumptions commonly adopted to deal with difficult scheduling problems. The follow-ing sections are dedicated to one special case that can be solved in polynomial time, and to the development of an approximation algorithm for the general case.

4. Fixed sequences

In this section, we consider a simplified situation where the sequences of both types of jobs are fixed, i.e., two independent pre-determined sequences are given. Under this assumption, the prob-lem reduces to finding an interleaved sequence on machine M0from

the two given fixed sequences. The problem is motivated as fol-lows. We have two sets of products to manufacture and each set has its processing sequence predetermined by some job characteristics or technological constraints. The products will be assigned a com-mon resource, i.e., machine time on the stage-one machine, so we need to construct an interleaved sequence of all the products on the common machine. In most scheduling problems, schedules are im-plicitly implied from sequences. For some problems, it is neverthe-less hard to determine an optimal schedule from fixed sequences. In two-machine flowshop scheduling with batch processing, it is not trivial to determine an optimal batching policy of a given sequence

to minimize the total completion time. To minimize the makespan in the F(1, 2) model, Hermann and Lee[9]reduced the problem to the problem to minimize the maximum lateness, which is solvable in O(n log n) time. Shafransky and Strusevich[24]studied open shop scheduling to minimize the makespan, subject to a given job se-quence on one machine. They investigated several cases and showed them to be NP-hard or polynomially solvable.

We denote the F(1, 2) problem with two fixed sequences by

F(1, 2)_WMTfixed_seq. Consider the two schedules in Fig. 2 as an example. Both schedules are obtained from interleaving sequences (J1, J2) and (J3, J4); however, the weighted sums of machine

com-pletion times of the two interleaved sequences (J3, J1, J2, J4) and

(J1, J3, J4, J2) are different. For notational simplicity, let (J1, J2, . . . , Jn1) and (Jn1+1, Jn1+2, . . . , Jn) denote the given sequences. Given the two sequences, there are (n1+ n2)!/(n1)!(n2)! possible interleaved

se-quences. The problem is to develop a solution algorithm to deter-mine an interleaved sequence from such a solution space with an exponential size. Preliminary investigation suggests that the prob-lem is not trivial to solve. At least, no simple method or dispatching rule has been found to deliver optimal solutions.

In this section we explore several structural properties that can help identify promising candidates. The main algorithm will con-struct at most n1(n2+ 1) schedules, among which an optimal

sched-ule is identified. In particular, for each job Jiof type 1, and for each

l= 0, ... , n2, the algorithm finds a feasible schedule (if such exists)

to minimize the completion time of machine M2. The schedule must

satisfy the following two requirements: there are exactly l jobs of type 2 preceding job Jion M0and job Jiis the first job of type 1 after which machine M1has no idle time.

Let

∗be an optimal sequence. We examine the processing of type-1 jobs on machine M1. In the following discussion, when index

i_{= 0 refers to a job, a dummy job with zero processing time is}

assumed; when index i= 0 refers to job completion times, C0,0= 0

and C1,0= 0 are assumed. First, we find the largest index i such

that C0i

>

C1,i−1and C0r

C1,r−1for any r, i

<

r

n1. In other words,

on machine M1, a non-zero idle time exists before job Jiand jobs

Ji, Ji+1, . . . , Jn1are processed consecutively without idle time inserted. We further assume that exactly j type-2 jobs Jn1+1, . . . , Jn1+jprecede job Jiin

∗. Therefore, job Jihas i+ j − 1 predecessors. The above arrangement implies that the completion time of machine M1is

C(1)(

∗)= i r=1 p0r+ n1+j r=n1+1 p0r+ n1 r=i p1r.

Given the above arrangement with the fixed completion time C(1)₍

∗₎

of machine M1, we then seek to minimize the completion time

C(2)₍

∗_{) of machine M}₂_{. The above discussion leads to the following}

notion.

Definition. The ordered pair (i, j), 1

i

n1, 0

j

n2is admissible if

there exists an interleaved sequence

that satisfies the following conditions:

(a) Job Ji is preceded by type-1 jobs J1, . . . , Ji−1 and type-2 jobs

Jn1+1, . . . , Jn1+j.

(b) C_1,i−1(

)

<

C0,i(

)=ir=1p0r+nr=n1+j1+1p0r. (c) For i

<

r

n1, C0,r(

)

C1,r−1(

).

Condition (b) specifies that there is a non-zero idle time on ma-chine M1before Ji. Condition (c) dictates that job Jiis the last job of set

J

1to have idle time before its second-stage operation, i.e., the

processing of Ji, . . . , Jn1 on machine M1 is consecutive without idle time inserted. An illustration of the configuration with two sched-ules is shown inFig. 4.

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Ji i-1 type-1 jobs j type-2 jobs M₀

M₁ M2

Ji i-1 type-1 jobs j type-2 jobs M0

M₁ M₂

Fig. 4. Tow schedules for an admissible ordered pair (i, j).

Definition. For the ordered pair (i, j), 1

i

n1, 0

j

n2, define the

basic sequence

(i, j) as (J1, . . . , Ji−1, Jn1+1, . . . , Jn1+j, Ji, . . . , Jn1, Jn1+j+1, . . . , Jn).

Lemma 2. If the basic sequence

(i, j) violates condition (b) or condition (c), then the ordered pair (i, j) cannot be admissible.

Proof. The starting timei_r₌₁p0r+n_r_=n1+j₁₊₁p0rof job Jiis fixed for the pair (i, j). The subsequence (J1, . . . , Ji−1, Jn1+1, . . . , Jn1+j) clearly pro-vides the smallest completion time of Ji−1on M1for all the possible

interleaved sequences derived from (J1, . . . , Ji−1) and (Jn1+1, . . . , Jn1+j). Therefore, if the subsequence (J1, . . . , Ji−1, Jn1+1, . . . , Jn1+j) violates dition (b), then so will all the other interleaved sequences. If con-dition (c) fails, then on M1a non-zero idle time exists before some

type-1 job Ji, i

>

i, in the basic sequence. The idle time before job Ji can be eliminated only if the idle time before Jivanishes, which vio-lates condition (b). Therefore, if condition (c) fails, the ordered pair (i, j) cannot be admissible. The proof is complete.

The theme of our solution algorithm now becomes one of deter-mining all the admissible ordered pairs (i, j) using Lemma 2. For each admissible ordered pair, we find an interleaved sequence that mini-mizes the completion time of machine M2. Finally, we take the

mini-mum of the weighted sum of machine completion times over all the admissible pairs. The minimization of weighted sum of machine-two completion times of the basic sequence consists of two subproblems: Problem X: Minimize the completion time of Jn1+jon machine M2 with condition (b) satisfied, and;

Problem Y: Subject to the solution for Problem X, minimize the completion time of job Jnon machine M2with condition (c) satisfied.

For Problem X, it suffices to consider the partial sequence

¯(i, j)= (J1, . . . , Ji−1, Jn1+1, . . . , Jn1+j). For 1

r

i− 1, define ¯

I

ras the subse-quence of type-2 jobs of the subset_{Jn₁₊₁, . . . , Jn1+j} that are sched-uled between jobs Jrand Jr+1. In addition, define ¯

I

0as the

subse-quence of type-2 jobs of{Jn1+1, . . . , Jn1+j} that are scheduled before

J1. The set of jobs in sequence ¯

I

r, 0

r

i− 1, is denoted by { ¯

I

r}.

Initially, in subsequence

¯(i, j), we have ¯

I

0= ¯

I

1= · · · = ¯

I

i−2=

null, and ¯

I

i−1= (Jn1+1, . . . , Jn1+j). The following procedure trans-forms subsequence

¯(i, j) so that the completion time of job Jn1+jis minimized, subject to condition (b).

Let

¯_{(i, j) be a sequence derived by interleaving the two types}

of jobs from the partial sequence

¯(i, j)= (J1, . . . , Ji−1, Jn1+1, . . . , Jn1+j) for Problem X. Sequence

¯_{(i, j) is called feasible if condition (b) is}

satisfied. Given two feasible sequences

¯_{(i, j) and}

_¯_{(i, j), we say that}

¯

_{(i, j) dominates}

_¯_{(i, j) if}r

l=0{ ¯

I

l} ⊇rl=0{ ¯

I

l} for all 0

r

i− 1. The schedule ofFig. 4(a) dominates that ofFig. 4(b).

Observation 1. Giventwo feasible sequences

¯(i, j) and

¯(i, j) for

Prob-lem X, if

¯_{(i, j) dominates}

_¯_{(i, j), then the completion times of J}

n1+j

on machine M0and machine M2with respect to

¯(i, j) are no greater

than those with respect to

¯(i, j), i.e., C0,n1+j(

¯(i, j))

C0,n1+j(

¯(i, j))

and C2,n1+j(

¯

_{(i, j))}

_C 2,n1+j(

¯

_{(i, j)).}

The observation reveals that Problem X reduces to finding a fea-sible sequence that dominates all the feafea-sible ones. The following procedure is designed to find such a sequence. In the algorithm, the symbols “⊕” and “

” denote sequence concatenation and sequence deletion, respectively.

PROCEDUREX(

¯(i, j))

Step 1: Set ¯

I

0= ¯

I

1= · · · = ¯

I

i−2= null, and ¯

I

i−1=(Jn1+1, . . . , Jn1+j).

Step 2: Set r= 0; s = n1+ 1.

Step 3: While (r

<

i− 1) and (s

n1+ j) do

if moving Js∈ ¯

I

i−1into ¯

I

rwill not violate condition (b)

then ¯

I

r:= ¯

I

r⊕ (Js); ¯

I

i−1:= ¯

I

i−1

(Js); s := s + 1;

else r := r + 1.

Step 4: Return sequence

¯(i, j)= ( ¯

I

0, J1, ¯

I

1, . . . , ¯

I

i−2, Ji−1, ¯

I

i−1).

Lemma 3. PROCEDUREX for Problem X produces a feasible sequence that

dominates all the feasible sequences.

Proof. First, note that during the course of execution of PROCEDURE

X, the produced sequence

¯(i, j) satisfies condition (b), so feasibility is maintained.

Let

¯_{(i, j)} _{= ( ¯}

_I

0, J1, ¯

I

1, . . . , ¯

I

i−2, Ji−1, ¯

I

i−1) be a feasible

se-quence not dominated by sese-quence

¯(i, j). There must be a small-est r such that r_l₌₀{ ¯

I

l} ⊂ rl=0{ ¯

I

l}. Let Jx be the type-2 job of

r l=0{ ¯

I

l}\rl=0{ ¯

I

l} with the smallest index. By the logic of the if

test in Step 3, augmenting ¯

I

rwith Jx will cause infeasibility, and thus

¯(i, j) cannot be feasible. Therefore, there cannot be such a feasible sequence

¯_{(i, j).}

With regard to the computing time, it is easy to see that con-dition (b) is examined for at most O(n1+ n2)= O(n) times. A na¨ve

approach can be deployed to examine condition (b) for the current sequence under consideration by a simple O(n) loop, thus, resulting in an overall O(n2_{) time. The concept of composite jobs, which was}

proposed by Kurisu[14], can be used to reduce to constant time the time required by checking condition (b) for a single sequence. Appli-cations of composite jobs can be found in, e.g., Monma[18], Sidney [25], and Cheng and Lin[3].

Assume that the string of jobs (J1, . . . , Ji−1) is to be processed on

machines M0and M1as in the two-machine flowshop without any

other jobs inserted in the string. By the concept of composite job, the processing of the job string (J1, . . . , Ji−1) can be replaced by a single

composite job, denoted by J[1:i−1]in the sense that the string and its

corresponding composite job have the same total idle time on M1.

To define composite job J[1:i−1], we first consider jobs Ji−2and Ji−1.

Composite job J[i−2:i−1]is defined by letting

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M₀ M 1 J₁ J₁ J₂ J₂ J₃ J₄ J₃ J₄ M 0 M 1 J_[1:2] J_[1:2] J₃ J₄ J₃ J₄ M 0 M 1 J_[1:3] J_[1:3] J₄ J₄ M 0 M 1 J_[1:4] J_[1:4]

Fig. 5. Formation of composite jobs.

and

p1,[i−2:i−1]= max{0, p1,i−2− p0,i−1} + p1,i−1.

Combining Ji−3and J[i−2:i−1], we can then define J[i−3:i−1]by letting

p0,[i−3:i−1]= p0,i−3+ max{0, p0,[i−2:i−1]− p1,i−3},

and

p1,[i−3:i−1]= max{0, p1,i−3− p0,[i−2:i−1]} + p1,[i−2:i−1].

Therefore, composite jobs J[i−2:i−1], J[i−3:i−1], . . . , J[1:i−1]can be

succes-sively found in a backward manner in O(n) time. Composite jobs

J[r:s], 1

r

<

s

i− 1, can be similarly derived in O(n2) time. This

preprocessing step will be elaborated in the analysis of another al-gorithm to be discussed later.

Lemma 4 (Kurisu [14]). Sequences (J1, . . . , Ji−1) and (J1, . . . , Jr−1, J[r:s],

Js+1, . . . , Ji−1), 1

r

<

s

i−1, have the same total idle time on machine

M1.

The schedules shown inFig. 5demonstrate how composite jobs are formed and the total idle time on the stage-two machine remains unchanged. Assume that all the composite jobs J[r:i−1], 1

r

i− 1

are derived when the input instance is given. The required time for the preprocessing step is O(n2_{), which will not be included in the}

computing time of PROCEDUREX.

Lemma 5. Given the composite jobs defined, PROCEDUREX solves

Prob-lem X in O(n) time.

Proof. Lemma 3 confirms that PROCEDUREX produces a feasible sched-ule that minimizes the completion times of job Jn1+jon machines

M0and M2. To complete the proof, it suffices to show that checking

condition (b) for each iteration of Step 3 requires only O(1) time. Let T0 and T1 be the completion times of machines M0 and M1

immediately after consideration of appending a particular type-2 job

Js– ¯

I

r. If not terminated, depending on whether or not the insertion

of Js into ¯

I

ris feasible, the algorithm will proceed to examining either Js₊₁ and ¯

I

ror Jsand ¯

I

r+1. For the former case, job Js+1and

composite job J[r+1:i−1] are scheduled from T0and T1. On machine

M1, the idle time immediately preceding J[r+1:i−1]is

= max{0, T0+

p0,s+1+ p0,[r+1:i−1]− T1}. Therefore, the completion time of job Ji−1is

C1,i−1= T1+

+

i−1

l=r+1

p1l.

If, on the other hand, Jsand ¯

I

r+1are considered in the next iteration,

then

= max{0, T0+ p0,s+ p0,[r+2:i−1]− T1} and

C1,i−1= T1+

+

i−1

l=r+2

p1l.

Having the value of C1,i−1, we can readily examine condition (b) in

constant time. The proof is complete.

Based upon the sequence

¯(i, j) produced by PROCEDUREX, we can then proceed to the second part to deal with Problem Y. We start with the sequence

(i, j)= ¯

(i, j)⊕ (Ji, . . . , Jn1, Jn1+j+1, . . . , Jn).

For i

r

n1− 1, define

I

ras the subsequence of type-2 jobs of {Jn1+j+1, . . . , Jn} that are scheduled between jobs Jrand Jr+1. The se-quence of type-2 jobs arranged after Jn1is denoted by

I

n1. Initially, in the subsequence

(i, j), we have

I

i=

I

i+1= · · · =

I

n1−1= null, and

I

n1= (Jn1+j+1, . . . , Jn). As the development and analysis are sim-ilar to those for Problem X, we omit the details of the proofs.

PROCEDUREY

Step 1: Set

I

i=

I

i+1= · · · =

I

n1−1=null, and

I

n1=(Jn1+j+1, . . . , Jn).

Step 2: Set

(i, j)= ¯

(i, j)⊕ (Ji,

I

i, . . . ,

I

n1−1, Jn1,

I

n1).

Step 3: Set r= i; s = n1+ j + 1.

Step 4: While (r

<

n1) and (s

n) do

if moving Js∈

I

n1into

I

rdoes not violate condition (c)

then

I

r:=

I

r⊕ (Js);

I

n1:=

I

n1

(Js); s :=s + 1;

else r :_{= r + 1.}

Step 5: Return sequence

(i, j)= ¯

(i, j)⊕ (Ji,

I

i, . . . ,

I

n1−1, Jn1,

I

n1).

Lemma 6. PROCEDUREY for Problem Y produces a feasible sequence that

dominates all the feasible sequences.

Lemma 7. Given the defined composite jobs, PROCEDUREY solves

Prob-lem Y in O(n) time.

The result concerning the determination of the earliest comple-tion time of machine M2 for an admissible pair (i, j) is summarized

in the following.

Lemma 8. Given two fixed sequences and an admissible pair (i, j), a schedule yielding the smallest completion time of machine M2 can be

found in O(n) time.

Proof. If the pair (i, j) is admissible, then the basic schedule (i, j)

is obtained by sequence concatenation. PROCEDURE X is invoked on the prefix partial sequence

¯(i, j), which is initially given by (J1, . . . , Ji−1, Jn1+1, . . . , Jn1+j). Then we invoke PROCEDUREY on sequence

(i, j)= ¯

(i, j)⊕ (Ji, . . . , Jn1, Jn1+j+1, . . . , Jn). After execution of the two procedures, we obtain a sequence in which the type-2 jobs are scheduled as early as possible such that the completion time of M2

is minimum while maintaining conditions (b) and (c). The execution of PROCEDURESX and Y takes O(n) time. This completes the proof.

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Consider the following instance with five jobs of each type.

n1= n2= 5

Jobs J1 J2 J3 J4 J5 J6 J7 J8 J9 J10

pi0 3 5 9 8 2 2 4 11 3 2

pi1, pi2 5 6 8 7 3 6 5 13 2 1 Subsequences (J1, J2, J3, J4, J5) and (J6, J7, J8, J9, J10) are given.

Ad-missible ordered pairs are (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5).

All the other ordered pairs are inadmissible. Taking or-dered pair (3, 3) as an example, we have the basic sequence (J1, J2, J6, J7, J8, J3, J4, J5, J9, J10). Through PROCEDURE X, jobs J6, J7, J8 are

moved with condition (b) preserved. The sequence output by PROCEDURE X is (J6, J7, J8, J1, J2, J3, J4, J5, J9, J10). Through PROCEDURE Y,

jobs J9, J10 are moved with condition (c) preserved. The final

out-put is (J6, J7, J8, J1, J2, J3, J4, J9, J10, J5). The Gantt charts of the above

sequences are shown inFig. 6.

The following algorithm integrates the above ingredients to de-termine the interleaved sequence that minimizes the weighted sum of machine completion times.

ALGORITHMFIXED-SEQUENCES

Input: Two sequences (J1, J2, . . . , Jn1) and (Jn1+1, Jn1+2, . . . , Jn).

Output: Interleaved sequence

∗that minimizes the weighted sum of machine completion times.

Step 1: Set Z∗= ∞;

Step 2: Derivei_r₌₁p0iandnr=i1p0ifor 1

i

n1, andir=n1+1p0i for 1

i

n2.

Step 3: For i_{= 1 to n}1define J[i:i]= Ji.

Step 4: For s= n1down to 2 do

for r= s − 1 down to 1 do

Derive composite job J[r:s]from Jrand J[r+1:s].

Step 5: For each pair (i, j), 1

i

n1, 0

j

n2, do

if basic sequence

(i, j) is admissible, then Call PROCEDUREX and PROCEDUREY;

if Z(

(i, j))

<

Z∗,then

∗₌

(i, j); Z∗_{= Z(}

∗). Step 6: Return

∗and Z∗.

Theorem 2. ALGORITHMFIXED-SEQUENCESsolves problem F(1, 2)_WMTfixed_seq in O(n3_{) time.}

Proof. The algorithm inspects all the possible ordered pairs, for

each of which PROCEDURESX and Y produce a sequence with the ear-liest completion time of M2. Taking the minimum weighted sum

of machine completion times among all the feasible ordered pairs, ALGORITHMFIXED-SEQUENCESreturns the optimal solution. As for the computing time, the preprocessing part, Steps 1-4, of ALGORITHMFIXED -SEQUENCEStakes O(n2_{) time. There are O(n}2_{) ordered pairs (i, j) to}

ex-amine in Step 5 and examining a pair requires O(n) time. Therefore, the overall running time is O(n3_).

In the following we extend the result given in Theorem 2. Recall the agreeable condition defined in Section 3. We consider the reverse of the agreeable condition:

rev_agr: For any jobs Ji, Jj∈

J

1, if p0i

<

p0j, then p1i

p1j, and for

any jobs Ji, Jj∈

J

2, if p0i

<

p0j, then p2i

p2j.

The case satisfying condition rev_agr is denoted by

F(1, 2)_WMTrev_agr. To solve this case, we re-index the type-1 jobs such that if i

<

j, then “p0i

<

p0j” or “p0i= p0j and p1i

p1j”. Break

ties arbitrarily. Type-2 jobs are similarly re-indexed.

Lemma 9. For problem F(1, 2)_WMTrev_agr, there is an optimal

sched-ule in which the jobs of each type are sequenced in increasing order of their indices. J₁, J₂, J₆, J₇, J₈, J₃, J₄, J₅, J₉, J₁₀ M0 M1 M₂ J₆, J₇, J₈, J₁, J₂, J₃, J₄, J₅, J₉, J₁₀ M₀ M₁ M₂ J₆, J₇, J₈, J₁, J₂, J₃, J₄, J₉, J₁₀, J₅ M0 M₁ M2

Fig. 6. Execution of PROCEDURESX and Y on ordered pair (3,3).

Proof. Assume that in some optimal solution for F(1, 2)_WMTrev_agr, there are jobs not sequenced as specified. Let Ji, Jj, 1

i

<

j

n1, be

the first two type-1 jobs such that job Jjprecedes job Ji. We ex-change the positions of the two jobs. The standard job-interex-change

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argument is applied in two phases, based on operations rather than on jobs. First, we interchange their stage-one operations and leave their stage-two operations unchanged. That is, p0iand p1jconstitute

one job, and p0jand p1iconstitute another job. It is clear that the

completion of any job of

J

\{Ji, Jj} will not be deferred. Next we in-terchange the positions of the dedicated operations of the two jobs. Similarly, no completion time of any other jobs will increase. There-fore, the schedule derived from swapping the positions of jobs Jiand

Jjwill not increase the makespan. Continuing the swapping process, if necessary, we can obtain a schedule in which the type-1 jobs are sequenced in increasing order of their indices. The same line of rea-soning can be applied to type-2 jobs. This completes the proof.

By Theorem 2, the following result immediately follows.

Corollary 1. Problem F(1, 2)_WMTrev_agrcan be solved in O(n3_{) time.}

5. Approximation algorithm

This section is devoted to the development of a heuristic and analysis of its worst-case performance. Given a minimization prob-lem and an approximation algorithm H, the performance ratio of the algorithm on a given instance is defined as the ratio of the solution value ZHgiven by the algorithm to the optimal value Z∗, i.e., ZH/Z∗. If for any instance, an algorithm attains a performance ratio less than or equal to r, then it is called an r-approximate algorithm for the studied problem. We give a 4₃-approximate algorithm for the prob-lem F(1,2)_WMT in this section.

First, we derive a lower bound on the objective value for analysis. Jobs of

J

1and jobs of

J

2are independently sequenced by Johnson's

rule. A lower bound is then obtained from the resulting weighted sum of completion times:

LB_{= w}1C(1)JS + w2C(2)JS ,

where C(1)_JS and C(2)_JS are the makespan (completion times of the two dedicated machines) of the two Johnson's sequences. An extra term can be added to the bound if we consider the interaction between the jobs of two different types. Let P1=ni=11 p0iand P2=ni=n1+1p0i. Assume that the last job on machine M0 belongs to type 1. If we

consider only type-1 jobs, then for a particular sequence

, the dif-ference between the completion times of machines M1 and M0 is

no less than C(1)_JS − P1. The insertion of the processing of type-2 jobs

on M0 will delay the completion of the last type-1 job in

by P2.

Consequently, the makespan of sequence

will increase by at least max{0, P2−(C_JS(1)−P1)}. Adding this term to the previous lower bound,

we get

LB1= w1C(1)JS + w2CJS(2)+ w1max{0, P1+ P2− CJS(1)} = w2C(2)JS + w1max{CJS(1), P1+ P2}.

On the other hand, if the last job on machine M0belongs to type 2,

then we have the following lower bound:

LB2= w1C_JS(1)+ w2max{C(2)_JS , P1+ P2}.

The heuristic algorithm presented below is mainly based upon application of Johnson's algorithm to the two job sets

J

1 and

J

2,

which we will show to be a4

3-approximate algorithm for the problem

F(1,2)_WMT.

HEURISTICH:

Step 1: Let

(

J

1) (resp.

(

J

2)) be the Johnson's sequence for

J

1

(resp.

J

2).

Step 2: If w2P1

w1P2, then return

(

J

1)⊕

(

J

2); otherwise,

return

(

J

2)⊕

(

J

1).

The computing time required by HEURISTICH is dominated by the sorting stage for deriving the two Johnson's sequences. Therefore, the running time is O(n log n). Denote by ZH the weighted sum of machine completion times of the schedule produced by HEURISTICH. In the following, we analyze the performance ratio of HEURISTICH. Without loss of generality, we assume that w2P1

w1P2. The case

of w2P1

w1P2 is symmetric. By the algorithm, all the type-1 jobs

sequenced by Johnson's rule are scheduled first as a block. The ma-chine completion times of the schedule produced by HEURISTICH are

C_JS(1)and P1+ C(2)JS , respectively. Therefore, the weighted sum is given by ZH= w1CJS(1)+ w2(P1+ C(2)JS ).

Theorem 3. HEURISTICH is a4

3-approximate algorithm for F(1,2)_WMT and

the bound is tight.

Proof. To establish the correctness of the result, we focus on

esti-mating max_{ZH/LB1, ZH/LB2} because

min{LB1, LB2}

Z∗.

Let P1=

P2for some positive

.

Case 1: LB1

LB2

In this case, ZH/LB1 is considered. The assumption that

w2P1

w1P2 implies the inequality

w2

w1. Because w2P1 =

w2

P2

w2CJS(2), we also have C (2) JS

P1/

. Therefore, ZH LB1 = w1C(1)JS + w2P1+ w2C(2)JS w2C(2)_JS + w1max{C_JS(1), P1+ P2}

1+ w2P1 w2CJS(2)+ w1max{C(1)JS , P1+ P2} (because C(1)_JS

max{C(1) JS , P1+ P2})

1+ w2P1 w2CJS(2)+ w1P1+ w1P2 (because max_{C(1)_JS , P1+ P2}

P1+ P2)

1₊ w2P1 w2P1/

+

w2P1+ w2P1 (because C(2)_JS

P1/

, w1

w2, w1P2

w2P1) = 1 +

2₊

_{+ 1}. Case 2: LB1

LB2 Considering ZH/LB2, we have ZH LB2 = w1C(1)_JS + w2P1+ w2C(2)_JS w1C(1)JS + w2max{CJS(2), P1+ P2}

1₊ w2P1 w1CJS(1)+ w2max{C(2)JS , P1+ P2} (because C(2)_JS

max_{C(2)_JS , P1+ P2})

1+ w2P1 w1CJS(1)+ w2P1+ w2P2 (because max{C(2) JS , P1+ P2}

P1+ P2).

We relate w1C_JS(1)and w2P2in the denominator to w2P1 in the

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2n-1 5n+1 M₂ M₀ M₁ J₁ J₂ J₁ J₂ 6n-1 4n-3 J₃,J₄, ..., J_n J₃,J₄, ..., J_n 2n+2 3n+3 M₂ M₀ M₁ J₁ J₂ J₁ J 3,J4, ..., Jn J₂ 4n+1 J₃,J₄, ..., J_n 4n-3

Fig. 7. (a): Schedule produced by HEURISTICH. (b) Optimal schedule.

Therefore, ZH LB2

1+ w2P1 w1C(1)_JS + w2P1+ w2P2

1+ w2P1

w2P1+ w2P1+ w2P1/

= 1 +

2₊

_{+ 1}.

From the discussion of Cases 1 and 2, a performance ratio of

/(

2 ₊

_{+ 1) is obtained. We further note that the}

func-tion

/(

2₊

_{+ 1) attains its maximum of} 1

3 when

= 1. Thus,

ZH/Z∗

ZH/min{LB1, LB2}

43.

Consider the following instance of n jobs to establish the asymp-totic tightness of this bound. In the instance n1= 1 and n2= n − 1.

Assume that the stage-two machines are equally weighted. The sin-gle type-1 job is defined as p01= 2n − 2, p11= 1. The type-2 jobs are

defined as p02= 3, p22= 3n, and p0i= 2, p2i= 1 for 3

i

n. Because

P1= 2n − 2

<

P2= 2(n − 2) + 3, HEURISTICH produces the sequence

(J1, J2, J3, . . . , Jn) (Fig. 7(a)) with ZH= (2n − 1) + (6n − 1) = 8n − 2. The optimal sequence for the instance is (J2, J1, J3, . . . , Jn) (Fig. 7(b)) with

Z∗_{= (2n + 2) + (4n + 1) = 6n + 3. As n approaches infinity, the ratio} ZH/Z∗approaches43, implying that 43 is, indeed, a tight bound.

Considering the results of Section 4, we know that a better heuris-tic can be designed by applying ALGORITHMFIXED-SEQUENCESto the two independent Johnson's sequences for the two types of jobs. How-ever, there exist no clear structural properties for the analysis of the performance ratio.

6. Conclusion

In this paper, we investigated a scheduling problem in a two-stage flowshop that has a common critical machine in two-stage one and two independent dedicated machines in stage two. The objective is to minimize the weighted sum of machine completion times. We showed that the problem is strongly NP-hard even under three com-mon assumptions on the input instances. Given two fixed sequences for the two types of jobs, we developed an O(n3_{) algorithm to}

deter-mine an optimal interleaved sequence. We also designed a heuristic for the problem and analyzed its performance ratio as 4

3.

For further research, it will be of interest to generalize our results to a constant number (

>

2) of stage-two machines. The case with a variable number of stage-two machines is also interesting. A third possible research issue is analysis of the performance ratio addressed at the end of Section 5. Under some circumstances, the machines could have different processing modes, e.g., batch machines can be considered. Incorporating batch machines, with continuous batches or simultaneous batches, into the differentiation model will give rise to several interesting research problems.

Acknowledgments

We are grateful to the reviewers for their helpful comments on an earlier version of the paper. Lin and Tian were supported in part by the National Science Council of Taiwan under Grant number 97-2923-H-009-001-MY3.

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