Johnson's rule, composite jobs and the relocation problem

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Short Communication

Johnson’s rule, composite jobs and the relocation problem

T.C.E. Cheng

a,*

, B.M.T. Lin

b

a_{Department of Logistics, The Hong Kong Polytechnic University, Kowloon, Hong Kong}

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

Received 26 April 2007; accepted 5 November 2007 Available online 22 November 2007

Abstract

Two-machine flowshop scheduling to minimize makespan is one of the most well-known classical scheduling problems. Johnson’s rule for solving this problem has been widely cited in the literature. We introduce in this paper the concept of composite job, which is an artificially constructed job with processing times such that it will incur the same amount of idle time on the second machine as that incurred by a chain of jobs in a given processing sequence. This concept due to Kurisu first appeared in 1976 to deal with the two-machine flowshop scheduling problem involving precedence constraints among the jobs. We show that this concept can be applied to reduce the computational time to solve some related scheduling problems. We also establish a link between solving the two-machine flow-shop makespan minimization problem using Johnson’s rule and the relocation problem introduced by Kaplan. We present an intuitive interpretation of Johnson’s rule in the context of the relocation problem.

Keywords: Flowshop; Makespan; Johnson’s rule; Composite job; Relocation problem

1. Introduction

Johnson’s seminal work on makespan minimization in a two-machine flowshop (Johnson, 1954) is one of the pio-neering papers in the scheduling literature. In the past half a century, the flowshop model and its solution algorithms have been included in most text books on operations man-agement, especially in books on scheduling theory (Brucker, 2007; Pinedo, 2001). Numerous research papers have been published on extensions of the flowshop model, and on exploration of the computational complexity issues and development of solution algorithms for variants of the flow-shop model. The focus of this paper is not to develop new models or algorithms, but to discuss the concept of ‘‘com-posite job”, which was proposed by Kurisu (1976). While the concept of composite job is very useful and has peda-gogical value, it has not been widely applied in flowshop scheduling. The second goal of this paper is to introduce

the relocation problem, which was first studied byKaplan (1986), and present an intuitive interpretation of Johnson’s rule for solving the two-machine flowshop makespan mini-mization problem in the context of the relocation problem. The two-machine flowshop to minimize makespan can be formally defined as follows. Let N ¼ fJ1; J2; . . . ; Jng

be a set of n jobs to be processed in a two-machine ﬂow-shop. Each job consists of two operations, which must be processed on the ﬁrst and second machine in that order, respectively. The processing times of job Ji on the

two-machines are denoted by p_iand q_i, respectively. The prob-lem, denoted by the three-ﬁeld notation F 2jjCmax(Graham et al., 1979), is to ﬁnd a schedule that completes all the jobs in the shortest time, i.e., minimizing the makespan. It can be shown that any schedule can be transformed into a sche-dule in which the processing sequences on the two-machines are identical without increasing the makespan. Consequently, solution schedules are permutation sequences, i.e., the two-machines have the same job sequence. As a result, we use sequence, instead of schedule, throughout this paper for simplicity of presentation.

*

Corresponding author. Tel.: +852 27665216; fax: +852 23645245. E-mail address:lgtcheng@inet.polyu.edu.hk(T.C.E. Cheng).

www.elsevier.com/locate/ejor European Journal of Operational Research 192 (2009) 1008–1013

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To solve the F 2jjCmax problem, Johnson (1954) gave a

decision rule, now popularly called Johnson’s rule: For any two jobs Ji; Jj2 N , if minfpi; qjg 6 minfqi; pjg, then

schedule job Jiearlier than job Jj. Based on this rule, a

solu-tion algorithm, known as Johnson’s algorithm, can be designed, which has a time complexity of Oðn log nÞ. Diﬀer-ent forms of Johnson’s algorithm have been presDiﬀer-ented in the literature. The following two forms are probably the most commonly adopted:

Form 1: Select the shortest processing time amongst all the unscheduled operations. If the operation belongs to machine one, then schedule the job of that operation in the earliest open position; else schedule the job of that operation in the last open position. Remove the job from the job set. Repeat the process until all the jobs are scheduled.

Form 2: Partition the job set N into Nþ _{¼ fJ} ijpi6

q_i; Ji2 N g and N ¼ fJijpi> qi; Ji2 N g. Schedule

the jobs in Nþ _{in non-decreasing order of p}

i, and

sche-dule the jobs in N _{in non-increasing order of q} i.

Con-catenate the two sequences.

Most operations management textbooks adopt Form 1 of the algorithm. It is clear that the time complexity of Johnson’s algorithm is dominated by the sorting procedure because we need to ﬁnd an ordered list of the jobs or oper-ations. Throughout this paper we call any algorithm John-son’s algorithm that is based on JohnJohn-son’s rule, regardless of the way in which it is presented, and denote by JiJ Jjif job Ji precedes Jjby Johnson’s rule.

The rest of this paper is organized into four sections. In Section2 we introduce the concept of composite job, fol-lowed by an illustrative example. We also give examples of applications of the concept of composite job. In Section

3we discuss the relocation problem, which has been shown to be equivalent to the two-machine ﬂowshop scheduling problem to minimize makespan. We present an intuitive

interpretation of Johnson’s rule in the context of the relo-cation problem. We conclude the paper and suggest some directions for future research in Section 4.

2. Composite jobs

The makespan of a given job sequence for F 2jjCmax is

the sum of all machine-two processing times plus the sum of all idle times incurred on the second machine. Therefore, minimizing the makespan is equivalent to finding a job sequence that minimizes the sum of idle times on the sec-ond machine. In studying the two-machine flowshop sched-uling problem with precedence constraints in chains among the jobs,Kurisu (1976)introduced the concept of a chain of jobs. That is, given a chain of jobs that will be processed consecutively without interruption, Kurisu defined a com-posite job as a job that will cause the same amount of idle time as the sum of idle times incurred by the jobs following the processing sequence defined by the chain. The reader is referred toKurisu (1976)for details of the proof. Kambu-rowski (2000) established the definition of composite job via considering flowshop scheduling with time lags.

For notational simplicity, consider the natural sequence S ¼ ðJ1; J2; . . . ; JnÞ of the jobs. We can replace a chain or a

subsequence of jobs by a composite job. Construct com-posite job J½1:2from jobs J1 and J2 by letting

p_½1:2¼ p1þ maxf0; p2 q1g; and

q_½1:2¼ maxf0; q1 p2g þ q2:

It is easy to see that schedule S and schedule S0¼ ðJ½1:2; J3; . . . ; JnÞ have the same total idle time. With

J_½1:2 and J3, we can similarly construct composite job

J½1:3 by letting p½1:3¼ p½1:2þ maxf0; p3 q½1:2g and

q_½1:3¼ maxf0; q_½1:2 p3g þ q3. Following this line of

rea-soning, for any 1 < k 6 n, composite job J½1:kis recursively

deﬁned by p_½1:k¼ p_½1:k1þ maxf0; pk q½1:k1g and q½1:k¼

maxf0; q_½1:ðk1Þ pkg þ qk. Indeed, p½1:kcan be interpreted

M M₁ M₂ M11 M M2 M1 M22 M₁ Eliminate overlapping M₁ M M2

a

b

c

d

Collect idle time on M2

Identify overlapping operations on machines

operations

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as the sum of machine-two idle times in the subsequence ðJ1; J2; . . . ; JkÞ (seeFig. 1). As shown inFig. 1, a sequence

of jobs has overlapping machine-one and machine-two operations. The overlapping intervals can be removed without changing the total idle time on machine-two. A composite job is deﬁned by trimming the overlapping inter-vals. In the backward direction, composite job J½k:ncan be

similarly deﬁned for any k; 1 6 k < n. The computational time of the above procedure for constructing composite jobs is OðnÞ.

Consider the example shown inFig. 2. The sum of idle times of the optimal schedule is 6. The idle time of compos-ite job J½1:5 is 6, too. If we combine two composite jobs

‘‘J½1:3and J½4:5” or ‘‘J½1:2and J½3:5”, then we have the same

result.

The above procedure constructs composite jobs J½1:kand

J½k:nfor 1 6 k 6 n in OðnÞ time. We can extend the

proce-dure to construct J½i:j for 1 6 i < j 6 n in the

follow-ing. Composite jobs J½1:2; J½1:3; . . . ; J½1:n are successively

determined in OðnÞ time. Similarly, J½2:3; J½2:4; . . . ; J½2:n;

J½3:4; . . . ; J½3:n; . . . ; J½n1:n are successively determined. The

total running time is Oðn2_Þ.

In the following we present some examples to demon-strate applications of the concept of composite job.Birman and Mosheiov (2004) studied a two-machine ﬂowshop scheduling problem, which is to determine the due date and schedule the jobs so as to minimize the following objective function: Zðq; dÞ ¼ maxfWE_max

16j6NEj; WT

max16j6NTj; Wddg, where Ej and Tj are, respectively the

earliness and tardiness of job Jj; WE, WT and Wd are

respectively the unit penalty of earliness and tardiness, and for setting the due date to be d. They proposed an Oðn2_{log nÞ algorithm that can optimally solve the problem.}

The time complexity results from OðnÞ repetitions of putt-ing a job in the ﬁrst position and applyputt-ing Johnson’s algo-rithm for the remaining jobs. For each job Jk and

Johnson’s sequence for the remaining n 1 jobs, we can compute the makespan of the sequence ðJk; J1; J2; . . . ;

Jk1; Jkþ1; . . . ; JnÞ by considering job Jk and two composite

jobs J½1:k1and J½kþ1:n. This takes Oð1Þ time. Therefore, the

overall time complexity is reduced to Oðn log nÞ time. The second example concerns the problems F 2jf jCmax

and F 2jljCmax, which were investigated by Saadani et al. (2005). In the F 2jf jCmaxproblem, there is a proper subset

A N of the jobs that cannot be scheduled in the ﬁrst posi-tion in a job sequence. The counterpart problem F 2jljCmax

has a subset B N of the jobs that cannot be scheduled in the last position in a job sequence. Both problems have been shown to be solvable in Oðn2_{Þ time by ﬁrst ﬁnding}

Johnson’s sequence for the jobs in N, and then testing each job in Nn A (respectively, N n B) in the first (respectively, last) position. Because the makespan of each sequence can be calculated in OðnÞ time, the overall time complexity is Oðn2_{Þ. To search for a simple rule, like Johnson’s rule,} Saadani et al. (2005) proposed four intuitive rules, and gave counter examples of these rules. They also developed several interesting but complicated properties for these two problems. Here we do not provide a simple rule, but show that the two problems can be solved in Oðn log nÞ time. As in the due date assignment problem in the first example, we select a job in Nn A out of sequence S and schedule it first. The total testing time is thus OðnÞ, which is dominated by the Oðn log nÞ time required for constructing the initial sequence S. An extended problem is F 2jf ; ljCmax, in which

the forbidden constraints are simultaneously applied to the ﬁrst and last positions. This problem is solvable in polyno-mial time too, but requires Oðn3_{Þ time (}_{Saadani et al.,} 2005). To apply the concept of composite job, we construct J½i:j for 1 6 i < j 6 n by a two-level nested loop. Assume

jobs Ji2 N n A and Jj2 N n B, i 6¼ j, are selected to be

positioned in the ﬁrst and last positions. The makespan of the test sequence can be calculated in constant time using two regular jobs and three composite jobs: Ji; J½1:i1; J½iþ1:j1; J½jþ1:n; Jj. To construct all the

com-posite jobs J½i:j requires Oðn2Þ time and the above testing

procedure takes Oðn2_{Þ time, too. Therefore, the overall}

time complexity reduces from Oðn3_{Þ to Oðn}2_Þ.

3. Applications in proofs

To prove the correctness of Johnson’s rule, we usually assume that in some optimal schedule there exist two con-secutive jobs Jiand Jjsuch that JiJ Jjand Jjprecedes Ji.

Subject to this assumption, we then show that the make-span of the schedule S0_{obtained by swapping J}

i and Jjis

no larger than that of the original schedule S. In some ﬂow-shop scheduling problems, we however need to establish the validity of the job-interchange argument for two jobs that are not necessarily scheduled in consecutive positions. Consider as an example the following problem (Sekiguchi, 1983; Cheng et al., 2000). The jobs are categorized into groups. When the processing of a job follows some job from another group, a setup time depending on the group to which the job belongs is required. In this problem

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setting, Sekiguchi (1983) showed that jobs from the same group need to be sequenced by Johnson’s rule. To prove this property, the job-interchange argument may not be applied for two consecutive jobs in the assumed optimal schedule because the two jobs may belong to diﬀerent groups. To make the jobs of a group follow the sequence determined by Johnson’s rule, we may encounter the situa-tion where two jobs from the same group are not scheduled consecutively and they violate Johnson’s rule.

In the following we demonstrate the use of the concept of composite job to show the correctness of Johnson’s rule by considering two jobs that are not necessarily consecu-tive. Assume JiJ Jj and Jj precedes Ji in some optimal

schedule. Let ðJj1; Jj2; . . . ; JjkÞ be the sequence of jobs

scheduled between Jjand Ji. We show that either moving

job Jj forward to the position immediately following Ji

or moving job Ji backward to the position immediately

preceding Jj will not increase the makespan. For

conve-nience in discussion, we partition the jobs in N into Nþ ¼ fJijpi6qi; Ji2 N g and N¼ fJijpi> qi; Ji2 N g.

Jobs in N _{(respectively, N}þ_{) are called negative}

(respec-tively, non-negative). The meanings of the terms ‘‘negative” and ‘‘non-negative” will be elaborated in the next section on the relocation problem. Moreover, the following prop-erty of the composite job will be used.

Property 1. Given job sequenceðJj; Jjþ1; . . . ; JiÞ, there are

the following inequalities: pj6p½j:i and qi6q½j:i.

To complete the proof of swapping two jobs that are not necessarily consecutive, we consider the following three cases:

Case 1: Jj; Ji2 Nþ: Consider the composite job J~jformed

by the sequenceðJj; Jj1; . . . ; JjkÞ. If J~j is negative,

then we swap J_~j and Ji and come up with the

sequenceðJi; Jj; Jj1; . . . ; JjkÞ, in which Ji precedes

Jj. On the other hand, if J~jis non-negative, then

by Property 1, p_~jP p_jP p_i. Therefore, we can also swap J~jand Ji, and have the same sequence.

Case 2: Jj; Ji2 N: If the composite job J~i deﬁned by

sequenceðJj1; . . . ; Jjk; JiÞ is non-negative, then we

can swap J~jand Jjand come up with the sequence

ðJj1; . . . ; Jjk; Ji; JjÞ, in which Ji precedes Jj. If J~iis

negative, thenProperty 1implies qj6qi6q~i, and

we can swap Jjand J~i.

Case 3: Jj2 N; Ji2 Nþ: If the composite job deﬁned by

ðJj1; . . . ; JjkÞ is negative, then we swap Ji and the

composite job, followed by swapping Ji and Jj.

The derived sequence isðJi; Jj; Jj1; . . . ; JjkÞ. If the

composite job is non-negative, then we swap Jj

and the composite job, followed by swapping Jj

and Ji. The derived sequence isðJj1; . . . ; Jjk; Ji; JjÞ.

Now, job Jiprecedes Jj.

The reasoning used above can be slightly modiﬁed (or simpliﬁed) to construct a proof of Form 2 of Johnson’s algorithm.

4. Proof of Form 2 of Johnson’s algorithm

Let S be an optimal schedule diﬀerent from the schedule produced by Form 2 of Johnson’s algorithm. Let Ji; Jjbe

the ﬁrst pair of jobs in schedule S such that the relation JiJ Jj holds but Jj precedes Ji. Recall that

ðJj1; Jj2; . . . ; JjkÞ is the sequence of jobs scheduled between

Jjand Ji. We show that eliminating the out-of-order jobs

between Ji and Jj will not increase the makespan.

More-over, we need to guarantee that no two jobs that are orig-inally in order will become out of order. Similarly, we consider the same three cases:

Case 1: Jj; Ji2 Nþ: The analysis of the makespan is the

same as above. Because Jj Jj0 for j₁6j06j_k,

by transitivity, JiJ JjJ Jj0 for j₁6j06j_k.

More pairs of out-of-order jobs can possibly be eliminated and no extra pair of jobs become out of order.

Case 2: Jj; Ji2 N: The analysis of the makespan is the

same as above. Transforming the original sequence ðJj; Jj1; . . . ; Jjk; JiÞ into ðJj1; . . . ; Jjk; Ji; JjÞ makes

jobs Jiand Jjin order without causing a new pair

of out-of-order jobs.

Case 3: Jj2 N; Ji2 Nþ: The reasoning for the new

subse-quence ðJi; Jj; Jj1; . . . ; JjkÞ (respectively,

subse-quence ðJj1; . . . ; Jjk; Ji; JjÞ) is the same as that

used for proving Case 1 (respectively, Case 2). Continuing the above swapping procedure, if necessary, we will ﬁnally come up with a schedule as speciﬁed by Johnson’s algorithm. The proof is completed.

5. The relocation problem

Alternative proofs of Johnson’s rule have been proposed in the literature in order to simplify presentation, e.g., Kam-burowski (1997). In this section we give a proof of John-son’s rule in the context of the relocation problem, which is a resource-constrained single-machine scheduling prob-lem formulated and studied in connection with a redevelop-ment project in Boston (Kaplan, 1986; Kaplan and Berman, 1988; PHRG, 1986). A pool of v0units of a single-type of

resource is provided for processing the jobs. Job Jiacquires

and consumes aiunits of the resource from the pool for its

processing and returns bi units of the resource to the pool

upon its completion. A schedule or sequence is said to be feasible if each job following the sequence can be success-fully processed. The objective is to determine the minimum initial resource level v0 that guarantees the existence of a

feasible schedule. Kaplan and Amir (1988) showed that the relocation problem is equivalent to F 2jjCmax. The

equiv-alence is based upon the fact that given a job sequence the sum of idle times on the second machine in F 2jjCmaxis equal

to the required initial resource level guaranteeing the feasi-bility of the processing sequence for the relocation problem. For recent research developments on the relocation

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problem, the reader is referred toKononov and Lin (2006, submitted for publication).

If we consider the relocation problem from the perspec-tive of investment planning, then we have the setting in which n projects are required to be ﬁnished and each pro-ject Ji requires an investment of ai units of resources and

will provide a return of b_iunits of resources. To determine a sequence to execute the projects so as to minimize the ini-tial capital, we need to work out the whole plan as follows: we execute the projects having non-negative proﬁts bi ai

ﬁrst and arrange them in non-decreasing order of invest-ments ai. The projects having negative proﬁts follow and

are arranged in non-increasing order of returns bi. This

set-ting provides an intuitive interpretation of Form 2 of John-son’s algorithm given in Section1.

We apply the concept of composite job to the relocation

problem. Consider the subsequence of jobs

ðJi; Jiþ1; . . . ; JjÞ; 1 6 i < j 6 n. We make the following

def-inition for composite job J½i:j:

a½i:j¼ a½i:j1þ maxf0; aj b½i:j1g;

b½i:j¼ maxf0; b½i:j1 ajg þ bj:

In the relocation problem, a job (project) Jiin N is called

negative because its contribution is bi ai<0. On the other

hand, non-negative jobs have non-negative contributions, i.e., bi aiP0. Therefore, the processing of a negative

job reduces the resource level in the pool while the process-ing of a positive job produces extra units of the resource.

Assume S is some optimal sequence and its resource requirement is v0. Let Ji and Jj be two consecutive jobs

in S such that JiJ Jj, but Jj precedes Ji. We consider

the following three cases to show that another schedule that is feasible with respect to v0can be obtained by

swap-ping Jj and Ji.

Case 1: Jj; Ji2 Nþ: Because ai6ajand bi aiP0,

swap-ping the two jobs preserves the feasibility of the sequence.

Case 2: Jj2 N; Ji2 Nþ: Because Jj is negative and Ji is

non-negative with respect to the resource level at the start of their processing, swapping the two jobs preserves the feasibility of the sequence.

Case 3: Jj; Ji2 N: Let vt be the resource level before the

start of job Jjin sequence S. Because Jjis negative

and consumes the resource, the resource level becomes vt ajþ bj, which is no less than ai to

ensure the feasibility of job Ji. After the swap, vtis

suﬃcient for Ji. The remaining issue we need to

address is to ensure the feasibility of job Jj

after the swap. Consider the following chain of derivations:

ðvt aiþ biÞ ajPðvt aiþ bjÞ aj

ðbecause biP bjÞ ¼ ðvt ajþ bjÞ aiP0:

Therefore, job Jjcan be processed after the swap.

6. Conclusion

In this paper, we introduced the concept of composite job and the relocation problem, which were ﬁrst proposed in 1976 and 1986, respectively, but have not been widely deployed or studied in the scheduling literature. We dem-onstrated their potential application in reducing the com-putational time required to solve related scheduling problems or simplifying the presentation of solution algo-rithms and their proofs. There is a considerable scope in applying the concept of composite job in scheduling. Studying the relocation problem with temporal constraints, such as processing times, release dates and due dates, is a worthwhile future research direction. From our experience, theoretical challenges arise whenever temporal parameters are involved in the relocation problem.

Acknowledgements

This research was supported in part by The Hong Kong Polytechnic University under a grant from the Area of Strategic Development in Chinese Business Services. The authors are grateful to the anonymous reviewers for their constructive comments.

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