European Journal of Operational Research 227 (2013) 76–80

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European Journal of Operational Research

journal homepage: www.elsevier .com/locate /e jor

Discrete Optimization

Single-machine scheduling problems with actual time-dependent andjob-dependent learning effect

Zhongyi Jiang ⇑, Fangfang Chen, Huiyan KangSchool of Mathematics and Physics, Changzhou University, Changzhou, Jiangsu 213164, PR China

a r t i c l e i n f o

Article history:Received 22 June 2011Accepted 6 December 2012Available online 20 December 2012

Keywords:SchedulingLearning effectActual time-dependentJob-dependentNP-hard

0377-2217/$ - see front matter � 2012 Elsevier B.V. Ahttp://dx.doi.org/10.1016/j.ejor.2012.12.007

⇑ Corresponding author. Tel.: +86 519 86330297; faE-mail address: jiangzhongyi2008@hotmail.com (Z

a b s t r a c t

In this study, we introduce an actual time-dependent and job-dependent learning effect into single-machine scheduling problems. We show that the complexity results of the makespan minimization prob-lem and the sum of weighted completion time minimization problem are all NP-hard. The complexityresult of the maximum lateness minimization problem is NP-hard in the strong sense. We also providethree special cases which can be solved by polynomial time algorithms.

� 2012 Elsevier B.V. All rights reserved.

1. Introduction

In classical scheduling problems the processing time of a job isassumed to be known and fixed. But in many modern industrialprocesses, producers (machines, plants, workers, etc.) can improvetheir efficiency continuously with time. As a result, the processingtime of a given job is shorter if it is scheduled later, rather than ear-lier in the sequence. This phenomenon is known as ‘learning effect’in the literature by Badiru (1992).

To the best of our knowledge, Biskup (1999) and Cheng andWang (2000) were among the pioneers that brought the learningeffect into the field of scheduling. Biskup (1999) modeled thelearning effect by the job processing time expressed as a non-increasing function of its position in a sequence, i.e. if job Ji isscheduled in position r, its actual processing time is pir = pir

a,where pi is the normal processing time of job Ji, a 6 0 is a constantlearning coefficient. Biskup showed that the single-machine prob-lems to minimize total deviations of job completion time from acommon due date and to minimize the sum of job completion timeare polynomially solvable. A different shape of the learning func-tion was considered by Cheng and Wang (2000). In their modelthe actual processing time was given as follows: pir = ai � bi -min{r � 1,gi}, i,r = 1, . . ., n where bi > 0 denotes the linear rationof job Ji and gi is its learning threshold. The objective is to minimizethe maximum lateness. They showed that the problem is NP-hardin the strong sense and then identified two special cases that arepolynomially solvable. They also proposed two heuristics and ana-

ll rights reserved.

x: +86 519 86330301.. Jiang).

lyzed their worst-case performance. A more general learning effectmodel was considered by Mosheiov and Sidney (2003). In theirmodel the actual processing time of job Ji in position r ispir ¼ pirai , where ai is the learning coefficient associated with jobJi. This model is based on the realistic situation that the learning ef-fect of some jobs may be better than that of others in a schedulingsequence, i.e. the learning effect is job-dependent. They showedthat the makespan minimization problem, the total flow time min-imization problem, a due date assignment problem, and total flowtime minimization on unrelated parallel machines remain polyno-mially solvable. Mosheiov and Sidney (2005) also introduced ageneral single machine scheduling problem with position learningeffect and showed that the problem can be solved in polynomialtime if the jobs have a common due date. Lin (2007) proved thatthe problem is NP-hard.

Kuo and Yang (2006) introduced a time-dependent learning ef-fect model that the learning ‘‘curve’’ depends on the total normalprocessing time of jobs previously executed, i.e. pir = (1 + p[1] +p[2] + � � � + p[r�1])api, where a 6 0 is a constant learning coefficientand p[k] is the normal processing time of a job if it is scheduledin position k in a sequence. They showed that the SPT-sequenceis the optimal sequence for the objective of minimizing the totalcompletion time.

Yang and Kuo (2007)introduced a learning effect that is basedon the sum of actual processing time. This model is intended toovercome the shortcoming caused by the assumption that theactual processing time depends on the sum of the normal process-ing time of the preceding jobs. In their model, the actual processing

time of job Ji in position r is: pir ¼ 1þ pA½1� þ pA

½2� þ � � � þ pA½r�1�

� �api,

Z. Jiang et al. / European Journal of Operational Research 227 (2013) 76–80 77

where a 6 0 is a constant learning coefficient and pA½k� is the actual

processing time of the job scheduled in position k in a sequence.They showed that the SPT-sequence is optimal for the objectivesof minimizing the makespan, the total completion time and thesum of the kth power of completion time. They also provide a poly-nomial time algorithm to minimize the sum of the weighted com-pletion time if jobs have agreeable weights.

Some other scheduling problems with time-dependent learn-ing effects can be found in Koulamas and Kyparisis (2007), Wanget al. (2008), Gawiejnowicz and Lin (2010). Extensive surveysof different scheduling models and problems involving jobswith learning effect can be found in Cheng et al. (2004),Bachman and Janiak (2004), Gawiejnowicz (2008) and Biskup(2008).

The learning model introduced by Yang and Kuo (2007) is basedon the assumption that the sum of actual processing time of thepreviously finished jobs should add to the workers experienceand cause ‘‘learning effect’’. However in many situations, a workermust deal with different types of jobs which have different charac-teristics. And the conditions of the same type of jobs are different.For example, a mechanical engineer in a factory or a repairman inan automobile repair plant must repair or maintain different typesof equipments or cars. The conditions of the same type of equip-ments or cars are different. The actual processing time of previousworks certainly could add to the workers experience, improve theproductivity and cause ‘‘learning effect’’. However the improve-ment of the productivity of some jobs may be slower than that ofothers because of their characteristics, which need more experi-ences and more time to learn. Therefore the actual processing timeof a job is not only affected by the the actual processing time ofjobs previously finished, but also the job itself. Based on this, weintroduce a new model where the learning effect is actual process-ing time dependent and job dependent in this paper. This model ismore general. We focus on classical single-machine objective func-tions such as makespan, total weighted completion time, and max-imum lateness. We show that relaxing the assumption of thelearning coefficient significantly changes the complexity of theproblem.

The remaining part of this paper is organized as follows. In Sec-tion 2 we formulate the model. In Section 3 we present the com-plexity status of several single machine scheduling problems. Thelast section presents the conclusions.

2. Problem formulation

There are given a single machine and n non-preemptive jobsthat are immediately available for processing at time zero. The ma-chine can handle one job at a time. Associated with each job Ji

(i = 1, . . . ,n) there is a normal processing time pi, a learning coeffi-cient ai a due date di and a weight wi. In addition, let p[r] and pA

½r�be the normal and the actual processing time of a job if it is sched-uled in the rth position in a sequence respectively. In this paper, weconsider a new learning effect model, i.e.,

pir ¼ 1þ pA½1� þ pA

½2� þ � � � þ pA½r�1�

� �aipi ð1Þ

where pA½1� ¼ p½1�; pA

½s� ¼ 1þ pA½1� þ pA

½2� þ � � � þ pA½s�1�

� �aips; s ¼ 2; . . . ;

r � 1 and ai 6 0 is a constant learning coefficient associated withjob Ji. For convenience, we denote this actual time-dependent andjob-dependent learning effect in Eq. (1) by LEatj.

For a given schedule p = (J1, J2, . . . , Jn). Let Ci = Ci(p) represent thecompletion time of job Ji. Let Cmax ¼maxfCiji ¼ 1; . . . ;ng;P

wiCi ¼Pn

i¼1wiCi and Lmax = max{Ci � diji = 1, . . . ,n} representthe makespan, the sum of weighted completion time andthe maximum lateness of a given permutation respectively.

Thus, using the three-field notation schema ajbjc introducedby Graham et al. (1979), the problems to minimize the maxi-mum completion time, the sum of weighted completion timeand the maximum lateness in a single-machine scheduling aredenoted by 1jLEatjjCmax, 1jLEatjj

PwiCi, 1jLEatjjLmax respectively.

3. Computational complexity

In this section, we present the complexity results of the sched-uling problems under consideration. The decision versions of theproblems can be stated as follows:

PROBLEM-Q1. Given a set J = {J1, . . . , Jn} of jobs, a normalprocessing time pi P 0, a learning coefficient ai 6 0, an LEatj

actual processing time function for each job Ji, and a positiveinteger y1, is there a non-preemptive schedule such thatCmax 6 y1?PROBLEM-Q2. Given a set J = {J1, . . . , Jn} of jobs, a normalprocessing time pi P 0, a learning coefficient ai 6 0, an LEatj

actual processing time function, a weight wi P 0 for each jobJi, and a positive integer y2, is there a non-preemptive schedulesuch that

PwiCi 6 y2?

PROBLEM-Q3. Given a set J = {J1, . . . , Jn} of jobs, a normalprocessing time pi P 0, a learning coefficient ai 6 0, an LEatj

actual processing time function, a due date di P 0 for each jobJi, and a positive integer y3, is there a non-preemptive schedulesuch that Lmax 6 y3?

It is clear that the three decision problems are all in NP.Firstly, we give a lemma which is useful for the following

theorems.

Lemma 1. Given a function f ðxÞ ¼ aþ xþ ð1þaþbÞ21þaþx , and a P 0, b is

non-negative integer, then minf(x) = 2a + 2b + 1, if and only if x = b.

Proof. f 0ðxÞ ¼ 1� ð1þaþbÞ2

ð1þaþxÞ2, it is obviously that

f 0ðxÞ > 0; when x > b; f 0ðxÞ < 0; when x < b; f 0ðxÞ¼ 0; when x ¼ b

So, minf(x) = 2a + 2b + 1, if and only if x = b. h

We now show that PARTITION, which has been shown to beNP-complete (Garey and Johnson (1979)), can be transformed toProblem-Q1 in polynomial time.

PARTITION. Given a finite set A = {x1,x2, . . . ,xn} and a ‘‘size’’s(xi) 2 Z+ for every xi such that

Pxi2AsðxiÞ ¼ 2B, is there a partition

A1 and A2 such thatP

xi2A1sðxiÞ ¼

Pxi2A2

sðxiÞ ¼ B?

Theorem 1. The decision version of the 1jLEatjjCmax problem is NP-complete.

Proof. Given an instance of PARTITION, we can construct aninstance of n + 2 jobs for the 1jLEatjjCmax problem as follows:

For every element xi 2 A, create a common job Ji (1 6 i 6 n) suchthat pi = s(xi), ai = 0.

Create two special jobs Jn+1 and Jn+2. The normal processingtimes and learning coefficients of the two special jobs arepn+1 = (B + 1)2, an+1 = �1 and pn+2 = (2 + 3B)2, an+2 = �1.

We claim that there is a solution of PARTITION if and only ifthere is a schedule that Cmax 6 y1 (y1 = 3 + 6B).

(IF) Given a partition A1 and A2 such thatPxi2A1

sðxiÞ ¼P

xi2A2sðxiÞ ¼ B. Without loss of generality, let A1 -

= {x1,x2, . . . ,xm}, and A2 = {xm+1,xm+2, . . . ,xn}

78 Z. Jiang et al. / European Journal of Operational Research 227 (2013) 76–80

We construct a schedule as shown below:

J1; J2; . . . ; Jm; Jnþ1; Jmþ1; Jmþ2; . . . ; Jn; Jnþ2

Then

Cmax ¼Xm

i¼1

sðxiÞ þðBþ 1Þ2

1þPm

i¼1sðxiÞþXn

j¼mþ1

sðxjÞ

þ ð2þ 3BÞ2

1þPm

i¼1sðxiÞ þ ðBþ1Þ2

1þPm

i¼1sðxiÞþPn

j¼mþ1sðxjÞ

¼ Bþ ð1þ BÞ þ Bþ ð2þ 3BÞ ¼ 3þ 6B ¼ y1

(Only IF) Assume there is a schedule for the scheduling problemin which Cmax 6 y1. Then we’ll derive a solution to the PARTITIONproblem.

Suppose the actual processing time of the special job Jn+1 andJn+2 are x and y in the schedule respectively. Firstly we prove thatthe special job Jn+2 cannot be placed before the special job Jn+1

when Cmax 6 y1 in the schedule. Conversely, if the special job Jn+2 isplaced before the special job Jn+1, without loss of generality, theschedule can be shown as below:

J1; J2; . . . ; Jl; Jnþ2; Jlþ1; . . . Jk; Jnþ1; Jkþ1; . . . ; Jn

Then

Cmax ¼Xl

i¼1

sðxiÞ þ yþXk

i¼lþ1

sðxiÞ þ xþXn

i¼kþ1

sðxiÞ ¼ 2Bþ xþ y

It is easy to see thatXl

i¼1

sðxiÞ 6Xn

i¼1

sðxiÞ andXk

i¼1

sðxiÞ 6Xn

i¼1

sðxiÞ

Thus

y ¼ pnþ2 1þXl

i¼1

sðxiÞ !�1

P pnþ2 1þXn

i¼1

sðxiÞ !�1

ð2Þ

x ¼ pnþ1 1þXk

i¼1

sðxiÞ þ y

!�1

P pnþ1 1þXn

i¼1

sðxiÞ þ y

!�1

ð3Þ

Let

y� ¼ pnþ2 1þXn

i¼1

sðxiÞ !�1

¼ ð2þ 3BÞ2

1þ 2B

Thus

y P y� > 0 ð4Þ

Cmax P 2Bþ yþ pnþ1 1þXn

i¼1

sðxiÞ þ y

!�1

ð5Þ

¼ 2Bþ yþ ð1þ BÞ2

1þ yþ 2Bð6Þ

Given a function

gðyÞ ¼ 2Bþ yþ ð1þ BÞ2

1þ yþ 2B

thus

g0ðyÞ ¼ 1� ð1þ BÞ2

ð1þ yþ 2BÞ2> 0; when y > 0 ð7Þ

From (4)–(7), we can get that

gðyÞP gðy�Þ ¼ 2Bþ ð2þ 3BÞ2

1þ 2Bþ ð1þ BÞ2

1þ 2Bþ ð2þ3BÞ21þ2B

and

Cmax � y1 P gðyÞ � y1 P gðy�Þ � y1

¼ 2Bþ ð2þ 3BÞ2

1þ 2Bþ ð1þ BÞ2

1þ 2Bþ ð2þ3BÞ21þ2B

� 3� 6B

¼ ð2þ 3BÞ2

1þ 2Bþ ð1þ BÞ2

1þ 2Bþ ð2þ3BÞ21þ2B

� ð1þ BÞ � ð2þ 3BÞ

¼ ð2þ 3BÞ2 � ð1þ 2BÞð2þ 3BÞ1þ 2B

� ð1þ BÞð1þ 2BÞ1þ 2B

þ ð1þ BÞ2

1þ 2Bþ ð2þ3BÞ21þ2B

¼ ð1þ BÞð1þ BÞ1þ 2B

þ ð1þ BÞ2

1þ 2Bþ ð2þ3BÞ21þ2B

> 0

Thus the special job Jn+2 must be placed after the special job Jn+1 inthe schedule that Cmax 6 y1. Without loss of generality, the schedul-ing that Cmax 6 y1 can be shown as below:

J1; J2; . . . ; Jr; Jnþ1; Jrþ1; . . . ; Jt ; Jnþ2; Jtþ1; . . . ; Jn

and

x ¼ pnþ1 1þXr

i¼1

sðxiÞ !�1

¼ ð1þ BÞ2 1þXr

i¼1

sðxiÞ !�1

y ¼ pnþ2 1þXt

i¼1

sðxiÞ þ x

!�1

P pnþ2 1þXn

i¼1

sðxiÞ þ x

!�1

¼ ð2þ 3BÞ2

1þ xþ 2B

Cmax ¼ 2Bþ xþ y P 2Bþ xþ ð2þ 3BÞ2

1þ xþ 2B

It is easy to see that y ¼ ð2þ3BÞ21þxþ2B when Jn+2 was placed in the final po-

sition as shown bellow

J1; J2; . . . ; Jr; Jnþ1; Jrþ1; . . . ; Jn; Jnþ2

Given a function g2ðxÞ ¼ 2Bþ xþ ð2þ3BÞ21þxþ2B. From the Lemma 1, ming2(-

x) = 3 + 6B if and only if x = 1 + B, i.e. min Cmax = 3 + 6B = y1 if andonly if x = 1 + B. Therefore the actual processing time of the specialjob Jn+1 is

x ¼ ð1þ BÞ2

1þXr

i¼1

sðxiÞ¼ 1þ B) 1þ

Xr

i¼1

sðxiÞ ¼ 1þ B)Xr

i¼1

sðxiÞ ¼ B

we get a solution to the PARTITION problem.The transformation we performed is polynomial. Thus we

complete the proof of the NP-completeness result. h

Next, we show that PARTITION can be transformed to Problem Q2in polynomial time.

Theorem 2. The decision version of the 1jLEatjjP

wiCi problem is NP-complete.

Proof. Given an instance of PARTITION, we can construct aninstance of n + 1 jobs for the 1jLEatjj

PwiCi problem as follows:

For every elements xi 2 A, create a common jobs Ji(1 6 i 6 2m)such that pi = s(xi), ai = 0, wi = 0.

Create a special jobs Jn+1. The normal processing time, learningcoefficient and the weight of the special job are pn+1 = (B + 1)2,an+1 = �1 and wn+1 = 1.

We claim that there is a solution of PARTITION if and only ifthere is a schedule that

PwiCi 6 y2ðy2 ¼ 1þ 2BÞ.

Z. Jiang et al. / European Journal of Operational Research 227 (2013) 76–80 79

(IF) Given a partition A1 and A2 such thatPxi2A1

sðxiÞ ¼P

xi2A2sðxiÞ ¼ B. Without loss of generality, let A1 -

= {x1,x2, . . . ,xm}, and A2 = {xm+1,xm+2, . . . ,xn}We construct a schedule as shown below:

J1; J2; . . . ; Jm; Jnþ1; Jmþ1; . . . ; Jn

ThenXwiCi ¼ Cnþ1 ¼

Xm

i¼1

sðxiÞ þðBþ 1Þ2

1þPm

i¼1sðxiÞ¼ Bþ ð1þ BÞ ¼ y2

(Only IF) Assume there is a schedule for the scheduling problemin which

PwiCi 6 y2. Then we’ll derive a solution to the PARTI-

TION problem.Suppose the special job Jn+1 is placed in the position r + 1, and

the actual completion time of the job just before it is x. Thenwithout loss of generality, the schedule can be shown as below:

J1; J2; . . . ; Jr; Jnþ1; Jrþ1; . . . ; Jn

Thus x ¼Pr

i¼1sðxiÞ andP

wiCi ¼ Cnþ1 ¼ xþ pnþ1ð1þ xÞ�1 ¼ xþð1þ BÞ2ð1þ xÞ�1. From Lemma 1, min

PwiCi ¼ 1þ 2B ¼ y2, if and

only if x = B, i.e.Xr

i¼1

sðxiÞ ¼ B

we get a solution of the PARTITION problem.The transformation we performed is polynomial. Thus we

complete the proof of the NP-completeness result. h

Finally, we show that 3-PARTITION, which has been shown tobe NP-complete in the strong sense (Garey and Johnson (1979)),can be transformed to Problem-Q3 in pseudo-polynomial time.

3-PARTITION. Given non-negative integer B and a finite set of3m elements A = {x1,x2, . . . ,x3m}, a ‘‘size’’ s(xi) 2 Z+ for each xi 2 A

such that s(xi) satisfies B/4 < s(xi) < B/2 andP3m

i¼1sðxiÞ ¼ mB, is therea partition A1, A2, . . ., Am of set A such that for each subsetAk;P

xi2AksðxiÞ ¼ B?

Theorem 3. The decision version of the 1jLEatjjLmax problem isstrongly NP-complete.

Proof. For an instance of 3-PARTITION, we construct an instanceof 4m jobs as follows:

For every elements xi 2 A, we create a common jobs Ji

(i = 1, . . . ,3m) such that pi = s(xi), ai = 0. These jobs have a samedue-date d = 2m�1 � 1 + (2m � 1)B.

Create m special jobs J3m+j(j = 1, . . . ,m) such that p3m+j = (2-j�1 + (2j � 1)B)2, d3m+j = 2j � 1 + (2j+1 � 2)B, a3m+j = �1.

We claim that there is a partition of 3-PARTITION if and only ifthere is a schedule that Lmax 6 0(y3 = 0).

(IF) If the partition is a solution to 3-PARTITION, without loss ofgenerality, let Ai = {x3i�2,x3i�1,x3i} and

Pxj2Ai

sðxjÞ ¼ Bði ¼ 1; . . . ;nÞ.we construct a schedule as shown below:

hbA1iJ3mþ1hbA2iJ3mþ2; . . . ; hbAmiJ3mþm ð8Þ

where hbAki; ð1 6 k 6 mÞ denotes the common jobs correspondingto the elements in Ak.

Firstly, we prove that the actual completion time of the specialjob J3m+k(1 6 k 6m) is

C3mþk ¼ ð2kþ1 � 2ÞBþ 2k � 1

by induction on k. Consider the special job J3m+1

C3mþ1 ¼ Bþ p3mþ1ð1þ BÞ�1 ¼ Bþ ð1þ BÞ2ð1þ BÞ�1 ¼ 1þ 2B

Suppose the actual completion time of the special job J3m+k is

C3mþk ¼ ð2kþ1 � 2ÞBþ 2k � 1

then the completion time of the special job J3m+k+1 is

C3mþkþ1 ¼ ð2kþ1 � 2ÞBþ 2k � 1þ Bþ p3mþkþ1ð1þ ð2kþ1 � 2ÞBþ 2k

� 1þ BÞ�1

¼ ð2kþ1 � 2ÞBþ 2k � 1þ Bþ ð2k þ ð2kþ1 � 1ÞBÞ2ð1þ ð2kþ1

� 2ÞBþ 2k � 1þ BÞ�1

¼ ð2kþ1 � 2ÞBþ 2k � 1þ Bþ 2k þ ð2kþ1 � 1ÞB

¼ ð2kþ2 � 2ÞBþ 2kþ1 � 1

Thus the completion time of the special job J3m+l(1 6 l 6m) is

C3mþl ¼ 2l � 1þ ð2lþ1 � 2ÞB ¼ d3mþl

The completion time of the last common job in Am is the completiontime of special job J3m+m minus its actual processing time

C ¼ C3mþm � p3mþmð1þ C3mþm�1 þ BÞ�1

¼ 2m � 1þ ð2mþ1 � 2ÞB� p3mþmð1þ ð2m � 2ÞBþ 2m�1 � 1þ BÞ�1

¼ 2m � 1þ ð2mþ1 � 2ÞB� ð2m�1 þ ð2m � 1ÞBÞ2ð1þ ð2m � 2ÞB

þ 2m�1 � 1þ BÞ�1

¼ 2m�1 � 1þ ð2m � 1ÞB ¼ d

Therefore all jobs are completed by their due dates in the schedule,i.e. Lmax 6 0. h

(ONLY IF) Assume there is a schedule for the scheduling prob-lem in which Lmax 6 0. From this schedule, we will derive a solutionto the 3-PARTITION problem through the follow Claims:

Claim 3. The actual completion time of the special jobpA½3mþk� > 1þ Bð1 6 k 6 mÞ in any schedule.

Proof. Suppose the actual processing time of the job just beforethe special job J3m+k (1 6 k 6m) is x (x P 0). Then

pA½3mþk� ¼ xþ p3mþk

1þ xP xþ p3mþ1

1þ x¼ xþ ð1þ BÞ2

1þ xP 1þ 2B > 1þ B

from Lemma 1. h

Claim 4. In the schedule Lmax 6 0, there must be that:

(i). The completion time of the special job J3m+1 must beC3m+1 = 1 + 2B. And before the special job, there must be threecommon jobs (says J1, J2, J3, without loss of generality), andp1 + p2 + p3 = s(x1) + s(x2) + s(x3) = B.

(ii). The completion time of the special job J3m+k, (2 6 k 6m) mustbe C3m+k = 2k � 1 + (2k+1 � 2)B. The jobs between the specialjob J3m+k�1 and J3m+k must be three common jobs (says J3k�2,J3k�1, J3k, without loss of generality), andp3k�2 + p3k�1 + p3k = s(x3k�2) + s(x3k�1) + s(x3k) = B

Proof. we prove the property by induction on k. Firstly, we con-sider the special job J3m+1. Suppose the actual completion time ofthe job just before the special job J3m+1 is x(x P 0). Then the com-pletion time of the special job J3m+1 will be

C3mþ1 ¼ xþ ð1þ BÞ2

1þ x

From Lemma 1, minC3m+1 = 2B + 1, if and only if x = B. The specialjob J3m+1 will not be late if and only if x = B. The other special jobs

80 Z. Jiang et al. / European Journal of Operational Research 227 (2013) 76–80

cannot be settled before the special job J3m+1, otherwise there willbe C3m+1 > 2B + 1 from Claim 3 and Lemma 1. The jobs before jobJ3m+1 must be three common jobs (says J1, J2, J3, without loss ofgenerality), for s(xi) 2 Z+ and B

4 < sðxiÞ < B2. These three common jobs

construct a set A1, andP

xi2A1xi ¼ B.

Suppose the completion time of the special job J3m+k(k P 2) isC3m+k = 2k � 1 + (2k+1 � 2)B. Consider the special job J3m+k+1, let

C3mþkþ1 ¼ 2k � 1þ ð2kþ1 � 2ÞBþ xþ ð2k þ ð2kþ1 � 1ÞBÞ2

1þ 2k � 1þ ð2kþ1 � 2ÞBþ x

where x denotes the sum of the actual processing time of the jobsjust between J3m+k and J3m+k+1. From Lemma 1, min C3m+k+1 =2k+1 � 1 + 2(2k+1 � 1)B = d3m+k+1, if and only if x = B. So the otherspecial jobs cannot be placed between the special job J3m+k andJ3m+k+1, otherwise there will be C3m+k+1 > 2k+1 � 1 + 2(2k+1 � 1)Bfrom Claim 3 and Lemma 1. And there must be three common jobs(says J3k�2, J3k�1, J3k, without loss of generality) between J3m+k andJ3m+k+1 for s(xi) 2 Z+ and B

4 < sðxiÞ < B2. These three common jobs con-

struct a partition Ak, andP

xi2AksðxiÞ ¼ B. h

From Claim 4, we get a solution to the 3-PARTITION problem.The transformation we performed is pseudo-polynomial. Thus wecomplete the proof of the strong NP-completeness result. h

4. Polynomially solvable cases

Although the general problems have been shown to be NP-hardor NP-hard in the strong sense, there are some special cases whichcan be solved in polynomial time.

If all jobs have the same learning coefficient, i.e. ai = a, i = 1, . . .,n, the learning model LEatj will be the same as the model LEat intro-duced by Yang and Kuo (2007). So from Yang and Kuo (2007), wecan easily get the following theorems.

Theorem 4. For the problem 1jLEatjjCmax, if all jobs have the commonlearning coefficient, i.e. ai = a, i = 1, . . ., n, there exist an optimalschedule in which the jobs sequence is determined by the SPT rule.

Theorem 5. For the problem 1jLEatjjP

wiCi, if all jobs have the com-mon learning coefficient and agreeable weights, i.e. ai = a, i = 1, . . ., n,and pi 6 pj implies wi P wj for all jobs Ji and Jj, there exists an optimalschedule in which jobs are sequenced in non-decreasing order of pi/wi

(SWPT rule).If all jobs have a common due date, i.e. di = d, i = 1, . . ., n, it is

obvious that the maximum lateness is minimize by minimizingthe maximum completion time. Then from Theorem 4, we can eas-ily get the following theorem:

Theorem 6. For the problem 1jLEatjjLmax, if all jobs have the commonlearning coefficient and common due date, i.e. ai = a and di = d, i = 1,. . ., n, there exist an optimal schedule in which the jobs sequence isdetermined by the SPT rule.

5. Conclusions

In this paper, we have established the complexity status of the1jLEatjjCmax problem and 1jLEatjj

PwiCi problem by performing

transformation from the PARTITION problem, and establishedthe complexity status of the 1jLEatjjLmax problem by performingtransformation from the 3-PARTITION problem. We showed thatthe makespan minimization problem and the sum of weightedcompletion time minimization problem are all NP-hard, and themaximum lateness minimization problem is NP-hard in the strongsense. We also provide three special cases which can be solved bypolynomial time algorithms. Whether the makespan minimizationproblem and the sum of weighted completion time minimizationproblem can be solved by pseudo-polynomial time algorithmsare still unknown. For further research on these problems, wemay consider the development of pseudo-polynomial time algo-rithms for them. We will also develop exact algorithms, such asbranch-and-bound, or approximation methods.

Acknowledgements

We are grateful to the anonymous referees for their valuablecomments on an earlier version of this paper.

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