European Journal of Operational Research 174 (2006) 1184–1190

www.elsevier.com/locate/ejor

Discrete Optimization

Minimizing the total completion timein a single-machine scheduling problem

with a time-dependent learning effect

Wen-Hung Kuo a, Dar-Li Yang b,*

a Department of Industrial Engineering and Technology Management, Da-Yeh University, Chang-Hwa 515, Taiwan, ROCb Department of Information Management, National Formosa University, Yun-Lin 632, Taiwan, ROC

Received 17 August 2004; accepted 8 March 2005Available online 13 June 2005

Abstract

In this study, we introduce a time-dependent learning effect into a single-machine scheduling problem. The time-dependent learning effect of a job is assumed to be a function of total normal processing time of jobs scheduled in frontof it. We introduce it into a single-machine scheduling problem and we show that it remains polynomially solvable forthe objective, i.e., minimizing the total completion time on a single machine. Moreover, we show that the SPT-sequenceis the optimal sequence in this problem.� 2005 Elsevier B.V. All rights reserved.

Keywords: Scheduling; Time-dependent; Learning effect; Single machine

1. Introduction

A scheduling problem is very important in a manufacturing system. Hence, numerous scheduling prob-lems have been studied for many years. For most of them, the processing time of a job is independent of itsposition in a scheduling sequence. However, in many realistic situations, because the firms and employeesperform the same task repeatedly, they learn how to perform more efficiently. Therefore, the actual process-ing time of a job is shorter if it is scheduled later, rather than earlier in the sequence. This phenomenon isknown as the ‘‘learning effect’’ in the literature. Although different types of learning effects have been

0377-2217/$ - see front matter � 2005 Elsevier B.V. All rights reserved.doi:10.1016/j.ejor.2005.03.020

* Corresponding author. Fax: +886 5 633 8302.E-mail address: dlyang@nfu.edu.tw (D.-L. Yang).

W.-H. Kuo, D.-L. Yang / European Journal of Operational Research 174 (2006) 1184–1190 1185

studied extensively in various areas (see Nadler and Smith, 1963; Yelle, 1979), it has rarely been studied inthe context of scheduling.

Biskup (1999) was the first to investigate the learning effect in scheduling problems. He assumed that thetime needed to perform an operation decreases by the number of repetitions, i.e., the processing time of ajob is a function of the job position in a sequence. In this job-independent learning effect model, the actualprocessing time of job j if scheduled in position r, is given by

pjr ¼ pjra; j; r ¼ 1; . . . ; n; ð1Þ

where pj is the normal (sequence-independent) processing time of job j, a 6 0 is a constant learning indexand n is the total number of jobs. Under this assumption, the learning effect of a job only depends on thenumber of jobs that are scheduled in front of it. It implies the learning rates of all jobs are all the same.There are several studies in this context. Biskup showed that single-machine scheduling problems with alearning effect still remain polynomially solvable if the objective is to minimize the deviation from a com-mon due date or to minimize the sum of flow times. Mosheiov (2001a,b) applied similar solution techniquesto some other single-machine scheduling problems, and minimum total flow time on parallel identical ma-chines. Lee et al. (2004) considered the learning effect in a bi-criterion single-machine scheduling problemand proposed a heuristic algorithm to search for optimal or near-optimal solutions. Lee and Wu (2004)proposed a heuristic algorithm to solve a total completion time minimization problem in a two-machineflowshop with a learning effect.

However, in some other practical situations, different jobs usually have different processing times due tothe number of operations. Thus, the firms and employees will learn more if they perform a job with longerprocessing time. Therefore, Mosheiov and Sidney (2003) further considered the learning in the productionprocess of some jobs to be faster than those of others, i.e., the learning is job-dependent. They showed thatthe makespan and the total flow time minimization problems on a single machine, a due-date assignmentproblem, and total flow time minimization on unrelated parallel machines remain polynomially solvable. Inthis job-dependent learning effect model, the actual processing time of job j if scheduled in position r, is givenby

pjr ¼ pjraj ; j; r ¼ 1; . . . ; n; ð2Þ

where aj is a negative job-dependent parameter. It proposes a concept that different jobs have differentlearning effects. The job-dependent learning effect model is more general than the job-independent one.

On the other hand, Biskup and Simons (2004) further considered both autonomous and induced learningeffects in a common due date scheduling problem. They focus on a scheduling problem where the processingtimes decrease according to a learning rate, which can be influenced by an initial cost-inducing investment.They derived some structural properties of the problem and presented a polynomially bound solution pro-cedure to search an optimal solution.

In this paper, we will introduce another viewpoint of a learning effect, i.e., the more time you practice,the better performance you obtain. Thus, we propose a time-dependent learning effect model and incorpo-rate it into a single-machine scheduling problem. The objective is to minimize the total completion time ofall jobs.

2. Notations and assumptions

There are n jobs to be processed on a single machine. Each of them is available at time zero. Let pj be thenormal (sequence-independent) processing time of job j (Jj, j = 1, 2, . . ., n) in a sequence and p[k] be thenormal processing time of a job if scheduled in the kth position in a sequence. The normal processing time

1186 W.-H. Kuo, D.-L. Yang / European Journal of Operational Research 174 (2006) 1184–1190

of a job is incurred if the job is scheduled first in a sequence. The processing times of the following jobs aresmaller than their normal processing times because of the learning effect. To introduce the notion that themore you practice, the better you learn, we define a time-dependent learning effect as follows. Let pir be theactual processing time of Ji if it is scheduled in position r in a sequence. Then

pir ¼�1þ p½1� þ p½2� þ � � � þ p½r�1�

�api ¼ 1þ

Xr�1

k¼1

p½k�

!a

pi; ð3Þ

where a 6 0 is a constant learning index.In such a learning effect model, we can see that the actual processing time of Ji is affected by the total

normal processing time of the previous (r � 1) jobs schedules. It implies that the more normal processingtime a job has, the more operations the job needs. In this situation, because the number of operations in ajob cannot be decreased when there exists a learning effect in the production process, the number of practicesin the job cannot be decreased. Therefore, in this study, the time-dependent learning effect of a job is as-sumed to be a function of total normal processing time of the previous (r � 1) jobs schedules. The number

1 in Eq. (3) is the modifying term to guarantee it is a learning effect, i.e., 0 < 1þPr�1

k¼1p½k�� �a

6 1. It is

apparent that the learning effect model given in Eq. (3) is very practical in many realistic situations. Seemore detailed explanations in Appendix A.

3. Minimum total completion time

In this section, we study a single-machine scheduling problem with a time-dependent learning effect. Fora given schedule q, let Ci = Ci(q) represent the completion time of Ji. The total completion time is denotedbyPn

i¼1Ci. For convenience, we denote the time-dependent learning effect mentioned in the previous sec-tion by LEt. Thus, using the conventional notation, the problem of total completion time minimization on asingle machine is denoted by 1=LEt=

Pni¼1Ci.

The classical scheduling problem of minimizing the total completion time on a single machine is opti-mized by the SPT policy. In addition, Biskup (1999) showed that the total completion time minimizationproblem with a job-independent learning effect (1=LE=

Pni¼1Ci) was also optimized by sequencing jobs

according to the SPT rule. In this section, we show that when a time-dependent learning effect is assumedfor each job, 1=LEt=

Pni¼1Ci is still optimized by the SPT rule.

Theorem 1. For the total completion time minimization problem of 1=LEt=Pn

i¼1Ci, there exists an optimal

schedule in which the job sequence is determined by the SPT rule.

Before proving Theorem 1, we need the following two lemmas.

Lemma 1. 1 � (1 + t)a + at(1 + t)a�1 P 0 if a 6 0 and t P 0.

Proof. Let h(t) = 1 � (1 + t)a + at(1 + t)a�1. Then we have h 0(t) = a(a � 1)t(1 + t)a�2 P 0 for a 6 0 andt P 0. Hence h(t) is increasing on a 6 0 and t P 0 for h(t) P h(0) = 0. Thus, we have 1 � (1 + t)a +at(1 + t)a�1 P 0 for a 6 0 and t P 0. This completes the proof. h

Lemma 2. k(1 � (1 + t)a) � (1 � (1 + kt)a) P 0 if k P 1, t P 0 and a 6 0.

Proof. Let

f ðkÞ ¼ kð1� ð1þ tÞaÞ � ð1� ð1þ ktÞaÞ. ð4Þ

W.-H. Kuo, D.-L. Yang / European Journal of Operational Research 174 (2006) 1184–1190 1187

To take the first and second derivates of Eq. (4) with respect to k, we obtain

f 0ðkÞ ¼ 1� ð1þ tÞa þ atð1þ ktÞa�1 ð5Þ

and

f 00ðkÞ ¼ aða� 1Þt2ð1þ ktÞa�2. ð6Þ

Hence, f 0(k) is increasing on k P 1, t P 0 and a 6 0 for f00(k) P 0. In addition, from Lemma 1, we have

f 0ð1Þ ¼ 1� ð1þ tÞa þ atð1þ tÞa�1 P 0. ð7Þ

Therefore, f 0(k) P f 0(1) P 0 for k P 1, t P 0 and a 6 0.Hence, f(k) is increasing on k P 1, t P 0 and a 6 0. Also, f(k) P f(1) = 0 for k P 1, t P 0 and a 6 0.Therefore, we have k(1 � (1 + t)a) � (1 � (1 + kt)a) P 0 for k P 1, t P 0 and a 6 0. This completes

the proof. h

Proof of Theorem 1. Let S1 = (p1, Jh, Ji, Jj,p2) denote a sequence where job i (Ji) and job j (Jj) are scheduledin the rth and the (r + 1)th positions, and let S2 denote the same sequence with Ji and Jj in opposite posi-tions. Moreover, let p1 and p2 be the partial sequences of S1 (or S2) before and after Jh, Ji and Jj, respec-tively. p1 or p2 may be empty. The S1 and S2 sequences are shown in Fig. 1.

Let Cl(S1) denote the completion time of Jl in sequence S1 and Cl(S2) denote the completion time of Jl insequence S2. In order to prove that the total completion time of 1=LEt=

Pni¼1Ci is minimized by the SPT-

sequence (that is pi 6 pj), It is sufficient to show that (a) Cj(S1) 6 Ci(S2) and (b) Ci(S1) + Cj(S1) 6 Cj(S2) +Ci(S2). Part (a) guarantees that all jobs scheduled in S1 after the pair of Ji and Jj, have completion times notlarger than their completion times in S2. Part (b) guarantees that the contribution to the total completiontime of Ji and Jj in sequence S1 is less than or equal to their contribution in sequence S2.

First, the proof of part (a) is given as follows.Since

CjðS1Þ ¼ ChðS1Þ þ pi 1þXr�1

k¼1

p½k�

!a

þ pj 1þ pi þXr�1

k¼1

p½k�

!a

ð8Þ

and

CiðS2Þ ¼ ChðS2Þ þ pj 1þXr�1

k¼1

p½k�

!a

þ pi 1þ pj þXr�1

k¼1

p½k�

!a

; ð9Þ

Fig. 1. A pairwise interchange of adjacent jobs.

1188 W.-H. Kuo, D.-L. Yang / European Journal of Operational Research 174 (2006) 1184–1190

we have

CiðS2Þ � CjðS1Þ ¼ ðpj � piÞ þ pið1þ pjÞa � pjð1þ piÞ

a if r ¼ 1 ð10Þ

or

CiðS2Þ �CjðS1Þ ¼ ðpj� piÞ 1þXr�1

k¼1

p½k�

!a

þ pi 1þ pjþXr�1

k¼1

p½k�

!a

� pj 1þ piþXr�1

k¼1

p½k�

!a

if r P 2;

ð11Þ

since Ch(S1) = Ch(S2).

Case 1. r = 1.Let k = pj/pi. Then Eq. (10) can be rewritten as

CiðS2Þ � CjðS1Þ ¼ kpið1� ð1þ piÞaÞ � pið1� ð1þ kpiÞ

aÞ.

Let t = pi > 0. Then from Lemma 2, if k = pj/pi P 1, we have

CiðS2Þ � CjðS1Þ ¼ piðkð1� ð1þ tÞaÞ � ð1� ð1þ ktÞaÞÞP 0.

Consequently, Cj(S1) 6 Ci(S2) if pi 6 pj.Case 2. r P 2.Let x ¼ 1þ

Pr�1k¼1p½k�.

Then

CiðS2Þ � CjðS1Þxa

¼ ðpj � piÞ þ pi 1þpj

x

� �a

� pj 1þ pi

x

� �a. ð12Þ

Because processing time of each job is greater than zero, x ¼ 1þPr�1

k¼1p½k� > 0. Thus, to prove thatCj(S1) 6 Ci(S2), it suffices to show that the right term of Eq. (12) is greater than or equal to zero. Letk = pj/pi.

Then Eq. (12) can be rewritten as

CiðS2Þ � CjðS1Þxa

¼ kpi 1� 1þ pi

x

� �ah i� pi 1� 1þ kpi

x

� �a� . ð13Þ

Let t = pi/x > 0. Then from Lemma 2, if k = pj/pi P 1, we have

CiðS2Þ � CjðS1Þxa

¼ piðkð1� ð1þ tÞaÞ � ð1� ð1þ ktÞaÞÞP 0. ð14Þ

Consequently, Cj(S1) 6 Ci(S2) if pi 6 pj.Note the proof of part (a) also shows that the makespan of all jobs is minimized by the SPT rule.Furthermore, the proof of part (b) is given as follows.Since

CiðS1Þ þ CjðS1Þ ¼ ChðS1Þ þ pi 1þXr�1

k¼1

p½k�

!a

þ ChðS1Þ þ pi 1þXr�1

k¼1

p½k�

!a

þ pj 1þ pi þXr�1

k¼1

p½k�

!a

ð15Þ

and

CjðS2Þ þ CiðS2Þ ¼ ChðS2Þ þ pj 1þXr�1

k¼1

p½k�

!a

þ ChðS2Þ þ pj 1þXr�1

k¼1

p½k�

!a

þ pi 1þ pj þXr�1

k¼1

p½k�

!a

ð16Þ

W.-H. Kuo, D.-L. Yang / European Journal of Operational Research 174 (2006) 1184–1190 1189

then

CjðS2Þ þ CiðS2Þ � ðCiðS1Þ þ CjðS1ÞÞ ¼ ðpj � piÞ þ ðpj � piÞ þ pið1þ pjÞa � pjð1þ piÞ

a if r ¼ 1 ð17Þ

or

CjðS2Þ þ CiðS2Þ � ðCiðS1Þ þ CjðS1ÞÞ

¼ ðpj � piÞ 1þXr�1

k¼1

p½k�

!a

þ ðpj � piÞ 1þXr�1

k¼1

p½k�

!a

þ pi 1þ pj þXr�1

k¼1

p½k�

!a

� pj 1þ pi þXr�1

k¼1

p½k�

!a

if r P 2. ð18Þ

Since pj � pi P 0, then the first term of Eq. (17) is non-negative. In addition, we can see that the rest ofthe terms of Eq. (17) are the same as Eq. (10). Therefore, from the proof of part (a), the sum of the rest ofthe terms of Eq. (17) is also non-negative, so we obtain

CiðS1Þ þ CjðS1Þ 6 CjðS2Þ þ CiðS2Þ for pi 6 pj; if r ¼ 1.

Similarly, since pj � pi P 0 and 1þPr�1

k¼1p½k� > 0, then the first term of Eq. (18) is also non-negative. Again,we can see that the rest of the terms of Eq. (18) are the same as Eq. (11). Therefore, from the proof of part(a), the sum of the rest of the terms of Eq. (18) is also non-negative, we obtain

CiðS1Þ þ CjðS1Þ 6 CjðS2Þ þ CiðS2Þ for pi 6 pj; if r P 2.

This completes the proof of (b) and thus of the theorem. h

Hence, the optimal job-sequence of the scheduling problem of 1=LEt=Pn

i¼1Ci can be obtained by a sort-ing algorithm. That is, the problem of 1=LEt=

Pni¼1Ci can be solved in O(n log n) time. Furthermore, we

demonstrate the result of the theorem in the following example.

Example 1. n = 5, p1 = 7, p2 = 3, p3 = 5, p4 = 4 and p5 = 6. The learning index a = �0.5. The optimaljob-sequence is (2, 4, 3, 5, 1) which is according to the SPT rule. In addition, the total completion time iscalculated as follows:

X5

i¼1

Ci ¼X5

i¼1

Xi

j¼1

p½j� 1þXj�1

k¼1

p½k�

!a

¼ 3þ 5þ 6.77þ 8.43þ 10.04 ¼ 30.24;

where p[1] = p2, p[2] = p4, p[3] = p3, p[4] = p5, and p[5] = p1.

4. Conclusions

In this study, we introduce a time-dependent learning effect into a single-machine scheduling problem.The time-dependent learning effect of a job is assumed to be a function of total processing time of jobsscheduled in front of it. It is very practical in many realistic situations. Furthermore, we show that it re-mains polynomially solvable for the objective, i.e., minimizing the total completion time on a single ma-chine. In addition, we show that the SPT-sequence is still the optimal sequence in this problem. It isclearly worthwhile for future research to investigate the time-dependent learning effect in the context ofadditional scheduling problems, including multi-machine and job-shops settings.

1190 W.-H. Kuo, D.-L. Yang / European Journal of Operational Research 174 (2006) 1184–1190

Acknowledgements

The authors would like to thank the anonymous referees for their helpful comments and suggestions.This research was supported in part by the National Science Council of Taiwan, Republic of China, undergrant number NSC-93-2213-E-150-019.

Appendix A

In fact, Eq. (3) is a concise representation of the time-dependent learning effect. However, there is oneproblem of scale-dependent. That is, we do not have one-to-one correspondence between the learning indexand the measure unit. Thus, the time-dependent learning effect model can be modified slightly by adding ascaling constant c > 0 as follows:

pir ¼ 1þ cXr�1

k¼1

p½k�

!a

pi.

If the constant c is 1 when the scale is ‘‘hour’’, say, then c is 1/60 if the scale is changed to ‘‘minute’’.In this way, the scale problem is disappeared. Also, we can see that the proof of Theorem 1 under theabove learning effect model is similar to that in Section 3 if we let t = cpi > 0 in Case 1 (r = 1) andx ¼ 1þ c

Pr�1k¼1p½k� and t = cpi/x > 0 in Case 2 (r P 2). Thus, the SPT rule is still optimal for the total

completion time under the above model. However, we focus on the concept of the time-dependent learningeffect and we use the same measure unit in this study. Thus, we still use the concise representation (Eq. (3))of the time-dependent learning effect.

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