14
Computers & Operations Research 33 (2006) 2099 – 2112 www.elsevier.com/locate/cor Single-machine group scheduling 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, Taiwan 515, ROC b Department of Information Management, National Formosa University,Yun-Lin,Taiwan 632, ROC Available online 7 January 2005 Abstract In many realistic situations, the more time you practice, the better learning effect you obtain. Thus, we propose a time-dependent learning effect and introduce it into the single-machine group scheduling problems. The two objectives of scheduling problems are to minimize the makespan and the total completion time, respectively.We also provide two polynomial time algorithms to solve these problems. 2004 Elsevier Ltd. All rights reserved. Keywords: Group scheduling; Time-dependent; Learning effect 1. Introduction In classical scheduling problems, the processing times of jobs are assumed to be constant values. However, this assumption is not appropriate for the modeling of many modern industrial processes where very often a job, executed in the same or almost the same conditions, has a varied processing time. For example, a worker has to assemble a large number of similar products. The processing time for the worker to assemble one product depends on his knowledge, skills, organization of the working place, and other aspects. With repetition the worker learns how to produce the product more efficiently. Consequently, he is better skilled, his knowledge has increased, and the working place is better organized. As a result, the processing time for the worker to assemble one product decreases. This phenomenon is known as the “learning effect” in the literature. The learning effect was first applied to the industry more than 60 years Corresponding author. Tel.: +886 5 631 5004; fax: +886 5 633 8302. E-mail addresses: [email protected] (W.-H. Kuo), [email protected] (D.-L. Yang). 0305-0548/$ - see front matter 2004 Elsevier Ltd. All rights reserved. doi:10.1016/j.cor.2004.11.024

Single-machine group scheduling with a time-dependent learning effect

Embed Size (px)

Citation preview

Page 1: Single-machine group scheduling with a time-dependent learning effect

Computers & Operations Research 33 (2006) 2099–2112www.elsevier.com/locate/cor

Single-machine group scheduling with a time-dependent learningeffect

Wen-Hung Kuoa, Dar-Li Yangb,∗aDepartment of Industrial Engineering and Technology Management, Da-Yeh University, Chang-Hwa, Taiwan 515, ROC

bDepartment of Information Management, National Formosa University, Yun-Lin, Taiwan 632, ROC

Available online 7 January 2005

Abstract

In many realistic situations, the more time you practice, the better learning effect you obtain. Thus, we proposea time-dependent learning effect and introduce it into the single-machine group scheduling problems. The twoobjectives of scheduling problems are to minimize the makespan and the total completion time, respectively. Wealso provide two polynomial time algorithms to solve these problems.� 2004 Elsevier Ltd. All rights reserved.

Keywords: Group scheduling; Time-dependent; Learning effect

1. Introduction

In classical scheduling problems, the processing times of jobs are assumed to be constant values.However, this assumption is not appropriate for the modeling of many modern industrial processes wherevery often a job, executed in the same or almost the same conditions, has a varied processing time. Forexample, a worker has to assemble a large number of similar products. The processing time for the workerto assemble one product depends on his knowledge, skills, organization of the working place, and otheraspects. With repetition the worker learns how to produce the product more efficiently. Consequently, heis better skilled, his knowledge has increased, and the working place is better organized. As a result, theprocessing time for the worker to assemble one product decreases. This phenomenon is known as the“learning effect” in the literature. The learning effect was first applied to the industry more than 60 years

∗ Corresponding author. Tel.: +886 5 631 5004; fax: +886 5 633 8302.E-mail addresses: [email protected] (W.-H. Kuo), [email protected] (D.-L. Yang).

0305-0548/$ - see front matter � 2004 Elsevier Ltd. All rights reserved.doi:10.1016/j.cor.2004.11.024

Page 2: Single-machine group scheduling with a time-dependent learning effect

2100 W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112

ago (Wright [1]). Since then it has been studied extensively in a number of areas (Nadler and Smith [2],Yelle [3]). However, the learning effect has not been studied in the context of scheduling until recently.

Biskup [4] was the first to investigate the learning effect in scheduling problems. He assumed a learningprocess that reflects a decrease in the process time as a function of the number of repetitions i.e. as afunction of the job position in the sequence. Biskup showed that single-machine scheduling problemswith a learning effect still remain polynomially solvable if the objective is to minimize the deviation froma common due date or to minimize the sum of flow times. Later, Mosheiov [5] continued to study Biskup’smodel and introduced a polynomial solution to the single-machine makespan minimization problem. Inaddition, he showed that the classical solutions in some other scheduling problems do not hold when thelearning effect is taken into consideration. Mosheiov [6] further considered the scheduling problem ofminimizing flow time on parallel identical machines. He showed that the problem has a solution whichis polynomial in the number of jobs. Lee et al. [7] considered the learning effect in a bi-criterion single-machine scheduling problem. The objective is to find a sequence that minimizes a linear combination ofthe total completion time and the maximum tardiness. Lee et al. proposed a heuristic algorithm to searchfor optimal or near-optimal solutions. Lee and Wu [8] also proposed a heuristic algorithm to solve the totalcompletion time minimization problem in a two-machine flowshop with a learning effect. The assumptionof these papers is that the learning effect is job-independent, i.e. a common learning curve for all the jobsis assumed. However, in many realistic situations, learning of some jobs may be better than that of others.Therefore, Mosheiov and Sidney [9] further considered learning in the production process of some jobs tobe faster than that of others, i.e. the learning is job-dependent. They showed that the scheduling problemsof makespan and total flow time minimization on a single-machine, a due-date assignment problem, andtotal flow time minimization on unrelated parallel machines remain polynomially solvable.

Previous studies, mentioned above, assumed that the learning is a function of job repetitions. In job-independent situations, the processing time of job j if scheduled in position r, is given by

prj = pjr

a, j, r = 1, . . . , n, (1)

where pj is the normal (sequence-independent) processing time of job j, a�0 is a constant learning indexand n is the total number of jobs. Under this assumption, the learning effect will be the same if a job isscheduled in the same position in a sequence. It implies that the learning effect of a job only dependson the number of jobs that are scheduled in front of it. However, different jobs usually have differentprocessing times due to various quantities of the products. Theoretically, it is intuitive that the firms andemployees will learn more if they perform a job with longer processing time. However, in the assumptionmentioned above, no matter how long the processing times of the previous jobs are, the following jobwill attain the same learning effect if it is scheduled in the same position in a sequence. This propositionis really unreasonable.

On the other hand, in job-dependent situations, the processing time of job j if scheduled in position r,is given by

prj = pjr

aj , j, r = 1, . . . , n, (2)

where aj is a negative job-dependent parameter. It proposes a concept that different jobs have differentlearning effects and this approach is more reasonable. However, it still cannot clearly describe time-dependent learning situations mentioned above. That is, the more time you practice, the better learningperformance you will obtain.

Page 3: Single-machine group scheduling with a time-dependent learning effect

W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112 2101

In addition, a number of basic scheduling problems are extended to the group scheduling problemsrecently (see Baker [10], Vickson and Alfredsson [11], Yang and Chern [12]). In these problems, thejobs are classified into groups. There is no setup time between two consecutive jobs in the same group.However, each group requires a group setup time to set up the tools, jigs, and fixtures on machines. Thus,in order to increase the efficiency of operations and reduce the setup time of jobs, it is always betterto arrange the jobs of the same group to be processed in succession. This is also the concept for grouptechnology.

In this paper, we will introduce a time-dependent learning effect and incorporate it into two single-machine group scheduling problems i.e. a makespan minimization problem, and a total completion timeminimization problem.

2. Notations and assumptions

The problem is developed using the following notations. Additional notations will be introduced whenneeded throughout the paper.

m the number of groups (m�2)Gi the ith group, i = 1, 2, . . . , m

ni the number of jobs in group Gi , i = 1, 2, . . . , m

n the total number of jobs (i.e. n1 + n2 + · · · + nm = n)Jij the jth job in group Gi , j = 1, 2, . . . , ni

si the group setup time of group Gi

pij the normal processing time of Jij in the original sequencepr

ij the actual processing time of Jij which is scheduled in the rth position in a sequence ingroup Gi

pi[k] the normal processing time of Ji[k] which is scheduled in the kth position in a sequence ingroup Gi

pki[k] the actual processing time of Ji[k] which is scheduled in the kth position in a sequence in

group Gi

Cij the completion time of Jij

Ci[k] the completion time of Ji[k] which is scheduled in the kth position in a sequence in groupGi

Cmax the makespan of all jobs∑Cij total completion time of all jobs

There are n jobs to be classified into m groups and to be processed on a single machine. All jobs areavailable at time zero. It is assumed that there is no setup time between two consecutive jobs in the samegroup. However, a group setup time is required to process a group. We assume that the group setup time issequence-independent. Moreover, the normal processing time of a job is incurred if the job is scheduledfirst in a sequence of a certain group. The actual processing times of the following jobs are smaller thantheir normal processing times because of the learning effect. To introduce the concept that the more youpractice, the better you learn, we define a time-dependent learning effect as follows:

prij = (1 + pi[1] + pi[2] + · · · + pi[r−1])aipij , (3)

Page 4: Single-machine group scheduling with a time-dependent learning effect

2102 W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112

where ai �0 is a constant learning index in a certain group Gi . From Eq. (3), we can see that thelearning effect of a job is affected by the total normal processing time of the previous (r − 1) jobsschedules. The number 1 in Eq. (3) is the modifying term to guarantee that it is a learning effect, i.e.

0 <(

1 +∑r−1k=1 pi[k]

)ai

�1. It is apparent that the learning effect model given in Eq. (3) is more reasonable

than the learning effect models defined in Eqs. (1) and (2) in realistic situations.

3. Makespan minimization

In this section, we study a single-machine group scheduling problem with a time-dependent learningeffect. The objective is to minimize the makespan of all jobs. For convenience, we denote the time-dependent learning effect mentioned in the previous section by LEt . In addition, let G denote that theproblem is a group scheduling problem and S denote the existence of a sequence-independent groupsetup time. Therefore, using the conventional notation, the single-machine group scheduling problemwith a time-dependent learning effect and the sequence-independent group setup time is denoted by1/G, S, LEt/Cmax.

In order to solve the proposed problem, the optimal job sequence in each group and the optimal groupsequence are to be determined.

Theorem 1. For the makespan minimization problem of 1/G, S, LEt/Cmax, the optimal schedule satis-fies (a) the job sequence in each group is the smallest processing time (SPT) first, and (b) the groups canbe sequenced in any order.

Proof. We assume that group Gi is processed in the ith position in a group sequence. Hence, the makespanof the scheduling problem 1/G, S, LEt/Cmax is given as follows:

Cmax =(s1 + p1

1[1] + p21[2] + · · · + p

n11[n1]

)+(s2 + p1

2[1] + p22[2] + · · · + p

n22[n2]

)+ · · · +

(sm + p1

m[1] + p2m[2] + · · · + p

nm

m[nm])

=m∑

i=1

⎛⎝si +

ni∑j=1

pi[j ]

⎛⎝1 +

j−1∑k=1

pi[k]

⎞⎠

ai⎞⎠

=m∑

i=1

si +m∑

i=1

ni∑j=1

⎛⎝pi[j ]

⎛⎝1 +

j−1∑k=1

pi[k]

⎞⎠

ai⎞⎠ . (4)

The first term of Eq. (4) is a constant. Therefore, to prove that the makespan of the problem 1/G, S,

LEt/Cmax is minimum, we only have to show that the second term is minimum. Note that the secondterm is concerned with the job sequence in each group and is not influenced by the group sequence.Hence, for the problem 1/G, S, LEt/Cmax, the optimal group sequence can be scheduled in any orderand then part(b) follows.

Next, we will determine the job sequence in each group to minimize the second term of Eq. (4). Thatis, we have to show that the optimal job sequence is the smallest processing time first (part(a)). We focuson a certain group Gi , then determining the job sequence will reduce the problem to 1/LEt/Cmax. It

Page 5: Single-machine group scheduling with a time-dependent learning effect

W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112 2103

is proved that the optimal job sequence is the smallest processing time first. See the detailed proof inAppendix A. �

Using Theorem 1, a simple algorithm to determine an optimal group schedule of problem 1/G, S,

LEt/Cmax is developed as follows.

Algorithm 1.Step 1: Arrange jobs of each group in a non-decreasing order of pij ,

i.e. pi[1]�pi[2]� · · · �pi[ni ], i = 1, 2, . . . , m.

Step 2: Groups are scheduled in any order.

Obviously, the optimal job sequence within a certain group Gi can be obtained in O(ni log ni).Therefore, in Algorithm 1, the complexity of Step 1 is

∑mi=1 O(ni log ni), and that of Step 2 is O(1). It

is easy to show that∑m

i=1 O(ni log ni)�O(n log n). We can see that

m∑i=1

ni log ni �m∑

i=1

ni log nk �(log nk)

(m∑

i=1

ni

)= n log nk �n log n

for n = (∑mi=1 ni

)and nk = max{n1, n2, . . . , nm}. Thus,

∑mi=1 O(ni log ni)�O(n log n). Hence, the

complexity of Algorithm 1 is O(n log n). In addition, we demonstrate the result of Theorem 1 in thefollowing example.

Example 1. m= 2, G1 : {J11, J12}, s1 = 2, p11 = 8, p12 = 5, a1 =−0.5, G2 : {J21, J22}, s2 = 3, p21 = 6,p22 = 9, a2 = −0.3.

Solution. According to Algorithm 1, we solve Example 1 as follows:Step 1: In group G1, the optimal job sequence is J12 → J11.

In group G2, the optimal job sequence is J21 → J22.Step 2: The optimal schedule is [J12 → J11] → [J21 → J22] or [J21 → J22] → [J12 → J11].Therefore, the optimal value of the minimum makespan is calculated as follows:

Cmax = s1 + p11[1] + p2

1[2] + s2 + p12[1] + p2

2[2]= 2 + 5 + (1 + 5)−0.5(8) + 3 + 6 + (1 + 6)−0.3(9) = 24.29,

where p1[1] = p12, p1[2] = p11, p2[1] = p21, and p2[2] = p22.

4. Total completion time minimization

In the classical single-machine grouping scheduling problem 1/G, S/∑

Cij , the sum of completiontimes of all jobs in each group is minimized by using the SPT rule, and by arranging groups in a non-

decreasing order of(si +∑ni

j=1 pij

)/ni (Ham et al. [13]). Similarly, for the total completion time

minimization problem of 1/G, S, LE/∑

Cij , we will show that the optimal job sequence in each groupand the optimal group sequence can be obtained by Theorem 2.

Page 6: Single-machine group scheduling with a time-dependent learning effect

2104 W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112

Theorem 2. For the total completion time minimization problem of 1/G, S, LEt/∑

Cij , the optimalschedule satisfies (a) the job sequence in each group is the smallest processing time first, and (b) the

groups are sequenced in a non-decreasing order of(si +∑ni

j=1 pji[j ])

/ni .

Proof. We assume that group Gi is processed in the ith position in a group sequence. Hence, the totalcompletion time of all jobs is given as follows:

The first group G1:

C1[1] = s1 + p1[1],C1[2] = s1 + p1[1] + p1[2](1 + p1[1])a1,

. . .

C1[n1] = s1 + p1[1] + p1[2](1 + p1[1])a1 + · · · + p1[n1](1 + p1[1] + p1[2] + · · · + p1[n1−1])a1 .

The second group G2:

C2[1] = C1[n1] + s2 + p2[1],C2[2] = C1[n1] + s2 + p2[1] + p2[2](1 + p2[1])a2,

. . .

C2[n2] = C1[n1] + s2 + p2[1] + p2[2](1 + p2[1])a2

+ · · · + p2[n2](1 + p2[1] + p2[2] + · · · + p2[n2−1])a2 .

. . .

. . .

. . .

The last group Gm:

Cm[1] = Cm−1[nm−1] + sm + pm[1],Cm[2] = Cm−1[nm−1] + sm + pm[1] + pm[2](1 + pm[1])am,

. . .

Cm[nm] = Cm−1[nm−1] + sm + pm[1] + pm[2](1 + pm[1])am

+ · · · + pm[nm](1 + pm[1] + pm[2] + · · · + pm[nm−1])am

Therefore, the total completion time of all jobs is

m∑i=1

ni∑j=1

Ci[j ] =m∑

i=1

nisi +m∑

i=2

niCi−1[ni−1]

+m∑

i=1

⎛⎝nipi[1] +

ni∑j=2

(ni − j + 1)pi[j ]

⎛⎝1 +

j−1∑k=1

pi[k]

⎞⎠

ai⎞⎠ . (5)

The first term of Eq. (5) is a constant. Therefore, to prove the total completion time is minimum, wehave to show that both the second term and the last term of Eq. (5) are also minimum.

Page 7: Single-machine group scheduling with a time-dependent learning effect

W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112 2105

Note that the second term can be expressed as

m∑i=2

niCi−1[ni−1] =m∑

i=2

ni

i−1∑j=1

⎛⎝sj +

nj∑k=1

pj [k]

(1 +

k−1∑l=1

pj [l]

)aj⎞⎠

which is concerned with both the job sequence in each group and the group sequence. By applying part(a)of Theorem 1,

sj +nj∑

k=1

pj [k]

(1 +

k−1∑l=1

pj [l]

)aj

is minimized by the SPT rule. This means the optimal job sequence is the smallest processing time first.In addition,

m∑i=2

ni

i−1∑j=1

⎛⎝sj +

nj∑k=1

pj [k]

(1 +

k−1∑l=1

pj [l]

)aj⎞⎠

is minimized in a non-decreasing order of⎛⎝si +

ni∑j=1

pi[j ]

⎛⎝1 +

j−1∑l=1

pi[l]

⎞⎠

ai⎞⎠/ ni

(see Ham et al. [13]). Thus, the optimal group sequence is scheduled in a non-decreasing order of(si +∑ni

j=1 pji[j ])/ni .

The last term

m∑i=1

⎛⎝nipi[1] +

ni∑j=2

(ni − j + 1)pi[j ]

⎛⎝1 +

j−1∑k=1

pi[k]

⎞⎠

ai⎞⎠

is concerned with the job sequence in each group and is not influenced by the group sequence. Hence,we only need to show that

nipi[1] +ni∑

j=2

(ni − j + 1)pi[j ]

⎛⎝1 +

j−1∑k=1

pi[k]

⎞⎠

ai

is minimum. Similarly, we focus on a certain group Gi , then determining the job sequence will reducethe problem to 1/LEt/

∑Cij . Again, it will be proved that the optimal job sequence is the smallest

processing time first. See the detailed proof in Appendix B. �

Based on Theorem 2, a simple algorithm to determine the optimal group schedule of the problem1/G, S, LEt/

∑Cij is developed as follows:

Page 8: Single-machine group scheduling with a time-dependent learning effect

2106 W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112

Algorithm 2.Step 1: Arrange jobs of each group in a non-decreasing order of pij ,

i.e. pi[1]�pi[2]� · · · �pi[ni ], i = 1, 2, . . . , m.

Step 2: Groups are scheduled in a non-decreasing order of(si +∑ni

j=1 pji[j ])

/ni .

Obviously, the optimal job sequence in a certain group Gi can be obtained in O(ni log ni) and theoptimal group sequence can be obtained in O(m log m). Therefore, in Algorithm 2, the complexity ofStep 1 is

∑mi=1 O(ni log ni), and that of Step 2 is O(m log m). Hence, the complexity of Algorithm 2

is O(n log n). Also, we demonstrate the result of Theorem 2 in the following example.

Example 2. The same data in Example 1 is used, that is, m= 2, G1 : {J11, J12}, s1 = 2, p11 = 8, p12 = 5,a1 = −0.5, n1 = 2; G2 : {J21, J22}, s2 = 3, p21 = 6, p22 = 9, a2 = −0.3, n2 = 2.

Solution. According to Algorithm 2, we solve Example 2 as follows.Step 1: In group G1, the optimal job sequence is J12 → J11.

In group G2, the optimal job sequence is J21 → J22.Step 2:⎛

⎝s1 +2∑

j=1

pj1[j ]

⎞⎠/ n1 = 5.13 <

⎛⎝s2 +

2∑j=1

pj2[j ]

⎞⎠/ n2 = 7.01.

Hence, the optimal group sequence is G1 → G2. Therefore, the optimal schedule is [J12 → J11] →[J21 → J22].

Consequently, the optimal value of the total completion time is calculated as follows.∑Cij = (s1 + p1

1[1]) + (s1 + p11[1] + p2

1[2]) + (s1 + p11[1] + p2

1[2] + s2 + p12[1])

+ (s1 + p11[1] + p2

1[2] + s2 + p12[1] + p2

2[2])= (2 + 5) + (2 + 5 + (1 + 5)−0.5(8)) + (2 + 5 + (1 + 5)−0.5(8) + 3 + 6)

+ (2 + 5 + (1 + 5)−0.5(8) + 3 + 6 + (1 + 6)−0.3(9)) = 60.82,

where p1[1] = p12, p1[2] = p11, p2[1] = p21, and p2[2] = p22.

5. Conclusions

In this study, we introduce a time-dependent learning effect into single-machine group schedulingproblems. The time-dependent learning effect of a job is assumed to be a function of total processingtimes of jobs scheduled in front of it. It is more reasonable than the job-independent or the job-dependentlearning effect. Furthermore, we show that the single-machine group scheduling problem with a time-dependent learning effect remains polynomially solvable for two objectives, i.e. minimizing the makespanand the total completion time. In addition, we propose two algorithms to solve the two problems.

Finally, in this study, the time-dependent learning effect of a job is assumed to be a function of totalnormal processing time of jobs scheduled in front of it. It implies that the more normal processing time

Page 9: Single-machine group scheduling with a time-dependent learning effect

W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112 2107

a job has, the more operations the job needs. In this situation, because the number of operations in a jobcannot be decreased when there is a learning effect in the production process, the number of the practicesin the job cannot be decreased, too. Thus, in our study, the time-dependent learning effect of a job isassumed to be a function of total normal processing time of jobs scheduled in front of it. However, insome other situations, a learning effect may actually depend on time. Therefore, it is worthwhile for futureresearch to investigate a time-dependent learning effect which is a function of total actual processingtime of jobs in scheduling problems.

Acknowledgements

The authors would like to thank the anonymous referee for his helpful comments and suggestions. Thisresearch was supported in part by the National Science Council of Taiwan, Republic of China, underGrant no. NSC-93-2213-E-150-019.

Appendix A

Before proving part (a) in Theorem 1, we need the following lemmas.

Lemma A.1. (1 + t) ln(1 + t) − t �0 if t �0.

Proof. Let k(t)=(1+ t) ln(1+ t)− t . Then we have k′(t)= ln(1+ t)�0 if t �0. Hence, k(t) is increasingon t �0 and k(t)�k(0)=0 for all t �0. Therefore, we have (1+ t) ln(1+ t)− t �0 if t �0. This completesthe proof. �

Lemma A.2. (1 + t)x − (1 + tx)�0 if x�1, t �0.

Proof. Let h(x) = (1 + t)x − (1 + tx). Then we have h′(x) = (1 + t)x ln(1 + t) − t . From Lemma A.1,we have h′(1)= (1 + t) ln(1 + t)− t �0 for all t �0. In addition, h′′(x)= (1 + t)x(ln(1 + t))2 �0. Hence,h′(x) is increasing on x�1, t �0 and h′(x)�h′(1)�0 for all x�1, t �0. Therefore, h(x) is increasingas well. Again, since h(x)�h(1) = 0, we have (1 + t)x − (1 + tx)�0 if x�1, t �0. This completes theproof. �

Lemma A.3. �(1 − (1 + t)a) − (1 − (1 + �t)a)�0 if ��1, t �0 and a�0.

Proof. Lemma A.3 is true when t = 0. For the rest of the proof, assume t > 0.

Let g(�) = � − f (�), (A.1)

where

f (�) = 1 − (1 + �t)a

1 − (1 + t)a. (A.2)

To take the first and second derivates of Eq. (A.1) with respect to �, we obtain

g′(�) = 1 + at(1 + �t)a−1

1 − (1 + t)a(A.3)

Page 10: Single-machine group scheduling with a time-dependent learning effect

2108 W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112

and

g′′(�) = a(a − 1)(1 + �t)a−2t2

1 − (1 + t)a. (A.4)

Hence g′(�) is increasing on ��1, t > 0 and a�0 for g′′(�)�0. In addition,

g′(1) = 1 + a(1 + t)a−1t

1 − (1 + t)a= 1 − (1 + t)a−1(1 + (1 − a)t)

1 − (1 + t)a. (A.5)

From Lemma A.2, we have

(1 + t)1−a �(1 + (1 − a)t) for (1 − a)�1 and t > 0.

Then

(1 + t)a−1(1 + (1 − a)t)�1. (A.6)

Therefore, g′(1)�0. Consequently g′(�)�0 for ��1, t > 0 and a�0.Hence, g(�) is increasing on ��1, t > 0 and a�0. Also, g(�)�g(1)�0 for ��1, t > 0 and a�0.Thus,

g(�) = � − f (�) = � − 1 − (1 + �t)a

1 − (1 + t)a= �(1 − (1 + t)a) − (1 − (1 + �t)a)

1 − (1 + t)a�0.

Therefore, we have �(1 − (1 + t)a) − (1 − (1 + �t)a)�0 for ��1, t > 0, and a�0. This completes theproof. �

Proof of part(a) in Theorem 1. Let Si1 = (�i1, Jij , Jik, Jil, �i2) denote a sequence in group Gi whereJik and Jil are scheduled in the rth and the (r + 1)th positions, and let Si2 denote the same sequence ingroup Gi with Jik and Jil in opposite positions. Moreover, let �i1 and �i2 denote the partial sequencesof Si1 (or Si2) before and after Jij , Jik , and Jil , respectively. �i1 or �i2 may be empty. The Si1 and Si2sequences are shown in Fig. 1.

For a certain group Gi , let Cip(Si1) and Cip(Si2) denote the completion times of Jip in sequence Si1and Si2, respectively. Then we have

Cil(Si1) = Cij (Si1) + pik

⎛⎝1 +

r−1∑q=1

pi[q]

⎞⎠

ai

+ pil

⎛⎝1 + pik +

r−1∑q=1

pi[q]

⎞⎠

ai

(A.7)

and

Cik(Si2) = Cij (Si2) + pil

⎛⎝1 +

r−1∑q=1

pi[q]

⎞⎠

ai

+ pik

⎛⎝1 + pil +

r−1∑q=1

pi[q]

⎞⎠

ai

. (A.8)

Page 11: Single-machine group scheduling with a time-dependent learning effect

W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112 2109

Captions for Figures

�i1 JijSi1

Pik Pil

Si2

Jik Jil �i2

�i1 Jij Jil Jik �i2

r(r-1) (r+1)

r(r-1) (r+1)

1+Σ Pi[q] 1+Pik+Σ Pi[q]

r-1

q=1

r-1

q=1

ai ai

Pil Pik1+Σ Pi[q] 1+Pil+Σ Pi[q]

r-1

q=1

r-1

q=1

ai ai

Fig. 1. A pairwise interchange of adjacent jobs in a certain group Gi .

In order to prove that the makespan of 1/LEt/Cmax is minimized by the SPT-sequence (that is pik �pil),we have to show that Cil(Si1)�Cik(Si2). First, we compute

Cik(Si2) − Cil(Si1) = (pil − pik) + pik(1 + pil)ai − pil(1 + pik)

ai if r = 1 (A.9)

or

Cik(Si2) − Cil(Si1) = (pil − pik)

⎛⎝1 +

r−1∑q=1

pi[q]

⎞⎠

ai

+ pik

⎛⎝1 + pil +

r−1∑q=1

pi[q]

⎞⎠

ai

− pil

⎛⎝1 + pik +

r−1∑q=1

pi[q]

⎞⎠

ai

if r �2 (A.10)

since Cij (Si1) = Cij (Si2).Case 1: r = 1Let � = pil/pik . Then we can rewrite Eq. (A.9) as

Cik(Si2) − Cil(Si1) = �pik(1 − (1 + pik)ai ) − pik(1 − (1 + �pik)

ai ).

Let t = pik > 0. Then from Lemma A.3, if � = pil/pik �1, we have

Cik(Si2) − Cil(Si1) = pik(�(1 − (1 + t)ai ) − (1 − (1 + �t)ai ))�0.

Consequently, Cil(Si1)�Cik(Si2) if pik �pil .Case 2: r �2Let x = 1 +∑r−1

q=1 pi[q].

Page 12: Single-machine group scheduling with a time-dependent learning effect

2110 W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112

Then

Cik(Si2) − Cil(Si1)

xai= (pil − pik) + pik

(1 + pil

x

)ai − pil

(1 + pik

x

)ai

. (A.11)

Because processing time of each job is greater than zero, x = 1 + ∑r−1q=1 pi[q] > 1. Hence, to prove

that Cil(Si1)�Cik(Si2), we only have to show that the right term of Eq. (A.11) is greater than or equal tozero. Let � = pil/pik .

Then we can rewrite Eq. (A.11) as

Cik(Si2) − Cil(Si1)

xai= �pik

[1 −

(1 + pik

x

)ai]

− pik

[1 −

(1 + �pik

x

)ai]

. (A.12)

Let t = pik/x > 0.Then from Lemma A.3, if � = pil/pik �1, we have

Cik(Si2) − Cil(Si1)

xai= pik(�(1 − (1 + t)ai ) − (1 − (1 + �t)ai ))�0. (A.13)

Consequently, Cil(Si1)�Cik(Si2) if pik �pil .Therefore, from the results of Cases 1 and 2, repeating this interchange argument for all jobs not

sequenced according to the SPT rule yields part (a) in Theorem 1. �

Appendix B

Proof of part (a) in Theorem 2. Let Si1 = (�i1, Jij , Jik, Jil, �i2) denote a sequence in group Gi , whereJik and Jil are scheduled in the rth and the (r + 1)th positions, and let Si2 denote the same sequence ingroup Gi with Jik and Jil in opposite positions. Moreover, let �i1 and �i2 denote the partial sequences ofSi1 (or Si2) before and after Jij , Jik , and Jil , respectively. �i1 or �i2 may be empty. Si1 and Si2 sequencesare shown in Fig. 1.

For a certain group Gi , let Cip(Si1) and Cip(Si2) denote the completion times of Jip in sequence Si1and Si2, respectively. In order to prove that the total completion time of 1/LEt/

∑Cij is minimized by

the SPT-sequence (that is pik �pil), It suffices to show that (1) Cil(Si1)�Cik(Si2), and (2) Cik(Si1) +Cil(Si1)�Cil(Si2) + Cik(Si2). Part (1) guarantees that all the jobs scheduled in Si1 after the pair of Jik

and Jil have completion times not larger than their completion times in Si2. Part (2) guarantees that thecontribution to the total completion time of Jik and Jil in sequence Si1 is less than or equal to theircontribution in sequence Si2.

We can see that part (1) is given in Appendix A. Therefore, we only have to prove part (2). The proofof part (2) is given as follows.

Since

Cik(Si1) + Cil(Si1) = Cij (Si1) + pik

⎛⎝1 +

r−1∑q=1

pi[q])ai

⎞⎠

+ Cij (Si1) + pik

⎛⎝1 +

r−1∑q=1

pi[q]

⎞⎠

ai

+ pil

⎛⎝1 + pik +

r−1∑q=1

pi[q]

⎞⎠

ai

Page 13: Single-machine group scheduling with a time-dependent learning effect

W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112 2111

and

Cil(Si2) + Cik(Si2) = Cij (Si2) + pil

⎛⎝1 +

r−1∑q=1

pi[q]

⎞⎠

ai

+ Cij (Si2) + pil

⎛⎝1 +

r−1∑q=1

pi[q]

⎞⎠

ai

+ pik

⎛⎝1 + pil +

r−1∑q=1

pi[q]

⎞⎠

ai

,

then

Cil(Si2) + Cik(Si2) − (Cik(Si1) + Cil(Si1))

= (pil − pik) + (pil − pik) + pik(1 + pil)ai − pil(1 + pik)

ai , if r = 1 (B.1)

or

Cil(Si2) + Cik(Si2) − (Cik(Si1) + Cil(Si1))

= (pil − pik)

⎛⎝1 +

r−1∑q=1

pi[q]

⎞⎠

ai

+ (pil − pik)

⎛⎝1 +

r−1∑q=1

pi[q]

⎞⎠

ai

+ pik

⎛⎝1 + pil +

r−1∑q=1

pi[q]

⎞⎠

ai

− pil

⎛⎝1 + pik +

r−1∑q=1

pi[q]

⎞⎠

ai

if r �2. (B.2)

Since pil −pik �0, then the first term of Eq. (B.1) is non-negative. In addition, we can see that the restof the terms of Eq. (B.1) are the same as Eq. (A.9). Therefore, from the proof of part (1), the sum of therest of the terms of Eq. (B.1) is also non-negative, so we obtain

Cik(Si1) + Cil(Si1)�Cil(Si2) + Cik(Si2) if r = 1.

Similarly, since pil − pik �0 and 1 + ∑r−1q=1 pi[q] > 0, then the first term of Eq. (B.2) is also non-

negative. Again, we can see that the rest of the terms of Eq. (B.2) are the same as Eq. (A.10). Therefore,from the proof of part (1), the sum of the rest of the terms of Eq. (B.2) is also non-negative, so we obtain

Cik(Si1) + Cil(Si1)�Cil(Si2) + Cik(Si2) if r �2

Thus, repeating this interchange argument for all jobs not sequenced according to the SPT rule completesthe proof of part (a) in Theorem 2. �

References

[1] Wright TP. Factors affecting the cost of airplanes. Journal of Aeronautical Sciences 1936;3:122–8.[2] Nadler G, SmithWD. Manufacturing progress functions for types of processes. International Journal of Production Research

1963;2:115–35.[3] Yelle LE. The learning curve: historical review and comprehensive survey. Decision Science 1979;10:302–28.[4] Biskup D. Single-machine scheduling with learning considerations. European Journal of Operational Research

1999;115:173–8.[5] Mosheiov G. Scheduling problems with learning effect. European Journal of Operational Research 2001;132:687–93.[6] Mosheiov G. Parallel machine scheduling with learning effect. Journal of the Operational Research Society 2001;52:1–5.

Page 14: Single-machine group scheduling with a time-dependent learning effect

2112 W.-H. Kuo, D.-L. Yang / Computers & Operations Research 33 (2006) 2099–2112

[7] LeeWC,Wu CC, Sung HJ.A bi-criterion single-machine scheduling problem with learning considerations.Acta Informatica2004;40:303–15.

[8] Lee WC, Wu CC. Minimizing total completion time in a two-machine flowshop with a learning effect. International Journalof Production Economics 2004;88:85–93.

[9] Mosheiov G, Sidney JB. Scheduling with general job-dependent learning curves. European Journal of Operational Research2003;147:665–70.

[10] Baker KR. Scheduling groups of jobs in the two-machine flow shop. Mathematical and Computer Modeling 1990;13:29–36.

[11] Vickson RG, Alfredsson BE. Two- and three-machine flow shop scheduling problems with equal sized transfer batches.International Journal of Production Research 1992;30:1551–74.

[12] Yang DL, Chern MS. Two-machine flowshop group scheduling problem. Computers & Operations Research 2000;27:975–85.

[13] Ham I, Hitomi K, Yoshida T. Group technology: application to production management. Boston: Kluwer Nijhoff, 1985. p.100–7.