Information Sciences 274 (2014) 303–309

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Information Sciences

journal homepage: www.elsevier .com/locate / ins

Single machine scheduling withsum-of-logarithm-processing-times based deterioration

http://dx.doi.org/10.1016/j.ins.2014.03.0040020-0255/� 2014 Elsevier Inc. All rights reserved.

⇑ Corresponding author. Tel.: +86 21 66135652.E-mail addresses: shenyinyin@126.com (N. Yin), lykang@shu.edu.cn (L. Kang), wangjibo75@163.com (J.-B. Wang).

Na Yin a,b, Liying Kang a,⇑, Ping Ji c, Ji-Bo Wang b,c,d

a Department of Mathematics, Shanghai University, Shanghai 200444, Chinab School of Science, Shenyang Aerospace University, Shenyang 110136, Chinac Department of Industrial and Systems Engineering, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong, Chinad State Key Laboratory for Manufacturing Systems Engineering, Xi’an Jiaotong University, Xi’an 710053, China

a r t i c l e i n f o

Article history:Received 30 June 2009Received in revised form 4 January 2011Accepted 3 March 2014Available online 15 March 2014

Keywords:SchedulingSingle machineDeteriorating job

a b s t r a c t

In this study we consider the single machine scheduling problems with sum-of-logarithm-processing-times based deterioration, i.e., the actual job processing time is a function of thesum of the logarithm of the processing times of the jobs already processed. We show thateven with the introduction of the sum-of-logarithm-processing-times based deteriorationto job processing times, single machine makespan minimization problem remain polyno-mially solvable. But for the total completion time minimization problem, we show thatthe optimal schedule is not always V-shaped with respect to job normal processing times.Heuristic algorithms and computational results are presented for the total completion timeminimization problem.

� 2014 Elsevier Inc. All rights reserved.

1. Introduction

In classical scheduling problems, the processing times of jobs are assumed to be constant values. However, in many/various practical/real life settings/applications, job processing times are an increasing function of their starting times or theirpositions in the sequence. This phenomenon, known as deterioration, has been extensively studied in the last decade invarious machine settings and performance measures. Extensive surveys of research related to scheduling deteriorating jobscan be found in Alidaee and Womer [1], and Cheng et al. [3]. We refer the reader to book Gawiejnowicz [7] for more detailson scheduling problems with time-dependent processing times.

Recently, Wu et al. [20] considered single-machine total weighted completion time scheduling problem under lineardeterioration. They proposed a branch-and-bound method and several heuristic algorithms to solve the problem. Toksarand Guner [17] considered the parallel machine earliness/tardiness (ET) scheduling with simultaneous effects of learningand linear deterioration, sequence-dependent setups, and a common due-date for all jobs. They introduced a mixed nonlin-ear integer programming formulation for the problem. Wu and Lee [19] considered single-machine group schedulingproblems with deteriorating setup times and job processing times. Cheng et al. [4] considered some scheduling problemswith the phenomena of job deterioration and learning exist simultaneously. They showed that the single-machine problemsare polynomially solvable if the performance criterion is makespan, total completion time, total weighted completion time,or maximum lateness. Cheng et al. [5] considered some scheduling problems with the actual job processing time is a function

304 N. Yin et al. / Information Sciences 274 (2014) 303–309

of jobs already processed. Lee et al. [10] considered a total completion time scheduling problem in the m-machine permu-tation flow shop with deteriorating jobs. They proposed a dominance rule and an efficient lower bound to speed up thesearching for the optimal solution. Lee et al. [11] considered a single machine scheduling with a new deterioration modelwhere the actual job processing time is a function of jobs already processed. They showed that the makespan problemremains polynomially solvable under the proposed model. Tang and Liu [16] considered two scheduling problems for atwo-machine flowshop where a single machine is followed by a batching machine. The first problem is that there is a trans-porter to carry the jobs between machines. The second problem is that there are deteriorating jobs to be processed on thesingle machine. For the first problem with minimizing the makespan, they formulate it as a mixed integer programmingmodel and then proved that it is strongly NP-hard. A heuristic algorithm is proposed for solving this problem and its worstcase performance is analyzed. For the second problem, they derived the optimal algorithms with polynomial time forminimizing the makespan, the total completion time and the maximum lateness, respectively. Wang et al. [18] consideredsingle machine scheduling problem with time-dependent deterioration. They showed that, even with the introduction oftime-dependent deterioration to job processing times, the single-machine makespan minimization problem remainspolynomially solvable. In general, the deterioration models can be classified into three types, namely time-dependent dete-rioration (see, e.g., [1,3,10,12,16,17,19,20]), position-dependent deterioration (see, e.g., [2,13,14]) and sum-of-processing-times-based deterioration (see, e.g., [4,11,15,18]).

However, the actual processing time of a given job drops to very large precipitously as the number of jobs increases in theposition-based deterioration model and when the normal job processing times are large in the sum-of-processing-times-based deterioration model. Motivated by this observation, we propose a new deterioration model where the actual jobprocessing time is a function of the sum of the logarithm of the processing times of the jobs already processed. This modelis adopted from Cheng et al. [6] and Wang et al. [18]. The sum-of-processing-times-based deterioration model can bedescribed by the following example. There are some products that need to be processed by a cutting tool. Because of wearof the cutting tool, the time required for processing a single product increases with respect to the processing time of productsalready executed.

The remaining part of this study is organized as follows. In Section 2, we formulate the model. In Section 3, we considerseveral single machine scheduling problems. In Section 4, we give some heuristics and computational results for the totalcompletion time problem. The last section presents the conclusions.

2. Problems description

There are given a single machine and n independent and non-preemptive jobs J ¼ fJ1; J2; . . . ; Jng that are immediatelyavailable for processing. The machine can handle one job at a time and preemption is not allowed. Each job j has a normal(basic) processing time pj with ln pj P 1. Cheng et al. [6] considered the following model, i.e., the actual processing time of

job Jj if it is scheduled in the rth position in a sequence is pjr ¼ pj 1þPr�1

i¼1 ln p½i�� �a

, where p½i� denotes the normal processing

time of job scheduled in the ith position in the sequence, and a � 0 is the learning index. Wang et al. [17] considered the

model where processing time of job Jj is pjr ¼ pj 1þPr�1

i¼1 p½i�� �a

, where a P 0 denote deterioration index. In this paper, we

propose a new model stem from Cheng et al. [6] and Wang et al. [18]. Specially, the actual processing time of job Jj if it isscheduled in the rth position in a sequence is

pjr ¼ pj 1þXr�1

i¼1

ln p½i�

!a

;

where p½i� denotes the normal processing time of job scheduled in the ith position in the sequence, and a denote deteriorationindex with 0 6 a � 1. In the remaining part of the paper, all the problems considered will be denoted using the three-fieldnotation schema a j b j c introduced by Graham et al. [8].

Suppose that p and p0 are two job schedules. The difference between p and p0 is a pairwise interchange of two adjacentjobs Ji and Jj, i.e., p ¼ ½S1; Ji; Jj; S2� and p0 ¼ ½S1; Jj; Ji; S2�, where S1 and S2 each denote a partial sequence. Furthermore, we as-sume that there are r � 1 jobs in S1. Thus, jobs Ji and Jj are the rth and ðr þ 1Þth job in p, whereas jobs Jj and Ji are scheduled inthe rth and ðr þ 1Þth position in p0. In addition, let A denote the completion time of the last job in S1. Under p, the completiontimes of jobs Ji and Jj are respectively

CiðpÞ ¼ Aþ pi 1þXr�1

l¼1

ln p½l�

!a

ð1Þ

and

CjðpÞ ¼ Aþ pi 1þXr�1

l¼1

ln p½l�

!a

þ pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

: ð2Þ

Similarly, the completion times of jobs Jj and Ji in p0 are respectively

N. Yin et al. / Information Sciences 274 (2014) 303–309 305

Cjðp0Þ ¼ Aþ pj 1þXr�1

l¼1

ln p½l�

!a

ð3Þ

and

Ciðp0Þ ¼ Aþ pj 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

: ð4Þ

3. Several single machine scheduling problems

First, we give some lemmas, they are useful for the following theorems (the proofs of the lemmas are given in theAppendix).

Lemma 1. ð1þ dÞa � adð1þ dÞa�1 � 1 P 0 if 0 < d � 1 and 0 6 a 6 1.

Lemma 2. ð1þ dxÞa � adð1þ dxÞa�1 � 1 P 0 if 0 < d � 1;0 6 a 6 1 and x P 1.

Lemma 3. ð1� kÞ þ kð1þ dxÞa � ð1þ d ln kþ dxÞa P 0 for k P 1;0 < d � 1;0 6 a 6 1 and x P 1.

Theorem 1. For the problem 1 j pjr ¼ pj 1þPr�1

i¼1 ln p½i�� �a

;0 6 a 6 1 j Cmax, the optimal schedule can be obtained by sequencingthe jobs in non-increasing order of pj (the LPT rule).

Proof. (By contradiction). Let us suppose an optimal schedule p0 that is not ordered in LPT order, i.e., pi P pj. To show that pdominates p0, it suffices to show that CjðpÞ 6 Ciðp0Þ. The p and p0 sequences are shown in Fig. 1. Based on Eqs. (2) and (4), wehave

Ciðp0Þ � CjðpÞ ¼ ðpj � piÞ 1þXr�1

l¼1

ln p½l�

!a

þ pi 1þXr�1

l¼1

ln p½l� þ ln pj

!a

� pj 1þXr�1

l¼1

ln p½l� þ ln pi

!a

:

Then,

Ciðp0Þ � CjðpÞ

pj 1þPr�1

l¼1 ln p½l�� �a ¼ ð1�

pi

pjÞ þ pi

pj1þ

ln pj

1þPr�1

l¼1 ln p½l�

!a

� 1þ ln pi

1þPr�1

l¼1 ln p½l�

!a

:

Let k ¼ pipj; x ¼ ln pj; d ¼ 1

1þPr�1

l¼1p½l�

, we have

Ciðp0Þ � CjðpÞ

pj 1þPr�1

l¼1 ln p½l�� �a ¼ ð1� kÞ þ kð1þ dxÞa � ð1þ d ln kþ dxÞa:

Since k ¼ pipj

P 1, by Lemma 3, we have

Ciðp0Þ � CjðpÞ

pj 1þPr�1

l¼1 p½l�� �a P 0:

So we obtain

CjðpÞ 6 Ciðp0Þ: ð5Þ

Therefore p0, which is not scheduled in LPT, cannot be optimal. Hence, the optimal schedule will be in the LPT order. �

Fig. 1. A pairwise interchange of adjacent jobs.

306 N. Yin et al. / Information Sciences 274 (2014) 303–309

In the classical problem 1kP

Cj, the optimal schedule can be obtained by the SPT rule. However, we will show that the

problem 1 j pjr ¼ pj 1þPr�1

i¼1 ln p½i�� �a

;0 6 a 6 1 jP

Cj cannot always be solved optimally by sequencing jobs in non-decreas-

ing order of pj (the SPT rule) or by sequencing jobs in non-increasing order of pj (the LPT rule), the example is as follows:

Example 1. n ¼ 3; p1 ¼ 10; p2 ¼ 11; p3 ¼ 12. When the deterioration index a ¼ 0:5. The SPT sequence is ½J1; J2; J3�,PCjðSPTÞ ¼ 94:1138. The LPT sequence is ½J3; J2; J1�;

PCjðLPTÞ ¼ 97:4053. Obviously, the optimal sequence is

½J2; J1; J3�;P

CjðOPTÞ ¼ 94:0000.From Example 1, we know that the classical SPT order or LPT order cannot always give an optimal solution for the prob-

lem 1 j pjr ¼ pj 1þPr�1

i¼1 ln p½i�� �a

; 0 6 a 6 1 jP

Cj. It remains an open problem.

For the problem 1 j pjr ¼ pjra; a P 0 j

PCj, Mosheiov [14] proved that the optimal schedule of the problem is V-shaped

with respect to the job normal processing times. It means that the jobs scheduled before the job with the smallest normalprocessing time and the jobs scheduled after it are arranged in non-increasing and non-decreasing order of their normal pro-cessing times, respectively. However, we prove by a counter-example that the optimal schedule of problem

1 j pjr ¼ pj 1þPr�1

i¼1 ln p½i�� �a

;0 6 a 6 1 jP

Cj is not V-shaped with respect to the job normal processing times.

Example 2. n ¼ 10; p1 ¼ 4; p2 ¼ 24; p3 ¼ 29; p4 ¼ 47; p5 ¼ 49; p6 ¼ 51; p7 ¼ 64; p8 ¼ 66; p9 ¼ 87; p10 ¼ 96. When the deteri-oration index a ¼ 0:5, the optimal sequence (by enumerative algorithm) is ½J8; J1; J3; J2; J4; J5; J6; J7; J9; J10�;P

CjðOPTÞ ¼ 7276:6107, which is not V-shaped with respect to the job normal processing times.

4. Heuristic algorithms and computational experiments

For the classical scheduling 1kP

Cj, the optimal schedule can be obtained by the SPT rule. In order to solve the problem

1 j pjr ¼ pj 1þPr�1

i¼1 ln p½i�� �a

;0 6 a 6 1 jP

Cj approximately, we can use the SPT rule as a heuristic algorithm.

For the classical completion time variance minimization problem, Kanet [9] made use of the V-shaped property and pre-sented a heuristic algorithm to solve it. In this section, we propose a heuristic algorithm to solve the problem

1 j pjr ¼ pj 1þPr�1

i¼1 ln p½i�� �a

;0 6 a 6 1 jP

Cj. The procedure of heuristic algorithm is adopted from Kanet [9] idea to obtain

a near optimal solution. In summary, the procedure of the heuristic algorithm is stated as follows:

Heuristic Algorithm 1 (HA1)

Step 1. Choose the job Jj with the smallest pj in J and label it as job Jk.Step 2. Remove the largest job from J and label it job Ji, and create a partial schedule S with only job Jk.Step 3. Insert job Ji to the immediate left of job Jk and create a partial schedule S0. Insert job Ji to the immediate right of job Jk

and create a partial schedule S00. From S0 and S00, choose the schedule with a smaller total completion time as the newschedule S. Delete job Ji from J.

Step 4. If J is not empty, go to step 2. Otherwise, stop.

From the Heuristic Algorithm 1, we can propose the following heuristic algorithm:

Heuristic Algorithm 2 (HA2)

Step 1. Choose and remove the job Jj with the smallest pj in J and label it as job Jk.Step 2. Remove the smallest last job from J and label it job Ji, and create a partial schedule S with only job Jk.Step 3. Insert job Ji to the immediate left of partial schedule S and create a partial schedule S0. Insert job Ji to the immediate

right of partial schedule S and create a partial schedule S00. From S0 and S00, choose the schedule with a smaller totalcompletion time as the new schedule S. Delete job Ji from J.

Step 4. If J is not empty, go to step 2. Otherwise, stop.

To evaluate the performance of SPT rule and the heuristic algorithms, a computational experiment was conducted. The SPTrule and the heuristic algorithms were coded in VC++ 6.0 and the computational experiments were run on a Pentium 4 personalcomputer with a RAM size of 1G. The test problems were generated as follows. For each job Jj, the job normal processing time pj

were generated from a uniform distribution over the integers between 3 and 100. For each HA, 6 different job sizes (n = 8, 9, 10,11, 12 and 13) and 5 different deterioration rates (a ¼ 0:1;0:3; 0:5;0:7 and 0:9) were used. As a consequence, 25 experimentalconditions were examined and 50 replications were randomly generated for each condition. A total of 1500 problems weretested.

In order to test the performance of the heuristic algorithms, relative to the optimal solutions where the optimal solution isobtained by enumerative algorithm. We did not try to optimize the running time of the enumerative algorithm, since our

Table 1The error percentages of the heuristic algorithms.

n a SPT rule HA1 HA2

Mean Max Mean Max Mean Max

0.1 0.000000 0.000000 0.000000 0.000000 0.000000 0.0000008 0.3 0.000449 0.003861 0.000000 0.000000 0.000407 0.003861

0.5 0.027267 0.087489 0.003985 0.032070 0.010692 0.0363170.7 0.072478 0.158826 0.004007 0.103467 0.017611 0.1034670.9 0.151665 0.290924 0.014884 0.290924 0.056587 0.290924

0.1 0.000005 0.000164 0.000000 0.000000 0.000007 0.0001729 0.3 0.005704 0.022609 0.002118 0.034973 0.005633 0.022609

0.5 0.038610 0.083096 0.003531 0.056550 0.011656 0.0565500.7 0.091656 0.161490 0.004107 0.091896 0.015928 0.0918960.9 0.149655 0.273301 0.014433 0.223471 0.031235 0.114711

0.1 0.000021 0.000945 0.000000 0.000000 0.000023 0.00035210 0.3 0.009261 0.028564 0.003011 0.038623 0.008437 0.028564

0.5 0.045329 0.077645 0.004555 0.072740 0.013972 0.0727400.7 0.092990 0.149730 0.008439 0.120673 0.009192 0.0903940.9 0.139937 0.254457 0.006893 0.230339 0.026391 0.222771

0.1 0.000134 0.000561 0.000005 0.000124 0.000056 0.00035411 0.3 0.011817 0.032236 0.003177 0.019373 0.009925 0.023576

0.5 0.050061 0.082859 0.001979 0.035537 0.008646 0.0399180.7 0.096228 0.159513 0.011699 0.144805 0.013594 0.0643260.9 0.135233 0.210275 0.010239 0.182645 0.020748 0.076857

0.1 0.000356 0.000986 0.000012 0.000142 0.000241 0.00085212 0.3 0.013231 0.030321 0.005834 0.030385 0.009731 0.023491

0.5 0.052697 0.082411 0.006786 0.077680 0.011853 0.0689440.7 0.092207 0.146228 0.015748 0.125466 0.012819 0.0456290.9 0.138278 0.199631 0.009253 0.174867 0.014733 0.065859

0.1 0.000687 0.001542 0.000354 0.001245 0.000567 0.00135413 0.3 0.014102 0.033512 0.006311 0.025931 0.009385 0.026231

0.5 0.055642 0.075859 0.004208 0.065127 0.005012 0.0367010.7 0.104231 0.134727 0.021165 0.140947 0.014707 0.0669280.9 0.126116 0.175812 0.007131 0.138523 0.014551 0.054104

N. Yin et al. / Information Sciences 274 (2014) 303–309 307

main goal was to evaluate the performance of the heuristic algorithm by comparing the heuristic solutions with the optimalsolutions. For the heuristic algorithms, the mean and maximum percentage deviation of the heuristic algorithm from theoptimal solution, i.e., (Heur�OPT)/OPT, are reported, where Heur is the the value of total completion time by using theheuristic methods HA1 (HA2), OPT is the value of total completion time by using the optimal schedule. From Table 1, wesee that for all cases the average error percentages of heuristic algorithms HA1 and HA2 are always less than 3%. Moreover,it can be observed that the performance of heuristic algorithms increase as the value of a decreases.

5. Conclusions

In this study we considered scheduling problems with sum-of-logarithm-processing-times based deterioration on asingle machine. We showed that the makespan minimization problem can be solved in polynomial time. However, thecomputational complexity of the total completion time minimization problem is still open. We also proposed heuristicalgorithms and showed by computational experiments that Heuristic Algorithm 1 performs effectively and efficiently inobtaining near-optimal solutions. It is suggested for future research to investigate this open problem, consider the sum-of-logarithm-processing-times based deterioration effect in the context of other scheduling problems or propose moresophisticated and efficient heuristic algorithms.

Acknowledgements

We are grateful to three anonymous referees for their constructive comments on an earlier versions of our paper. Thework described in this paper was partially supported by a grant from the Research Grants Council of the Hong Kong SpecialAdministrative Region, China (Project No. PolyU 517011), The Hong Kong Polytechnic University (Project G-YM99), and theopen project of The State Key Laboratory for Manufacturing Systems Engineering (Xi’an Jiaotong University) (Grant No.sklms201306). This research was also partially supported by the National Natural Science Foundation of China (Nos.11171207, 91130032).

Appendix A

308 N. Yin et al. / Information Sciences 274 (2014) 303–309

Lemma 1. ð1þ dÞa � adð1þ dÞa�1 � 1 P 0 if 0 < d � 1 and 0 6 a 6 1.

Proof. Let f ðdÞ ¼ ð1þ dÞa � adð1þ dÞa�1 � 1. Taking the first derivative of f ðdÞ with respect to d, we have

f 0ðdÞ ¼ �aða� 1Þdð1þ dÞa�2 P 0

for 0 < d � 1 and 0 6 a 6 1. Thus, this implies that f ðdÞ is a non-decreasing function for d. Hence, f ðdÞP f ð0Þ ¼ 0. This com-pletes the proof. �

Lemma 2. ð1þ dxÞa � adð1þ dxÞa�1 � 1 P 0 if 0 < d � 1; 0 6 a 6 1 and x P 1.

Proof. Let gðxÞ ¼ ð1þ dxÞa � adð1þ dxÞa�1 � 1. Taking the first derivative of gðxÞ with respect to x, we have

g0ðxÞ ¼ adð1þ dxÞa�1 � d2aða� 1Þð1þ dxÞa�2 P 0

for 0 < d � 1;0 6 a 6 1 and x P 1. Thus, this implies that gðxÞ is an increasing function for x P 1. From Lemma 1, we havegðxÞP gð1Þ ¼ ð1þ dÞa � adð1þ dÞa�1 � 1 P 0 for 0 < d � 1;0 6 a 6 1 and x P 1. This completes the proof. �

Lemma 3. ð1� kÞ þ kð1þ dxÞa � ð1þ d ln kþ dxÞa P 0 for k P 1;0 < d � 1;0 6 a 6 1 and x P 1.

Proof. Let hðkÞ ¼ ð1� kÞ þ kð1þ dxÞa � ð1þ d ln kþ dxÞa. Taking the first and the second derivative of hðkÞ with respect to k,we have

h0ðkÞ ¼ �1þ ð1þ dxÞa � adð1þ d ln kþ dxÞa�1

k

and

h00ðkÞ ¼ adð1þ d ln kþ dxÞa�1 � aða� 1Þd2ð1þ d ln kþ dxÞa�2

k2

¼ adð1þ d ln kþ dxÞa�2½ð1þ d ln kþ dxÞ � ða� 1Þd�k2 :

Since 0 6 a � 1; x P 1; k P 1, it implies that h00ðkÞP 0. Therefore, h0ðkÞ is an increasing function for k P 1. From Lemma 1, wehave

h0ð1Þ ¼ �1þ ð1þ dxÞa � adð1þ dxÞa�1 P 0:

Using the fact that h0ðkÞ is an increasing function for k P 1, this implies that

h0ðkÞP h0ð1ÞP 0:

Therefore, it also implies that hðkÞ is an increasing function for k P 1. Since hð1Þ ¼ 0, we have

hðkÞP 0

for k P 1;0 < h � 1; a > 1;0 < a � 1 and x > 0. This completes the proof. �

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