Research on scheduling problems with general effects of deterioration and learning

Research on scheduling problems with general effects of deterioration and learning

Information Sciences 307 (2015) 89–94 Contents lists available at ScienceDirect Information Sciences journal homepage: www.elsevier.com/locate/ins ...

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Information Sciences 307 (2015) 89–94

Contents lists available at ScienceDirect

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

Research on scheduling problems with general effects of deterioration and learning Limin Wang a,⇑, Jian Jin b, Ji-Bo Wang c, Ping Ji d a

College of Science, Liaoning Shihua University, Fushun 113001, China School of Government, Beijing Normal University, Beijing 100875, China c School of Science, Shenyang Aerospace University, Shenyang 110136, China d Department of Industrial and Systems Engineering, The Hong Kong Polytechnic University, Hung Hom, Kowloon, Hong Kong b

a r t i c l e

i n f o

Article history: Received 22 May 2014 Received in revised form 9 February 2015 Accepted 14 February 2015 Available online 21 February 2015 Keywords: Scheduling Single machine Deteriorating job Learning effect

a b s t r a c t Jobs with deterioration and learning effect co-exist in many industry engineering and logistics management situations. In this note, the general effects of deterioration and learning are considered. It is proved that some single machine scheduling problems are still polynomially solvable under the proposed model. The note also shows that some results in Lee and Lai’s recent paper (Lee and Lai, 2011) are incorrect by a counterexample. Ó 2015 Elsevier Inc. All rights reserved.

1. Introduction In classical scheduling it is assumed that the processing times of the jobs are constant. However, in many realistic settings, because firms and employees perform a task over and over again, they learn how to perform more efficiently. The production facility (a machine, a worker) improves continuously over time. As a result, the production time of a given product is shorter after it is scheduled, rather than the earlier process in the sequence. This phenomenon is known as the ‘‘learning effect’’ in the literature [1,5,10,12,14]. On the other hand, job deterioration appears, e.g., in scheduling maintenance jobs or cleaning assignments [6,8,9,11,13], where any delay in processing a job is penalized by incurring additional time for accomplishing the job. Extensive surveys of different scheduling models and problems involving jobs with start time dependent processing times can be found in Cheng et al. [3] and Gawiejnowicz [4]. An extensive survey of research related to scheduling with learning effects was provided by Biskup [2]. The recent paper ‘‘Scheduling problems with general effects of deterioration and learning’’ [7] addresses the single machine scheduling problems with general effects of deterioration and learning in which the actual job processing time is a general function on the processing times of the jobs already processed and its scheduled position. For the single-machine case, the authors showed that the scheduling problems to minimize makespan, total completion time and sum of the lth (l > 0) power of completion times can be solved in polynomial time. In addition, they also showed that the problems of minimizing the total weighted completion time, the maximum lateness, the maximum tardiness and the total tardiness are polynomially solvable under certain agreeable conditions. In this note, some counter-examples are given to show the ⇑ Corresponding author. E-mail addresses: [email protected] (L. Wang), [email protected] (J.-B. Wang). http://dx.doi.org/10.1016/j.ins.2015.02.021 0020-0255/Ó 2015 Elsevier Inc. All rights reserved.

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L. Wang et al. / Information Sciences 307 (2015) 89–94

incorrectness of the main results in Lee and Lai [7]. Besides, a revised model of Lee and Lai [7] is also provided. We show that some single machine scheduling problems are still polynomially solvable under the revised model. We shall follow the notation and terminology given in Lee and Lai [7]. There are n jobs J 1 ; J 2 ; . . . ; J n given to be processed on a single-machine. Each job J j has a due-date dj and a weight wj . All jobs are non-preemptive and are ready for processing at time 0. The machine can handle at most one job at a time and cannot stand idle until the last job assigned to it has finished processing. The actual processing time of job J i ði ¼ 1; 2; . . . ; nÞ if scheduled in position r is A pj½r

! r1 X ¼ pj f bk p½k ; r ;

r; j ¼ 1; 2; . . . ; n;

ð1Þ

k¼1

where pj is the normal processing time of job J j , p½k is the normal processing time of a job if it is scheduled in the kth position in a sequence, and bi > 0 for all i. It is assumed that f is a general function satisfying the following conditions: (1) f : ½0; 1Þ  ½1; 1Þ ! ð0; 1Þ is a differentiable non-decreasing function with respect to the first variable x, non-increasing with @ f ðx; y0 Þ is non-decreasing with respect to x for every fixed y0 and f ð0; 1Þ ¼ 1. respect to the second variable y; (2) f x ðx; y0 Þ ¼ @x The first condition implies that the actual job processing times might be prolonged due to the deterioration effect which depends on the processing times of the jobs that are already processed. The second condition implies that the actual job processing times might be shortened due to the learning effect which depends on its scheduled position. For a given schedule p, let C j ¼ C j ðpÞ denote the completion time of job J j . The objective functions to be minimized are P makespan C max ¼ maxfC j jj ¼ 1; 2; . . . ; ng, total completion time C j , total completion time and sum of the lth (l > 0) power P l P of completion times C j , total weighted completion time wj C j , maximum lateness Lmax ¼ maxfLj jj ¼ 1; 2; . . . ; ng, where P Tj. Lj ¼ C j  dj , maximum tardiness T max ¼ maxfT j jj ¼ 1; 2; . . . ; ng, where T j ¼ maxfC j  dj ; 0g, and total tardiness Lee and Lai [7] arrived at the following results in the single-machine scheduling problems: 0

A Theorem 1 (Theorem 1, [7]). For the problem 1jpj½r ¼ pj f

P r1

k¼1 bk p½k ; r

 jC max , the optimal schedule is obtained by the shortest

processing time (SPT) rule (i.e., by sequencing jobs in non-decreasing order of pj ).

0

A Theorem 2 (Theorem 2, [7]). For the problem 1jpj½r ¼ pj f

weighted shortest processing time (WSPT) rule if   p i:e:; by sequencing jobs in non-decreasing order of wj .

P

r1 k¼1 bk p½k ; r

the

 P j wi C i , the optimal schedule is obtained by the

processing

times

and

the

weights

are

agreeable

j

0

A Theorem 3 (Theorem 3, [7]). For the problem 1jpj½r ¼ pj f

P

r1 k¼1 bk p½k ; r



jLmax , the optimal schedule is obtained by the earliest

due date (EDD) rule if the job processing times and the due dates are agreeable (i.e., by sequencing jobs in non-decreasing order of dj ). 0

A Theorem 4 (Theorem 4, [7]). For the 1jpj½r ¼ pj f

P r1

k¼1 bk p½k ; r

 P j T i problem, the optimal schedule is obtained by the EDD rule

if the job processing times and the due dates are agreeable.

2. Counter-examples In the following, we will show that Theorems 10 –40 are not correct by a counter-example. Counter-Example 1. Let f

P r1

k¼1 bk p½k ; r



 a P ¼ 1 þ r1 , where a P 1. n ¼ 3; p1 ¼ 1; p2 ¼ 2; p3 ¼ 200; w1 ¼ 3; w2 ¼ 2; k¼1 bk p½k

w3 ¼ 1; d1 ¼ 4; d2 ¼ 10; d3 ¼ 200; b1 ¼ 2; b2 ¼ 3, and a ¼ 2. 0

According to the result of Theorem 1 , we can obtain the schedule ½J 1 ; J 2 ; J 3 , then we have

C 1 ¼ 1; C 2 ¼ 1 þ 2  ð1 þ 2  1Þ2 ¼ 19; C 3 ¼ 1 þ 2  ð1 þ 2  1Þ2 þ 200  ð1 þ 2  1 þ 3  2Þ2 ¼ 16219: Hence, C max ¼ C 3 ¼ 16219. However, if the schedule is ½J 2 ; J 1 ; J 3 , then we have

C 2 ¼ 2; C 1 ¼ 2 þ 1  ð1 þ 2  2Þ2 ¼ 27; C 3 ¼ 2 þ 1  ð1 þ 2  2Þ2 þ 200  ð1 þ 2  2 þ 3  1Þ2 ¼ 12827: Hence, C max ¼ C 3 ¼ 12827.

L. Wang et al. / Information Sciences 307 (2015) 89–94

P

r1 k¼1 bk p½k ; r

A Obviously, for the problem 1jpj½r ¼ pj f

91

 jC max , the SPT sequence is not the optimal schedule. Hence,

0

Theorem 1 is not correct. P 0 wi C i ¼ 3  1 þ 2  19þ Similarly, according to the result of Theorem 2 , we can obtain the schedule ½J 1 ; J 2 ; J 3 , and P 0 1  16219 ¼ 16260. However, if the schedule is ½J 2 ; J 1 ; J 3 , wi C i ¼ 2  2 þ 3  27 þ 1  12827 ¼ 12912. Hence, Theorem 2 is not correct. 0 According to the result of Theorem 3 , we can obtain the schedule ½J 1 ; J 2 ; J 3 , and Lmax ¼ maxf1  4; 19  10; 16219 0 200g ¼ 16019. However, if the schedule is ½J 2 ; J 1 ; J 3 , Lmax ¼ maxf2  10; 27  1; 12827  200g ¼ 12627. Hence, Theorem 3 is not correct. P 0 According to the result of Theorem 4 , we can obtain the schedule ½J 1 ; J 2 ; J 3 , and T i ¼ maxf1  4; 0g þ maxf19  10; 0gþ P maxf16219  200; 0g ¼ 16028. However, if the schedule is ½J 2 ; J 1 ; J 3 , T i ¼ maxf2  10; 0g þ maxf27  4; 0gþ 0 maxf12827  200; 0g ¼ 12650. Hence, Theorem 4 is not correct. 3. A revised form of job processing times In this section, we still use the same notations as in the paper Lee and Lai [7], i.e., the model (1) with b1 P b2 P    P bn > 0. The difference between Lee and Lai [7] and this paper is that Lee and Lai [7] considered the model P  P  r1 r1 A A pj½r ¼ pj f k¼1 bk p½k ; r with bi > 0, and we consider the model pj½r ¼ pj f k¼1 bk p½k ; r with b1 P b2 P    P bn > 0. A Theorem 1. For the problem 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r



jC max , the optimal schedule is obtained by the shortest processing time

(SPT) rule. Proof. By the standard job interchange method, let S ¼ ½p1 ; J j ; J k ; p2 ; S0 ¼ ½p1 ; J k ; J j ; p2 , where p1 and p2 are partial sequences, and p1 or p2 may be empty. Furthermore, we assume that there are r  1 jobs in p1 . In addition, let A denote the completion time of the last job in p1 . Suppose pj 6 pk . To show S dominates S0 , it suffices to show that the ðr þ 1Þth jobs in S and S0 satisfy the condition that (i) C k ðSÞ 6 C j ðS0 Þ and (ii) C u ðSÞ 6 C u ðS0 Þ for any J u in p2 . From the proof of Theorem 1 in Lee and Lai [7], we have

C j ðS0 Þ  C k ðSÞ P 0:

ð2Þ

This completes the proof of part (i). Now let J h denote the first job schedule in

C h ðSÞ ¼ C k ðSÞ þ ph f

p2 . Then we have

r1 X

!

bl p½l þ br pj þ brþ1 pk ; r þ 1

ð3Þ

! r1 X bl p½l þ br pk þ brþ1 pj ; r þ 1 :

ð4Þ

l¼1

and

C h ðS0 Þ ¼ C j ðS0 Þ þ ph f

l¼1

From pj 6 pk and br P brþ1 , we have

br pj þ brþ1 pk  br pk  brþ1 pj ¼ ðbr  brþ1 Þðpj  pk Þ  0:

ð5Þ

From (2)–(5), we have C h ðS0 Þ P C h ðSÞ since f is a non-decreasing function with respect to the first variable x, i.e., we have showed that the first job J h in p2 , which starts earlier in S than S0 , completes earlier in S. Similarly, we have C u ðS0 Þ P C u ðSÞ for any J u in p2 . A ¼ Hence, repeating this argument will lead to the optimality of the SPT-sequence for the problem 1jpj½r P  r1 pj f l¼1 bl p½l ; r jC max . h Remark 1. The incorrectness of Theorems 1–4 in [7] is that they did not consider the case (ii) C u ðSÞ 6 C u ðS0 Þ for any J u in p2 (proof of case (ii) using the condition br P brþ1 , see Theorem 1 in Lee and Lai [7] and this paper). A Theorem 2. For the problem 1jpj½r ¼ pj f

P

r1 l¼1 bl p½l ; r

(SPT) rule. Proof. Similarly to the proof of Theorem 1. h

 P j C j , the optimal schedule is obtained by the shortest processing time

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L. Wang et al. / Information Sciences 307 (2015) 89–94

A Theorem 3. For the problem 1jpj½r ¼ pj f

P

r1 l¼1 bl p½l ; r

 P j C lj , the optimal schedule is obtained by the shortest processing time

(SPT) rule. Proof. Similarly to the proof of Theorem 1. h

A Theorem 4. For the problem 1jpj½r ¼ pj f

P

r1 l¼1 bl p½l ; r

 P j wj C j , the optimal schedule is obtained by the weighted shortest process-

ing time (WSPT) rule if the processing times and the weights are agreeable. Proof. Similarly to the proof of Theorem 1 and the proof of Theorem 2 in Lee and Lai [7]. h If the normal processing times of all jobs are equal, i.e., pj ¼ p for 1 6 j 6 n, then we have the following corollary. A Corollary 1. For the problem 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r

 P ; pj ¼ pj wj C j , the optimal schedule can be obtained by sequencing the

jobs in non-increasing order of wj . If the weights of all jobs are equal, i.e., wj ¼ w for 1 6 j 6 n, then we have the following corollary. A Corollary 2. For the problem 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r

 P ; wj ¼ wj wj C j , the optimal schedule can be obtained by sequencing the

jobs in non-decreasing order of pj (the SPT rule). Let h be a positive real number. If wj pj ¼ h, then jobs have reversely agreeable weights. Hence, we have the following corollary. A Corollary 3. For the problem 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r

 P ; wj pj ¼ hj wj C j , the optimal schedule can be obtained by sequencing the

jobs in non-decreasing order of pj (the SPT rule).

A Theorem 5. For the problems 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r



A jLmax and 1jpj½r ¼ pj f

P

r1 l¼1 bl p½l ; r



jT max , the optimal schedule is obtained

by the EDD rule if the job processing times and the due dates are agreeable. Proof. Similarly to the proof of Theorem 1 and the proof of Theorem 3 in Lee and Lai [7]. h By using the similar method of Theorem 5, the following corollaries can be easily obtained. A Corollary 4. For the problems 1jpj½r ¼ pj f

P

r1 l¼1 bl p½l ; r

 P  r1 A ; pj ¼ pjLmax and 1jpj½r ¼ pj f l¼1 bl p½l ; r ; pj ¼ pjT max , an optimal

schedule can be obtained by sequencing the jobs in non-decreasing order of dj (the EDD rule).

A Corollary 5. For the problems 1jpj½r ¼ pj f

P

r1 l¼1 bl p½l ; r



A ; dj ¼ djLmax and 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r



; dj ¼ djT max , an optimal sched-

ule can be obtained by sequencing the jobs in non-decreasing order of pj (the SPT rule).

A Corollary 6. For the problems 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r



A ; dj ¼ kpj jLmax and 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r



; dj ¼ kpj jT max , the optimal

schedule can be obtained by sequencing the jobs in non-decreasing order of dj (the EDD rule).

A Theorem 6. For the problem 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r

 P j Lj , the optimal schedule can be obtained by sequencing the jobs in non-

decreasing order of pj , (the SPT rule). P P P P P P Proof. For any given schedule, the total lateness is Lj ¼ nj¼1 ðC j  dj Þ ¼ nj¼1 C j  nj¼1 dj . Since nj¼1 dj is a constant, Lj is Pn Pn minimized if j¼1 C j is minimized. By Corollary 1, j¼1 C j is minimized by the SPT rule and so the SPT rule leads to an optimal P  P r1 A schedule for the problem 1jpj½r ¼ pj f Lj . h l¼1 bl p½l ; r ; spsd j P  P P r1 A The single-machine total tardiness problem 1jj T j is NP-hard. Hence the problem 1jpj½r ¼ pj f T j must be l¼1 bl p½l ; r j NP-hard, too. But it is still optimal if the job processing times and the due dates are agreeable.

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L. Wang et al. / Information Sciences 307 (2015) 89–94 A Theorem 7. For the problem 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r

 P j T j , the optimal schedule is obtained by the EDD rule if the job

processing times and the due dates are agreeable.

Proof. It is similar to the proof of Theorem 1 and the proof of Theorem 4 in Lee and Lai [7]. h Using the similar method of Theorem 7, the following corollaries can be easily obtained. A Corollary 7. For the problem 1jpj½r ¼ pj f

P r1

l¼1 bl p½l ; r

 P ; pj ¼ pj T j , the optimal schedule can be obtained by sequencing the jobs

in non-decreasing order of dj (the EDD rule).

A Corollary 8. For the problem 1jpj½r ¼ pj f

P

r1 l¼1 bl p½l ; r

 P ; dj ¼ kpj j T j , the optimal schedule can be obtained by sequencing the

jobs in non-decreasing order of dj (the EDD rule). Remark 2. Obviously, the complexity of obtaining the SPT order (the WSPT order, and the EDD order) is Oðn log nÞ. Hence the problems of Theorems 1–7 are polynomially solvable scheduling problems. In the following, we will demonstrate the results of the theorems (i.e., Theorems 1–7) by the following example: Example

1. Let

f

P

r1 l¼1 bl p½l ; r



 a P ¼ 1 þ r1 rb , l¼1 bl p½l

where

a P 1; b  0.

n ¼ 4; p1 ¼ 1; p2 ¼ 2; p3 ¼ 10; p4 ¼ 12;

b1 ¼ 0:4; b2 ¼ 0:3; b3 ¼ 0:2; b4 ¼ 0:1; w1 ¼ 5; w2 ¼ 4; w3 ¼ 3; w4 ¼ 2; d1 ¼ 15; d2 ¼ 14; d3 ¼ 13; d4 ¼ 12, a ¼ 1:1 and b ¼ 0:2. According to Theorems 1–7, we know that the optimal schedule is ½J 1 ; J 2 ; J 3 ; J 4  for the following objective functions: P P P P C j ; C 2j , wj C j ; Lmax , T max and T j . In addition, the completion times are calculated as follows:

C max ;

C 1 ¼ 1; C 2 ¼ 1 þ 2  ð1 þ 0:4  1Þ1:1  20:2 ¼ 3:5210; C 3 ¼ 3:5210 þ 10  ð1 þ 0:4  1 þ 0:3  2Þ1:1  30:2 ¼ 20:7281; C 4 ¼ 20:7281 þ 12  ð1 þ 0:4  1 þ 0:3  2 þ 0:2  10Þ1:1  30:2 ¼ 62:5145; C max ¼ C 4 ¼ 62:5145; X X X

C j ¼ 1 þ 3:5210 þ 20:7281 þ 62:5145 ¼ 87:7636; C 2j ¼ 12 þ 3:52102 þ 20:72812 þ 62:51452 ¼ 4351:114; wj C j ¼ 5  1 þ 4  3:5210 þ 3  20:7281 þ 2  62:5145 ¼ 206:2973;

Lmax ¼ maxf1  15; 3:5210  14; 20:7281  13; 62:5145  12g ¼ 50:5145; T max ¼ maxfmaxf1  15; 0g; maxf3:5210  14; 0g; maxf20:7281  13; 0g; maxf62:5145  12; 0gg ¼ 50:5145; X X

Lj ¼ 1  15 þ 3:5210  14 þ 20:7281  13 þ 62:5145  12 ¼ 33:7636; T j ¼ maxf1  15; 0g þ maxf3:5210  14; 0g þ maxf20:7281  13; 0g þ maxf62:5145  12; 0g ¼ 58:2426:

4. Conclusion The main contribution of this note is to notice that results of [7] will hold if coefficients b1 ; b2 ; . . . ; bn are ordered as b1 P b2 P    P bn . Such more general effects of deterioration and learning model happen in many realistic situations. Hence, we gave a revised model with the general effects of deterioration and learning. We showed that some single machine scheduling problems are still polynomially solvable under the revised model. Further research may focus on other objective functions, consider multi machines scheduling or propose more general and practical models (for example, abrupt (delta function) deterioration events, such as an accident and/or unforeseen machine breakdown).

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Acknowledgements The authors are grateful to the anonymous referees for their helpful comments on earlier version of the article. The work described in this paper was partially supported by a grant from The Hong Kong Polytechnic University (Project 4-BCBJ), the National Natural Science Foundation of China (Grant Nos. 71471120 and 61104058), the China Postdoctoral Science Foundation (2014M560972) and Institutions of Higher Learning in Liaoning Province Outstanding Young Scholars Growth Plans (Grant No. LJQ2014039). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]

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