Machine scheduling with deteriorating jobs and DeJong’s learning effect

Computers & Industrial Engineering 91 (2016) 42–47

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Machine scheduling with deteriorating jobs and DeJong’s learning effect Min Ji a,⇑, Xiaoying Tang a, Xin Zhang a, T.C.E. Cheng b a b

School of Computer Science and Information Engineering, Contemporary Business and Trade Research Center, Zhejiang Gongshang University, Hangzhou 310018, PR China Department of Logistics and Maritime Studies, The Hong Kong Polytechnic University, Kowloon, Hong Kong

a r t i c l e

i n f o

Article history: Received 22 March 2015 Received in revised form 18 October 2015 Accepted 23 October 2015 Available online 28 October 2015

a b s t r a c t We consider parallel-machine scheduling with deteriorating jobs and DeJong’s learning effect. We focus on the problems to minimize the total completion time and the makespan. We show that the former is polynomially solvable, while the latter is NP-hard, for which we provide a fully polynomial-time approximation scheme. Ó 2015 Elsevier Ltd. All rights reserved.

Keywords: DeJong’s learning effect Job deterioration Scheduling FPTAS

1. Introduction Scheduling with deteriorating jobs was first considered by Gupta and Gupta (1988) and Brown and Yechiali (1990) independently. Later, Mosheiov (1994) studied a special case proposed by Brown and Yechiali (1990) (referred to as simple linear deterioration) and gave polynomial-time algorithms for several classical scheduling criteria. Cheng and Ding (1998) derived some relationships between scheduling problems involving time-dependent jobs with increasing/decreasing linear processing times. General results on the relations between scheduling problems involving jobs with time-dependent processing times can be found in Gawiejnowicz, Kurc, and Pankowska (2009a, 2009b), Gawiejnowicz and Kononov (2014), and Rustogi and Strusevich (2012, 2014). Focusing on scheduling problems involving jobs with time-dependent processing times, Gawiejnowicz (2008) gave a detailed survey of such scheduling problems and discussed advanced topics such as application of matrix methods; for the latest results on such scheduling problems, the reader may refer to Yin, Cheng, and Wu (2014, 2015). Furthermore, recent papers that consider scheduling with deteriorating jobs include Ji and Cheng (2009, 2010), Lee, Wu, and Liu (2009), Lee, Wang, Shiau, and Wu (2010), Gawiejnowicz and Kononov (2010), Gawiejnowicz and Lin (2010), Wang, Wang, and Ji (2011), Li, Ng, Cheng, and Yuan (2011), Ji, Hsu, and Yang (2013), and Yin, Wu, Cheng, and Wu

⇑ Corresponding author. Tel./fax: +86 571 28008303. E-mail addresses: [email protected] (M. Ji), [email protected] (T.C.E. Cheng). http://dx.doi.org/10.1016/j.cie.2015.10.015 0360-8352/Ó 2015 Elsevier Ltd. All rights reserved.

(2015), among others. There are many applications of scheduling models where the processing time of a job is an increasing function of its start time. These include repayment of multiple loans, derusting of operations, control of queues in communication systems, etc. For a list of applications of scheduling models with deteriorating jobs, the reader may refer to Gawiejnowicz (2008). On the other hand, it is necessary and reasonable to consider the learning effect in scheduling. Motivated by observations in the aircraft industry, Wright (1936) initiated research on the learning effect in manufacturing. Gawiejnowicz (1996) first considered learning in scheduling research, which was later popularized by Dondeti and Mohanty (1998), Biskup (1999), and Cheng and Wang (2000). Subsequently, many scheduling researchers have devoted a great deal of effort to this stream of research and proposed a large variety of position-based learning effect models (see, e.g., Agnetis, Billaut, Gawiejnowicz, Pacciarelli, & Soukhal, 2014; Cheng, Kuo, & Yang, 2013; Low & Lin, 2011; Lu, Wei, & Wang, 2012; Wang & Wang, 2013; Yin, Xu, & Wang, 2010). Biskup (2008) and Agnetis et al. (2014) provided comprehensive reviews of this line of research. Extending the work to the multiagent scheduling context, Agnetis et al. (2014) provided details on multi-agent scheduling research. Almost all of these models considering the position-dependent learning effect suffer a common drawback that when a job’s position is large enough in a schedule, its actual processing time is close to zero. Following Wright’s (1936) notion of learning, DeJong (1957) proposed a new learning model T s ¼ T 1 ðM þ ð1  MÞ=sm Þ, where T s is the time required for the sth cycle of the batch, T 1 is the time required for the first cycle of a batch, s stands for the sth cycle, M represents

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M. Ji et al. / Computers & Industrial Engineering 91 (2016) 42–47

the ‘‘factor of incompressibility” ð0 6 M 6 1Þ, and m is the exponent of reduction ð0 < m < 1Þ. It is clear that when s increases, the time required for the sth cycle will naturally fall and approach a certain limit, i.e., MT 1 . So DeJong’s learning model overcomes the drawback associated with the log-linear learning model. Subsequent studies have empirically validated DeJong’s learning model, e.g., Yelle (1979), Badiru (1992), Okolowski and Gawiejnowicz (2010), Ji, Yao, Yang, and Cheng (2014), etc. Scheduling problems considering the effects of learning and job deterioration at the same time has been extensively studied in the literature (see, e.g., Huang, Wang, & Ji, 2014; Kuo & Yang, 2011; Toksari & Güner, 2010; Wang, 2009a, 2009b; Wang, Huang, Wang, Yin, & Wang, 2009; Wang & Wang, 2014; Wu, Wang, & Wang, 2011; Yang & Kuo, 2010). However, to the best of our knowledge, most studies assume that the position-dependent learning curve approaches 0 when a job’s position is large enough in a schedule, which is unrealistic because this implies no further improvement can be made after some amount of production in manufacturing. DeJong’s learning curve can overcome this drawback, but it has seldom been combined with deteriorating jobs despite its relevance in practice. The exception is Wang (2009a), who considered two models that combine DeJong’s learning effect with deteriorating jobs, i.e., pjr ¼ pj aðtÞðM þ ð1  MÞr a Þ and pjr ¼ pj ðaðtÞ þ M þ ð1  MÞra Þ, where 0 6 M 6 1 is an incompressible factor. He provided polynomial-time algorithms to solve some single-machine problems. The phenomenon of learning and job deterioration occurring simultaneously can be found in many real-life situations. For example, as manufacturing becomes increasingly competitive, in order to provide customers with greater product variety, organizations are moving towards shorter production runs and frequent product changes, which give rise to the phenomenon of learning and deterioration in performing operational activities. Considering both the learning and forgetting effects in measuring productivity should be helpful in improving the accuracy of production planning and productivity estimation (see, e.g., Nembhard & Osothsilp, 2002). In this paper we combine DeJong’s learning effect with deteriorating jobs for scheduling as follows: pj ¼ pj ½M þ ð1  MÞr a  þ at ð0 6 M 6 1Þ, where a is the non-positive learning index and a is the non-negative job deterioration rate. Note that if M ¼ 0, the model reduces to the model pjr ¼ pj ra þ at considered by Yang and Kuo (2010). The rest of the paper is organized as follows: In Section 2 we provide the problem description. In Section 3 we present some useful preliminary results. In Section 4 we provide a fully polynomial-time approximation scheme (FPTAS) to minimize the makespan, and propose polynomial-time algorithm to minimize the total completion time in the parallel-machine setting. In Section 5 we conclude the paper and suggest topics for future research.

a P 0 is the common job deterioration rate, and t is the starting time of job J j in the schedule. The objectives are to minimize the makespan C max ¼ Pn maxj¼1;2;...;n C j and the total completion time j¼1 C j , where C j denotes the completion time of job J j in a given schedule. Using the three-field notation of Graham, Lawler, Lenstra, and Rinnooy Kan (1979) for describing scheduling problems, we denote the j ½M þ ð1  MÞra  þ atjC max and problems under study as P m jpj ¼ p P a j ½M þ ð1  MÞr  þ atj C j . Pm jpj ¼ p 3. Preliminary results In this section we give some useful preliminary results for solving the scheduling problems under consideration. Given m parallel machines, we assume that the number of jobs allocated to machine i is ni ði ¼ 1; 2; . . . ; mÞ. Then the allocation of n jobs to m machines can be expressed as Pðn; mÞ ¼ ðn1 ; n2 ; . . . ; nm Þ with Pm i¼1 ni ¼ n. Let ½ij denote the job that occupies the jth position ½ij denote the on machine i in a given schedule, and p½ij and p actual processing time and the normal processing time of the job, respectively. Lemma 1. The completion time of the job scheduled in the jth position on machine i, i.e., C ½ij , for i ¼ 1; 2; . . . ; m; j ¼ 1; 2; . . . ; ni , is equal to Pj  Q , where Q jk ¼ ð1 þ aÞjk ½M þ ð1  MÞka  for k ¼ 1; p k¼1 ½ik jk 2; . . . ; j. ½i1 ½Mþ Proof (By mathematical induction). When j ¼ 1; C ½i1 ¼ p ð1  MÞ1a , so the proposition is true for j ¼ 1. Assume that the P ½ik Q lk , where Q lk ¼ proposition is true for j ¼ l, i.e., C ½il ¼ lk¼1 p a

ð1 þ aÞlk ½M þ ð1  MÞk  for k ¼ 1; 2; . . . ; l. When j ¼ l þ 1,

½i;lþ1 ½M þ ð1  MÞðl þ 1Þ  þ aC ½il C ½i;lþ1 ¼C ½il þ p a

¼

l X ½ik ð1 þ aÞlþ1k ½M þ ð1  MÞka  p k¼1

½i;lþ1 ð1 þ aÞlþ1ðlþ1Þ ½M þ ð1  MÞðl þ 1Þ  þp a

¼

lþ1 lþ1 X X ½ik ð1 þ aÞlþ1k ½M þ ð1  MÞka  ¼ ½ik Q lþ1;k ; p p k¼1

k¼1 lþ1k

a

where Q lþ1;k ¼ ð1 þ aÞ ½M þ ð1  MÞk  for k ¼ 1; 2; . . . ; l þ 1. Hence, the proposition is also true for j ¼ l þ 1. By the principle of mathematical induction, the lemma is established. h Lemma 2. If jza1  za2 j 6 d minfza1 ; za2 g, for 0 < d 6 1; a 6 0, and are positive integers, then jðz1 þ 1Þa  ðz2 þ 1Þa j z1 ; z2 a a 6 d minfðz1 þ 1Þ ; ðz2 þ 1Þ g.

2. Notation and problem formulation We describe the scheduling problem as follows. There is a set of n independent jobs fJ 1 ; J 2 ; . . . ; J n g waiting to be processed on m parallel machines. Each machine can handle only one job at a time and the processing of a job cannot be interrupted. All the jobs are available for processing at time zero. In this paper we study a scheduling model that considers both learning and job deterioration simultaneously as follows:

j ½M þ ð1  MÞr a  þ at; pj ¼ p

ð1Þ

j is the where, given a schedule, pj is the actual processing time and p normal processing time of job Jj ; 0 6 M 6 1 is the incompressible factor, r is the position of Jj ; a 6 0 is the common learning index,

Proof. Without loss of generality, we assume z1 6 z2 . Noting that za1  za2 6 dza2 , we have

 a  a z1 þ 1 z1 6 6 ð1 þ dÞða 6 0Þ: z2 þ 1 z2

ð2Þ

i.e.,

ðz1 þ 1Þa  ðz2 þ 1Þa 6 dðz2 þ 1Þa :

ð3Þ

Similarly, if z1 > z2 , we have

ðz2 þ 1Þa  ðz1 þ 1Þa 6 dðz1 þ 1Þa :

ð4Þ a

a

From (3) and (4), we conclude that jðz1 þ 1Þ  ðz2 þ 1Þ j 6 d minfðz1 þ 1Þa ; ðz2 þ 1Þa g. h

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M. Ji et al. / Computers & Industrial Engineering 91 (2016) 42–47

We introduce the main Properties 1 and 2 associated with the Partition procedure provided by Kovalyov and Kubiak (1998, 1999). For details on the Partition (A; e; d) procedure and the notation we use below, we refer the reader to Kovalyov and Kubiak (1998, 1999). Property 1. jeðxÞ  eðx0 Þj 6 d minfeðxÞ; eðx0 Þg for any x; x0 2 Aek ; k ¼ 1; 2; . . . ; ke .

r i0 ðxÞ ¼ 0; x r j j ðxÞ

¼

i ¼ 1; 2; . . . ; m;

xj r j1 ðxÞ

þ 1;

r ij ðxÞ ¼ r ij1 ðxÞ; Ri0 ðxÞ

¼ 0;

x

for i – xj ;

i ¼ 1; 2; . . . ; m; a

x

Rij ðxÞ ¼ Rij1 ðxÞ; jAj

jAj

Property 2. ke 6 log eðx Þ=d þ 2 for 0 < d 6 1 and 1 6 eðx Þ.  3. ke 6 C log e xði1 þ1Þ dþ2 ð Þ

i f 0 ðxÞ



1

Property 0 < eðx

i1 þ1

Þ 6    6 ðx

Proof. Suppose

ðjAjÞ

0
for

and

that

ðike 1 þ1Þ

x

; x

ðike 1 þ2Þ

ðjAjÞ

; ... ; x

2

Aeke

and

eðxðike 1 þ1Þ Þ 6 eðxðike 1 þ2Þ Þ 6    6 eðxðjAjÞ Þ. Then for ke > 2, we have 1 1 1 1 : 16 < <  < 6 k 2 eðxðjAjÞ Þ eðxðike 1 þ1Þ Þ ð1 þ dÞeðxðike2 þ1Þ Þ ð1 þ dÞ e eðxði1 þ1Þ Þ

Thus, from (5), we have



ðke  2Þd 6 ðke  2Þd þ log

1 eðxðjAjÞ Þ

! 6 C log

For ke 6 2, we have ke 6 C log



¼ 0; ¼

i

i ¼ 1; 2; . . . ; m;

xj f j1 ðxÞ

x

x

j j ½M þ ð1  MÞRj j ðxÞ þ af j1 þp ðxÞ;

i

for i – xj :

Thus, given a fixed number of machines m, problem Pm jpj ¼ pj ½M þ ð1  MÞra  þ atjC max reduces to the following problem:

 min

 i max f n ðxÞjx 2 X :

i¼1;2;...;m

Based on the above analysis, given a fixed number of machines m, we provide a formal description of the FPTAS, namely Algorithm Am e , as follows:



1 eðxðjAjÞ Þ



where C ¼ maxf1= log 2; 1g.

ð5Þ



6 Cðke  2Þ logð1 þ dÞ þ C log ð1 þ dÞke 2 ¼ C log eðxðjAjÞ Þ

x f j j ðxÞ

for i – xj ;

f j ðxÞ ¼ f j1 ðxÞ;

Þ 6 1, where C ¼ maxf1= log 2; 1g.

a

x

j ðxÞ þ 1Þ ; Rj j ðxÞ , ðr j j ðxÞÞ ¼ ðrj1



1

eðxði1 þ1Þ Þ

1 eðxði1 þ1Þ Þ

 ;

. d þ 2.

This completes the proof of the property. h

Algorithm Am e Step 1. (Initialization) Number the jobs in non-decreasing order of j (Property 4). Set Y 0 ¼ fð0; 0; . . . ; 0Þg and j ¼ 1. p Step 2. (Generation of Y 1 ; Y 2 ; . . . ; Y n ) For set Y j1 , generate Y 0j by adding xj ; xj ¼ 1; 2; . . . ; m, in position j of each vector from Y j1 . Calculate the following for any x 2 Y 0j . x

x

j r j j ðxÞ ¼ rj1 ðxÞ þ 1;

r ij ðxÞ ¼ rij1 ðxÞ; 4. Parallel-machine scheduling problems In this section we study two problems under our scheduling model to minimize the makespan and the total completion time, respectively.

x Rj j ðxÞ

¼

for i – xj ;

a x ðr j j ðxÞÞ ;

Rij ðxÞ ¼ Rij1 ðxÞ; x xj f j j ðxÞ ¼ f j1 ðxÞ þ i

i

f j ðxÞ ¼ f j1 ðxÞ;

for i – xj ; x

x

j j ½M þ ð1  MÞRj j ðxÞ þ af j1 p ðxÞ;

for i – xj :

j ½M þ ð1  MÞr a  þ atjC max 4.1. Problem P m jpj ¼ p The makespan minimization problem under study is NP-hard trivially as it is a special case of the NP-hard problem Pm kC max when a ¼ 0 and a ¼ 0 or M ¼ 1. Below we give a fully polynomial-time approximation scheme (FPTAS) for it. j ½M þ ð1  MÞr a þ For the single-machine problem 1jpj ¼ p atjC max , it is not hard to verify that there exists an optimal solution in which the jobs are sequenced in the shortest processing time (SPT) order of their normal processing times, which leads to the following property.

If j ¼ n, then set Y n ¼ Y 0n and go to Step 3. If j < n, then set d ¼ e=ð2ðn þ 1ÞÞ and perform the following computation.

Property 4. There exists an optimal solution for problem Pm jpj ¼ pj ½M þ ð1  MÞra  þ atjC max in which the jobs on each machine are sequenced in the SPT order of their normal processing times.

Divide

By Property 4, we re-index the jobs in non-decreasing order of their normal processing times, if necessary. We then introduce the variable xj to denote the machine on which job J j is processed. Let X be the set of all the vectors x ¼ ðx1 ; x2 ; . . . ; xn Þ with xj ¼ 1; 2; . . . ; m and j ¼ 1; 2; . . . ; n. We define the following initial and recursive functions on X:

Call Partition ðY 0j ; Rij ; dÞ ði ¼ 1; 2; . . . ; mÞ to partition set Y 0j into Ri

Ri

Ri

disjoint subsets Y 1j ; Y 2j ; . . . ; Y k j i . R j

Call Partition

i ðY 0j ; f j ; dÞ

joint subsets

fi Y 1j ;

set

Y a1 ...am b1 ...bm ¼

Y 0j f1 Y aj1

ði ¼ 1; 2; . . . ; mÞ to partition set Y 0j into dis-

fi Y 2j ;

fi

. . . ; Y kj i . f

into

\

f2 Y aj2

j

the

\  \

fm Y ajm

following R1j

R2j

disjoint

subsets

Rm j

\ Y b1 \ Y b2 \    \ Y bm ,

a1 ¼ 1; 2;. .. ;kf 1 ; a2 ¼ 1; 2;. .. ;kf 2 ; .. .; am ¼ 1;2; .. .; kf mj ; j

j

b1 ¼ 1;2; .. .; kR1 ; b2 ¼ 1; 2;. .. ;kR2 ; .. .; bm ¼ 1; 2;. .. ;kRmj ; j

j

For each non-empty subset Y a1 ...am b1 ...bm , choose a vector xða1 ...am b1 ...bm Þ

such

i fmaxi¼1;2;...;m ðf j ðxÞjx

that

max

2 Y a1 ...am b1 ...bm g.

i ða1 ...am b1 ...bm Þ Þ i¼1;2;...;m f j ðx

¼ min

45

M. Ji et al. / Computers & Industrial Engineering 91 (2016) 42–47

Set

n h i o k k k ða1 ...am b1 ...bm Þ  k k x f jþ1 ðx Þ  f jþ1 e ¼ f j ðx Þ þ pjþ1 M þ ð1  MÞRjþ1 ðx Þ þ af j ðx Þ

n Y j ¼ xða1 ...am b1 ...bm Þ ja1 ¼ 1;2;. .. ;kf 1 ; .. . ; am ¼ 1;2; .. .; kf mj ;

n

h 

ða ...am b ...bm Þ i o k k

1  f j xða1 ...am b1 ...bm Þ þ pjþ1 M þ ð1  MÞRkjþ1 e þ af j xða1 ...am b1 ...bm Þ x 1

j

b1 ¼ 1;2; .. .; kR1 ; .. . ; bm ¼ 1;2;. .. ;kRmj ; and Y a1 ...am b1 ...bm j  R1 R2 Rm f 1j f 2j fm ¼ Y a1 \ Y a2 \   \ Y ajm \ Y b1j \ Y b2j \   \ Y bmj – Ø and j ¼ j þ 1:



ða ...am b ...bm Þ  k k

1 6 ð1 þ aÞ f j ðx Þ  f j xða1 ...am b1 ...bm Þ þ ð1  MÞpjþ1 Rkjþ1 ðx Þ  Rkjþ1 e x 1 h i k k 6 d1 ð1 þ aÞf j ðx Þ þ ð1  MÞpjþ1 Rkjþ1 ðx Þ 6 d1 f jþ1 ðx Þ:

Repeat Step 2. For i – k, we have

i

Step 3. (Solution) Set Q ðxÞ ¼ maxi¼1;2;...m f n ðxÞ, x 2 Y n . Select vector x0 2 Y n such that Q ðx0 Þ ¼ minfQ ðxÞg. Let x ¼ ðx1 ; x2 ; . . . ; xn Þ be an optimal solution for problem j ½M þ ð1  MÞr a  þ atjC max and L ¼ logðmaxfn; 1=e; p max ; P m jpj ¼ p ð1 þ aÞgÞ, where pmax ¼ maxj¼1;2;...;n pj . We have the following result. Theorem 1. Given a fixed number of machines m, Algorithm Am e finds x0 2 X for problem Pm jpj ¼ pj ½M þ ð1  MÞra  þ atjC max such that Q ðx0 Þ 6 ð1 þ eÞQ ðx Þ in Oðn3mþ1 L2mþ1 =e2m Þ.

 i i i ða1 ...am b1 ...bm Þ  i

i x f jþ1 ðx Þ  f jþ1 e ¼ f j ðx Þ  f j xða1 ...am b1 ...bm Þ 6 d1 f jþ1 ðx Þ: Then we conclude that

i i ða1 ...am b1 ...bm Þ  i x f jþ1 ðx Þ  f jþ1 e 6 d1 f jþ1 ðx Þ; i ða1 ...am b1 ...bm Þ  i 6 ð1 þ d1 Þf jþ1 ðx Þ; x f jþ1 e

i ¼ 1; 2; . . . ; m:

i ¼ 1; 2; . . . ; m:

j and a1 ; . . . ; am ; b1 ; . . . ; bm . By the definition of Ae , such an j always   exists, e.g., j ¼ 1. Algorithm Am e may not choose ðx1 ; . . . ; xj ; 0; . . . ; 0Þ for further construction; however, for a vector xða1 ...am b1 ...bm Þ chosen instead of it, due to Property 1, we have

 i eða1 ...am b1 ...bm Þ  i

i ða1 ...am b1 ...bm Þ  x  f jþ1 xðc1 ...cm d1 ...dm Þ 6 df jþ1 e f jþ1 x

m

i  i i Rj ðx Þ  Rj ðxða1 ...am b1 ...bm Þ Þ 6 dRj ðx Þ;

i

6 dð1 þ d1 Þf jþ1 ðx Þ:

i ¼ 1; 2; . . . ; m;

and

 i  i

i f j ðx Þ  f j xða1 ...am b1 ...bm Þ 6 df j ðx Þ;

i ¼ 1; 2; . . . ; m:

Set d1 ¼ d. We consider vector ðx1 ; . . . ; xj ; xjþ1 ; 0; . . . ; 0Þ and   e Without x ða1 ...am b1 ...bm Þ ¼ xa11 ...am b1 ...bm ; . . . ; xaj 1 ...am b1 ...bm ; xjþ1 0; . . . ; 0 . n  loss of generality, we assume xjþ1 ¼ k. If min Rkj ðx Þ; Rkj ðxða1 ...am b1 ...bm Þ Þg ¼ 0, then we have



ða ...am b ...bm Þ  k

ða ...am b ...bm Þ a a k k 1 1 x 1 x 1 Rjþ1 ðx Þ  Rjþ1 e ¼ ðrjþ1 ðx ÞÞ  r kjþ1 e

n

o and if min Rkj ðx Þ; Rkj xða1 ...am b1 ...bm Þ > 0, by Lemma 2, we have

For i ¼ 1; 2; . . . ; m, from (8) and (11), we obtain  i i i

i ða1 ...am b1 ...bm Þ  x f jþ1 ðx Þ  f jþ1 xðc1 ...cm d1 ...dm Þ 6 f jþ1 ðx Þ  f jþ1 e i eða1 ...am b1 ...bm Þ  i ðc1 ...cm d1 ...dm Þ  i i   f jþ1 x þ f jþ1 x 6 d1 f jþ1 ðx Þ þ dð1 þ d1 Þf jþ1 ðx Þ i

¼ ðd þ d1 ð1 þ dÞÞf jþ1 ðx Þ;

Similarly, from (6) and (10), we have

i

i

i ¼ 1; 2; . . . ; m:

dnjþ1 ¼d þ dð1 þ d1 Þ þ dð1 þ d2 Þ þ    þ dð1 þ dnj Þ n X 6 d ð1 þ dÞ j j¼0

For i – k, we have

ða ...a b ...b Þ  i 

 i i x 1 m 1 m ¼ Rj ðx Þ  Rij xða1 ...am b1 ...bm Þ 6 d1 Rijþ1 ðx Þ: Rjþ1 ðx Þ  Rjþ1 e

¼ð1 þ dÞnþ1  1 nþ1 X ðn þ 1Þn . . . ðn  j þ 2Þ  e  j ¼ 2 j!ðn þ 1Þ j j¼1

Then, we conclude that

Similarly, assuming xjþ1 ¼ k, we have

ð13Þ

Set dl ¼ d þ dl1 ð1 þ dÞ; l ¼ 2; 3; . . . ; n  j þ 1. From (12), we obtain  i i

i f jþ1 ðx Þ  f jþ1 xðc1 ...cm d1 ...dm Þ 6 d2 f jþ1 ðx Þ.

Since

h ia 6d1 r kj ðx Þ þ 1 ¼ d1 Rkjþ1 ðx Þ:

i ¼ 1; 2; . . . ; m:

ð12Þ



 i i i Rjþ1 ðx Þ  Rjþ1 xðc1 ...:cm d1 ...dm Þ 6 ðd þ d1 ð1 þ dÞÞRjþ1 ðx Þ; i ¼ 1; 2;. . .; m:

i



a  a ¼ ðr kj ðx Þ þ 1Þ  rkj xða1 ...am b1 ...bm Þ þ 1

ða ...am b ...bm Þ  1 6 ð1 þ d1 ÞRijþ1 ðx Þ; x 1 Rijþ1 e

ð11Þ

jf n ðx Þ  f n ðx0 Þj 6 dnjþ1 f n ðx Þ;



ða ...am b ...bm Þ  k

ða ...am b ...bm Þ a a k k 1 1 x 1 x 1 ¼ ðr jþ1 ðx ÞÞ  r kjþ1 e Rjþ1 ðx Þ  Rjþ1 e

i ¼ 1; 2; . . . ; m:

ð10Þ

Repeating the above argument for j þ 2; . . . ; n, we show that there exists x0 2 Y n such that

¼j1a  1a j 6 d1 Rkjþ1 ðx Þ;



ða ...am b ...bm Þ  i i i 1 x 1 Rjþ1 ðx Þ  Rjþ1 e 6 d1 Rjþ1 ðx Þ;

ð9Þ

Suppose that ~ xða1 ...am b1 ...bm Þ 2 Y c1 ...cm d1 ...dm # Y 0jþ1 and algorithm Am e ðc1 ...cm d1 ...dm Þ chooses x 2 Y c1 ...cm d1 ...dm instead of e x ða1 ...am b1 ...bm Þ in the (j þ 1)th iteration. Combining (7) and (9), for i ¼ 1; 2; . . . ; m, we obtain



 i eða1 ...am b1 ...bm Þ   Rijþ1 xðc1 ...cm d1 ...dm Þ Rjþ1 x

ða ...am b ...bm Þ  1 6 dRijþ1 e x 1 6 dð1 þ d1 ÞRijþ1 ðx Þ;

Proof. Suppose that ðx1 ; . . . ; xj ; 0; . . . ; 0Þ 2 Y a1 ...am b1 ...bm # Y 0j for some

ð8Þ

ð6Þ

6

nþ1 X 1 e j j¼1

ð7Þ

j! 2 i

6 e; i

i

then we have jf n ðx Þ  f n ðx0 Þj 6 ef n ðx Þ; Setting

k f n ðx0 Þ

¼

i maxi¼1;2;...m f n ðx0 Þ,

i ¼ 1; 2; . . . ; m.

we have

46

M. Ji et al. / Computers & Industrial Engineering 91 (2016) 42–47

max f i ðx0 Þ  max f i ðx Þ ¼ f k ðx0 Þ  max f i ðx Þ n n n i¼1;2;...m n i¼1;2;...m i¼1;2;...m k k 6 f n ðx0 Þ  f n ðx Þ k

5. Conclusions

i

6 ef n ðx Þ 6 e max f n ðx Þ: i¼1;2;...m

i Then in Step 3, vector x0 will be chosen such that maxi¼1;2;...m f n ðx0 Þ i i i maxi¼1;2;...m f n ðx Þj 6 maxi¼1;2;...m f n ðx0 Þ  maxi¼1;2;...m f n ðx Þ 6 e

We consider parallel-machine scheduling of deteriorating jobs that are subject to DeJong’s learning effect. We focus on the problems to minimize the total completion time and the makespan. We show that the former is polynomially solvable, while the latter is NP-hard, for which we provide a fully polynomial-time approximation scheme. For future research, it is worth considering more general models of learning and job deterioration, and/or in other machine environments such as the flowshop.

i

maxi¼1;2;...m f n ðx Þ. Therefore, we have Q ðx0 Þ 6 ð1 þ eÞQ ðx Þ. We analyze the time complexity of Algorithm Am e by noting that the most time-consuming operation is iteration j in Step 2, i.e., a call of the procedure Partition, which requires OðjY 0j j log jY 0j jÞ time to complete. To estimate

jY 0j j,

recall that

jY 0jþ1 j

¼ mjY j j 6 mkf 1    j

kf mj kR1    kRmj . By Properties 2 and 3, we have kf i 6 2ðn þ 1Þ j j

 max ð1 þ aÞn1 Þ=e þ 2 6 6ðn2  1ÞL=e þ 2, kRi 6 2Cðn þ 1Þlog n1a logðnp j

e þ 2 6 2aCðn þ 1ÞL=e þ 2, i ¼ 1;2;...;m. So jY 0j j ¼ Oðn3m L2m =e2m Þ, then jY 0j jlogjY 0j j ¼ Oðn3m L2mþ1 =e2m Þ. Hence, the time complexity of the algorithm is Oðn3mþ1 L2mþ1 =e2m Þ. h j ½M þ ð1  MÞr a  þ atj 4.2. Problem P m jpjr ¼ p By Lemma 1, we have

Pn

j¼1 C j

¼

P

Cj

Pm Pni nPni i¼1

j¼1

k¼j ð1

þ aÞkj

ni m X X ½ij ; Z ½ij p i¼1 j¼1

where

" Z ½ij ¼

ni X

# kj

ð1 þ aÞ

a

ðM þ ð1  MÞj Þ;

We thank the anonymous referees for their many helpful comments on earlier versions of our paper. Ji was supported in part by the National Social Science Foundation of China (Grant No. 14CGL071), Zhejiang Provincial Natural Science Foundation of China (Grant No. LR15G010001), and the Contemporary Business and Trade Research Center of Zhejiang Gongshang University, which is a key Research Institute of Social Sciences and Humanities of the Ministry of Education. Cheng was supported in part by The Hong Kong Polytechnic University under the Fung Yiu King – Wing Hang Bank Endowed Professorship in Business Administration and the National Natural Science Foundation of China (Grant No. 71390334).



a ½ij g. Thus, given the vector Pðn; mÞ ¼ ðn1 ; n2 ; ðM þ ð1  MÞj Þp . . . ; nm Þ, the problem reduces to the following matching problem which is polynomially solvable (see page 261 in Hardy, Littlewood, & Polya, 1967):

min

Acknowledgements

j ¼ 1; 2; . . . ni ;

k¼j

i ¼ 1; 2; . . . m: Based on the above analysis, we propose an algorithm to solve the problem as follows: Algorithm 1. Step 1. Generate all the possible allocations Pðn; mÞ ¼ ðn1 ; n2 ; . . . ; nm Þ. Step 2. For each Pðn; mÞ ¼ ðn1 ; n2 ; . . . ; nm Þ, calculate the positional weights Z ½ij ; match the job with the largest normal processing time to the position with the smallest value of Z ½ij , the job with the second largest normal processing time to the position with the second smallest value of Z ½ij , and P so on; then calculate nj¼1 C j . Pn Step 3. Choose the smallest j¼1 C j and find the corresponding optimal schedule. In a similar way as the proof of Theorem 5 in Ji et al. (2014), one can prove the following result. Theorem 2. Given the number of machines m, problem P j ½M þ ð1  MÞra  þ atj C j can be solved in Oðnm log nÞ Pm jpjr ¼ p time.

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