Computers & Industrial Engineering 68 (2014) 35–41
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Group scheduling and job-dependent due window assignment based on a common flow allowance q Min Ji a,⇑, Ke Chen a, Jiaojiao Ge 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 8 March 2013 Received in revised form 9 September 2013 Accepted 23 November 2013 Available online 1 December 2013 Keywords: Single-machine scheduling Group technology Common flow allowance Job-dependent due window Earliness/tardiness
a b s t r a c t We study a single-machine group scheduling and job-dependent due window assignment problem in which each job is assigned an individual due window based on a common flow allowance. In the group technology environment, the jobs are divided into groups in advance according to their processing similarities and all the jobs of the same group are processed consecutively in order to improve production efficiency. A sequence-independent machine setup time precedes the processing of the first job of each group. A job completed earlier (later) than its due window will incur an earliness (tardiness) penalty. Our goal is to find the optimal sequence for both the groups and jobs, together with the optimal due window assignment, to minimize the total cost that comprises the earliness and tardiness penalties, and the due window starting time and due window size costs. We give an O(n log n)time algorithm to solve this problem. Ó 2013 Elsevier Ltd. All rights reserved.
1. Introduction We study a single-machine scheduling and job-dependent due window assignment problem in a group technology (GT) environment. In the manufacturing industry, it is well known that firms can increase production efficiency by adopting GT. GT is an approach to manufacturing that seeks to improve efficiency in high-volume production by exploiting the similarities of different products and activities in their production. Through decades of application, people have found many advantages of GT. For instance, changeover between different jobs in the same group is simplified, reducing the costs or time involved; jobs in the same group spend less time waiting, which results in less work-in-process inventory; jobs in the same group tend to move through production in a direct route, reducing the manufacturing lead time (Li, Ng, and Yuan, 2011). As an example of GT in practice, Conway, Maxwell, and Miller (1967) consider paint production on a single machine. Customer orders vary in colors, but they can be divided into major color groups, such as red, blue, and green. Within a color group, e.g., red, colors may range from very light to dark red. The times to set up the machine to produce paint in colors of the same group are small, since it is natural to produce paint from lighter to darker colors, so these setup times are negligible. On the other
q
This manuscript was processed by Area Editor Subhash C. Sarin.
⇑ Corresponding author. Tel./fax: +86 571 28008303.
E-mail addresses:
[email protected] (M. Ji),
[email protected] (T.C.E. Cheng). 0360-8352/$ - see front matter Ó 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.cie.2013.11.017
hand, the time to switch the machine from the production of paint of one color group to another color group is substantial, which may include the times to clean the machine and change the tools, and this setup time is generally color-independent, i.e., the setup time is sequence-independent. In view of these observations, it is clear that dividing the orders into groups according to their processing similarities can significantly increase production efficiency. While GT research was originated by Mitrofanov (1966) and Opitz (1970), scheduling studies taking GT into consideration include Chen, Li, and Tang (1997), Ng, Cheng, Janiak, and Kovalyov (2005), Logendran, Carson, and Hanson (2005), Allahverdi, Ng, Cheng, and Kovalyov (2008), Cheng, Kovalyov, Ng, and Lam (2008), Shabtay, Itskovich, Yedidsion, and Oron (2010), Behnamian, Zandieh, and Ghomi (2010), Li et al. (2011), and Bai, Li, and Huang (2012), among others. While there is a large body of literature on due date scheduling involving the constant due date, i.e., each job is given the same due date, only a few studies focus on the common flow allowance, i.e., each job is given a due date that is equal to the sum of its processing time and a common flow allowance. The common flow allowance that reflects equal waiting time is a decision variable. Based on the common flow allowance, we set a due date for each job, the value of which has different implications. While a small value of the common flow allowance (leading to earlier due dates) makes the supplier more attractive to customers, a large value (i.e. late due dates) increases the flexibility of production but decreases attractiveness to customers. Depending on the value of the common flow allowance, a total cost is incurred, which the system
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M. Ji et al. / Computers & Industrial Engineering 68 (2014) 35–41
seeks to minimize. The system intends to inform each customer of a time that they have to wait until their job is completed through the optimal value of the common flow allowance. The delivery date announced will not generally be realized. However, the cost to the system will be minimized (Adamopoulos and Pappis, 1996). Due date scheduling research involving the common flow allowance includes Cheng (1986), Alidaee (1991), Cheng, Oguz, and Qi (1996), Wang (2006), Shabtay and Steiner (2007), and Wang and Wang (2010). In a recent paper, Mosheiov and Oron (2010) extend the flow allowance notion to the due window assignment setting in which each job is assigned an individual due window. Following Liman, Panwalker, and Thongmee (1998), they consider that a job completed earlier (later) than its due window will incur an earliness (tardiness) penalty, while a job completed within its due window is considered as on time that will incur no penalty. In this new due window assignment model, in addition to the standard job scheduling decisions, one has to determine the job-dependent 1 due windows. The due window starting time of a job Jj is dj , which is equal to the sum of its processing time pj and a common flow allowance q1, which is a job-independent constant, i.e., 1 dj ¼ pj þ q1 . Similarly, the due window completion time of the 2 job Jj is dj , which is equal to the sum of its processing time pj 2 and a larger common flow allowance q2, i.e., dj ¼ pj þ q2 . Mosheiov and Oron (2010) provide an O(n log n) solution for this problem. Mor and Mosheiov (2012) and Chen, Ji, and Ge (2013) extend this problem to the case including a maintenance activity. They consider several versions of this problem, including the duration of the maintenance activity is (i) a constant time, (ii) an increasing function of its starting time, and (iii) position-dependent. They also investigate extensions of the position-dependent processing times. They solve all these problems by converting them into an Assignment Problem. To the best of our knowledge, there is no research considering GT and due window assignment based on a common flow allowance simultaneously. The published papers in this area either study a job-dependent due window based on a common flow allowance without group technology or a job-dependent due date based on a common flow allowance with group technology. There is no paper studying a job-dependent due window based on a common flow allowance with group technology. Compared with the due date assignment problem based on a common flow allowance, we need to determine two values of the common flow allowance. Compared with the due window assignment problem based on a common flow allowance without group technology, we need to determine where the common flow allowance is, i.e., the common flow allowance may exist in other groups. Therefore our model is more general and covers the results of several papers, such as Li et al. (2011) and Mosheiov and Oron (2010). Shabtay et al. (2010) provide some practical examples of scheduling multiple groups of similar products with an assignable common due date and resource allocation on a single machine. They point out that GT implementation is especially important when the specifications of products destined to different markets, different market segments, or even different customers vary. This may be due to specific packaging, labelling, or other processes. In this paper we extend the notion of a due date to a job-dependent due window based on a common flow allowance. Generally, in real-life production, a set of independent jobs divided into groups are given, where jobs of the same group must be processed consecutively and groups are not allowed to be interweaved so that GT can be used to take advantage of their processing similarities. Mosheiov and Oron (2010) first introduced job-dependent due windows based on a common flow allowance. In our paper each job is assigned an individual due window, but all the jobs within the same group share a common due window size. We adopt the common flow allowance
procedure provided by Adamopoulos and Pappis (1996) to assign the due windows. We seek to determine the optimal group sequence, the optimal job sequence, and the optimal due window assignment to minimize the total cost that comprises the earliness and tardiness penalties, and the due window starting time and due window size costs. The remainder of this paper is organized as follows: In Section 2 we introduce the notation and formulate the problem. In Section 3 we present some properties of the optimal solution. In Section 4 we provide an O(n log n) algorithm to solve the problem. In Section 5 we conclude the paper and suggest future research topics. 2. Problem formulation The scheduling problem under study is formulated as follows: We are given a set of n independent jobs. In order to take advantage of GT in production, all the jobs are divided into m groups {G1, G2, . . ., Gm} in advance according to their processing similarities. All the jobs are available at time zero, and job preemption is not allowed. A single machine is available to process all the jobs. Each group Gi has ni jobs fJ i1 ; J i2 ; . . . ; J ini g, where Jij denotes the job P Jj of group Gi and m i¼1 ni ¼ n. Jobs of the same group must be processed consecutively. Each job has a normal processing time pij, and pi[j] means the processing time of the job in the jth position of Gi. There is a sequence-independent machine setup time si preceding the processing of the first job of group Gi. Each job has a jobh i 1 2 1 2 dependent due window dij ; dij , where dij 6 dij . The window size of the jobs of the same group is the same. Let Cij denote the completion time of job Jij. The earliness and tardiness of job Jij is given by n o 1 2 Eij ¼ max dij C ij ; 0 and T ij ¼ maxfC ij dij ; 0g, respectively, for i ¼ 1; . . . ; m; j ¼ 1; . . . ; ni . The job-dependent due window is determined by using the common flow allowance procedure as follows: The due window starting time of job Jij in Gi is given by 1
dij ¼ pij þ q1i ;
ð1Þ
and its due window completion time is given by 2
dij ¼ pij þ q2i ;
ð2Þ
where q1i , q2i are group-dependent decision variables and q1i < q2i . So the window size of group Gi is Di ¼ q2i q1i . Similar to Liman et al. (1998), our goal is to determine the optimal group sequence, the optimal job sequence, and the optimal due window assignment to minimize an objective function that comprises the earliness and tardiness penalties, and the due window starting time and due window size costs. So for a given schedule p, the objective function is given by
zðpÞ ¼
ni m X X
ai Eij þ bi T ij þ ci d1ij þ di Di ;
ð3Þ
i¼1 j¼1
where ai, bi, ci, di are non-negative parameters representing the unit costs of earliness, tardiness, due window starting time, and due window size, respectively. Using the three-field notation of Graham, Lawler, Lenstra, and Rinnooy Kan (1979) for describing scheduling problems, we can denote the problem under study as P Pni 1 1jGT; JDj m i¼1 j¼1 ai Eij þ bi T ij þ ci dij þ di Di , where JD means the job-dependent due window. 3. Preliminary results In this section, some useful preliminary results are given to solve the scheduling problem. It is clear that an optimal schedule exists that starts at time zero and contains no machine idle time.
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M. Ji et al. / Computers & Industrial Engineering 68 (2014) 35–41
Lemma 1.
ð2Þ bi 1
ni ni X X 2 T ij ¼ bi ðC i½j di½j Þ j¼1
1
If C ij 6 dij ) C iðj1Þ 6 diðj1Þ ; for i ¼ 1; . . . ; m; j ¼ 1; . . . ; ni : 2 2 If C ij P dij ) C iðjþ1Þ P diðjþ1Þ ; for i ¼ 1; . . . ; m; j ¼ 1; . . . ; ni :
j¼li þ1
"
Si þ si þ
¼ bi
! !! li þ1 li 1 X X pi½j Si þ si þ pi½j þ D2 þ piðli þ1Þ j¼1
Proof. For a given schedule p, if C ij 6
1 dij ,
then we have
þ
Si þ si þ
2
P q2i () C ij P q2i () C ij þ piðjþ1Þ P q2i þ piðjþ1Þ () C iðjþ1Þ
j¼1
Adamopoulos and Pappis (1996) proved the following result about the common flow allowance procedure, which Mosheiov and Oron (2010) extended to the due window setting. In this paper we prove that the result remains valid in the GT scheduling environment. Lemma 2. For a given schedule p, if the values of the common flow allowance q1i and q2i are within the starting and ending times of group Gi, then the optimal values of q1i and q2i coincide with the job completion times of group Gi for i = 1, . . ., m. Proof. Assume that ki 1 X pi½j
ð4Þ
j¼1 li 1 X pi½j
ð5Þ
j¼1
where Si denotes the starting time of Gi, si is the setup time, and ki and li mean the kith and lith positions of group Gi respectively (ki < li). Thus the completion time of the kith job is equal to its due window starting time. So according to Lemma 1, all the jobs before it are early jobs. Similarly, all the jobs after the lith job are tardy jobs. Consider that there exists an optimal schedule without the staPki 1 ted property, i.e., q1i ¼ Si þ si þ j¼1 pi½j þ D1 ; 0 6 D1 6 pi½ki , and P l 1 i q2i ¼ Si þ si þ j¼1 pi½j þ D2 ; 0 6 D2 6 pi½li . We only consider Gi here as an example to demonstrate that the result is correct. The total cost of Gi as a function of D1 and D2 is given by ni ni ni ni X X X X 1 Eij þ bi T ij þ ci dij þ di ðq2i q1i Þ: j¼1
j¼1
ni X
Eij ¼ ai
j¼1
j¼1
!
j¼1
j¼1
¼ ci ðni þ 1Þ
ki 1 X
ni X pi½j þ ci ni ðSi þ si Þ þ ci ni D1
j¼1
j¼ki
pi½j þ ci
" ! !# ni ni li 1 ki 1 X X X X 2 1 ð4Þ di ðqi qi Þ ¼ di Si þ s i þ pi½j þ D2 Si þ si þ pi½j þ D1 j¼1
j¼1
j¼1
j¼1
li 1 X ¼ di ni pi½j þ di ni ðD2 D1 Þ j¼ki
We see that each component is a linear function of D1 or D2, so the total cost of Gi is a linear function of both arguments. We can either decrease D1 and D2 to zero, or increase them to pi½ki and pi½li , respectively, to obtain a lower cost. In any case, we can see that q1i and q2i coincide with some jobs’ completion times, respectively h. Lemma 3. For a given schedule p, if the values of the common flow allowance q1i and q2i are within the starting and ending times of group Gi, then there exists an optimal schedule in which Pki 1 Pli 1 pi½j and q2i ¼ Si þ si þ j¼1 pi½j , where q1i ¼ Si þ si þ j¼1 ki ¼ dni ðdi ci Þ=ai e and li ¼ dni ðbi di Þ=bi e. Proof. This lemma can be easily proved by the standard technique of small perturbations. As given in Lemma 2, there is an optimal Pki 1 pi½j and q2i ¼ Si þ si þ schedule in which q1i ¼ Si þ si þ j¼1 Pli 1 1 j¼1 pi½j . The effect of moving qi D units of time to the left is
ai ðki 1ÞD ci ni D þ di ni D:
ð6Þ
The effect of moving q1i D units of time to the right is
ai ki D þ ci ni D di ni D:
j¼1
Both (6) and (7) are non-negative due to the optimality of the original schedule. Then from ai ðki 1ÞD ci ni D þ di ni D P 0 and
j¼1
"
¼ ai
Si þ s i þ
ki 1 X
!
Si þ s i þ
!
ki 1 X pi½j þ D1 þ pi½2 ðSi þ si þ pi½1 þ pi½2 Þ j¼1
þ... þ
!
Si þ s i þ
ki 1 X
! pi½j þ D1 þ pi½ki
j¼1
ki X pi½j Si þ s i þ j¼1
!!# ¼ ai
ki 1 X jpi½j þ ai ki D1 j¼1
ð7Þ
ai ki D þ ci ni D di ni D P 0, we obtain ki ¼ dni ðdai i ci Þe. Using the same
pi½j þ D1 þ pi½1 ðSi þ si þ pi½1 Þ
j¼1
þ
!!#
ki ki X X 1 Ei½j ¼ ai ðdi½j C i½j Þ
1 1 1 i ½ðdi½1 C i½1 Þ þ ðdi½2 C i½2 Þ þ ... þ ðdi½ki C i½ki Þ
¼a
li 1 X Si þ si þ pi½j þ D2 þ pi½ni
j¼1
We consider each cost component separately as follows:
ð1Þ ai
pi½j
j¼1
ni ni ki 1 X X X 1 ð3Þ ci dij ¼ ci Si þ si þ pi½j þ D1 þ pi½j
j¼1
!
j¼li
C ij ¼ C iðj1Þ þ pij P dij ¼ q2i þ pij () C iðj1Þ P q2i ) C iðj1Þ þ pij
Z i ðD1 ; D2 Þ ¼ ai
ni X
!!
ni X ¼ bi ðni jÞpi½j bi ðni li ÞD2
2
q2i ¼ Si þ si þ
j¼1 li 1 X Si þ si þ pi½j þ D2 þ pi½ðli þ2Þ
j¼1
If C ij P dij , the we have
q1i ¼ Si þ si þ
pi½j
Si þ si þ
þ .. . þ
1
6 q1i þ piðj1Þ () C iðj1Þ 6 diðj1Þ :
2
!
j¼1
1
C ij ¼ C iðj1Þ þ pij 6 dij ¼ q1i þ pij () C iðj1Þ 6 q1i ) C iðj1Þ
P diðjþ1Þ :
li þ2 X
!
iÞ e. h method, we obtain li ¼ dni ðbbi d i
Remarks. For a given schedule p, if 0 < dni ðdi ci Þ=ai e < dni ðbi di Þ=bi e, we can determine the optimal common flow allowance for a group by Lemma 2 and Lemma 3. But the ratios may not meet the above inequality, e.g., for given parameters, there may exist inequality dni ðbi di Þ=bi e 6 dni ðdi ci Þ=ai e, so we need further analysis. Note that dni ðbi di Þ=bi e 6 dni ðdi ci Þ=ai e means ni ðdi ci Þ=ai P ni ðbi di Þ=bi .
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M. Ji et al. / Computers & Industrial Engineering 68 (2014) 35–41
(i) With a shift of q1i (without changing q2i ) by D units of time to the right, resulting in a change of total cost of DZ = Dkiai + Dnici Dnidi. With a shift of q2i (without changing q1i ) by D units of time to the right, resulting in a change of DZ = D(ni li)bi + Dnidi. If li < ni(bi di)/bi, a shift of q2i by D units of time to the right till li ¼ dni ðbi di Þ=bi e can only reduce the total cost. If ki < ni(bi di)/bi, then ki < ni(bi di)/bi < ni(di ci)/ai, so a shift of q1i by D units of time to the right till ki ¼ dni ðbi di Þ=bi e can only reduce the total cost. (ii) With a shift of q1i (without changing q2i ) by D units of time to the left, resulting in a change of total cost of DZ = D(ki 1)ai Dnici + Dnidi. With a shift of q2i (without changing q1i ) by D units of time to the left, resulting in a change of DZ = D(ni li + 1)bi Dnidi. If ki > ni(di ci)/ai + 1, a shift of q1i by D units of time to the left till ki ¼ dni ðdi ci Þ=ai e can only reduce the total cost. If li > ni(di ci)/ai + 1, then li > ni(di ci)/ai + 1 > ni(bi di)/bi + 1, so a shift of q2i by D units of time to the left till li ¼ dni ðdi ci Þ=ai e can only reduce the total cost. From (i) and (ii), we can see that dni ðbi di Þ=bi e 6 ki 6 li 6 dni ðdi ci Þ=ai e, if ki 6 li , so a further shift of q1i to the right and a further shift of q2i to the left till ki = li can only reduce the total cost, which means that q1i ¼ q2i , and the due window assignment problem reduces to a due date assignment problem. So in this case, the due date of group Gi is given by
dij ¼ pij þ qi q1i
ð8Þ
Proof. If dni ðbi di Þ=bi e 6 0, which means bi 6 di , then a shift of q2i by D units of time to the left until q1i will result in a change of the total cost DZ 6 0, so q2i can be shifted to q1i , and the due window assignment problem reduces to a due date assignment problem. Using the same analytical method of Li et al. (2011), we can get the result. h From Lemmas 2–6, according to the ratio of dni ðdi ci Þ=ai e and dni ðbi di Þ=bi e, there exist three cases: Case I: 0 < dni ðdi ci Þ=ai e < dni ðbi di Þ=bi e. This will be the normal due window assignment problem based on a common flow allowance, and the common flow allowance can be determined by Lemma 2 and lemma 3. Case II: dni ðdi ci Þ=ai e 6 0 < dni ðbi di Þ=bi e. This is a special due window assignment problem and the common flow allowance can be determined by Lemma 5. Case III: dni ðbi di Þ=bi e 6 dni ðdi ci Þ=ai e or dni ðdi ci Þ=ai e < dni ðbi di Þ=bi e 6 0. The problem reduces to a due date assignment problem based on a common flow allowance. The common flow allowance can be determined by Lemmas 4 and 6. The following lemma is useful for solving the problems. Lemma 7. Given two sequences of non-negative numbers xi and yi, P the sum of the products of the corresponding elements ni¼1 xi yi is minimized if the sequences are monotonic in the opposite way. Proof. See Hardy, Littlewood, and Polya (1967). h
4. Optimal properties of the problem 4.1. Analysis of case I
q2i
If and are the same, and the due window becomes a due date. Li et al. (2011) study such a model. According to their results, we have the following Lemma. Lemma 4. If dni ðbi di Þ=bi e 6 dni ðdi ci Þ=ai e, then the due window assignment problem reduces to a due date assignment problem. According to Li et al. (2011), the optimal common flow allowance is as follows:
8 ki 1 > X > ci i i qi ¼ ; j¼1 > > : 0 otherwise
Note that the values of ki and li given in Lemma 3 are independent of the job processing times and the job sequence. Therefore, it is optimal for any job sequence within each group of case I. Without loss of generality, we assume Gi is case I. For group Gi. The optimal ki and li can be observed from Lemma 3, so does the Pki 1 optimal values of q1i and q2i , which q1i ¼ Si þ si þ j¼1 pi½j and P l 1 i q2i ¼ Si þ si þ j¼1 pi½j . Recall that Si denotes the starting time of Gi and si is the sequence-independent setup time. For a given schedule p, the total cost of all the jobs within Gi is given by Z i ðpÞ ¼ ai
ni ni ni ni X X X X 1 Eij þ bi T ij þ ci dij þ di ðq2i q1i Þ j¼1
where ki ¼ dni ðbi ci Þ=ðai þ bi Þe . From Lemma 4, without loss of generality, we only consider the case dni ðbi di Þ=bi e > dni ðdi ci Þ=ai e in the following Lemmas 5 and 6.
"j¼1 ¼ ai
j¼1
j¼1
j¼1 j¼li ! ! j¼1 ki 1 X Si þ s i þ pi½j þ pi½1 ðSi þ si þ pi½1 Þ þ
j¼1 ! !!# ki 1 ki X X Si þ si þ pi½j þ pi½ki Si þ si þ pi½j
þ
Lemma 5. If dni ðdi ci Þ=ai e 6 0 < dni ðbi di Þ=bi e, the problem becomes a special due window assignment problem, with the common Pli 1 flow allowance q1i ¼ 0, q2i ¼ Si þ si þ j¼1 pi½j , and li ¼ dni ðbi di Þ=bi e.
j¼1
ki ni ni ni X X X X 1 2 1 ¼ ai ðdi½j C i½j Þ þ bi ðC i½j di½j Þ þ ci di½j þ di ðq2i q1i Þ
j¼1
j¼1 ! !! li li 1 X X Si þ s i þ pi½j Si þ si þ pi½j þ pi½li þ
" þbi
j¼1 j¼1 ! !!# ni li 1 X X pi½j Si þ si þ pi½j þ pi½ni Si þ si þ
þ
j¼1
q1i
Proof. If dni ðdi ci Þ=ai e 6 0, meaning di 6 ci , then a shift of by D units of time to the left will result in a change of the total cost DZ 6 0, so result q1i can be shifted to the left until it equals time zero. Use the same analytical method of Lemmas 2 and 3, we can get the result about q2i . Lemma 6. If dni ðdi ci Þ=ai e < dni ðbi di Þ=bi e 6 0, then the due window assignment problem also reduces to a due date assignment problem. the optimal common flow allowance is qi = 0.
þci
ni X
Si þ s i þ
!
k i 1 X
pi½j þ pi½j þ di
j¼1
j¼1
j¼1 " ni X
ki 1 X Si þ si þ pi½j
!#
j¼1
Si þ si þ
li 1 X pi½j
!
j¼1
ki 1 ni X X ¼ ai jpi½j þ bi ðni jÞpi½j
j¼1
j¼1
j¼li
k ni li 1 i 1 X X X þci ðni þ 1Þ pi½j þ ci pi½j þ ci ni ðSi þ si Þ þ di ni pi½j j¼1
j¼ki
ni X ¼ wij pi½j þ ci ni ðSi þ si Þ j¼1
j¼ki
ð9Þ
39
M. Ji et al. / Computers & Industrial Engineering 68 (2014) 35–41
where -i = min {bi, ci}
Where
8 > < ai j þ ci ðni þ 1Þ; 1 6 j 6 ki 1 wij ¼ di ni þ ci ki 6 j 6 li 1 > : bi ðni jÞ þ ci li 6 j 6 ni
ð10Þ
Pni Note that the first term j¼1 wij pi½j of the total cost is only concerned with the job processing sequence within a group, and is not influenced by the processing orders of the other groups. The second term of the total cost is only concerned with the starting time of the group, and is independent of the internal job sequence of each group. The following lemma is immediately obtained from this property. Lemma 8. The optimal job sequence within Gi can be obtained by matching the smallest wij value to the largest pij value, the second smallest wij value to the second largest pij value, and so on.
4.2. Analysis of case II Using the same method as Section 4.1 and the result of Lemma 5, we get that for a given schedule p, the total cost of all the jobs within Gi is given by:
¼ bi
ni ni ni ni X X X X 1 Eij þ bi T ij þ ci dij þ di ðq2i q1i Þ j¼1 ni X
j¼1 2
ðC i½j di½j Þ þ c
j¼li
"
Si þ si þ
¼ bi
j¼1 ni X 1 di½j i j¼1
þ di
þ
ðq2i 0Þ
j¼1
! !! li li 1 X X þ pi½j Si þ si þ pi½j þ pi½li j¼1
! !!# ni li 1 X X pi½j Si þ si þ pi½j þ pi½ni Si þ s i þ j¼1
j¼1
ni ni li 1 X X X Si þ si þ pi½j þci ð0 þ pi½j Þ þ di j¼1
j¼1
!
j¼1
j¼1
ci þ di ni ;
1 6 j 6 li 1
bi ðni jÞ þ ci ; li 6 j 6 ni
j¼li
ð11Þ
j¼1
ð12Þ
Pni Note that the first term j¼1 wij pi½j of the total cost is only concerned with the job processing sequence within a group and is not influenced by the processing orders of the other groups. The second term of the total cost is only concerned with the starting time of the group and is independent of the internal job sequence of each group. The same as Section 4.1, we can use Lemma 8 to assign the job sequence within a group. 4.3. Analysis of case III If dni ðbi di Þ=bi e 6 dni ðdi ci Þ=ai e or dni ðdi ci Þ=ai e < dni ðbi di Þ=bi e 6 0, then the due window assignment problem based on a common flow allowance reduces to a due date assignment problem based on a common flow allowance. We can get the following results from Li et al. (2011), who study a similar model. Lemma 9. For a given schedule p, the cost function of all the jobs within group Gi can be expressed as follows:
Z i ðpÞ ¼
ni X wij pi½j þ ni -i ðSi þ si Þ; j¼1
ki 6 j 6 n i
for ci < bi
ð14Þ
and
wij ¼ ci þ ðni jÞbi ;
1 6 j 6 ni
for c P b :
ð15Þ
i i Pni Note that the first term j¼1 wij pi½j of the total cost is only concerned with the job processing sequence within a group and is not influenced by the processing orders of the other groups. The second term of the total cost is only concerned with the starting time of the group and is independent of the internal job sequence of each group. Using the same method of Section 4.1, we can determine the job sequence according to Lemma 8.
By the analysis of Sections 3 and 4, the optimal internal job sequence is observed. And the following lemma enables us to determine the processing orders of the groups. Lemma 10. For a given schedule p, there is an optimal group processing order in which all the groups are scheduled in nondecreasing order of ui,
j¼1
Where
wij ¼
ci þ ðni jÞbi ;
ni X ¼ bi ðni jÞpi½j
ni li 1 ni X X X þci pi½j þ di ni ðSi þ si Þ þ di ni pi½j ¼ wij pi½j þ di ni ðSi þ si Þ j¼1
ðni þ 1Þci þ jai ; 1 6 j 6 ki 1
! 8 ni X > > > si þ pij =ni ci ; if group Gi is case I > > > > j¼1 > > ! > > ni < X si þ pij =ni di ; if group Gi is case II where /i ¼ > > j¼1 > > ! > > ni > X > > > pij =ni -i ; if group Gi is case III > : si þ
j¼1 ni X
j¼1
5. An optimal solution for the group sequence
Proof. It can be easily proved by Lemma 7. h
Z i ðpÞ ¼ ai
wij ¼
ð13Þ
Proof. We prove the result by contradiction. Let p be an optimal schedule that does not satisfy Lemma 10. In this schedule there must be at least two adjacent groups, say Gl followed by Gk, such that ul > uk. Without loss of generality, we assume that Gl is case I, and Gk is case II. That is:
sl þ
Pnl
j¼1 plj
nl cl
>
sk þ
Pnk
j¼1 pkj
nk dk
:
ð16Þ
Swapping Gl and Gk, while leaving the other groups in their ori~ . Recall that a change in the ginal order, we obtain a new schedule p processing order of the groups does not affect the schedule of the job sequence within a group. Thus the difference in the values of ~ is only due to the starting the objectives between schedule p and p time of Gl and Gk. Letting Sl(p) denote the starting time of Gl in schedule p, we have
Sk ðpÞ ¼ Sl ðpÞ þ sl þ
nl X
plj ;
ð17Þ
j¼1
~ Þ ¼ Sl ðpÞ; Sk ð p ~ Þ ¼ Sl ðpÞ þ sk þ Sl ð p
ð18Þ nk X
pkj :
ð19Þ
j¼1
Substituting (17)–(19) into (9) and (11), the difference between the two schedules is as follows:
40
M. Ji et al. / Computers & Industrial Engineering 68 (2014) 35–41
~ Þ ¼ cl nl ðSl ðpÞ þ sl Þ þ dk nk ZðpÞ Zðp
nl X Sl ðpÞ þ sl þ plj þ sk
!
j¼1
dk nk ðSl ðpÞ þ sk Þ cl nl ¼ dk nk sl þ
nl X
! plj
cl nl
nk X Sl ðpÞ þ sk þ pkj þ sl nk X sk þ pkj
j¼1
!
Table 2 The wij values of each group.
! G1 G2 G3
j¼1
wi1
wi2
wi3
wi4
30 29 36
33 23 28
21 14 18
5 5 8
>0
j¼1
This contradicts the optimality of schedule p and completes the proof of the lemma. h Based on the above analysis, now we can present the following P Pni 1 algorithm to solve the 1jGT; JDj m i¼1 j¼1 ðai Eij þ bi T ij þ ci dij þ di Di Þ problem. Algorithm 1 Step 1. Calculate the ratios of dni ðdi ci Þ=ai e and dni ðbi di Þ=bi e, and according to Lemmas 2–6, determine which case each group is. Step 2. Arrange the groups in non-decreasing order of ui by Lemma 10. Step 3. For the groups of case I, calculate wij according to Eq. (10), and ki and li according to Lemma 3. Sequence the jobs by Lemma 8. For the groups of case II, calculate wij according to Eq. (12) and li according to Lemma 5. Sequence the jobs by Lemma 8. For the groups of case III, calculate wij according to Eq. (14), (15) and ki according to Lemma 4 or 6. Sequence the jobs by Lemma 8. Step 4. For the groups of case I, assign the common flow allowance according Eqs. (4) and (5), and set the job dependent due windows by Eqs. (1) and (2). For the groups of case II, assign the due window according to Lemma 5, and Eqs. (1) and (2). For the groups of case III, assign the due window according to Lemma 4 or 6, and Eq. (8).
To determine the computational complexity of Algorithm 1, note that Step 1 can be performed in linear time. Step 2 needs P O(m log m) time, and Step 3 needs Oð m i¼1 ni log ni Þ time, while Step Pm 4 takes a linear time. Since i¼1 ni ¼ n and m < n, the overall time complexity of Algorithm 1 is O(n log n). Theorem 1. Algorithm 1 solves the 1jGT; JDj 1 þci dij þ di Di Þ problem in O(n log n) time.
Pm Pn i i¼1
j¼1
ðai Eij þ bi T ij
Table 3 The internal job sequence of each group.
G1 G2 G3
1
2
3
4
J12 J22 J33
J11 J21 J32
J13 J24 J31
J14 J23 J34
Table 4 Due windows (due dates) of each job.
G1 G2 G3
Ji1
Ji2
Ji3
Ji4
[43, 45] 80 [7, 17]
[44, 46] 78 [6, 16]
[46, 48] 83 [5, 15]
[47, 49] 81 [8, 18]
problem reduces to a due date assignment problem, and k2 ¼ dn2 ðb2 c2 Þ=ða2 þ b2 Þe ¼ d4ð9 5Þ=ð4 þ 9Þe ¼ 2, hence G2 is case III. For group G3, the given parameters are c3 > d3 and l3 ¼ dn3 ðb3 d3 Þ=b3 e ¼ d4ð10 7Þ=10e ¼ 2, so according to Lemma 5, this will be a special due window assignment problem, hence G3 is case II. Step 2. The ui values obtained from Lemma 10 are u1 = (7 + 2 + 3 + 5 + 6)/(4 ⁄ 5) = 1.15, u2 = (10 + 9 + 7 + 12 + 10)/ (4 ⁄ 5) = 2.4, and u3 = (5 + 7 + 6 + 5 + 8)/(4 ⁄ 7) = 1.107. So u3 < u1 < u2, hence the group processing sequence will be G3 first, G1 second, and G2 the last. Step 3. For G1, calculate wij according to Eq. (10). For G2, calculate wij according to Eq. (14). For G3, calculate wij according to Eq. (12). According to Lemma 8, assign the internal job sequence of each group. The results are shown in Tables 2 and 3. Step 4. For G1, the common flow allowance is determined by Pk1 1 p1½j ¼ 31 þ 7 þ 3 ¼ 41 and q21 ¼ S1 þ s1 þ q11 ¼ S1 þ s1 þ j¼1 Pl1 1 j¼1 p1½j ¼ 31 þ 7 þ 5 ¼ 43. For G2, the common flow allowance Pk2 1 p2½j ¼ 54 þ 10 þ 7 ¼ 71. is determined by q2 ¼ S2 þ s2 þ j¼1 For G3, the common flow allowance is determined by q13 ¼ 0, Pl3 1 q23 ¼ S3 þ s3 þ j¼1 p3½j ¼ 0 þ 5 þ 5 ¼ 10. The due windows or
6. A numerical example
due dates of each job are shown in Table 4. A three-group due window assignment problem based on a common flow allowance is solved here to illustrate the solution procedure. Each group has four jobs. The job processing times, cost parameters, and setup times are given is Table 1. Step 1. For group G1, k1 ¼ dn1 ðd1 c1 Þ=a1 e ¼ d4ð7 5Þ=5e ¼ 2 l1 ¼ dn1 ðb1 d1 Þ=b1 e ¼ d4ð16 7Þ=16e ¼ 3. According to the Remarks, G1 is case I. For group G2, the given parameters are b2 < d2, so according to Lemma 6, the due window assignment Table 1 Parameters of the numerical example.
G1 G2 G3
ai
bi
ci
di
pi1
pi2
pi3
pi4
si
5 4 5
16 9 10
5 5 8
7 10 7
2 9 7
3 7 6
5 12 5
6 10 8
7 10 5
Executing the four steps, we solve the problem and the total P cost is ZðpÞ ¼ 3i¼1 Z i ðpÞ ¼ 3619. 7. Conclusions We consider a single-machine scheduling problem that considers both group technology and job-dependent due window assignment simultaneously. The objective is to find the optimal job and group schedules, and the optimal job-dependent due windows to minimize an objective function that includes the earliness and tardiness penalties, and due window starting time and due window size costs. We show that the problem can be solved in polynomial time. Future research can consider the problem in other settings, such as the job processing times are position-dependent, the machine needs a maintenance activity, and the multi-machine environment.
M. Ji et al. / Computers & Industrial Engineering 68 (2014) 35–41
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