A note on “minimizing makespan in three machine flow shop with deteriorating jobs”

Computers & Operations Research 72 (2016) 93–96

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A note on “minimizing makespan in three machine flow shop with deteriorating jobs” Abbas-Ali Jafari a, Hassan Khademi-Zare a,n, M.M. Lotfi a, Reza Tavakkoli-Moghaddam b,c a b c

Department of Industrial Engineering, Faculty of Engineering, Yazd University, Yazd, Iran School of Industrial Engineering and Engineering Optimization Research Group, College of Engineering, University of Tehran, Tehran, Iran Universal Scientific Education and Research Network (USERN), Tehran, Iran

art ic l e i nf o

a b s t r a c t

Available online 23 February 2016

This paper is a note on “minimizing makespan in three machine flow shop with deteriorating jobs” [J.-B. Wang, M.-Z. Wang, minimizing makespan in three machine flow shop with deteriorating jobs, Computers & Operation Research 40 (2013) 547–557]. Wang and Wang presented a branch-and-bound algorithm with several dominance properties and a lower bound; however, we think that the dominance properties may not be true as they are neither necessary nor sufficient. We first show by means of a counter-example that the published dominance properties are incorrect, and then present a necessary and sufficient condition for them to be true. Moreover, a simplifying remark is provided for the above dominance properties. & 2016 Elsevier Ltd. All rights reserved.

Keywords: Scheduling Flow shop Makespan Linear deteriorating job Dominance property

1. Introduction In the recent paper, Wang and Wang [1] (called WW in the rest) considered three-machine flow shop scheduling with linear deteriorating jobs in which the actual processing time (Pij) of Jj (j¼1, 2, …, n) on machine Mi (i¼1, 2, 3), if it is started at time t in a sequence, is given as follows: P ij ¼ aij þbt where aij is the basic processing time of Jj on machine Mi and bZ 0 denotes the similar deterioration rate for all the jobs. The objective was to find a sequence that minimizes the makespan. Since the problem is NP-hard, they presented a branch and bound algorithm with several dominance properties as follows. The completion time of Jj and jth job on machine Mi under schedule S is denoted as C i j ðSÞ and C i ½j ðSÞ, respectively. Obviously, the completion time of Jj is Cj ¼ C3j. The problem classification may be represented as the following three-field notation: F3j prmu; P ij ¼ aij þ bt; bZ 0j C max : We consider αj ; β j and γ j instead of a1j ; a2j and a3j , respectively, for simplifying the relations. Let S1 ¼ ðπ ; J j ; J k ; π 0 Þ, and S2 ¼ ðπ ; J k ; J j ; π 0 Þ be obtained from S1 by a pairwise interchange of Jj and Jk, where π and π 0 are the partial sequences. In addition, it is assumed that the completion time of π on M1, M2 and M3 in S1 is A, B and C, respectively. It is obvious that A, B and C are fixed in S2 . WW expressed based on the following proposition that to confirm that S1 dominates S2 , it is sufficient to show that C k ðS1 Þ r C j ðS2 Þ. Proposition. Suppose that J ðiÞ

n

Either

k

and J j satisfy the following conditions :

αj ð1 þ bÞ3 þ αk ð1 þbÞ2 þ βk ð1 þ bÞ þ γ k r αk ð1 þ bÞ3 þ αj ð1 þ bÞ2 þ βj ð1 þbÞ þ γ j

or

αj ð1 þ bÞ3 þ αk ð1 þbÞ2 þ βk ð1 þ bÞ þ γ k r αk ð1 þ bÞ3 þ βk ð1 þ bÞ2 þ βj ð1 þbÞ þ γ j

or

αj ð1 þ bÞ3 þ αk ð1 þbÞ2 þ βk ð1 þ bÞ þ γ k r αk ð1 þ bÞ3 þ βk ð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

Corresponding author. Tel.: þ 98 35 31232423; fax: þ 98 35 38210699. E-mail addresses: [email protected] (A.-A. Jafari), [email protected] (H. Khademi-Zare), lotfi@yazd.ac.ir (M.M. Lotfi), [email protected] (R. Tavakkoli-Moghaddam).

http://dx.doi.org/10.1016/j.cor.2016.02.001 0305-0548/& 2016 Elsevier Ltd. All rights reserved.

94

A.-A. Jafari et al. / Computers & Operations Research 72 (2016) 93–96

or

Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ αk ð1 þbÞ2 þ β k ð1 þ bÞ þ γ k r Bð1 þ bÞ3 þ βk ð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

or

Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ αk ð1 þbÞ2 þ β k ð1 þ bÞ þ γ k r Bð1 þ bÞ3 þ βk ð1 þ bÞ2 þ β j ð1 þ bÞ þ γ j

or

Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ αk ð1 þbÞ2 þ β k ð1 þ bÞ þ γ k r Cð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

ðiiÞ

Either

αj ð1 þ bÞ3 þ βj ð1 þ bÞ2 þ βk ð1 þ bÞ þ γ k r αk ð1 þbÞ3 þ αj ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j

or

αj ð1 þ bÞ3 þ βj ð1 þbÞ2 þ βk ð1 þ bÞ þ γ k r αk ð1 þ bÞ3 þ βk ð1 þbÞ2 þ βj ð1 þ bÞ þ γ j

or

αj ð1 þ bÞ3 þ βj ð1 þbÞ2 þ βk ð1 þ bÞ þ γ k r αk ð1 þ bÞ3 þ βk ð1 þbÞ2 þ γ k ð1 þ bÞ þ γ j

or

Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ β j ð1 þbÞ2 þ β k ð1 þ bÞ þ γ k rBð1 þ bÞ3 þ βk ð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

or

Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ β j ð1 þbÞ2 þ β k ð1 þ bÞ þ γ k rBð1 þ bÞ3 þ βk ð1 þ bÞ2 þ β j ð1 þ bÞ þ γ j

or

Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ β j ð1 þbÞ2 þ β k ð1 þ bÞ þ γ k rCð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

ðiiiÞ

Either

αj ð1 þ bÞ3 þ βj ð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k r αk ð1 þ bÞ3 þ αj ð1 þ bÞ2 þ βj ð1 þbÞ þ γ j

or

αj ð1 þ bÞ3 þ βj ð1 þbÞ2 þ γ j ð1 þbÞ þ γ k r αk ð1 þ bÞ3 þ βk ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j

or

αj ð1 þ bÞ3 þ βj ð1 þbÞ2 þ γ j ð1 þbÞ þ γ k r αk ð1 þ bÞ3 þ βk ð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

or

Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ β j ð1 þbÞ2 þ γ j ð1 þ bÞ þ γ k r Bð1 þ bÞ3 þ βk ð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

or

Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ β j ð1 þbÞ2 þ γ j ð1 þ bÞ þ γ k r Bð1 þ bÞ3 þ βk ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j

or

Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ β j ð1 þbÞ2 þ γ j ð1 þ bÞ þ γ k r Cð1 þbÞ2 þ γ k ð1 þ bÞ þ γ j

ðivÞ Either

Bð1 þ bÞ3 þ β j ð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k r Að1 þ bÞ4 þ αk ð1 þ bÞ3 þ αj ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j

or

Bð1 þ bÞ3 þ βj ð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k r Að1 þbÞ4 þ αk ð1 þ bÞ3 þ β k ð1 þ bÞ2 þ β j ð1 þbÞ þ γ j

or

Bð1 þ bÞ3 þ βj ð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k r Að1 þbÞ4 þ αk ð1 þ bÞ3 þ β k ð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

or

βj ð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k r βk ð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

or

βj ð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k r βk ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j

or

Bð1 þ bÞ3 þ βj ð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k r Cð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

ðvÞ

Either

Bð1 þ bÞ3 þ β j ð1 þ bÞ2 þ βk ð1 þbÞ þ γ k r Að1 þ bÞ4 þ αk ð1 þ bÞ3 þ αj ð1 þbÞ2 þ β j ð1 þ bÞ þ γ j

or

Bð1 þ bÞ3 þ βj ð1 þ bÞ2 þ β k ð1 þ bÞ þ γ k rAð1 þ bÞ4 þ αk ð1 þbÞ3 þ β k ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j

or

Bð1 þ bÞ3 þ βj ð1 þ bÞ2 þ β k ð1 þ bÞ þ γ k rAð1 þ bÞ4 þ αk ð1 þbÞ3 þ β k ð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

or

Bð1 þ bÞ3 þ βj ð1 þ bÞ2 þ β k ð1 þ bÞ þ γ k rBð1 þ bÞ3 þ βk ð1 þbÞ2 þ γ k ð1 þ bÞ þ γ j

or

Bð1 þ bÞ3 þ βj ð1 þ bÞ2 þ β k ð1 þ bÞ þ γ k rBð1 þ bÞ3 þ βk ð1 þbÞ2 þ β j ð1 þ bÞ þ γ j

or

Bð1 þ bÞ3 þ βj ð1 þ bÞ2 þ β k ð1 þ bÞ þ γ k rCð1 þ bÞ2 þ γ k ð1 þbÞ þ γ j

ðviÞ Either

Cð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k r Að1 þ bÞ4 þ αk ð1 þ bÞ3 þ αj ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j

or

Cð1 þbÞ2 þ γ j ð1 þ bÞ þ γ k r Að1 þ bÞ4 þ αk ð1 þ bÞ3 þ βk ð1 þbÞ2 þ β j ð1 þ bÞ þ γ j

or

Cð1 þbÞ2 þ γ j ð1 þ bÞ þ γ k r Að1 þ bÞ4 þ αk ð1 þ bÞ3 þ βk ð1 þbÞ2 þ γ k ð1 þ bÞ þ γ j

or

Cð1 þbÞ2 þ γ j ð1 þ bÞ þ γ k r Bð1 þ bÞ3 þ βk ð1 þ bÞ2 þ γ k ð1 þ bÞ þ γ j

or

Cð1 þbÞ2 þ γ j ð1 þ bÞ þ γ k r Bð1 þ bÞ3 þ βk ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j

or

γ j ð1 þ bÞ þ γ k r γ k ð1 þbÞ þ γ j

Then, S1 dominates S2.

2. A counter-example The following example shows that the upper dominance rules in WW are incorrect. Example 1. Let n ¼5 and b¼0.5. The basic processing time of jobs on Mi are provided in Table 1. Let π ¼ 1  5, π 0 ¼ 2, Jj ¼3 and Jk ¼ 4; therefore, S1 ¼ 1  5  3 4  2 and S2 ¼ 1  5  4  3  2. According to Table 1, values of A, B and C will be 5, 13.50 and 26.25, respectively. The optimal sequence is S2 ¼ 1 5  4  3  2 with makespan 128.97 shown in Fig. 1 and Table 2. Notably, according to any arbitrary dominance rules in “WW” satisfied for this example, e.g., the following case, S1 with makespan 132.34 (Fig. 2 and Table 3) dominates S2 with makespan 128.97. Consequently, this leads to omit the optimal sequence S2 ; thus, the dominance

A.-A. Jafari et al. / Computers & Operations Research 72 (2016) 93–96

95

rules in “WW” are not correct. ði−1Þ

αj ð1 þ bÞ3 þ αk ð1 þ bÞ2 þ βk ð1 þ bÞ þ γ k ≤ αk ð1 þ bÞ3 þ αj ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j 6ð1 þ 0:5Þ3 þ 5ð1 þ 0:5Þ2 þ 7ð1 þ 0:5Þ þ 2 ¼ 44 ≤ 5ð1 þ 0:5Þ3 þ 6ð1 þ 0:5Þ2 þ 10ð1 þ 0:5Þ þ 9 ¼ 54:37

ðii−2Þ

αj ð1 þ bÞ3 þ βj ð1 þ bÞ2 þ βk ð1 þ bÞ þ γ k ≤ αk ð1 þ bÞ3 þ βk ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j 6ð1 þ 0:5Þ3 þ 10ð1 þ 0:5Þ2 þ 7ð1 þ 0:5Þ þ 2 ¼ 55:25 ≤ 5ð1 þ 0:5Þ3 þ 7ð1 þ 0:5Þ2 þ 10ð1 þ 0:5Þ þ 9 ¼ 56:63

ðiii−5Þ Að1 þ bÞ4 þ αj ð1 þ bÞ3 þ βj ð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k ≤ Bð1 þ bÞ3 þ βk ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j 5ð1 þ 0:5Þ4 þ 6ð1 þ 0:5Þ3 þ 10ð1 þ 0:5Þ2 þ 9ð1 þ 0:5Þ þ 2 ¼ 83:56 ≤ 13:5ð1 þ 0:5Þ3 þ 7ð1 þ 0:5Þ2 þ 10ð1 þ 0:5Þ þ 9 ¼ 85:31 ðiv−5Þ

βj ð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k ≤ βk ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j 10ð1 þ 0:5Þ2 þ 9ð1 þ 0:5Þ þ 2 ¼ 38 ≤ 7ð1 þ 0:5Þ2 þ 10ð1 þ 0:5Þ þ 9 ¼ 39:75

ðv−2Þ

Bð1 þ bÞ3 þ βj ð1 þ bÞ2 þ βk ð1 þ bÞ þ γ k ≤ Að1 þ bÞ4 þ αk ð1 þ bÞ3 þ βk ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j 13:5ð1 þ 0:5Þ3 þ 10ð1 þ 0:5Þ2 þ 7ð1 þ 0:5Þ þ 2 ¼ 80:56 ≤ 5ð1 þ 0:5Þ4 þ 5ð1 þ 0:5Þ3 þ 7ð1 þ 0:5Þ2 þ 10ð1 þ 0:5Þ þ 9 ¼ 81:93

ðvi−1Þ

Cð1 þ bÞ2 þ γ j ð1 þ bÞ þ γ k ≤ Að1 þ bÞ4 þ αk ð1 þ bÞ3 þ αj ð1 þ bÞ2 þ βj ð1 þ bÞ þ γ j 26:25ð1 þ 0:5Þ2 þ 9ð1 þ 0:5Þ þ 2 ¼ 74:56 ≤ 5ð1 þ 0:5Þ4 þ 5ð1 þ 0:5Þ3 þ 6ð1 þ 0:5Þ2 þ 10ð1 þ 0:5Þ þ 9 ¼ 79:69

The above erroneous Proposition will be corrected as follows: According to the above example, the condition C k ðS1 Þ r C j ðS2 Þ proposed by Wang and Wang [1] is neither necessary nor sufficient. This relation must be replaced with the condition t 3r ðS1 Þ ¼ MaxfC 2r ðS1 Þ; C k ðS1 Þg r t 3r ðS2 Þ ¼ MaxfC 2r ðS2 Þ; C j ðS2 Þg where r is the first job in π 0 and t 3r ðS1 Þ and t 3r ðS2 Þ are the starting time of Jr on M3 under schedules S1 and S2 , respectively. Since condition C k ðS1 Þ r C j ðS2 Þ in the Proposition is valid, if relation C 2r ðS1 Þ r MaxfC 2r ðS2 Þ; C j ðS2 Þg is also involved; then, it will be perfect although, the relation C k ðS1 Þ r C j ðS2 Þ is not necessary and if the relation MaxfC 2r ðS1 Þ; C k ðS1 Þg r MaxfC 2r ðS2 Þ; C j ðS2 Þg is established, then the relation C k ðS1 Þ r C j ðS2 Þ may not be indefeasible but S1 will dominate S2 . In the counter-example, we have r ¼ 2. Noteworthily, the relation C k ðS1 Þ ¼ 83:56 r C j ðS2 Þ ¼ 85:31 is satisfied; however, the relation C 2r ðS1 Þ ¼ 87:56 r MaxfC 2r ðS2 Þ; C j ðS2 Þg ¼ Maxf85:31; 85:31g ¼ 85:31 is not held; hence, S1 does not dominate S2 . In this example, S2 dominates S1 according to the relation MaxfC 2r ðS2 Þ; C j ðS2 Þg ¼ Maxf85:31; 85:31g ¼ 85:31 r MaxfC 2r ðS1 Þ; C k ðS1 Þg ¼ Maxf87:56; 83:56g ¼ 87:56, however, the relation C j ðS2 Þ ¼ 85:31 r C k ðS1 Þ ¼ 83:56 is not valid. Proof. Since the partial sequence π 0 is the same in S1 and S2 , therefore the makespan of both schedules depends on the starting time of Jr on M3 where r is the first job in π 0 as mentioned. On the other hand, each sequence with the smallest t 3r will have the smallest makespan. So, if the relation t 3r ðS1 Þ ¼ MaxfC 2r ðS1 Þ; C k ðS1 Þg r t 3r ðS2 Þ ¼ MaxfC 2r ðS2 Þ; C j ðS2 Þg is satisfied, then S1 will dominate S2 . Remark. To simplify proposition of WW, phrase Bð1 þ bÞ3 may be removed from both sides of the fourth and fifth equations of set (v). Table 1 Basic processing time of Jj on Mi (aij). job Machine

1

2

3

4

5

M1 M2 M3

2 2 4

9 9 1

6 10 9

5 7 2

2 6 6

M1

1

5

2 M2

M3

4

5 1

3

12.5 5

5

2

24.75 4

13.5

1

3

27.25

5

11.5

46.13

26.25

50.88

2

85.31

3

4

42.88

2

85.31

Fig. 1. Optimal sequence S2 ¼ 1  5  4  3 2.

128.97

96

A.-A. Jafari et al. / Computers & Operations Research 72 (2016) 93–96

Table 2 Starting, processing and completion time of S2 on Mi. Sequence M1

M2

M3

Starting time Processing time Completion time Starting time Processing time Completion time Starting time Processing time Completion time

1

5

4

3

2

0.00 2.00 2.00 2.00 3.00 5.00 5.00 6.50 11.50

2.00 3.00 5.00 5.00 8.50 13.50 13.50 12.75 26.25

5.00 7.50 12.50 13.50 13.75 27.25 27.25 15.63 42.88

12.50 12.25 24.75 27.25 23.63 50.88 50.88 34.44 85.31

24.75 21.38 46.13 50.88 34.44 85.31 85.31 43.66 128.97

Fig. 2. Non-optimal sequence S1 ¼ 1  5  3 4 2.

Table 3 Starting, processing and completion time of S1 on Mi. Sequence M1

M2

M3

Starting time Processing time Completion time Starting time Processing time Completion time Starting time Processing time Completion time

1

5

3

4

2

0.00 2.00 2.00 2.00 3.00 5.00 5.00 6.50 11.50

2.00 3.00 5.00 5.00 8.50 13.50 13.50 12.75 26.25

5.00 8.50 13.50 13.50 16.75 30.25 30.25 24.13 54.38

13.50 11.75 25.25 30.25 22.13 52.38 54.38 29.19 83.56

25.25 21.63 46.88 52.38 35.19 87.56 87.56 44.78 132.34

Reference [1] Wang J-B, Wang M-Z. Minimizing makespan in three machine flow shop with deteriorating jobs. Comput Oper Res 2013;40:547–57.