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Parallel Machine Scheduling with Batch Setup TimesAuthor(s): T. C. E. Cheng and Z.-L. ChenSource: Operations Research, Vol. 42, No. 6 (Nov. - Dec., 1994), pp. 1171-1174Published by: INFORMSStable URL: http://www.jstor.org/stable/171994 .Accessed: 09/05/2014 18:16

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CHENG AND CHEN / 1171

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WEIN, L. M., AND P. B. CHEVALIER. 1992. A Broader View of the Job-Shop Scheduling Problem. Mgmt. Sci. 38, 1018-1033.

PARALLEL MACHINE SCHEDULING WITH BATCH SETUP TIMES

T. C. E. CHENG Hong Kong Polytechnic, Hung Hom, Kowloon, Hong Kong

Z.-L. CHEN Shanghai Transportation Planning Institute, Shanghai, People's Republic of China

(Received September 1992; revisions received April, June 1993; accepted August 1993)

We consider a problem of scheduling several batches of jobs on two identical parallel machines to minimize the total completion time of jobs. A setup time is incurred whenever there is a switch from processing a job in one batch to a job in another batch. When the number of jobs is arbitrary, the computational complexity of the problem is posed as an open problem in the literature. We show in this note that the problem is binary NP-hard even when the setup times are sequence independent and all processing times are equal.

W e are given B batches of jobs. Each batch i for "Y 1 % i < B contains a set of Ni jobs: Ji = {Jil, Ji2, * ** JiNi. Let N = >fL1 Ni and J = UB 1 Ji. There are two identical parallel machines on which all the Njobs are to be processed. A batch of jobs cannot be processed by more than one machine simulta- neously. Each job Jii E J has a given processing time

pj E Z+, where Z+ is the set of positive integers. A setup time Si, E Z+ is incurred whenever a job from batch j is processed immediately after a job from batch i on the same machine. We let Sii = 0 for all 1 < i S B. Also, an initial setup time Soi E Z+ is incurred if a job from batchj is processed first on a machine. The setup times are called sequence independent if SJ = S, for 0 % i % B, where i 1 j; otherwise, the setup times are called sequence dependent.

Monma and Potts (1989) stated that when the num- ber of batches B is fixed there exist pseudopolynomial algorithms for the maximum com- pletion time, the maximum lateness, the total weighted completion

time, and the weighted number of late jobs problems. They also showed that when the number of batches B is arbitrary the maximum completion time, the max- imum lateness, the number of late jobs, and the total weighted completion time problems are all NP-hard. But the computational complexity of the total completion-time problem with an arbitrary number of batches was posed as an open problem in their paper.

We will study this open problem in this note. Spe- cifically, we will consider a special case of the prob- lem in which the setup times are sequence independent, i.e., Sij = Sj for all 0 S i S B with i X j, and all processing times are equal, i.e.,pij = p for all Ji1 E J. The objective is to find an optimal schedule to process the jobs on the two machines so that the total completion time f = -1--j ANi 1-i B Cij is mini- mized, where Ci1 is the completion time of job Jif. We call this special problem the two-parallel machine total completion time (P2TCT) problem and show that P2TCT is NP-hard by a reduction from PARTITION.

Subject classifications: Analysis of algorithms, computational complexity: NP-hardness. Production/scheduling: parallel machine scheduling. Area of review: MANUFACTURING, OPERATIONS AND SCHEDULING.

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1172 / CHENG AND CHEN

PRELIMINARY RESULT

The following lemma identifies the structure of an optimal schedule for P2TCT.

Lemma 1. Given an optimal schedule, suppose that a machine is to process ni, 0 S ni > Ni jobs from batch i, 1 > i > B. Then, on this machine, the following properties hold:

i. the jobs from the same batch are processed consecutively;

ii. the batches are sequenced in nondecreasing order of Si/ni-

Proof. Lemma 1 follows immediately from Proposi- tion 1 in Dobson, Karmarkar and Rummel (1987) when pij = p for all Jij E J.

NP-HARDNESS RESULT

Now, we will show that P2TCT is binary NP-hard by showing that its associated decision problem is NP- complete by a reduction from PARTITION, a known NP-complete problem (Garey and Johnson 1979).

The decision problems of PARTITION and P2TCT are stated as follows.

PARTITION. Given t positive integers a1, a2, ..., at such that -t=1 ai = 2A is even, does there exist a subset U C T = {1, 2, ..., t} such that >LEu ai = >ie\u ai - A?

P2TCT. Given two identical parallel machines, B batches of jobs, a set of Ni jobs: Ji = {Jil, Ji2, . . .* JiNi} a setup time Si for each batch i, a processing timep for each jobJij, 1 S i S B, 1 > j S Ni, and a threshold Y, does there exist a schedule such that

f = y- El --j --Ni l -_ Si --B C ij < Y?

Given any instance I of PARTITION, we can con- struct in polynomial time a corresponding instance I' of P2TCT as: B = t; Ni = ai, Si = 2Aai, 1 < i S B; p = 1;

Y=2A > aiaj-2A3+A2+A. 1

CHENG AND CHEN / 1173

Figure 1. A schedule with a batch split over both machines.

We now begin to show the NP-completeness of P2TCT.

Theorem. The decision version of P2TCT is binary NP-complete.

Proof

i. The proof that P2TCT E NP is trivial, and we omit it.

ii. PARTITION is polynomially reducible to P2TCT. To prove part ii, it suffices to show that there exists a solution to instance I' if and only if there exists a solution to instance I.

If Part

Suppose that there exists a subset U C T = {1, 2, ... , t} such that Lieu ai = >ie7\u ai = A. Without loss of generality, let U = {1, 2, ... , k} and T\U = {k + 1, k + 2, ... , t}. We now con- struct a schedule V for instance I' as follows: Machine one processes batches 1, 2,..., k and machine two pro- cesses batches k + 1, k + 2, ..., t; batches are processed in an arbitrary order. Then the total com- pletion time of schedule V is given by

k k N Ni+-+Nk

f (V)=I Si I Nj + I i i=l j=i i=l

B B Nk+1+ +NB

+ 2 Si ( Nj + i i=k+l j=l i-1

+2A I ai ( aj + 2 ai - 1 +a )

i=k1 \= 2i=1 \i=k 1

2A( aIai +?2 aj + ai) a2+

1 i=kl k+= 1 2 i=k+l

t ~

=2A I aiaj - + aiaj-2+

2 .i=1S i=l ik+l

= 2A I aiaj - 2A3+A2 +A = Y, 1

1174 / CHENG AND CHEN

andtwoare(1, 2, ..., k - 1, k) and (k + 1, k + 2, . .. , B, k), respectively. Then the total completion time of W is

k-1 k-1 k

PMW)= 2 Si 2 Nj) + nj 2 Si i=1 j=l i=1

N * *+Nk-I +nl

+ i=l

B B B + 2 Si(2 Nj) + (Nk -nl) 2 Si

i=k+l j=1 i=k

Nk +- * *+NB--nl

+ (1) i=l1

=2A( aiaj + aiaj) 1