parallel machine scheduling of machine-dependent jobs with unit-length

6
Short Communication Parallel machine scheduling of machine-dependent jobs with unit-length q Yixun Lin * , Wenhua Li Department of Mathematics, Zhengzhou University, Zhengzhou, Henan 450052, China Received 24 April 2002; accepted 5 December 2002 Abstract In this paper we study the parallel machine scheduling problem with unit-length jobs in which each job j is only allowed to be processed on a specified subset M j of machines. For the problem P jp j ¼ 1; M j jC max where the subset family fM j g is nested, Pinedo [Scheduling: Theory, Algorithms, and Systems, Prentice-Hall, Englewood Cliffs, NJ, 1995] established an least flexible job first algorithm. We further present polynomial algorithms for the problem with general subset family fM j g. As a generalization of the problem for nested family, we consider the problem for convex subset family fM j g and present an efficient algorithm for solving it. Ó 2003 Elsevier B.V. All rights reserved. Keywords: Scheduling; Parallel machine; Unit-length job; Machine eligibility constraint 1. Introduction The parallel machine scheduling problems have been extensively studied in the literature [3,4,7]. In a basic model, there is a set M ¼fM 1 ; M 2 ; ... ; M m g of identical machines, which are used to process n jobs J 1 ; J 2 ; ... ; J n . Each job J j has one operation which can be processed on any machine with processing time p j . The objective is to minimize the makespan C max . By the well-known three-field representation [3], this problem is denoted by P kC max . In some practical applications, the machines may have different capabilities (or eligibility re- strictions). This gives rise to a generalization that each job J j is only allowed to be processed on a specified subset M j of machines (M j M). This subset M j is called the admissible set of job J j (1 6 j 6 n). The generalized model can be denoted by P jM j jC max . In fact, it can be regarded as a special case of unrelated parallel machine sched- uling with processing times p ij ¼ p j ; if M i 2 M j ; 1; if M i 62 M j : Since P kC max is NP-hard, it is clear that the gen- eralization P jM j jC max is also NP-hard. Pinedo [9] proposed the problem P jp j ¼ 1; M j jC max , in which each job has a processing time of unit. When the admissible subset family fM j g is nested, i.e., q Project supported by NSFC (grant no. 10371112). * Corresponding author. E-mail address: [email protected] (Y. Lin). 0377-2217/$ - see front matter Ó 2003 Elsevier B.V. All rights reserved. doi:10.1016/S0377-2217(02)00914-1 European Journal of Operational Research 156 (2004) 261–266 www.elsevier.com/locate/dsw

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European Journal of Operational Research 156 (2004) 261–266

www.elsevier.com/locate/dsw

Short Communication

Parallel machine scheduling of machine-dependentjobs with unit-length q

Yixun Lin *, Wenhua Li

Department of Mathematics, Zhengzhou University, Zhengzhou, Henan 450052, China

Received 24 April 2002; accepted 5 December 2002

Abstract

In this paper we study the parallel machine scheduling problem with unit-length jobs in which each job j is only

allowed to be processed on a specified subset Mj of machines. For the problem P jpj ¼ 1;MjjCmax where the subset

family fMjg is nested, Pinedo [Scheduling: Theory, Algorithms, and Systems, Prentice-Hall, Englewood Cliffs, NJ,

1995] established an least flexible job first algorithm. We further present polynomial algorithms for the problem with

general subset family fMjg. As a generalization of the problem for nested family, we consider the problem for convex

subset family fMjg and present an efficient algorithm for solving it.

� 2003 Elsevier B.V. All rights reserved.

Keywords: Scheduling; Parallel machine; Unit-length job; Machine eligibility constraint

1. Introduction

The parallel machine scheduling problems have

been extensively studied in the literature [3,4,7]. In

a basic model, there is a set M ¼ fM1;M2; . . . ;Mmgof identical machines, which are used to process njobs J1; J2; . . . ; Jn. Each job Jj has one operation

which can be processed on any machine with

processing time pj. The objective is to minimize themakespan Cmax. By the well-known three-field

representation [3], this problem is denoted by

PkCmax.

In some practical applications, the machines

may have different capabilities (or eligibility re-

qProject supported by NSFC (grant no. 10371112).* Corresponding author.

E-mail address: [email protected] (Y. Lin).

0377-2217/$ - see front matter � 2003 Elsevier B.V. All rights reserv

doi:10.1016/S0377-2217(02)00914-1

strictions). This gives rise to a generalization thateach job Jj is only allowed to be processed on a

specified subset Mj of machines (Mj � M). This

subset Mj is called the admissible set of job Jj(16 j6 n). The generalized model can be denoted

by P jMjjCmax. In fact, it can be regarded as a

special case of unrelated parallel machine sched-

uling with processing times

pij ¼pj; if Mi 2 Mj;1; if Mi 62 Mj:

Since PkCmax is NP-hard, it is clear that the gen-

eralization P jMjjCmax is also NP-hard.

Pinedo [9] proposed the problem P jpj ¼ 1;MjjCmax, in which each job has a processing timeof unit. When the admissible subset family fMjg is

nested, i.e.,

ed.

Fig. 1. Network N .

262 Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266

Mi \Mj 6¼ ; ) Mi � Mj or

Mj � Mi for all i and j;

he established a least flexible job first (LFJ) algo-

rithm. The computational complexity of

P jpj ¼ 1;MjjCmax for general subset family fMjgwas left open. Later, Centeno and Armacost [2]

studied a special case of P jrj;MjjLmax, namely, the

problem of minimizing the maximum lateness Lmax

with release times rj and due dates dj ¼ rj þ c.Bhatia et al. [1] also considered the machine ca-

pability constraint, and they focused attention on

a more complicated model with precedence and

loading (setup) time constraints, which was shownto be NP-hard.

In this paper we mainly investigate the problem

P jpj ¼ 1;MjjCmax with general subset family

fMjg. First we present an Oðn3 log nÞ-time algo-

rithm, which can be naturally generalized to the

case where the machines have different speeds.

Then we give a more efficient algorithm for the

problem in which the family fMjg is convex,which is a generalization of nested family. Here, a

subset family fMjg is said to be convex if we can

order m machines in a sequence, say (M1;M2; . . . ;Mm), such that each subset Mj consists of consec-

utive terms of the sequence. This is similar to

the concept of ‘‘convex bipartite graph’’ [6]. In

the case of fMjg being convex, we obtain an

Oðmðmþ nÞÞ algorithm.The organization of the paper is as follows. In

Section 2 we establish the basic algorithm for

P jpj ¼ 1;MjjCmax with general constraint. Then,

we apply the algorithm to Qjpj ¼ 1;MjjCmax. Sec-

tion 3 is devoted to the case of convex subset

family fMjg and to develop a revised LFJ algo-

rithm. We provide some concluding remarks in

Section 4.

2. Binary search algorithm

Let us consider the problem P jpj ¼ 1;MjjCmax

for which M ¼ fM1;M2; . . . ;Mmg is the set of midentical machines; J1; J2; . . . ; Jn are the jobs with

unit-length (pj ¼ 1); the machine capability con-straint is represented by a family of admissible

subsets Mj ¼ fMi 2 MjJj can be processed on Mig(j ¼ 1; 2; . . . ; n); and the objective function is the

makespan Cmax (the final completion time of all jobs).

We will use a network flow based method on

the problem. The network flow theory can beconsulted in [6,8]. For any given subset family

fMjg we can construct a network N as follows.

Let X ¼ fx1; x2; . . . ; xmg correspond to the set of mmachines, Y ¼ fy1; y2; . . . ; yng to the set of n jobs.

The vertex set of N is V ¼ fs; tg [ X [ Y , where s isa source and t is a sink. The arc set A of N consists

of the following arcs:

ðs; xiÞ for all xi 2 X ði ¼ 1; . . . ;mÞ;ðxi; yjÞ for Mi 2 Mj ð16 i6m; 16 j6 nÞ;ðyj; tÞ for all yj 2 Y ðj ¼ 1; . . . ; nÞ:

All arcs ðxi; yjÞ and ðyj; tÞ have unit capacity; and

all arcs ðs; xiÞ have capacity c, a variable parame-

ter. Without loss of generality, we may assume

that c is a positive integer. We denote this network

by N ¼ ðV ;A; cÞ, as shown in Fig. 1.

Theorem 2.1. The problem P jpj ¼ 1;MjjCmax has afeasible schedule of makespan Cmax 6 c if and only ifthe value of max-flow in network N ¼ ðV ;A; cÞ is n.

Proof. If the scheduling problem has a feasible

schedule of Cmax 6 c, then at most c jobs are as-

signed on each machine. We may define a flow f in

network N as follows:

f ðxi; yjÞ ¼1; if Jj is assigned to Mi;

0; otherwise;

f ðyj; tÞ ¼Xmi¼1

f ðxi; yjÞ ¼ 1 for j ¼ 1; . . . ; n;

Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266 263

f ðs; xiÞ ¼Xnj¼1

f ðxi; yjÞ6 c for i ¼ 1; . . . ;m:

It is easy to see that this is a max-flow in N withvalue n. Conversely, if the value of a max-flow in Nis n, then there exists an integral max-flow f such

that all arcs ðyj; tÞ are saturated. This is because

that all these arcs have unit capacity and they form

a cut of capacity n. Thus the n unit flow f must

saturate this cut. So we can get a feasible schedule

by assigning job Jj to machine Mi if f ðxi; yjÞ ¼ 1.

Since each arc ðs; xiÞ has capacity c, it follows thatat most c jobs are assigned on each machine.

Therefore the makespan Cmax is not greater than c.This completes the proof. �

Based on this theorem, the problem P jpj ¼1;MjjCmax can be solved by searching for the

minimal parameter c. Thus we obtain the follow-

ing binary search algorithm.

BS Algorithm

Step 0: Construct the network N ¼ ðV ;A; cÞ fromthe subset family fMjg (where c is an un-

known parameter for the time being). Let

l ¼ 1 and u ¼ n.Step 1: If l ¼ u, then the optimal value is reached:

C�max ¼ l; return the schedule of makespan

l, stop.Step 2: Let c ¼ b1

2ðlþ uÞc. Find the max-flow in

network N ¼ ðV ;A; cÞ. If the value of

max-flow is exactly n, namely C�max 6 c,

then set u :¼ c and write down the sched-

ule of makespan c by means of the max-

flow. Otherwise, the value of max-flow is

less than n, namely, C�max > c, we set

l :¼ cþ 1. Go back to Step 1.

Fig. 2. Network N .

Theorem 2.2. The BS algorithm correctly solves theproblem P jpj ¼ 1;MjjCmax for any subset familyfMjg in Oðn3 log nÞ time.

Proof. By Theorem 2.1, we can assert that

l6C�max 6 u holds during the process of the algo-

rithm. As soon as the algorithm terminates,

l ¼ C�max ¼ u, an optimal solution is obtained. As

to the running time of the algorithm, it is clear that

the binary search has log n stages each of which is

a procedure of finding max-flow in network N . It is

well-known that there are Oðn3Þ algorithms for

the max-flow problem [6,8]. Therefore the time

bound of our algorithm is Oðn3 log nÞ. The theo-

rem follows. �

In fact, the network N considered in our algo-

rithm has special structure (it can be changed into

a �simple network� if each vertex xi is replaced by cvertices). So, the max-flow algorithm on it would

have complexity less than Oðn3Þ (see [8] for de-

tails). Therefore the complexity result in Theorem

2.2 can be improved slightly.

We can further consider the problemQjpj ¼ 1;MjjCmax, i.e., the case of uniform parallel

machines. In this case, m machines have different

speeds; the speed of machine Mi is denoted by vi.Accordingly, the actual processing time of job Jjon machine Mi is pj=vi (at the moment pj ¼ 1).

Without loss of generality we may assume that all

speeds v1; v2; . . . ; vm are positive integers (the ma-

chines with vi ¼ 1 are regarded as standard ma-chines). Let v be the least common multiple of

v1; v2; . . . ; vm. Then the completion time of every

job will be an integer if we take 1=v as a time unit.

We may construct the network N ¼ ðV ;A; cÞ asbefore, except that the capacity of arc ðs; xiÞ is

defined to be vic, as shown in Fig. 2.

Similar to Theorem 2.1, we have

Theorem 2.3. The problem Qjpj ¼ 1;MjjCmax has afeasible schedule of makespan Cmax 6 c if and only ifthe value of max-flow in network N ¼ ðV ;A; cÞ is n.

264 Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266

So we can use the binary search algorithm todetermine the minimal makespan C�

max and the

optimal schedule as before. However, the value of

parameter c now may not be integral. The range

½1=v; n� of c must be divided into vn� 1 sections by

the time unit 1=v and the binary search is executed

in these sections. Hence, there will be Oðlog nvÞstages in the BS algorithm. We then come to the

following conclusion.

Theorem 2.4. The BS algorithm correctly solves theproblem Qjpj ¼ 1;MjjCmax for any subset familyfMjg in Oðn3 log nvÞ time.

3. Convex subset family

We discuss a special case––the problem P jpj ¼1;MjjCmax where fMjg is a convex subset family.

We first address the problem of determining the

convexity of a subset family fMjg. In fact, a

family fMjg can be represented by a ð0; 1Þ-matrix

in which a row corresponds an element Mi and a

column corresponds a subset Mj, and the entry

ði; jÞ is 1 if and only if Mi 2 Mj. Then fMjg isconvex if and only if the matrix has �the consecu-

tive 1�s property for columns�, namely, the rows of

the matrix can be permuted in such a way that the

1�s in each column occur consecutively. This is also

equivalent to the interval hypergraph recognition

problem. It is well-known in the literature [5] that

the recognition of the consecutive 1�s property (as

well as that of interval graphs or interval hyper-graphs) can be solved in linear time. Hence, we

may assume here that the convexity of family

fMjg has been shown.

For the sake of simplicity, suppose that

M ¼ f1; 2; . . . ;mg. Moreover, by an integer inter-

val ½a; b�, we mean the set of integers

fi 2 Z j a6 i6 bg ¼ fa; aþ 1; . . . ; b� 1; bg (a6 b).For the case of convex family, we may assumethat m machines have been ordered as (1; 2; . . . ;m)such that each admissible set Mj is an integer in-

terval:

Mj ¼ ½aj; bj� ðaj; bj 2 M; aj 6 bjÞ:For any a; b 2 Mða6 bÞ, we define

J ½a; b� ¼ fJjja6 aj 6 bj 6 bgas a set of jobs whose admissible sets are included

in ½a; b�.

Theorem 3.1. The minimal makespan of the prob-lem is

C�max ¼ max

16 a6 b6m

jJ ½a; b�jb � a þ 1

� �:

Proof. Denote

k ¼ max16 a6b6m

jJ ½a; b�jb � a þ 1

� �:

We are going to prove C�max ¼ k. For any given

a and b (16 a6b6m), the jobs in J ½a; b� are onlyallowed to be processed on the b � a þ 1 machinesin ½a; b�. So, the final completion time c½a;b� of

these machines satisfies

c½a; b�P jJ ½a; b�jb � a þ 1

:

By the arbitrariness of a and b, we have C�max P k.

In order to show C�max 6 k, it suffices to con-

struct a feasible schedule of makespan k. We give

an algorithm as follows.

Revised LFJ Algorithm

Step 0: Let i ¼ 1; S ¼ ; (where S denotes the set of

jobs which have been assigned).

Step 1: Let Ti ¼ fJjjJj 62 S; aj 6 i6 bjg be the set

of jobs which can be assigned to machine

i. If Ti ¼ ;, go to the next step. If

0 < jTij6 k, then assign all jobs in Ti tomachine i; If jTij > k, then, according tothe non-decreasing order of fbjg (the

�LFJ� rule), assign the first k jobs in Ti tomachines i. Put all assigned jobs into S.

Step 2: Stop if i ¼ m; otherwise let i :¼ iþ 1, go

back to Step 1.

We proceed to show that the algorithm can

produce a feasible schedule of makespan k. If not,there must be a job Jj which cannot be assigned in

any machine. Suppose that this Jj is the first job

not being assigned (i.e., the one with the least bj

Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266 265

among those failed jobs). For brevity, the ma-chines which have received k jobs in the algorithm

are called �full machines�. Then, for aj 6 i6 bj, themachines i must be full, so that there is no room to

hold Jj. Besides, there may be more full machines.

Let b ¼ bj and let a be the least number such that

all machines in ½a; b� are full. Then 16 a6 aj. Note

that the machine ða � 1Þ is not full if a > 1. It can

be seen that Jj and all jobs assigned to the ma-chines in ½a; b� belong to J ½a; b�. In fact, for a job Jiassigned in ½a; b�, the rule in Step 1 implies that

bi 6 bj ¼ b. On the other hand, we have ai P a.Otherwise, ai 6 a � 1, Ji would be assigned to ma-

chine a � 1, as it is not full. So Ji 2 J ½a; b�. Hence

jJ ½a; b�jP ðb � a þ 1Þk þ 1;

thus

jJ ½a; b�jb � a þ 1

> k;

which contradicts the definition of k. Therefore

C�max ¼ k. The theorem follows. �

Theorem 3.2. The revised LFJ algorithm correctlysolves the problem P jpj ¼ 1;MjjCmax with convexsubset family fMjg in Oðmðmþ nÞÞ time.

Proof. The correctness has been proved in Theo-

rem 3.1. Let us see the time bound of the algo-

rithm. Note that we have to compute k in advance.For doing this, there are two steps as follows.

First, we should calculate an m� mmatrix kab

� �where

kab ¼ jJ ½a; b�j; if a6 b;0; otherwise:

This can be done by constructing the sets J ½a;�� ¼fJjja6 ajg and sets J ½�; b� ¼ fJjjbj 6 bg in OðmnÞtime (for each a or b, scan all n jobs); and com-

puting all jJ ½a; b�j ¼ jJ ½a;�� \ J ½�; b�j in Oðm2Þtime.

Second, compute

k ¼ max16 a6 b6m

kab

b � a þ 1

� �

in Oðm2Þ time.

After k has been computed, the revised LFJ

algorithm can be carried out in OðmnÞ time. In

fact, the algorithm has m stages and in each stagewe assign the jobs in Ti in OðnÞ time.

To summarize, the overall complexity of the

algorithm is Oðmðmþ nÞÞ. This completes the

proof. �

Pinedo [9] studied the case of nested subset

family fMjg (i.e., Mi \Mj 6¼ ; ) Mi � Mj or

Mj � Mi). It is easy to show by induction on nthat if a family fMjg is nested, then it is convex.

But the converse is not true. Theorem 3.2 implies

the following.

Corollary 3.3. For nested subset family fMjg, theproblem P jpj ¼ 1;MjjCmax has an Oðmðmþ nÞÞ al-gorithm.

The following example in [9] illustrates that the

LFJ algorithm may not work if fMjg is not nes-

ted:

M1 ¼ f1; 2g; M2 ¼ M3 ¼ f1; 3; 4g;M4 ¼ f2g; M5 ¼ M6 ¼ M7 ¼ M8 ¼ f3; 4g:

However, this fMjg is convex in viewing the order

(2,1,3,4) of machines. Our algorithm provides the

following solution:

4. Concluding remarks

In the foregoing sections we have investigated

the parallel machine scheduling problems withadmissible subset constraint––each job j is only

allowed to be processed on the machines in subset

Mj. These models come up in connection with

many application backgrounds. Meanwhile, they

have theoretic interest in the viewpoint of

266 Y. Lin, W. Li / European Journal of Operational Research 156 (2004) 261–266

combinatorics. In fact, they can be regarded asgeneralizations of the system of distinct represen-

tatives (SDR) problem: the existence of SDR is

equivalent to determining whether the problem

P jpj ¼ 1; MjjCmax (mP n) has a schedule of

Cmax ¼ 1.

There are many aspects worthwhile to further

study. For example, we have considered the ob-

jective Cmax. How about the others, e.g.,P

Cj,Lmax,

PUj, etc? In the case of ordinary processing

times (not necessarily unit-length), the problem

P jMjjCmax is NP-hard. It is significant to study the

approximation algorithms and other polynomially

solvable cases. As a generalization of nested subset

family, we have taken the convex subset family

fMjg into account. More results on other special

cases would be expected.

Acknowledgements

The authors would like to thank the referees for

their helpful comments on improving the repre-

sentation of the paper.

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