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An improved heuristic for two-machine owshop scheduling with

an availability constraint

T.C. Edwin Cheng

a;∗

_{, Guoqing Wang}

b;c

a_{Oce of the Vice-President (Research and Postgraduate Studies) The Hong Kong Polytechnic University,}

Kowloon, Hong Kong

b_{Department of Business Administration, Jinan University, Guangzhou, People’s Republic of China} c_{Department of Management, The Hong Kong Polytechnic University, Kowloon, Hong Kong}

Received 1 May 1998; received in revised form 1 March 2000

Abstract

In this paper we study the two-machine owshop scheduling problem with an availability constraint on the rst machine. We rst show that the worst-case error bound 1

2 of the heuristic provided by Lee [4] is tight. We then develop an improved heuristic with a worst-case error bound of 1

Keywords:Flowshop scheduling; Heuristics; Error bound

1. Introduction

In this paper, we address the two-machine owshop scheduling problem with an availability constraint on the rst machine. We are given two machinesM1and

M2, and a job setS={J1; : : : ; Jn}. Each job has to be

processed rst on M1 and then onM2. In our

prob-lem, M2 is always available, but M1 is unavailable

for job processing during a certain period of time and the time interval of unavailability is assumed to be known in advance. All jobs are assumedresumable, i.e. if a job cannot nish before the unavailable pe-riod of a machine, then its processing may be inter-rupted and resumed at a later time when the machine

∗_{Corresponding author. Fax: +852-2356-2682.}

E-mail address:[email protected] (T.C.E. Cheng).

becomes available again. The objective is to minimize the makespan.

The two-machine owshop scheduling problem with an availability constraint is rst studied by Lee [4]. He shows that the problem is NP-hard and de-velops a pseudo-polynomial dynamic programming algorithm to solve the problem optimally. When the availability constraint is imposed onM1, he provides

a heuristic H2 to tackle the problem and proves that its worst-case error bound is 1₂. But the problem of whether the bound is tight is left open. He also pro-vides another heuristic with a worst-case error bound of 1

3 for a variant of the problem, where the

availabil-ity constraint is imposed onM2. For other works on

owshop scheduling with availability constraints, we refer the interested reader to Kubiak et al. [3], Lee [5], and Cheng and Wang [1].

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224 T.C.E. Cheng, G. Wang / Operations Research Letters 26 (2000) 223–229

In this paper, we rst show that the bound1₂ is tight for H2. We then provide another heuristic to solve the problem and show that the heuristic has a worst-case error bound of 1₃.

2. Notation

For the problem under consideration, we use the following notation.

S={J1; : : : ; Jn}; the set of jobs,

M1,M2: machine 1 and machine 2,

s1, t1:M1 is unavailable from s1 to t1, where

06s16t1,

pi,qi: processing times forJi on M1 andM2,

re-spectively,

=hJ(1); : : : ; J(n)i: a permutation schedule, where

J(i)is theith job in,

Cmax(): the makespan of,

∗: an optimal schedule, C∗: the optimal makespan.

Following the notation of Lee [4], we denote the problem under study as F2=r−a(M1)=Cmax, i.e.

the makespan minimization problem in the two-machine owshop with a resumable availability con-straint onM1.

It is well-known that the classical two-machine owshop makespan minimization problem,F2==Cmax,

can be solved by Johnson’s rule [2]. Johnson’s rule can be implemented as follows:

Johnson’s algorithm. Divide S into two disjoint subsets A and B, where A={Ji|qi¿pi} and B= {Ji|qi¡ pi}. Order the jobs in A in nondecreasing

order ofpiand those jobs inBin nonincreasing order

ofqi. Sequence jobs inArst, followed by those inB.

LetC∗be the optimal makespan forF2==Cmax. It is

evident that

C∗6C∗: (1)

3. An improved heuristic and its worst-case error bound

The heuristic H2 proposed by Lee for F2=r −

a(M1)=Cmaxis as follows:

Heuristic H2

(i) Use Johnson’s algorithm to schedule the jobs and let the corresponding schedule be1

(ii) Sequence jobs in nonincreasing order of qi=pi

and let the corresponding schedule be 2. Let

t be the starting time of the last busy period onM2, namely M2 is idle immediately beforet.

Let Jk be the corresponding job that starts at t

onM2.

(iii) Sequence all jobs in the same order as that in Step (ii) except that we makeJk the rst job in

the sequence. Let the corresponding schedule be 3.

(iv) Choose the schedule with the minimum makespan from the above three schedules. Let CH2= min{Cmax(1); Cmax(2); Cmax(3)}.

It is shown in [4] that (CH2−C∗)=C∗61₂. Below,

we complete the analysis of the algorithm by demon-strating that this bound is tight. Consider a problem instance withn= 3. Letp1= 1,q1=a+ 1, p2=a,

q2 =a2, p3 =a + 2, and q3 = (a + 2)2, s1 =a,

t1=a2+a, and s2=t2= 0, wherea ¿1. It is clear

that Step (i) will result in sequence1=hJ1; J2; J3i

withCmax(1) = 3a2+ 5a+ 5 (see Fig. 1(a)). Steps

(ii) and (iii) will result in2=3=hJ3; J1; J2i, with

Cmax(2) =Cmax(3) = 3a2+ 6a+ 7 (see Fig. 1(b)).

Hence,CH2= 3a2+ 5a+ 5. However, the optimal

so-lution is∗=hJ2; J1; J3iwithC∗= 2a2+ 6a+ 7 (see

Fig. 1(c)). We see that (CH2−C∗)=C∗goes to 1₂ asa

approaches innity.

In what follows, we present an improved heuristic that nds a schedule forF2=r−a(M1)=Cmaxthat is at

most 4=3 times the optimal value.

Heuristic HI:

(i) LetJaandJbbe two jobs with the largest and the

second largest processing time on M2,

respec-tively, i.e. min{qa; qb}¿qifori= 1; : : : ; n,i6=a,

andi6=b.

(ii) Use Johnson’s algorithm to schedule the jobs and let the corresponding schedule be1.

(iii) Sequence jobs in nonincreasing order ofqi=pi.

Let the corresponding schedule be 2. Let t be

the starting time of the latest busy period onM2

in 2, namely M2 is idle immediately before t.

LetJkbe the corresponding job that starts atton

M2.

(iv) Sequence the jobs in the same sequence as that in Step (iii) except thatJaandJb are scheduled as

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Fig. 1. (a) Solution of Step (i); (b) solution of Steps (ii) and (iii); and (c) optimal solution.

(v) Ift ¿ s1andpk6s1, then construct4by

main-taining all jobs in the same order they appear in2, with the exception ofJk which is moved

to position u in which it nishes by time s1

and the job in position u+ 1 nishes after s1

onM1.

(vi) LetS1={Ji|_pqi_i¿1 andpi¡ pk}, andS2=S\

(S1∪ {Jk}). Let5=hS1; Jk; S2i, where all jobs

in S1 are scheduled in nonincreasing order of

qi=pi and jobs inS2 are scheduled according to

Johnson’s rule.

(vii) Choose the schedule with the minimum makespan from the above ve schedules. Let CHI = min{Cmax(1); Cmax(2); Cmax(3);

Cmax(4); Cmax(5)}.

Before we analyze the worst-case performance of heuristic HI, we rst establish the following three lem-mas which will be used in the subsequent analysis. The rst lemma is directly from Lee [4].

Lemma 1. There exists an optimal solution for

F2=r−a(M1)=Cmaxin which all jobs beforeJlfollow

Johnson’s rule and the remaining jobs(includingJl)

also follow Johnson’s rule; where Jl is the rst job

which nishes aftert1 onM1.

Lemma 2. LetT2(w) be the completion time of any

job J2(w) on M1 in 2. Given an arbitrary feasible

schedule;letJ(v) be the last job which nishes no

later than timeT2(w) onM1.We have

(i) Pv

j=1q(j)6Pwj=1q2(j),

(ii) Pn

j=v+1q(j)¿Pnj=w+1q2(j).

Proof It is clear that Pv

j=1p(j)6Pwj=1p2(j) and

Pn

j=v+1p(j)¿Pnj=w+1p2(j). Since all jobs are

se-quenced in nonincreasing order ofqi=piin2, we can

easily check the results.

Lemma 3. For schedule2and jobJkdened in Step

(iii)of Heuristic HI;the following inequalities hold: (i) Cmax(2)6C∗+qk,

(ii) Ift6s1ort¿t1+pk;thenCmax(2)6C∗+pk.

Proof Assume thatJ2(w)=Jk, thenCmax(2) =t+

Pn

j=wq2(j).

(i) LetJ∗(v)be the last job which nishes no later

than time t on M1 in ∗. From Lemma 2(ii), we

havePn

j=v+1q∗(j)¿Pnj=w+1q2(j). Hence, we have

Cmax(2)6C∗+qk.

(ii) LetL be the total idle time onM2 in2. It is

clear thatL=t−Pw−1

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Proof It suces to consider the case withP

pi¿ s1.

Note that Cmax(1)6C∗ + t1 − s1. Hence, if

t1−s16C∗=3 we are done. Therefore, in the

remain-der of the proof, we are only interested in the case with t1−s1¿ C∗=3. Let ˆS={Ji|qi¿ C∗=3; i= 1; : : : ; n}.

It is obvious that |Sˆ|62. When |Sˆ| = 0, from Lemma 3(i), we haveCmax(2)6C∗+qk¡4C∗=3.

Hence, we only need to consider the following two cases.

Case1:|Sˆ|= 2

In this case, we have ˆS={Ja; Jb}. Let us consider

Step (iv). Let C∗_{( ˆ}_S_{) denote the optimal completion}

time of the jobs in ˆSin3. It is clear thatC∗( ˆS)6C∗.

Lett′ _{be the starting time of the last busy period on}

M2 in3, and Jlbe the corresponding job that starts

for otherwise we are done. Now, we need to consider the following two subcases.

It is clear that schedule2 will result in the same

makespan for bothF2=r−a(M1)=CmaxandF2==Cmaxin

this subcase, and the optimal makespanC∗forF2=r−

a(M1)=Cmax is no less than the optimal makespanC∗

forF2==Cmax. LetA′={Ji|qi¿pi; i= 1; : : : ; n; i6=k}

andB′₌_S_\_(A′_{∪ {}_J_k}_{). It is not dicult to see that}

there exists an optimal solution forF2==Cmaxin which

all jobs inA′ _{are scheduled before}_J

k and all jobs in

B′_{are scheduled after}_J

kaccording to Johnson’s rule.

Thus, we have

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Now suppose thatqk¡ pk. LetN be the set of jobs

2.t ¿ s1. We consider the following two situations.

(a) There exists an optimal schedule∗ _{such that}_J k

nishes befores1onM1.

Let us focus on schedule 4 obtained in Step (v).

Lett′ be the starting time of the last busy period on M2;andJ4(v)be the corresponding job that starts att

′

onM2. Wheneverv ¡ uorv ¿ u, following the same

argument as that for Lemma 3, we can easily show thatCmax(4)−C∗6q4(v)6C

(b) There exists no optimal schedule such thatJk

n-ishes befores1 onM1.

It is clear thatpk¡ C∗=3 and sopk¡ qk;

other-wise, we haveC∗_¿_t

1−s1+pk+qk¿ C∗, a

contra-diction. Let us focus on schedule5 obtained in Step

(vi). We rst show that there exists an optimal solu-tion in which all jobs inS1 are scheduled beforeJk,

and all jobs in S2 are scheduled after Jk in this

sit-uation. Let∗ be an optimal schedule, and assume, without loss of generality, that∗satises Lemma 1. Then it is immediately clear that any jobJi∈S1which

nishes after t1 on M1 is scheduled beforeJk in∗.

This also means that all jobs inS1are scheduled

be-foreJk in∗. As it is also clear that any jobJi such

thatpi¿pk which nishes aftert1 can be scheduled

afterJk, now suppose that there is a jobJi such that

pi¿pk nishes before s1 in∗. Sinceqi¡ qk, it is

easy to see that interchangingJiandJkin∗will not

increase the makespan. This means that there exists an optimal solution in whichJk nishes befores1 on

M1, a contradiction. Again, as it is clear that any job

Ji such thatpi¿ qiandpi¡ pk which nishes after

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Fig. 2. Solution of Step (ii); (b) solution of Steps (iii) and (iv); (c) solution of Step (v); (d) solution of Step (vi) and (e) optimal solution.

some jobs such thatpi¿ qi andpi¡ pk which

n-ish befores1onM1in∗. LetJjbe the last job which

nishes befores1onM1in∗. It is clear thatpj¿ qj,

as∗_{satises Lemma 1. Since}_p

j¡ pk¡ qk, it is not

dicult to see that movingJj to the position

imme-diately after Jk will not increase the makespan.

Re-peating the same process, we see that there is an op-timal schedule in which all jobs in S2 are scheduled

afterJk.

AsJk nishes aftert1 onM1 in∗, and so all jobs

inS2 must also nish aftert1onM1, and it is optimal

to assign the jobs by Johnson’s rule. Let C∗(S1) be

the completion time of the last job inS1in the optimal

solution. Then we have C∗=C∗(S1) +L+qk+

X

Ji∈S2

qi; (8)

where L is the total idle time on M2 between the

last job of S1 and the last job of S2 in the optimal

solution.

LetC(S1) =C∗(S1) +be the completion time of

the last job inS1 in5. From Lemma 3(i), we know

that

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From (8) and (9), we have

Cmax(5) =C(S1) + max{0; L−}+qk+

X

Ji∈S2

qi

¡ C∗(S1) +L+qk+

X

Ji∈S2

qi+C∗=3

¡4C∗=3 The proof is complete.

Remark (i) The time complexity of the algorithm is O(nlogn).

(ii) Although we are not able to show that the bound is tight, the following instance with n = 3 shows that the bound cannot be better than 1

4.

Con-sider a problem withp1=a−1; q1= (a−1)2; p2=

a; q2=a2; p3=a+ 1, andq3= 2(a+ 1)2; s1= 2a,

and t1 = 2a2 + 2a, where a ¿¿1. It is clear that

Step (ii) will result in schedule1=hJ1; J2; J3iwith

Cmax(1)=5a2+7a+2 (see Fig. 2(a)). Both Steps (iii)

and (iv) will result in schedule2=3=hJ3; J2; J1i

withCmax(2) =Cmax(3) = 5a2+ 2 (see Fig. 2(b)).

Step (v) will result in schedule 4 = hJ2; J3; J1i

with Cmax(4) = 6a2 + 4a + 4 (see Fig. 2(c)).

Step (vi) will result in schedule 5 = hJ2; J1; J3i

with Cmax(5) = 5a2 + 7a + 2 (see Fig. 2(d)).

Hence, CHI = 5a2 + 2. However, the optimal

so-lution is ∗ ₌_h_J

3; J1; J2i with C∗ = 4a2 + 3a + 4

(see Fig. 2(d)). Hence, (CHI−C∗)=C∗goes to 1₄ asa

approaches innity.

Acknowledgements

This research was supported in part by The Hong Kong Polytechnic University under grant number G-YW16. We are grateful to an anonymous referee for his constructive comments on earlier versions of the paper.

References

[1] T.C.E. Cheng, G. Wang, Two-machine owshop scheduling with consecutive availability constraints, Inform. Process. Lett. 71 (1999) 49–54.

[2] S.M. Johnson, Optimal two- and three-stage production schedules with setup times included, Naval Res. Logist. Quart. 1 (1954) 61–68.

[3] W. Kubiak, J. Blazewicz, P. Formanowicz, G. Schmidt, A branch and bound algorithm for two machine ow shop with limited machine availability, The Abstracts of the Tenth Meeting of the European Chapter on Combinatorial Optimization 38 (1997).

[4] C.-Y. Lee, Minimizing the makespan in the two-machine owshop scheduling problem with an availability constraint, Oper. Res. Lett. 20 (1997) 129–139.