A generalized two-machine flowshop scheduling problem with processing time linearly dependent on job waiting-time

A generalized two-machine ¯owshop scheduling problemwith processing time linearly dependent on job waiting-time

Dar-Li Yanga,*, Maw-Sheng Chernb

aDepartment of Industrial Engineering and Management, Nankai College of Technology and Commerce, Nantou 54210,Taiwan, Republic of China

bDepartment of Industrial Engineering and Engineering Management, National Tsing Hua University, Hsinchu 30043,Taiwan, Republic of China

Abstract

We consider a two-machine ¯owshop scheduling problem where the processing times are linearlydependent on the waiting times of the jobs. The objective is to minimize the makespan. A 0±1 mixedinteger program and a heuristic algorithm are proposed. Some cases solved in polynomial time andcomputational experiments are also provided. # 1999 Elsevier Science Ltd. All rights reserved.

Keywords: Flowshop; Scheduling; Waiting time; NP-hard; 0±1 Mixed integer program

1. Introduction

In this paper, we consider a two-machine ¯owshop scheduling problem where the processingtimes are linearly dependent on the waiting times of the jobs prior to processing on the secondmachine. For a given set of jobs J={J1, J2, . . . , Jn}, let ai and bi be the regular processingtimes of job Ji on the ®rst machine M1 and the second machine M2, respectively. A tolerableinterval [li, ui ] on the waiting time of job Ji is given. For each job Ji which is processedcompletely on machine M1, let wi be the waiting time of Ji before its processing on machineM2. If li R wi R ui, then the processing time on machine M2 is bi. However, if wi < li orwi>ui, then the processing time on machine M2 is given by bi+a1(liÿwi ) or bi+a2(wiÿui ),

Computers & Industrial Engineering 36 (1999) 365±378

0360-8352/99/$ - see front matter # 1999 Elsevier Science Ltd. All rights reserved.PII: S0360-8352(99)00137-0

www.elsevier.com/locate/compindeng

* Corresponding author. Fax: +886-49-554412.E-mail address: [email protected] (D.L. Yang)

respectively, where a1 and a2 are nonnegative constants. The objective is to ®nd the optimalschedule minimizing the makespan. For example, there is a problem with a1=1, b1=6, a2=2,b2=4, a3=5, b3=2, l1=1, u1=2, l2=1, u2=3, l3=3, u3=5, a1=0.5 and a2=1. Fig. 1 illustratestwo feasible schedules with di�erent makespans, where the shady area is an additional time inaddition to the processing time required by the second machine.For the sake of convenience, we denote the function P(Ji )=[ai, bi+Wi ] as the processing

times of Ji on machines M1 and M2, respectively, where Wi=max{0, a1(liÿwi ), a2(wiÿui )}.Note that if ai=0 or bi=0, then it is natural to let Wi=0. Such a scheduling problem hasapplications in some ceramic, chemical [1], food [3,7], glass and metallurgical industries [6]. Forexample, in ceramic industries a waiting time between two processors is needed to ensure thesuitable operation on the next machine. If we reduce the required waiting time or increase it,then there always needs some additional time in addition to the processing time required bythe second machine. Sriskandarajah and Goyal [6] have considered problem for the case thatli=ui=0, i= 1, . . . , n. They presented a heuristic algorithm with worst-case analysis for such acase. Chern and Yang [1] have considered problems for the cases of li=0 for all i and ui=1for all i. They showed that for each case, the problem is NP-hard [2]. Therefore, our proposedproblem is NP-hard. For the one-machine case, Kunnathur and Gupta [4] consider a similarproblem where the processing time of a job is linearly dependent on its start time.In this paper, a 0±1 mixed integer program and a heuristic algorithm are proposed. Some

cases solved in polynomial time and computational experiments are also provided.

2. A 0±1 mixed integer programming formulation

In this section, we will show that the proposed problem can be formulated as 0±1 mixedinteger programming problem. We denote that

Ai � the starting time of job Ji on machine M1;

Fig. 1. A processing sequence (J1, J2, J3) with di�erent makespans.

D.-L. Yang, M.-S. Chern / Computers & Industrial Engineering 36 (1999) 365±378366

Bi � the starting time of job Ji on machine M2;

wi � the waiting time of job Ji before its processing on machine M2 � Bi ÿ Ai ÿ ai;

Wi � maxf0, a1�li ÿ wi �, a2�wi ÿ ui �g � maxf0, a1�li ÿ Bi � Ai � ai �, a2�Bi ÿ Ai ÿ ai ÿ ui �g;

yijk � 1, if job Ji precedes job Jj on machine Mk; � 0, otherwise;

Cmax � the makespan:

In the following, we will formulate the proposed problem as 0±1 mixed integer program.For each job i, it follows that

BirAi � ai: �1�In addition, it is necessary to assure that no two operations are processed simultaneously onthe same machine. If job Jj precedes job Ji on machine 1, then

AirAj � aj:

On the other hand, if job Ji precedes job Jj on machine 1, then

AjrAi � ai:

Hence, for each i and j, the following condition:

AirAj � aj or AjrAi � ai

must hold. It is equivalent to that

Ai �Myij1rAj � aj and Aj �M�1ÿ yij1�rAi � ai, 1Ri, jRn, �2�where M is a su�ciently large positive number.By the similar argument, for machine 2, the following condition:

Bi �Myij2rBj � bj �Wj and Bj �M�1ÿ yij2�rBi � bi �Wi, 1Ri, jRn: �3�must hold. We note that

Wi � maxf0, a1�li ÿ Bi � Ai � ai �, a2�Bi ÿ Ai ÿ ai ÿ ui �g: �4�If we de®ne yi1, yi2 and yi3 as

yi1 � 1, if li ÿ Bi � Ai � air0; � 0, otherwise,

yi2 � 1, if Bi ÿ Ai ÿ ai ÿ uir0; � 0, otherwise,

yi3 � 1, if liRBi ÿ Ai ÿ aiRui; � 0, otherwise,

D.-L. Yang, M.-S. Chern / Computers & Industrial Engineering 36 (1999) 365±378 367

then (4) is equivalent to:

a1�li ÿ Bi � Ai � ai �RWiRa1�li ÿ Bi � Ai � ai � �M�1ÿ yi1�, �5�

a2�Bi ÿ Ai ÿ ai ÿ ui �RWiRa2�Bi ÿ Ai ÿ ai ÿ ui � �M�1ÿ yi2�, �6�

0RWiRM�1ÿ yi3�, �7�

yi1 � yi2 � yi3 � 1, �8�where M represents a su�ciently large positive number.It is clear that if yi1=1 then Wi=a1(liÿBi+Ai+ai ); if yi2=1 then Wi=a2(BiÿAiÿaiÿui ); if

yi3=1 then Wi=0.For minimizing makespan, it is necessary to have

Cmax rBi � bi �Wi, 1RiRn: �9�Thus, an integer programming formulation of the proposed problem is given as follows:

minimize

Cmax

subject to

Cmax rBi � bi �Wi, 1RiRn,

BirAi � ai, 1RiRn,

Ai �Myij1rAj � aj, 1Ri, jRn,

Aj �M�1ÿ yij1�rAi � ai, 1Ri, jRn,

Bi �Myij2rBj � bj �Wj, 1Ri, jRn,

Bj �M�1ÿ yij2�rBi � bi �Wi, 1Ri, jRn,

Wira1�li ÿ Bi � Ai � ai �, 1RiRn,

WiRa1�li ÿ Bi � Ai � ai � �M�1ÿ yi1�, 1RiRn,

Wira2�Bi ÿ Ai ÿ ai ÿ ui �, 1RiRn,

WiRa2�Bi ÿ Ai ÿ ai ÿ ui � �M�1ÿ yi2�, 1RiRn,


WiRM�1ÿ yi3�, 1RiRn,

yi1 � yi2 � yi3 � 1, 1RiRn,

Cmax , Ai, Bi, Wir0, 1RiRn,

yi1, yi2, yi3, yijk � 0 or 1, 1Ri, jRn, 1RkR2:

Hence, the total number of constraints in (1)±(3), (5)±(8) and (9) is therefore n(2n + 6). Thereare 3n + 1 nonnegative variables of Cmax, Ai, Bi and Wi. We note that if yijk is in theformulation, then yjik need not be de®ned. Hence, there are n(n ÿ 1) 0±1 integer variables ofyijk. There are 3n 0±1 integer variables of yil, yi2 and yi3. The total number of variables istherefore n(n+ 5)+1.

3. A heuristic algorithm

In this section some solvable cases and a heuristic algorithm are proposed. In order todescribe the algorithm and the following theorems, we de®ne a `no-increase' schedule as eachjob can be early processed on machines M1 and M2 without increasing its processing time onmachine M2 and the processing order of jobs is the same on each machine. A Gantt diagramfor sequence (J1, J2, J3) is shown in Fig. 2. We note that if li=ui=0 for each i, then the no-increase schedule reduces to a no-wait schedule [5].

Theorem 1. If a1 > 1, max1RiRn

fai � li gR min1RiRn

fbi � li g and ai� � li� � min1RiRn

fai �li g, then �Ji� , A� is an optimal schedule, where A is an arbitrary subsequence of the jobs

without Ji � and we schedule the jobs with no-increase manner as shown in Fig. 2.

Proof. It is clear that a lower bound on the makespan is

min1RiRn

fai � li g �Xni�1

bi:

Fig. 2. A no-increase schedule with a1=1, b1=4, a2=2, b2=3, a3=6, b3=1, l1=2, u1=3, l2=2, u2=3, l3=1, u3=2and a1=a2=1.


If a1 > 1 and max1RiRn


fbi � li g, then the makespan of (Ji �, A ) with no-increase

manner is equal to the lower bound

ai� � li� �Xni�1

bi

(Fig. 3). Hence, (Ji �, A ) is an optimal schedule.

Theorem 2. If a1 > 1, min1RiRn

fai � li gr max1RiRn

fbi � li g and bi� � li� � min1RiRn

fbi �li g, then �A, Ji� � is an optimal schedule, where A is an arbitrary subsequence of the jobs

without Ji � and we schedule the jobs with no-increase manner.

Proof. This proof is similar to that of Theorem 1. It is clear that a lower bound on themakespan isXn

i�1ai � min

1RiRnfbi � li g:

If a1 > 1 and min1RiRn

fai � li gr max1RiRn

fbi � li g, then the makespan of (A, Ji �) is equal to thelower boundXn

i�1ai � bi� � li�

(Fig. 4). Hence, (A, Ji �) is an optimal schedule.

Theorem 3. The case is considered here with a1R1, max1RiRn


fbi � li g and ai� �a1li� � min

1RiRnfai � a1li g. If there is a job Jj such that aj+lj R bi �+a1li � then (Ji �, Jj, A ) is an

optimal schedule, where A is an arbitrary subsequence of the jobs without Ji �, Jj. In this casewe schedule the jobs (except Ji �) with no-increase manner as shown in Fig. 5.

Fig. 3. Optimal schedule for the case that a1 > 1, max1RiRn


fbi � li g and A � fJ1, . . . , Jng ÿ fJi� g.


Proof. It is clear that a lower bound on the makespan is

min1RiRn

fai � a1li g �Xni�1

bi:

If a1 R 1, max1RiRn


fbi � li g and aj+lj R bi �+a1li �, then the makespan of (Ji �, Jj,

A ) with no-increase manner (except Ji �) as shown in Fig. 5 is equal to the lower bound

ai� � a1li� �Xni�1

bi:

Hence, (Ji �, Jj, A ) is an optimal schedule.

Theorem 4. If a1 R 1, min1RiRn

fai gr max1RiRn

fbi � a1li g and bi� � a1li� � min1RiRn

fbi � a1li g, then (A,

Ji �) is an optimal schedule, where A is an arbitrary subsequence of the jobs without Ji �, and weschedule the jobs with no-wait manner as shown in Fig. 6.

Proof. It is clear that a lower bound on the makespan isXni�1

ai � min1RiRn

fbi � a1li g:

Fig. 4. Optimal schedule for the case that a1 > 1, max1RiRn

fbi � li gR min1RiRn

fai � li g and A � fJ1, . . . , Jng ÿ fJi� g.

Fig. 5. Optimal schedule for the case that a1 R 1, max1RiRn


fbi � li g, aj � ljRbi� � a1li� and A={J1,. . . , Jn }ÿ{Ji �, Jj }.


Fig. 6. Optimal schedule for the case that a1 R 1, min1RiRn

fai gr max1RiRn

fbi � a1li g and J[i ] is the job scheduled at theith position in the processing sequence.

Fig. 7. Gantt diagrams for four cases in Algorithm 1.


If a1 R 1, min1RiRn

fai gr max1RiRn

fbi � a1li g, then the makespan of (A, Ji �) with no-wait manner as

shown in Fig. 6 is equal to the lower bound

Xni�1

ai � bi� � a1li� :

Hence, (A, Ji �) is an optimal schedule.

In order to describe the algorithm, we let N={1, . . . , n } and T[i ],k be the completion timeof job J[i ] on machine Mk, k = 1, 2, where J[i ] is the job scheduled at the ith position in theprocessing sequence A=(J[1], J[2 ], . . . , J[n ]).

3.1. Algorithm 1

Step 0. (Initiation) Check those conditions stated in Theorems 1±4. If any one of conditionsholds, the optimal schedule is obtained. Otherwise, generate a heuristic solution in thefollowing steps.Step 1. (Determining the ®rst scheduled job in the processing sequence) Choose a jobJi �=J[1] such that ai� � min

i2Nfai g as the ®rst scheduled job of the processing sequence. Ties

may be broken arbitrarily. Set k = 1, T[0],1=T[0],2 =0 and N=Nÿ{i �}. Go to Step 3.Step 2. (Assigning the other n ÿ 1 jobs in the processing sequence) SetR1={i v ai+li R T[k ÿ 1],2ÿT[k ÿ 1],1 R ai+ui for i $ N}, R2={i v ai +li>T[k ÿ 1],2ÿT[k ÿ 1],1

for i $ N} and R3=NÿR1ÿR2. If R1$f, then go to Step 2.1. If R2$f, then go to Step2.2. Otherwise, go to Step 2.3.

2.1 Choose a job Jj such that aj � mini2R1

fai g. Ties may be broken arbitrarily. Place Jj in the

kth position of the processing sequence and set N=Nÿ{ j }. Go to Step 3.2.2 Choose a job Jj such that aj � lj � min

i2R2

fai � li g. Ties may be broken arbitrarily. Place

Jj in the kth position of the processing sequence and set N=Nÿ{ j }. Go to Step 3.2.3 Choose a job Jj such that aj � uj � max

i2R3

fai � ui g. Ties may be broken arbitrarily.

Place Jj in the kth position of the processing sequence and set N=Nÿ{ j }. Go to Step 3.

Step 3. (Determining the reduced period of waiting time and the additional time on machineM2 for each job) If T[k ÿ 1],2ÿT[k ÿ 1],1 < a[k ] (Fig. 7(a)), then go to Step 3.1. Ifa[k ] R T[k ÿ 1],2ÿT[k ÿ 1],1 R a[k ]+l[k ] (Fig. 7(b)), then go to Step 3.2. Ifa[k ]+l[k ] < T[k ÿ 1],2ÿT[k ÿ 1],1 R a[k ]+u[k ] (Fig. 7(c)), then go to Step 3.3. Ifa[k ]+u[k ] < T[k ÿ 1],2ÿT[k ÿ 1],1 (Fig. 7(d)), then go to Step 3.4.


3.1 If 0 < a1 < 1, then T[k ],1=T[k ÿ 1],1+ a[k ], T[k ],2=T[k ÿ 1],1+a[k ]+b[k ]+a1l[k ].If 1 R a1, then T[k ],1=T[k ÿ 1],1+a[k ], T[k ],2=T[k ÿ 1],1+a[k ]+l[k ]+ b[k ].If N$f, then set k=k + 1 and go to Step 2. Otherwise, stop.3.2 If 0 < a1 < 1, then T[k ],1=T[k ÿ 1],1+a[k ], T[k ],2=T[k ÿ 1],2+b[k ]+a1(a[k ]+l[k ]ÿT[k ÿ 1],2+T[k ÿ 1],1).If 1 R a1, then T[k ],1=T[k ÿ 1],1+a[k ], T[k ],2=T[k ÿ 1],1+a[k ]+l[k ] +b[k ].If N$f, then set k=k + 1 and go to Step 2. Otherwise, stop.3.3 T[k ],1=T[k ÿ 1],1+a[k ], T[k ],2=T[k ÿ 1],2 +b[k ]. If N$f, then set k=k+ 1 and go toStep 2. Otherwise, stop.3.4 T[k ],1=T[k ÿ 1],2ÿu[k ], T[k ],2=T[k ÿ 1],2+b[k ]. If N$f, then set k=k + 1 and go toStep 2. Otherwise, stop.

In the above algorithm, Step 1 chooses the ®rst scheduled jobs to minimize the idle time onmachine M2. Step 2 sequentially assigns the other jobs. Step 2.1 minimizes the idle time onmachines M1 and M2. Step 2.2 and Step 2.3 minimize the idle time on machine M2 and M1,respectively. Step 3 adjusts the reduced period of waiting time and the additional time onmachine M2 for each job. In Step 3.1 and Step 3.2, if a1r1, then it is useless to reduce the l[k ].However, if 1>a1>0, then it is useful to reduce the l[k ] as much as possible and thecalculation of the completion times on both machines is given. Step 3.3 depicts that ifa[k ]+l[k ] < T[kÿ1],2ÿT[kÿ1],1 R a[k ]+u[k ], then it is useless to reduce the l[k ] either a1r1 or1>a1>0. Step 3.4 depicts that if a[k ]+u[k ] < T[kÿ1],2ÿT[kÿ1],1, then we schedule the job J[k ]with no-increase manner.

Example 1. A three-job problem is given, where a1=2, b1=3, a2=1, b2=4, a3=3, b3=5,a1=a2=2, li=2, and ui=3 for i= 1, 2, 3. Since a1>1 and 5 � max

1R3fai � li gR min

1RiR3fbi �

li g � 5, by Theorem 1 the sequence (J2, J1, J3) with no-increase manner is an optimal schedule.The Gantt diagram is shown in Fig. 8 and the makespan is 15.

Example 2. A three-job problem is given, where a1=2, b1=3, a2=1, b2=4, a3=3, b3=5,a1=0.5, a2=2, li=2, and ui=3 for i = 1, 2, 3. Since a1R1, 5 � max

1RiR3fai � li gR min

1RiR3fbi �

li g � 5 and 4=a1+l1 R b2+a1l2=5, by Theorem 3 the sequence (J2, J1, J3) with no-increasemanner (except J2) is an optimal schedule. The Gantt diagram is shown in Fig. 9 and themakespan is 14.

Fig. 8. Gantt diagram for the sequence (J2, J1, J3).


Example 3. A three-job problem is given, where a1=3, b1=2, a2=4, b2=1, a3=4, b3=2,a1=0.5, a2=1, li=2, and ui=3 for i= 1, 2, 3. Since a1R1 and 3 � min

1RiR3fai gr max

1RiR3fbi �

a1li g � 3, by Theorem 4 the sequence (J1, J3, J2) with no-wait manner is an optimal schedule.The Gantt diagram is shown in Fig. 10 and the makespan is 13.

Example 4. A four-job problem is shown in Table 1, where a1=0.5 and a2=1. A near-optimalsolution (J4, J2, J3, J1) is determined by using the Algorithm 1. A Gantt diagram is shown inFig. 11 and the makespan is 21. An optimal solution solved by the 0±1 mixed integer programis shown in Table 2. An optimal schedule (J1, J4, J3, J2) is determined and its makespan is20.25. A Gantt diagram is given in Fig. 12.

4. Computational experiments

In order to evaluate the e�ciency of Algorithm 1, we generate several groups of randomproblems as follows:

1. n is equal to 10, 20, 30, 40, 50, 60, 70, 80, 90, or 100.2. ai is uniformly distributed over [1, 10].3. bi is uniformly distributed over [1, 10].4. li is uniformly distributed over [0, 5].5. ui is uniformly distributed over [5, 10].6. a1 is uniformly distributed over [0, 2].7. a2 is uniformly distributed over [0, 2].

The computational experiments are conducted on an IBM PC 586. The heuristic algorithm 1 iscoded in Pascal. A total of 200 test problems is generated. The computation times ofAlgorithm 1 for all the test problems are less than one second. For each of these random




Table 1Data for a four-job problem

i 1 2 3 4ai 2 4 8 1

bi 3 4 4 6li 1 1 2 2ui 2 4 3 4

Fig. 11. Gantt diagram for the sequence (J4, J2, J3, J1).

Table 2

Solution for a four-job problem

i 1 2 3 4

Ai 0 11.25 3 2Bi 2 16.25 11.5 5.5Wi 0.5 0 0.75 0

Fig. 12. Gantt diagram for the optimal sequence (J1, J4, J3, J2).


problems, a percentage error E=ChÿCo/Co � 100 is computed, where Ch and Co are themakespans of heuristic solution and optimal solution, respectively. The optimal solution iscomputed by the proposed integer program coding with STORM. We note that Co is replacedby Clow as n is greater than 20. Clow is the lower bound on the makespan and is estimated asfollows:

Clow � max

(Xni�1

ai � min1RiRn

f minfli � bi, a1li � bi gg, min1RiRn

f minfai � li,

ai � a1li gg �Xni�1

bi

),

where we assume that processing on each machine may be continuous.The results are given in Table 3. There are 20 test problems for each problem type. To

evaluate the overall performance of the heuristic algorithm, we compute the mean of all theaverage percentage errors reported in Table 3. The mean value is 1.06, which suggests that theheuristic algorithm, on the average, ®nds schedules that are within 1.06% of optimality. Basedon the computational experiments, as it can be seen from Table 3, the performance of thealgorithm 1 is quite satisfactory. We can also see that the average percentage errors aredecreased as the job number n is increased. In view of the NP-hardness of the problem, thisresult is quite encouraging since it provides an e�cient procedure for solving large-sizedproblems.

5. Conclusions

In this paper, we investigate a two-machine ¯owshop scheduling problem where theprocessing times are linearly dependent on the waiting times of the jobs. A 0±1 mixed integer

Table 3Computational results for Algorithm 1

n Observed in 20 test problems

Minimum percentage error Average percentage error Maximum percentage error

10 0 1.83 16.02

20 0 1.81 12.8530 0 1.72 10.1340 0 1.35 7.46

50 0 0.91 5.1160 0 0.82 5.8070 0 0.63 4.3880 0 0.61 4.31

90 0 0.52 4.01100 0 0.43 3.12


program and a heuristic algorithm are proposed. Some cases solved in polynomial time andcomputational experiments are also provided. From the computational experiments, theperformance of the proposed algorithm is quite satisfactory.In the future research, some heuristic algorithms or branch-and-bound algorithms may be

designed to solve the above proposed problem.

Acknowledgements

The authors would like to thank the anonymous referees for their helpful comments andsuggestions on improving the presentation of this paper. This work is supported by theNational Science Council of the Republic of China under grant NSC-85-2213-E007-007.

References

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[2] Garey MR, Johnson DS. Computers and intractability: a guide to the theory of NP-completeness. San

Francisco: Freeman, 1979.[3] Hodson A, Muhlemann AP, Price DHR. A microcomputer based solution to a practical scheduling problem. J

Opl Res Soc 1985;36:903±14.

[4] Kunnathur AS, Gupta SK. Minimizing the makespan with late start penalties added to processing times in asingle facility scheduling problem. Eur J Opl Res 1990;47:56±64.

[5] Reddi S, Ramamoorthy C. On the ¯owshop sequencing problem with no-wait in process. Opl Res Q1972;23:323±32.

[6] Sriskandarajah C, Goyal SK. Scheduling of a two-machine ¯owshop with processing time linearly dependent onjob waiting-time. J Opl Res Soc 1989;40:907±21.

[7] Yang D-L, Chern M-S. A two-machine ¯owshop sequencing problem with limited waiting time constraints.

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A generalized two-machine flowshop scheduling problem with processing time linearly dependent on job waiting-time