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European Journal of Operational Research 77 (1994) 415-428 415 North-Holland Theory and Methodology Scheduling a two-stage hybrid flowshop with separable setup and removal times Jatinder N.D. Gupta and Enar A. Tunc Department of Management, Ball State Uniuersity, Muncie, IN 47306, USA Received February 1992; revised September 1992 Abstract: This paper considers the two-stage flowshop scheduling problem where each stage consists of several identical parallel machines and the setup and removal times of each job at each stage are separated from the processing times. A polynomial optimization algorithm is developed for the special case where the first stage contains only one machine and the number of identical parallel machines at the second stage is equal to or greater than the total number of jobs. In view of the NP-completeness of this problem, four heuristic algorithms are developed for the case where there is one machine at stage 1 and the number of identical parallel machines at the second stage is less than the total number of jobs. The proposed heuristic algorithms are empirically tested to determine their effectiveness in finding an optimal schedule. I. Introduction Consider the two-stage hybrid flowshop sched- uling problem where n given jobs are to be pro- cessed on two stages in the same technological order, first on stage 1 and then on stage 2. At stage s, there are m s identical parallel machines. The processing time of job a at stage s is denoted by t(a, s). In addition to the processing time, each job requires a setup at stage s which takes S(a, s) amount of time. Similarly, once job a is completed at stage s, R(a, s) amount of time is required to remove the setup of job a at stage s before any other job can be processed on that particular machine. However, a job can be taken Correspondence to: Prof. J.N.D. Gupta, Department of Man- agement, Ball State University, Muncie, IN 47306, USA. to the second stage as soon as it completes its processing at the first stage without waiting for the removal operation at first stage. Unlimited intermediate storage space is available to hold partially completed jobs between the two stages. Transportation times are negligible and therefore are ignored. Similarly, once a job completes pro- cessing at the second stage, it is ready to be delivered. Given this scenario, it is desired to find a processing schedule which will minimize the total throughput time (called Makespan) in which all jobs complete processing on both stages. As a secondary criteria, it is desired to minimize the number of machines used at each stage. The hybrid flowshop scheduling problems of the type described above are quite common in practice, especially in the process industry where multiple servers (machines) are available at each stage (Salvador, 1973; Brah and Hunsucker, 0377-2217/94/$07.00 © 1994 - Elsevier Science B.V. All rights reserved SSD1 0377-2217(92)00419-H

Scheduling a two-stage hybrid flowshop with separable setup and removal times

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European Journal of Operational Research 77 (1994) 415-428 415 North-Holland

Theory and Methodology

Scheduling a two-stage hybrid flowshop with separable setup and removal times

Jatinder N.D. Gupta and Enar A. Tunc Department of Management, Ball State Uniuersity, Muncie, IN 47306, USA

Received February 1992; revised September 1992

Abstract: This paper considers the two-stage flowshop scheduling problem where each stage consists of several identical parallel machines and the setup and removal times of each job at each stage are separated from the processing times. A polynomial optimization algorithm is developed for the special case where the first stage contains only one machine and the number of identical parallel machines at the second stage is equal to or greater than the total number of jobs. In view of the NP-completeness of this problem, four heuristic algorithms are developed for the case where there is one machine at stage 1 and the number of identical parallel machines at the second stage is less than the total number of jobs. The proposed heuristic algorithms are empirically tested to determine their effectiveness in finding an optimal schedule.

I. Introduction

Consider the two-stage hybrid flowshop sched- uling problem where n given jobs are to be pro- cessed on two stages in the same technological order, first on stage 1 and then on stage 2. At stage s, there are m s identical parallel machines. The processing time of job a at stage s is denoted by t(a, s). In addition to the processing time, each job requires a setup at stage s which takes S(a, s) amount of time. Similarly, once job a is completed at stage s, R(a, s) amount of time is required to remove the setup of job a at stage s before any other job can be processed on that particular machine. However, a job can be taken

Correspondence to: Prof. J.N.D. Gupta, Department of Man- agement, Ball State University, Muncie, IN 47306, USA.

to the second stage as soon as it completes its processing at the first stage without waiting for the removal operation at first stage. Unlimited intermediate storage space is available to hold partially completed jobs between the two stages. Transportation times are negligible and therefore are ignored. Similarly, once a job completes pro- cessing at the second stage, it is ready to be delivered. Given this scenario, it is desired to find a processing schedule which will minimize the total throughput time (called Makespan) in which all jobs complete processing on both stages. As a secondary criteria, it is desired to minimize the number of machines used at each stage.

The hybrid flowshop scheduling problems of the type described above are quite common in practice, especially in the process industry where multiple servers (machines) are available at each stage (Salvador, 1973; Brah and Hunsucker,

0377-2217/94/$07.00 © 1994 - Elsevier Science B.V. All rights reserved SSD1 0377-2217(92)00419-H

416 J.N.D. Gupta, E.A. Tunc / Scheduling a two-stage hybrid flowshop

1991b) as well as in certain flexible manufactur- ing environments (Sawik, 1988; Zijm and Nelis- sen, 1990). Further, when processing times of second stage dominate those of the first stage, it is natural to increase the system capacity by adding another machine at the second stage (Gupta and Tunc, 1991). This is also true in the reverse case (Gupta, 1988).

If each of the two stages contains only one machine, the problem is polynomially solvable as shown by Sule (1982), Szwarc and Gupta (1987), and Proust, Gupta and Deschamps (1991) de- pending on the definition of makespan (see Sec- tion 2 below). Dileepan and Sen (1991) consider the two-stage flowshop problem with only one machine at each stage and describe a branch and bound algorithm for minimizing the maximum job lateness. Similarly, when the setup and removal times are ignored, the branch and bound algo- rithms of Salvador (1973) and Brah and Hun- sucker (1991a) or the mathematical programming formulation of Brah and Hunsucker (1991b) can be used to optimally solve small-sized problems. Heuristic algorithms are available to solve the problem under varying conditions as discussed by Deal and Hunsucker (1991), Gupta (1988), Gupta and Tunc (1991), and Hunsucker and Shah (1991). However, in the general case, nothing specific is known and following Garey and Johnson (1979), it has been shown that the problem is NP-hard (Gupta, 1988).

This paper considers the two-stage hybrid flowshop scheduling problems where the first stage contains only one machine, when the num- ber of machines at the second stage is greater than or equal to the total number of jobs, a polynomial optimization algorithm is developed to minimize makespan. Encouraged by earlier work where one of the two stages contains only one machine (Gupta, 1988; Gupta and Tunc, 1991) and the polynomial algorithm for the above special case (containing no more jobs than the number of identical parallel machines at the sec- ond stage), this paper proposes heuristic algo- rithms to solve the problem when the first stage contains only one machine. Computational expe- rience with several problems shows that the pro- posed algorithms are quite effective in solving these problems and that the effectiveness of the proposed algorithms increases as the problem size increases.

The paper is organized as follows: Section 2 discusses and defines two concepts of makespan used in the literature when separable setup and removal times are considered. For a special case where the number of jobs is no more than the number of identical parallel machines, a polyno- mial optimization algorithm is developed in Sec- tion 3. The global lower bounds that can also be used to develop a branch and bound algorithm are described in Section 4. Heuristic algorithms to solve the problem are developed in Section 5 and empirically tested for their effectiveness in finding minimal job-based makespan schedules in Section 6. The extensions of the proposed algo- rithms to solve problems when machine-based makespan concept is used are described in Sec- tion 7. This section also includes the results of computational experience to empirically test the effectiveness of two heuristic algorithms to mini- mize machine-based makespan. Finally, Section 8 summarizes the results, gives a few conclusions, and describes some fruitful areas for further re- search.

2. Definition of makespan

When setup and removal times are separable and are considered explicitly, the concept of makespan requires a reference point and an in- terpretation since the definition of makespan can change the optimal schedule. In order to com- pute the makespan of a schedule, consider a permutation

P = ( a , , a 2 . . . . . a , )

of n jobs. Let A l ( a i) be the time job a i com- pletes processing at stage 1 and Bl (a i ) be the time job a i completes its processing at stage 2. Let the immediately preceding job at the machine where job a i is processed at stage 2 be ag. Further, let A 2 ( a i) and B2(a i) be the specific machine availability times after processing job a i at stages 1 and 2, respectively. In addition, as- sume that all setup requirements are anticipatory implying that the setup for a job at a machine can be done in anticipation of the job arriving at that machine (Baker, 1990). Then, following the usual assumptions of the flowshop scheduling problem (Gupta, 1979) and using the physical conditions

J.N.D. Gupta, E.A. Tunc / Scheduling a two-stage hybrid flowshop 417

of the problem, the makespan can be computed using the following recursive relations:

3. Case with unlimited number of machines at second stage

Al(ai) =A2(a i_ , ) +S(a,, 1) +t(a i, 1), ( la)

A2( ai) = A,( ai) + R( a,, 1), ( lb)

B~( ai) = max{A,( ai); Be( ak) + S( ai, 2)}

+ t(a i, 2), (1c)

B2( ai) = Bl( ai) + R( a i, 2), ( ld)

where Al(qb) = A 2 ( ~ ) = Bl(CJS) = B2(q~) = 0 and q~ is a null schedule (job).

Now, the definition of makespan could either be based on the availability of any of the ma- chines in the shop after completing all jobs or availability of completed jobs. When the makespan is computed from the point of view of machine availability (called machine-based make- span), removal times of jobs are to be included since machines cannot process any other job dur- ing that time. This implies that the makespan of schedule P, M(P), will be defined as follows:

M(P) = max {A2(ai), B2(ai) }. (2) l <_i <~n

On the other hand, removal times should not be included when makespan is computed from the point of view of job availability. Thus, in this case, makespan (called job-based makespan), M(P), will be defined by the following equation:

M(P) = max {Bl(ai) }. (3) 1 <<_i<_n

Sule (1982) and Proust, Gupta and Deschamps (1991) use the machine-based makespan by con- sidering only the last stage and define the makespan of schedule P as follows:

M(P) = max {B2(ai) }. l <_i <_n

Szwarc and Gupta (1987), however, consider the job-based makespan and define the makespan as in (3). Depending on the specific definition of the makespan, the heuristic algorithm used in each case may be different. Unless otherwise stated, in this paper, the job-based makespan will be con- sidered and hence (1) and (3) will be used to compute the makespan.

When the number of machines at the second stage, m2, is unlimited (or m 2 > n) there is no possibility of any bottleneck at the second stage since each job can be assigned to a different machine. Since there is no preceding job at stage 2, the calculation of Bl(a i) in (1) simplifies as follows:

B,(ai)=max{A,(ai); S(a i ,2 )}+t (a i,2). (1')

Based on the above recursive relations (1) and (1'), the following theorem provides the polyno- mial optimization algorithm for this case.

Theorem 1. With unlimited number of machines at second stage, the schedule obtained by arranging jobs in descending order of h(a)= {t(a, 2 ) - R(a, 1)} values minimizes job-based makespan.

Proof. Consider two schedules P=trabTr and P' = o'baTr where

{t(a, 2 ) - R ( a , 1)} > {t(b, 2 ) - R ( b , 1)}.

The makespan of schedule P will be no greater than that of P ' if

max{Bl(a); Bl(b)} < max{B~(b); B~(a)}.

Using relations (1) and (1'), it follows that

max{Bl(a) ; Bl(b)}

=max{C + S(a, 1) +t(a, 1) +t(a, 2);

S(a, 2) + t ( a , 2);

C+S(a, 1 ) + t ( a , 1 ) + R ( a , 1 ) + S ( b , 1)

+t(b, 1) + t(b, 2); S(b, 2) + t(b, 2)}

where C equals the setup, completion and re- moval times of all jobs in ~r at stage 1.

In view of the condition that

(t(a, 2) - R ( a , 1)} > {t(b, 2) - R ( b , 1)},

the above equation leads to

max{B,(a) ; B,(b)}

< max{C + S(a, 1) + t(a, 1) + t(a, 2);

S(a, 2) + t(a, 2);

C+S(a, 1 ) + t ( a , 1 ) + R ( b , 1 ) + S ( b , 1)

+t(b, 1) +t(a,2); S(b,2) +t(b, 2)}.

418 J.N.D. Gupta, E.A. Tunc / Scheduling a two-stage hybrid flowshop

Since the first term under the maximum oper- ator in the right-hand side expression is less than the third term in this expression, the above ex- pression can be simplified as

max{Bl(a) ; Bl(b)}

< max{S(a, 2) +

+t(a, 1) + R ( b , 1)

S(b, 2)

t(a, 2); C+S(a, 1)

+S(b, 1) +t(b, 1) + t ( b , 2);

+ t(b, 2)}.

Further, in view of the above, it follows that

max{Bl(a) ; Bl(b)}

<max{C + S(b, 1) + t(b, 1) + t(b, 2);

S(b,2) + t(b, 2);

{C + S(b, 1) +t(b, 1) +R(b, 1)

+S(a, 1) + t(a, 1) + t(a, 2); S(a, 2) + t(a, 2)}

= max{B£(b); B~(a)}.

Hence if the conditions of Theorem 1 above are satisfied, M(P)<M(P') regardless of the jobs in or and ~'. In view of the transitive nature of the conditions, it follows that the schedule obtained by arranging jobs in descending order of the h(a) = {t(a, 2 ) - R(a, 1)} values minimizes makespan. Thus the proof of Theorem 1 is com- plete. []

4. Global lower bounds

The job-based makespan of any schedule can- not be less than the total completion time of all jobs at stage 1 plus the minimum processing time at stage 2. Thus, a lower bound on makespan, LBa, can be expressed as follows:

LB 1= ~ { S ( a , a ) + t ( a , 1 ) + R ( a , 1 ) } a - 1

- max {R(i, 1)} + min {t( i ,2)}. l <i <n l <_i <n

A second lower bound could be obtained by assuming that the second stage has unlimited number of machines. The makespan of the sched- ule obtained using Theorem 1 could be treated as a lower bound for the case where there are fewer than n machines at the second stage. Let M* denote this makespan. Then

LB z = M *.

Thus, the global lower bound, LB, on the minimum makespan for any schedule for the two-stage hybrid flowshop problem being consid- ered here can be stated by the following theorem:

Theorem 2. LB = max(LB1, LB2).

The above lower bounds can either be opera- tionalized to develop a branch and bound algo- rithm or can be used to test the effectiveness of the heuristic algorithms, when developing a branch and bound algorithm, additional work is required to modify the above lower bounds so that (a) initial partial schedules of already as- signed jobs are included in the definition of the lower bounds; and (b) lower bounds are effective in curtailing the search without too much compu- tational effort. However, for comparing the effec- tiveness of various heuristic algorithms, global lower bounds provided by Theorem 2 above, though not very tight, are sufficient.

5. Proposed heuristic algorithms

When the number of machines at the second stage is less than the total number of jobs, the problem has been shown to be NP-hard (Gupta, 1988) even when setup and removal times are not considered. Thus, the problem considered in this paper is NP-hard since a specific instance of this problem (with zero setup and removal times) is NP-hard. Therefore, it is desirable to develop efficient and effective heuristic algorithms to solve the problems.

The solution of the two-stage hybrid flowshop scheduling problem requires two aspects: se- quencing jobs on both stages and assignment of jobs to various machines at each stage. For the case with only one machine at stage 1, assume that the sequencing and assignment of jobs to machines at the second stage can be done inde- pendently. This assumption may produce subopti- mal results but saves considerable computational effort in finding an approximately optimal solu- tion.

The first part, sequencing of jobs, can be done by using one of the existing algorithms, perhaps Sule's (1982) rule or Szwarc and (Gupta's (1987) (henceforth called S&G) algorithm. The second part, one of assigning jobs to multiple machines

J.N.D. Gupta, E.A. Tunc / Scheduling a two-stage hybrid flowshop 419

Table 1

Data for an eight-job problem

Job a S(a. 1) t(a, 1) R ( a , l ) S(a, 2) t(a, 2) R(a, 2) p(a , 1) p ( a . 2 ) h(a)

1 10 22 5 7 26 8 25 29 21

2 5 13 3 5 12 7 13 16 9

3 7 36 2 6 18 2 37 18 16 4 6 18 8 3 39 3 21 34 31

5 5 26 7 9 13 5 22 11 6 6 3 12 11 10 20 6 5 15 9

7 9 9 3 2 9 3 16 9 6

8 2 22 6 9 17 11 15 22 11

at the second stage, is done by attempting to minimize the job-waiting (and machine idle) time at the second stage.

5.1. Sule's rule

To succinctly describe Sule's rule and the S & G algorithm, for each job a, define

p(a, 1 ) = S ( a , 1 ) - S ( a , 2 ) + t ( a , 1), (4a)

p ( a , 2 ) = t ( a , 2 ) + R ( a , 2 ) - R ( a , 1). (4b)

With these definitions, Sule's rule reduces to solving the two-machine flowshop scheduling problem using Johnson's (1954) algorithm with p(a, s) as the processing time of job a at stage s.

5.2. The S& G algorithm

The steps of the S&G algorithm can be de- scribed as follows:

Step 1. Let P - - ( a 1, a 2 . . . . . a n) be the sched- ule obtained by Sule's algorithm with processing times given by (4) above.

Step 2. For each i where R(ai, 2)> R(a~, 2), generate

P ' = ( a l , a2 . . . . ,ai l, a i+ l , . . . , a , , a i ) .

Let the total number of schedules thus generated be k - 1 .

Step 3. Calculate the makespan of each of the k schedules in Steps 1 and 2 above and select the one with the minimum makespan. This is the optimal schedule.

The above result does not extend to the hybrid flowshops. The following heuristic algorithms show some promise in finding approximately opti- mal solutions.

5.3. Heuristic algorithm 1

The first heuristic algorithm is based on the premise that Sule's rule coupled with an appro- priate assignment rule should produce an accept- able schedule for the problem.

The assignment rule used here is as follows: whenever a specific job at stage 2 experiences a waiting time because a specific machine is not

Table 2

Assignment of jobs to machines (Algorithm 1)

i Job a i Stage 1 Stage 2

Machine 1 Machine 2

Al(ai) A2(a i) Bl(ai) Bz(a,.) Bl(a i) B2(a ,) 1 6 15 26 35 41 -

2 2 44 47 - - 56

3 8 71 77 8 99 - 4 4 101 109 - - 140 5 1 141 146 167 175 - 6 3 189 191 207 209 -

7 5 222 229 235 240 - 8 7 247 250 256 259 -

63

143

420 J.N.D. Gupta, E.A. Tunc / Scheduling a two-stage hybrid flowshop

available, we assign that job to the machine that minimizes the job waiting time. The steps of heuristic algorithm 1, therefore, are as follows:

Step 1. (Sequence identification:) Let P = (a 1, a 2 . . . . . a n) be the schedule obtained using Sule's algorithm, breaking ties in favor of the job with lowest t(a, 1) or the greatest p(a , 2).

Step 2. (Assignment of jobs:) Starting with i = 1,successively assign each job a i to the latest available machine at stage 2 such that minimum job waiting is incurred at stage 2. If there is no job waiting, assign job a i to the latest available machine.

Step 3. (Approximate solution:) Accept the fi- nal assignment and schedule thus obtained as an approximate solution to the problem.

Step 3. From Table 2, it is seen that jobs (6, 8, 1, 3, 5, 7) are processed on machine 1 of stage 2 while jobs (2, 4) are processed on machine 2 of stage 2. The processing sequence at stage 1 is P = (6, 2, 4, 1, 8, 3, 5, 7). The makespan of this schedule, as found by using (1) and (3) is 256 time units.

The solution to the above problem is optimal as can easily be verified by computing the lower bound on the makespan. Further, the above ex- ample also illustrates that the proposed algorithm aims at minimizing the number of machines used at stage 2. Even if there were more than 2 ma- chines at stage 2, the proposed algorithm will use only 2 machines.

5.4. Heuristic algorithm 2

To illustrate the above algorithm, consider the eight-job problem of Table 1. There are two identical machines at stage 2 and only one ma- chine at stage 1.

The problem of Table 1 is solved as follows: Step 1. Use of Sule's algorithm results in the

schedule P = (6, 2, 8, 4, 1, 3, 5, 7). Step 2. The assignment of jobs is now done as

shown in Table 2. To start with, job 6 is assigned to first machine at stage 2. Job 6 completes its processing and removal at machine 1 of stage 2 at time 41 (as determined by using (1) and (3)). Now, job 2 is ready to be taken to second stage at 44 time units. However, if machine 1 at stage 2 is free only at time 41 and 8 units of time are needed for the setup of job 2 on stage 2, job 2 will have to wait for 5 time units if assigned to ma- chine 1 at stage 2. In order to avoid this delay in processing, job 2 is assigned to the second ma- chine at stage 2. Proceeding in this manner, jobs are assigned as shown in Table 2.

For the two-stage flowshop scheduling prob- lem with one machine at each stage, S&G algo- rithm was found to yield better results than Sule's rule (Szwarc and Gupta, 1987). Therefore, use of the improvement routine (Steps 2 and 3 in the S & G algorithm described above)will enhance the quality of the solution (in terms of decreasing the job-based makespan). With this modification, the steps of heuristic algorithm 2 can be stated as follows:

Step 1. (Sequence identification:)Let P = ( a l , a2 , . . . , a n) be the schedule obtained by Sule's algorithm, breaking ties in favor of the job with lowest t(a, 1) or the greatest p(a, 2).

Step 2. (Additional sequences generation:) For each i where R(ai , 2)> R(a n, 2), generate

e ' = ( a 1, a 2 . . . . . ai_ 1, a i + l , . . . , a n , ai) .

Let the total number of schedules thus generated be k - 1 .

Table 3 Schedules generated by algorithm 2

Schedule Stage 1 Stage 2

Machine 1 Machine 2

Makespan

Sule (6 ,2 ,8 ,4 ,1 ,3 ,5 ,7 ) 1 (2 ,8 ,4 ,1 ,3 ,5 ,7 ,6 ) 2 (6 ,8 ,4 ,1 ,3 ,5 ,7 ,2 ) 3 (6 ,2 ,4 ,1 ,3 ,5 ,7 ,8 ) 4 (6 ,2 ,8 ,4 ,3 ,5 ,7 ,1 ) 5 (6 ,2 ,8 ,4 ,1 ,3 ,7 ,5 )

(6, 8, 1, 3, 5, 7) (2, 4) 256 (2, 4, 6) (8, 1, 3, 5, 7) 259 (6, 8, 1, 3, 5, 7, 2) (4) 259 (6, 1, 3, 5, 7, 8) (2, 4) 261 (6, 8) (2, 4, 3, 5, 7, 1) 271 (6, 8, 1, 3) (2, 4, 7, 5) 256

J.N.D. Gupta, E.A. Tunc /Scheduling a two-stage hybrid flowshop 421

Step 3. (Assignment of jobs:) For each of the k schedules in Steps 1 and 2, starting with i = 1, successively assign each job a i to the latest avail- able machine at stage 2 such that minimum job waiting is incurred at stage 2. If there is no job waiting, assign job a i to the latest available ma- chine.

Step 4. (Approximate solut ion:)Among the schedules generated, accept the schedule and as- signment with minimum makespan as an approxi- mate solution to the problem.

To illustrate the above algorithm 2, reconsider the eight-job problem of Table 1 with two identi- cal machines at stage 2 and only one machine at stage 1. The first schedule generated will be as shown in Table 2 with job 7 as the last job. Since the removal times of jobs 1, 2, 5, 6, and 8 are greater than that of job 7, each of these jobs is assigned to the last sequence position. Table 3 shows the schedules, assignment to each machine at stage 2 and the makespans. This algorithm identifies two optimal schedules.

5.5. Heuristic Algorithm 3

The use of Sule's or S& G algorithm is advis- able since it tends to minimize the makespan when the number of machines at each stage is 1. However, there might be cases where such an algorithm performs poorly. Therefore, another sequencing rule may be desirable. Following Langston's (1987) analysis, it is rational to assume that a schedule generated by arranging jobs in a descending order of their processing times at the second stage (or the job with Longest Processing

time (LPT) at the second stage first) will tend to minimize makespan. In fact, for problems without any setup and removal times, application of LPT based algorithm was found to yield excellent re- sults (Gupta and Tunc, 1991). Therefore, pro- posed heuristic algorithm 3 uses this logic. The steps of this heuristic algorithm 3 are as follows:

Step 1. (Sequence ident if icat ion:)Let P = (a I, a 2 . . . . . a n) be the schedule obtained by ar- ranging jobs in descending order of their process- ing times at second stage, breaking ties in favor of the job with lowest t(a, 1).

Step 2. (Assignment of jobs:) Starting with i = 1,successively assign each job a i to the latest available machine at stage 2 such that minimum job waiting is incurred at stage 2. If there is no job waiting, assign job a i to the latest available machine.

Step 3. (Approximate solution:) Accept the fi- nal assignment and schedule thus obtained as an approximate solution to the problem.

As an illustration of the above algorithm, re- consider the eight-job problem of Table 1. As before, there are two identical machines at stage 2 and only one machine at stage 1. To solve the problem, we follow the steps of algorithm 3 above.

Step 1. Arranging the jobs in ascending values of t(a, 2) results in the schedule P = (4, 1, 6, 3, 8, 5, 2, 7).

Step 2. The assignment of jobs is now done as shown in Table 4. To start with, job 4 is assigned to first machine at stage 2. Job 4 completes its processing and removal at machine 1 of stage 2 at time 69. Now, job 1 is ready to be taken to second

T a b l e 4

A s s i g n m e n t o f j o b s to m a c h i n e s ( A l g o r i t h m 3)

i J o b a i S t a g e 1 S t a g e 2

M a c h i n e 1 M a c h i n e 2

A l ( a i ) Ae(a i) Bl(a i) Be(a i) Bl(ai) B2(ai)

1 4 24 32 63 66 - -

2 1 64 69 - - 0 98

3 6 84 95 104 110 - -

4 3 138 140 156 158 - -

5 8 164 170 - - 181 192

6 5 201 208 - - 214 219

7 2 226 229 - - 238 245

8 7 247 250 - - 256 259

422 J.N.D. Gupta, E.A. Tunc / Scheduling a two-stage hybrid flowshop

stage at 64 time units. However, if machine 1 at stage 2 is free only at time 69 and 7 units of time are needed for the setup of job 1 on stage 2, job 1 will have to wait for 12 time units if assigned to machine 1 at stage 2. In order to avoid this delay in processing, job 1 is assigned to second machine at stage 2. Proceeding in this manner, jobs are assigned as shown in Table 4.

Step 3. From Table 2, it is seen that jobs (4, 6, 3) are processed on machine 1 of stage 2 while jobs (1, 8, 5, 2, 7) are processed on machine 2 of stage 2. The processing sequence at stage 1 is P = (4, 1, 6, 3, 8, 5, 2, 7). The makespan of this schedule is 256.

As in the case of algorithm 1 and 2, the above example illustrates that the proposed algorithm 3 also aims at minimizing the number of machines used at stage 2. Even if there were more than 2 machines at stage 2, the proposed algorithm 3 will use only 2 machines.

5.6. Heuristic algorithm 4

Even though the number of machines at the second stage is less than the total number of jobs, the optimization algorithm for the unlimited ma- chine case can be used as a heuristic algorithm for problems with limited number of machines at the second stage (i.e. m 2 < n). For each job a, let

h ( a ) = { t (a , 2) - R ( a , 1)}.

Then, the steps of heuristic algorithm 4 are as follows:

ranging jobs in descending order of h(a), values breaking ties in favor of the job with lowest t(a, 1).

Step 2. (Assignment of Jobs:) Starting with i = 1, successively assign each job a i to the latest available machine at stage 2 such that minimum job waiting is incurred at stage 2. If there is no job waiting, assign job a i to the latest available machine.

Step 3. (Approximate Solution:) Accept the fi- nal assignment and schedule thus obtained as an approximate solution to the problem.

Consider once again, the eight-job problem of Table 1 with two identical machines at stage 2 and only one machine at stage 1. Below, the problem is solved using algorithm 4.

Step 1. Arranging the jobs in descending val- ues of h ( a ) = {t(a, 2 ) - R ( a , 1)} results in the schedule P = (4, 1, 3, 8, 6, 2, 7, 5).

Step 2. The assignment of jobs is now done as explained in Step 2 of algorithm 3. Proceeding in the same manner as in algorithm 3, jobs are assigned as shown in Table 5.

Step 3. From Table 5, it is seen that jobs (4, 8, 2, 7, 5) are processed on machine 1 of stage 2 while jobs (1, 3, 6) are processed on machine 2 of stage 2. The processing sequence at stage 1 is P = (4, 1, 6, 3, 8, 5, 2, 7). The makespan of this schedule is 256.

6. Computational experience

Step 1. (Sequence identification:)Let P = (al, a 2 . . . . . a n) be the schedule obtained by ar-

How good are the solutions obtained from the proposed heuristic algorithms? Attempts to de-

T a b l e 5

A s s i g n m e n t o f jobs to m a c h i n e s ( A l g o r i t h m 4)

i J o b a i S t a g e 1 S tage 2

M a c h i n e 1 M a c h i n e 2

A l (a i ) A 2(ai ) Bl(a i) n2(a i) Bl(ai) B2(ai)

1 4 24 32 63 66 - -

2 1 64 69 - - 90 98

3 3 112 114 - 130 132

4 8 138 144 155 166 - -

5 6 159 170 - - 179 185

6 2 188 191 200 207 - -

7 7 209 212 218 221 -

8 5 243 250 256 261 - -

J.N.D. Gupta, E.A. Tune / Scheduling a two-stage hybrid flowshop 423

velop the worst-case bounds on the deviation of the heuristic makespan from the optimal make- span were not successful. Further, no optimiza- tion algorithm is available to optimally solve this problem, even for moderately large number of jobs. Therefore, the effectiveness of the proposed heuristic algorithms in finding the minimum makespan schedules was empirically tested.

The heuristic algorithms presented in this pa- per also attempt to minimize the total number of machines used at the second stage. This can readily be seen from the fact that additional machines will be used only if job waiting occurs. Further, as the number of available machines at the second stage increases, makespan of a sched- ule found by any of the heuristic algorithms pre- sented in the previous section will decrease. Since the global lower bounds on job-based makespan presented in Section 3 are independent of the number of machines at the second stage, the percent deviation of the heuristic makespan from the global lower bound on the makespan will decrease as the number of machines at the sec- ond stage increases. Thus, the percentage devia- tion of heuristic makespan from global lower bound will be maximum when the second stage

Table 6 Percentage deviation from lower bounds for the heuristic job-based makespan using algorithm 1 (non-dominant data set)

n n I n 2 n 3 n 4 Avr. Min. Max.

5 9 2 4 5 4.33 0 26.67 6 11 4 2 3 3.29 0 23.94 7 I l 5 4 0 0.98 0 4.39 8 8 9 3 0 1.35 0 4.39 9 13 5 1 1 1.13 0 13.79

10 13 5 0 2 0.85 0 6.84 15 10 10 0 (I 0.65 0 2.23 20 15 5 0 0 0.29 0 2.84 25 11 9 0 1) 0.26 0 1.57 30 10 9 1 0 0.51 0 3.90 35 6 10 4 0 1.04 0 4.29 40 9 10 1 0 0.35 0 3.36 45 7 13 0 1) 0.42 0 2.07 50 4 14 2 0 0.74 0 3.17 60 5 15 0 0 0.31 0 2.02 70 7 13 0 0 0.18 0 1.45 80 5 15 0 0 0.21 0 1.19 90 5 15 0 0 0.22 0 1.11

100 1 19 0 0 0.24 0 0.48

contains only two machines. Therefore, the case with only two identical machines at second stage was examined in detail.

The proposed algorithms were programmed in FORTRAN to solve 760 problems ranging from 5 jobs to 100 jobs divided equally between two sets of problem data: In Data Set A (called Non- dominant data set), the processing times at both stages of the problems were generated from the same discrete uniform distribution in the range (1, 99). In Data Set B (called Dominant data set), the processing times at stage 1 were generated from the same discrete uniform distribution in the range (1, 49) while the processing times at stage 2 were generated from the same discrete uniform distribution in the range (1, 99). For both sets of problems, the removal times and setup times at both stages were generated from a discrete uniform distribution in the range (1, 9).

6.1. Efficiency of the proposed heuristic algorithms

One measure of the efficiency of the proposed algorithms is the computational time required to solve the problem. These computational times for the proposed algorithms were not measured since they are relatively small. The computational time to sort n jobs according to Sule's or LPT rule is O(n log n) while allocation of jobs to various machines and computation of makespan requires additional O(mn) computational time. Therefore, in the worst case, the number of computational time for algorithms 1, 3 and 4 will be O(n log n +nm) where n is the total number of jobs and m is the number of machines at stage 2. Algorithm 2, in the worst case, requires n additional job allocations and calculation of makespans. There- fore, the computational complexity of algorithm 2 is O{(n log n + nZm)}.

6.2. Effectil,eness of the proposed heuristic algo- rithms

For each problem, the percentage deviation of the makespan of the schedule generated by the heuristic algorithms from its global lower bound was calculated. Based on 20 problems of each size (n), the following statistics were collected: nl: Number of times heuristic makespan equalled its global lower bound. n2: Number of times the deviation of the heuris-

424 J.N.D. Gupta, E.A. Tunc / S c h e d u l i n g a two-stage hybrid flowshop

Table 7

Percentage deviation from lower bounds for the heuristic job-based makespan using algorithm 1 (dominant data set)

n n 1 n 2 n 3 rt 4 Avr. Min. Max.

5 1 2 0 17 21.55 0 58.93 6 3 5 0 12 13.13 0 47.09 7 5 5 2 8 6.82 0 22.78 8 4 3 2 11 9.58 0 33.50 9 5 5 1 9 5.87 0 20.31

10 5 5 2 8 5.37 0 17.91 15 7 9 1 3 3.38 0 31.38 20 12 6 1 1 0.85 0 10.99 25 13 5 1 1 0.73 0 6.95 30 10 10 0 0 0.40 0 2.72 35 14 4 1 1 0.52 0 6.35 40 10 7 1 2 0.99 0 6.47 45 14 6 0 0 0.09 0 0.69 50 13 7 0 0 0.07 0 0.29 60 14 6 0 0 0.03 0 0.33 70 11 8 1 0 0.28 0 4.45

80 13 7 0 0 0.04 0 0.19 90 9 11 0 0 0.06 0 0.19

100 9 11 0 0 0.04 0 0.14

Table 8

Percentage deviation from lower bounds for the heuristic job-based makespan using algorithm 2 (non-dominant data set)

n n I n 2 n 3 n 4 Avr. Min. Max.

5 10 4 2 4 2.70 0 12.92 6 14 4 1 1 1.47 0 16.99 7 18 2 0 0 0.20 0 2.69 8 14 4 2 0 0.77 0 4.39 9 15 3 1 1 1.03 0 13.79

10 17 2 0 1 0.46 0 6.84 15 14 6 0 0 0.39 0 2.23 20 16 4 0 0 0.15 0 2.02 25 13 7 0 0 0.16 0 0.82 30 17 3 0 0 0.05 0 0.57 35 10 10 0 0 0.21 0 0.80 40 16 3 1 0 0.18 0 3.36 45 14 6 0 0 0.08 0 1.09 50 12 8 0 0 0.11 0 0.83 60 11 9 0 0 0.08 0 0.82 70 16 4 0 0 0.01 0 0.09 80 15 5 0 0 0.02 0 0.17 90 14 6 0 0 0.02 0 0.26

100 20 0 0 0 0.00 0 0.00

tic makespan from its global lower bound was greater than zero but less than or equal to 3%. 1/3: Number of times the deviation of the heuristic makespan from its global lower bound was greater than 3% but less than or equal to 5%. n4: Number of times the deviation of the heuris- tic makespan from its lower bound was greater than 5%. Min.: Minimum percentage deviation of the heuristic makespan from its global lower bound. Avr.: Average percentage deviation of the heuris- tic makespan from its global lower bound. Max.: Maximum percentage deviation of the heuristic makespan from its global lower bound.

Tables 6 and 7 depict the summary statistics for 380 problems in each of the nondominant and dominant data sets for the proposed algorithm 1. Similar results for proposed algorithms 2, 3, and 4 are presented in Tables 8-13. From these re- sults, it is clear that the effectiveness of the proposed heuristic algorithms increases as the number of jobs increases. Further, the perfor- mance of the heuristic algorithms does not deteri- orate for large sized problems with dominant data set. Since the global lower bounds for prob- lems with dominant second stage are very weak, the percentage deviation of heuristic makespan from global lower bounds for small problems is quite large. In view of the fact that global lower

bounds were used instead of the optimal makespans, actual percent deviations from opti- mal makespan will be less than those reported in Tables 6-13.

Based on these computational results, it can be concluded that the proposed heuristic algo-

Table 9 Percentage deviation from lower bounds for the heuristic job-based makespan using algorithm 2 (dominant data set)

n n 1 n 2 n 3 n 4 Avr. Min. Max.

5 1 2 0 17 17.89 0 44.44 6 3 5 0 12 12.02 0 37.04 7 8 2 2 8 5.73 0 22.78 8 8 1 1 10 7.44 0 33.49 9 7 4 3 6 3.65 0 20.31

10 6 4 2 8 4.22 0 12.46 15 12 4 1 3 2.89 0 27.86 20 18 1 0 1 0.67 0 10.65 25 18 0 1 1 0.56 0 6.95 30 17 3 0 0 0.32 0 2.72 35 18 0 1 1 0.49 0 6.35 40 15 2 1 2 0.92 0 6.31 45 19 1 0 0 0.03 0 0.69 50 20 0 0 0 0.00 0 0.00 60 20 0 0 0 0.00 0 0.00 70 19 0 1 0 0.22 0 4.45 80 20 0 0 0 0.00 0 0.00 90 20 0 0 0 0.00 0 0.00

100 20 0 0 0 0.00 0 0.00

J.N.D. Gupta, E.A. Tunc / Scheduling a two-stage hybrid flowshop 425

Table 10 Percentage deviation from

job-based makespan using

set)

lower bounds for the heuristic algorithm 3 (non-dominant data

n n 1 n 2 n 3 t'/4 Avr. Min. Max.

5 15 3 0 2 0.91 0 7.87 6 16 1 1 2 1.32 0 15.44 7 13 6 1 0 0.36 0 4.55 8 17 2 0 1 0.66 0 11.63 9 16 3 0 1 0.44 0 7.76

10 19 1 0 0 0.09 0 1.87 15 19 1 0 0 0.05 0 1.04 20 18 2 0 0 0.02 0 0.22 25 14 6 0 0 0.07 0 0.35 30 16 4 0 0 0.04 0 0.33 35 14 6 0 0 0.05 0 0.34 40 13 7 0 0 0.04 0 0.25 45 15 5 0 0 0.02 0 0.22 50 15 5 0 0 0.02 0 0.14 60 16 4 0 0 0.01 0 0.11 70 11 9 0 0 0.03 0 0.10 80 10 10 0 0 0.03 0 0.13 90 12 8 0 0 0.02 0 0.13

100 9 11 0 0 0.04 0 0.08

Table 12 Percentage deviation from lower bounds for the heuristic

job-based makespan using algorithm 4 (non-dominant data

set)

n n t n 2 n 3 n 4 Avr. Min. Max.

5 17 1 0 2 0.87 0 7.87 6 16 1 1 2 1.31 0 15.44 7 19 0 1 0 0.23 0 4.55 8 19 0 0 1 0.39 0 7.76 9 18 1 0 1 0.45 0 8.91

10 19 1 0 0 0.09 0 1.87 15 19 1 0 0 0.02 0 0.31 20 20 0 0 0 0.00 0 0.00 25 20 0 0 0 0.00 0 0.00 30 20 0 0 0 0.00 0 0.00 35 20 0 0 0 0.00 0 0.00 40 20 0 0 0 0.00 0 0.00 45 20 0 0 0 0.00 0 0.00 50 20 0 0 0 0.00 0 0.00 60 20 0 0 0 0.00 0 0.00 70 20 0 0 0 0.00 0 0.00 80 20 0 0 0 0.00 0 0.00 90 20 0 0 0 0.00 0 0.00

100 20 0 0 0 0.00 0 0.00

rithms are quite effective in minimizing the makespan. In addition, the heuristic algorithm 4 outperforms the other three algorithms.

Table 11 Percentage deviation from lower bounds for the heuristic job-based makespan using algorithm 3 (non-dominant data

set)

n n 1 n 2 n 3 n 4 Avr. Min. Max.

5 0 2 2 16 17.16 2.15 47.41 6 4 5 2 9 10.82 0 47.09 7 8 4 1 7 5.14 0 23.53 8 5 5 2 8 6.72 0 31.86 9 6 1 4 9 5.28 0 17.62

10 9 2 0 9 5.65 0 22.03 15 9 5 0 6 5.71 0 40.76 20 13 5 0 2 1.39 0 13.75 25 11 6 0 3 1.42 0 9.06 30 11 6 1 2 1.01 0 7.92 35 14 4 0 2 0.95 0 10.38 40 13 3 0 4 1.70 0 9.91 45 15 5 0 (1 0.20 0 2.28 50 13 7 0 (1 0.08 0 0.35 60 9 11 0 0 0.11 0 0.35 70 10 9 0 1 0.34 0 5.63 80 11 9 0 0 0.05 0 0.19 90 8 12 0 (1 0.07 0 0.21

100 12 8 0 0 0.04 0 0.24

7. Minimizing machine-based makespan

Thus far, minimization of job-based makespan was considered. However, foregoing results can easily be extended to the machine-based

Table 13 Percentage deviation from lower bounds for the heuristic job-based makespan using algorithm 4 (dominant data set)

n n I n 2 n 3 n 4 Avr. Min. Max.

5 1 1 2 16 16.91 0 47.41 6 6 5 1 8 9.96 0 47.09 7 9 4 0 7 5.13 0 23.53 8 7 5 2 6 6.99 0 31.86 9 5 2 5 8 5.39 0 17.62

10 8 3 0 9 5.82 0 21.45 15 11 3 0 6 5.22 0 41.05 20 16 2 0 2 1.32 0 14.60 25 13 3 1 3 1.65 0 10.27 30 16 1 1 2 1.09 0 8.82 35 18 0 0 2 0.88 0 9.03 40 16 0 0 4 1.46 0 8.93 45 18 1 1 0 0.25 0 3.11 50 20 0 0 0 0.00 0 0.00 60 20 0 0 0 0.00 0 0.00 70 19 0 0 1 0.30 0 6.09 80 20 0 0 0 0.00 0 0.00 90 20 0 0 0 0.00 0 0.00

100 20 0 0 0 0.00 0 0.00

426 J.N.D. Gupta, E.A. Tune /Scheduling a two-stage hybrid flowshop

makespan provided the makespan of schedule

P = (a 1, a 2 , . . . , a , ) is def ined as follows:

M ( P ) = max{B2( a i)}.

As stated earl ier , this def ini t ion of makespan is

the same as that used by Sule (1982) and Proust,

Gup ta and Deschamps (1991). The following the-

o rem describes a polynomial a lgor i thm to mini-

mize the mach ine -based makespan for the case

where the n u m b e r of machines at the second

stage is g rea te r than or equal to the n u m b e r of

jobs.

Theorem 3. With an unlimited number of machines at the second stage, the schedule obtained by ar- ranging jobs in descending order of h(a) = ( t (a , 2)

+ R(a, 2) - R(a, 1)} values minimizes machine- based makespan.

Proof. The p roof of T h e o r e m 3 follows f rom

T h e o r e m 2 since t(a, 2) in T h e o r e m 2 can easily

be replaced by t(a, 2) + R(a, 2) wi thout causing

any change in the expressions. []

For problems where m 2 < n, def ine

h(a) = {t(a, 2) + R(a , 2) - R ( a , 1)}

Table 14 Percentage deviation from lower bounds for the heuristic machine-based makespan using algorithm 1 (non-dominant data set)

n n 1 n 2 n 3 n 4 Avr. Min. Max.

5 1 1 0 18 22.67 0 57.89 6 12 4 1 3 2.90 0 23.02 7 14 2 4 0 0.77 0 3.89 8 10 7 3 0 1.06 0 4.17 9 12 7 0 1 1.12 0 14.81

10 14 4 1 1 0.88 0 6.82 15 11 9 0 0 0.64 0 2.35 20 14 5 1 0 0.32 0 3.20 25 12 8 0 0 0.30 0 1.63 30 13 6 1 0 0.49 0 3.78 35 8 8 4 0 1.07 0 4.58 40 13 6 1 0 0.33 0 3.39 45 13 7 0 0 0.42 0 2.27 50 6 12 2 0 0.72 0 3.20 60 6 14 0 0 0.32 0 2.04 70 11 9 0 0 0.17 0 1.42 80 7 13 0 0 0.21 0 1.15 90 8 12 0 0 0.23 0 1.17

100 3 17 0 0 0.24 0 0.55

Table 15 Percentage deviation from lower bounds for the heuristic machine-based makespan using algorithm 1 (dominant data set)

n n I n 2 n 3 n 4 Avr. Min. Max.

5 1 1 0 18 22.67 0 57.89 6 3 5 0 12 13.59 0 47.92 7 6 4 2 8 6.85 0 33.16 8 4 2 2 12 9.74 0 31.56 9 6 2 3 9 6.88 0 20.63

10 4 4 1 11 6.95 0 20.38 15 8 5 1 6 6.10 0 46.80 20 11 4 2 3 1.82 0 10.92 25 9 4 3 4 2.16 0 9.89 30 10 4 3 3 2.07 0 9.68 35 7 9 1 3 1.57 0 7.12 40 8 4 5 3 2.62 0 13.12 45 11 6 2 1 0.87 0 5.30 50 4 13 1 2 1.13 0 5.09 60 7 11 2 0 1.06 0 4.44 70 5 14 0 1 0.79 0 6.38 80 3 15 2 0 1.14 0 4.66 90 5 15 0 0 0.91 0 2.75

100 3 17 0 0 0.68 0 2.44

and use heuris t ic a lgor i thm 4. Similarly, heuris t ic

a lgor i thm 1 can be used except that makespan is

found by using (2) ra ther than (3).

Tables 14-17 depict the results of heuris t ic

a lgor i thms 1 and 4 (with h(a) value as def ined

above) for the mach ine -based makespan where

lower bounds were also adjusted accordingly.

These results show that the p roposed algori thms

pe r fo rm equally well for this case as well even

though the lower bounds are not very tight.

8. Conclusions

This pape r has discussed the two-stage hybrid

f lowshop schedul ing p rob lem with separable setup

and removal t imes and p roposed solut ion proce-

dures for the case when there is only one ma-

chine at the first stage. Two concepts of makespan

were ident i f ied for schedul ing p rob lems with sep-

arable setup and removal times. For the case

where the n u m b e r of machines at the second

stage is less than the total n u m b e r of jobs, four

heuris t ic a lgor i thms have been p roposed and

tes ted as to thei r effect iveness in f inding a mini-

m u m makespan schedule. Computa t iona l results

indicate that the effect iveness of the p roposed

algori thms increases with the increase in the total

J.N.D. Gupta, E.A. Tunc / Schedufing a two-stage hybrid flowshop 427

Table 16 Percentage deviation from lower bounds machine-based makespan using algorithm data set)

for the heuristic 4 (non-dominant

n n l n e n 3 n 4 Avr. Min. Max.

5 16 1 0 3 1.26 0 11.40 6 16 1 1 2 1.25 0 14.72 7 19 0 1 0 0.18 0 3.69 8 19 0 0 1 0.30 0 6.01 9 18 1 0 1 0.39 0 7.69

10 19 1 0 0 0.10 0 2.05 15 19 1 0 0 0.02 0 0.31 20 20 0 0 0 0.00 0 0.00 25 20 0 0 0 0.00 0 0.00 30 20 0 0 0 0.00 0 0.00 35 20 0 0 0 0.00 0 0.00 40 20 0 0 0 0.00 0 0.00 45 20 0 0 0 0.00 0 0.00 50 20 0 0 0 0.00 0 0.00 60 20 0 0 0 0.00 0 0.00 70 20 0 0 0 0.00 0 0.00 80 20 0 0 0 0.00 0 0.00 90 20 0 0 0 0.00 0 0.00

100 20 0 0 0 0.00 0 0.00

n u m b e r of jobs. Since global lower bounds , which are not very tight, were used in testing the effec- t iveness of the proposed heurist ic algorithms, ac- tual per formance of these algori thms will be bet-

ter than that repor ted in this paper. Fur ther , the

logic used in the deve lopment of the heuristic

Table 17 Percentage deviation from lower bounds for the heuristic machine-based makespan using algorithm 4 (dominant data set)

n n x ne n 3 n 4 Avr. Min. Max.

5 1 1 1 17 17.45 0 47.52 6 5 3 4 8 10.49 0 44.79 7 10 2 1 7 5.11 0 25.91 8 7 4 1 8 6.67 0 31.56 9 5 2 2 11 5.59 0 19.78

10 8 3 1 8 5.67 0 22.35 15 10 3 1 6 5.61 0 39.53 20 16 2 0 2 1.31 0 14.51 25 14 2 2 2 1.58 0 9.89 30 16 1 1 2 1.11 0 8.18 35 18 0 0 2 0.76 0 9.19 40 16 0 0 4 1.71 0 10.91 45 18 2 0 0 0.21 0 2.41 50 20 0 0 0 0.00 0 0.00 60 19 1 0 0 0.01 0 0.19 70 19 0 0 1 0.28 0 5.61 80 20 0 0 0 0.00 0 0.00 90 20 0 0 0 0.00 0 0.00

100 20 0 0 0 0.00 0 0.00

algori thms tends to satisfy the secondary criteria,

namely the reduc t ion of the n u m b e r of machines

used at the second stage since addi t ional ma-

chines will be used only if necessary. In view of the NP-comple teness of the problem, these re-

sults are quite encouraging since large-sized prob-

lems can be efficiently solved. The excellent per formance of the proposed

heurist ic algori thms in optimally solving problems with a large n u m b e r of jobs can be partially

explained as follows. Af ter processing several jobs at the first machine, a queue of partially com-

pleted jobs is created be tween the two stages. This enables the second stage to operate inde-

pe nde n t of the first stage without any idle t ime

on any of the machines. Therefore , if the assign-

men t of jobs to machines is done so as to avoid machine idle and job wait ing times, the result ing makespan will tend to be min imum.

The proposed global lower bounds can be

modified and opera t ional ized to develop a branch and bound algori thm to solve the problem. Fur-

ther, in view of the relative efficiency of the proposed heuristic algorithms, the efficiency of such a branch and b o u n d algori thm can be im-

proved by augment ing the heurist ic a lgori thm at the f ront -end of a b ranch and bound algorithm.

This would decrease the computa t iona l effort

(both computa t iona l t ime to solve the problem and n u m b e r of end-nodes to be explored in find-

ing the opt imal schedule) required. This will also extend the size of the problem that can be solved

by b ranch and bound algorithms. This paper considered the case where there

are identical parallel machines at the second stage and only one machine at the first stage. It is possible to develop similar approximate solut ion

procedures for the reverse problem when the first

stage conta ins several identical parallel machines and the second stage conta ins only one machine. Fur ther , the logic used in the deve lopment of the

proposed heurist ic algori thms may be used for the solut ion of the general two-stage hybrid flow- shop problem where each stage conta ins identical

parallel machines.

Acknowledgement

Thought fu l comments of two anonymous refer- ees on an earl ier draft of the paper improved the

p resen ta t ion of results in this paper.

428 J.N.D. Gupta, E.A. Tunc / Scheduling a two-stage hybrid flowshop

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