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Scheduling a two-machineflowshop with travel timesto minimize maximumlatenessJ. W. Stevens & D.D. GemmillPublished online: 14 Nov 2010.

To cite this article: J. W. Stevens & D.D. Gemmill (1997) Schedulinga two-machine flowshop with travel times to minimize maximumlateness, International Journal of Production Research, 35:1, 1-15, DOI:10.1080/002075497195948

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Scheduling a two-machine ¯ owshop with travel times to minimizemaximum lateness

J. W. STEVENS² and D. D. GEMMILL³ *

As fully automated manufacturing becomes a realization, the problem of schedulingjobs to meet customer due dates come to the forefront. In this paper, possiblesequencing methods are generated for an automated two-machine ¯ owshop withnon-negligible transportation times and blocking of the second machine. Twoheuristics are developed to sequence a set of jobs with the objective of minimizingmaximum lateness. The cases of scheduling the ¯ owshop when it starts in a nullstate and a busy state are both investigated.

1. IntroductionScheduling is a topic that has received much attention from both OR researchers

and production practitioners. The problem undertaken in this paper involves thesequencing of jobs through a two-machine ¯ owshop with signi® cant travel timebetween the machines. Each machine is considered to be an entity that performs anoperation on a job. This operation can consist of any type of processing, such asassembly, machining, cleaning, or inspection. Similarly, the transportation device isconsidered to be an entity that moves a job from the ® rst machine to the second. Thisdevice could be a conveyor, AGV, crane, robot, or industrial truck, to name a fewexamples. The objective is to ® nd a sequence so that the maximum lateness isminimized for the entire set of jobs.

As more industrial technologies become available, the aim of fully automatedmanufacturing is becoming a realization . The innovation of ¯ exible manufacturingsystems (FMS) has been an important step in achieving this desired outcome. Alongwith this has come diŒerent automated equipment, such as robots, guided vehicles,and computer-controlled machining centres. As the application of these technologiesbecomes more widespread, appropriate scheduling rules for the system will be needed.It is the intent of this study to develop a sequencing method for a two-machine¯ owshop possessing an FMS characteristic, namely travel times.

The main objective is to create heuristics that provide good solutions to thetwo-stage ¯ owshop problem described above. While algorithms that produce optimalsolutions every time are desired, the di� culties presented by this particular problemmakes this ideal solution method unlikely. Though it does not guarantee an optimalresult, a decent heuristic will provide good or near-optimal solutions. It will serve as aguide to ® nding possible solutions to the problem being considered.

2. Literature reviewSome of the possible solution methods for this problem are complete enumeration,

0020±7543/96 $12.00 � 1997 Taylor & Francis Ltd.

INT. J. PROD. RES., 1997, VOL. 35, NO. 1, 1±15

Revision received November 1995.² The HON Company.³ Iowa State University, 205 Engineering Annex, Ames, IA 50011, USA.*To whom correspondence should be addressed.

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mixed integer and non-linear programming, branch-and-bound, dynamic programming,and heuristics. With the exception of heuristics, each of these methods has a timecomplexity function that cannot be bounded and thus are referred to as exponentialtime algorithms. It is impractica l to ® nd solutions to large problems with these typesof algorithms. If a problem is so hard that a polynomial time algorithm cannotpossibly solve it, it is referred to as intractable (Garey and Johnson 1979).

For the problem of concern in this paper, it is probable that a polynomial timealgorithm is unlikely to exist since several related problems have been proven to beintractable. MacCarthy and Liu (1993) state that the n=2=F= Tj, n=2=F=Lmax ,n=2=F=nT , n=2=F= Cj, n=3=F=Cmax , n=2=G=Lmax are all NP-hard. Therefore, itwill be concluded that the problem under consideration in this paper is likely to beintractable also. By no means does this indicate that a polynomial time algorithmcannot exist, but it does seem unlikely.

Single-stage sequencing has received the most attention by researchers. This isprobably due to the simplicity that a single-machine shop oŒers for evaluation, but itis surprising how applicable many of these results are to larger and more complexproblems. Smith (1956) states that when one schedules an n=1==F problem, the mean¯ ow time is minimized by sequencing the jobs in order of non-decreasing processingtime. The procedure of ordering jobs in this way is referred to as shortest processingtime (SPT) sequencing, and is considered the most important concept in the theory ofsequencing (Conway et al. 1967). When one schedules an n=1==Lmax problem, themaximum lateness is minimized by sequencing the jobs in order of non-decreasing duedates (Jackson 1955). (i.e. d[1] £ d[2] £ . . . £ d[n], where [i]= ith job to be processed.)Placing jobs in this type of order is referred to as earliest due date (EDD) sequencing.The EDD sequence will also provide an optimal solution for the n=1==Tmax problem(Conway et al. 1967).

Maybe the most signi® cant work done in the area of two-machine sequencing wasaccomplished forty years ago (Proust 1992). Johnson (1954) developed an algorithmto minimize total ¯ ow time in the n=2=F=Fmax problem, which is now commonlyreferred to as the J̀ohnson problem’. The shop setup for the scheduling problem ofconcern in this paper is presented by Panwalkar (1991). An algorithm is presentedwhich obtains an optimal solution for the n=2=F=Cmax problem with travel timesbetween machines included. The diŒerence between this problem and the workpresented here is that Panwalker’s measure of performance is to minimize makespaninstead of minimize maximum lateness. Stern and Vitner (1990) use a speciallystructured, travelling salesman problem formulation to obtain solutions to alsominimize makespan, with the exception that no buŒer storage exists before thesecond machine. In the case of Stern and Vitner (1990), only approximate solutionsare generated. Other similar two-stage problems are considered by Mitten (1959),Maggu et al. (1981), Maggu et al. (1982), Sule and Huang (1983) and CË etinkaya (1994).

3. Problem formulationConsider the problem of sequencing a set of n independent jobs to be processed on

two continuously available machines, called A and B. Each job is processed ® rst onmachine A and then on machine B. Associated with a job is a processing time for eachmachine, called Ai and Bi, and a corresponding due date di. The time it takes to loadand unload a job from a machine is considered to be part of the processing time, alongwith any setup or teardown time required for the machine to process the job. There is

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an in® nite amount of buŒer storage available before B and all jobs are ready forprocessing at time zero.

A single transporter is used to pickup a completed job from A and deliver it to B.The transporter takes time t to travel from A to B, and time t

9to travel from B to A.

When machine A ® nishes a job, it will remain on the machine until it is picked up. Ifthe transporter has not returned from machine B, A becomes blocked and no otherjobs can be processed until the completed job is removed. If B is busy when a jobarrives from A, the job is placed in the buŒer storage until it can be processed.

The objective is to ® nd a sequence to minimize the maximum lateness for the set ofjobs. Lateness, or Li, is simply the diŒerence between a job’s completion time and itsdue date. Thus, the maximum lateness, or Lmax is the largest Li obtained for the set ofn jobs. It should be noted that the objective function can return a positive or negativevalue, depending on whether the job corresponding to the Lmax result is late or earlyrespectively. Results will be analysed for the case where the ¯ owshop starts in the nullstate and also when it is in a busy state. A busy state refers to scheduling the set of jobsinto a ¯ owshop that is already operating.

The following list of assumptions is necessary to further describe the characteristicsof the ¯ owshop.

(1) Both machines are continuously available and ready at time zero.(2) The transporter is continuously available and ready at time zero, unless the

¯ owshop is in a busy state.(3) The machines may be idle.(4) The transporter may be idle.(5) Each machine can process only one job at any given time.(6) No job may be preempted once an operation has begun.(7) No job cancellation is allowed.(8) Each job has exactly two operations.(9) Each operation can be performed by only one machine.

(10) The technological constraints are known in advance.(11) The machine processing times are independent of sequence.(12) The travel times between machines are independent of sequence.(13) There is no randomness within the ¯ owshop (processing times are ® xed,

travel times are ® xed, number of jobs is ® xed, ready times are ® xed).

c[i]2 = . . . f (c[i- 1]2 and c[i]1) . . .

4. Heuristic developmentThe preliminary analysis begins by mentioning three scheduling algorithms that

contributed to the development of a possible heuristic solution. Because the EDDordering provides the most favourable solutions for the n=1==Lmax problem, itbecame a starting point for a heuristic presented later. The decision rule developedby Johnson (1954) states that job i precedes job j in an optimal sequence if

min{Ai , Bj} £ {Bi , Aj}.

It is assumed that all jobs are ready for processing at time zero. When scheduling ann=m=F=Fmax problem with all jobs ready for processing at time zero, one needs onlyconsider schedules with the same job order on the ® rst two machines because the rule

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is regular, as well as the last two machines (Conway et al. 1967). Hence, onlypermutation schedules need to be evaluated for the n=2=F=Fmax problem.

The ® nal algorithm to be mentioned was proposed by Panwalkar (1991) and is thepaper in which the ¯ owshop described here is based. A classical two-machine¯ owshop scheduling problem is considered with signi® cant travel times betweenmachines. One is referred back to the problem formulation given earlier for a detaileddescription of the ¯ owshop operation. The only diŒerence is that Panwalkar’sproblem uses a performance measure of minimizing makespan instead of maximumlateness. Panwalkar (1991) gives a constructive algorithm to minimize completiontime, or makespan. A short proof showing the optimality of the algorithm can befound in Panwalkar (1991). This result has dual importance for the problem presentedin this paper. F irst, the ¯ owshop setup and operation is exactly the same, though themeasure of performance is diŒerent. Secondly, the algorithm will provide optimalsolutions to a small subset of problems being considered here. Any problem in whichall due dates are identical will have the same solution for the objective of minimizingLmax , Fmax , or the makespan Cmax . Hence, the algorithm can be applied to theproblem considered in this paper to ® nd an optimal sequence to minimize Lmax whenall jobs have identical due dates. Two proposed heuristics will now be presented thatattempt to solve the n=2=F=Lmax problem with travel times between machines.

4.1. Heuristic notationDe® nitions of all variables that will be used in the proceeding heuristics are given

below:

n = number of jobs to be scheduled,[i ] = ith position in the sequence,m = position of the latest job in the sequence,t = travel time of the transporter from machine A to B,t9 = travel time of the transporter from machine B to A,

A[i] = processing time on machine A for the job in position i,B[i] = processing time on machine B for the job in position i,A9

[i] = maximum value between A[i] and the roundtrip travel time of thetransporter,

d[i] = due date for the job in position i,u[i] = delay at machine B for the job in position i due to the job in position i-1

still being processed,L[i] = lateness of the job in position i,Lm = maximum lateness for all jobs and m is its position,Lp = maximum lateness after a proposed sequence change and p is its position,

Lmax = maximum lateness for the ® nal sequence.

4.2. Proposed Heuristic 1 ( H1)The ® rst proposed heuristic uses several diŒerent ideas discussed earlier in this

section. The seven-step heuristic is stated below and an example using this heuristic isgiven in Table 1 for a problem containing seven jobs.

Step 1. Determine the due date range for the job set. If the range is less than 22time units, initially sequence the jobs according to Johnson’s algorithm;else initially sequence the jobs according to EDD order.

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Step 2. Create a tableau with the following columns: Job, A, A 9 , B, u, d, L .Let A9

[i]= max{A[i], t + t 9 }, where i = 1, . . ., n,u[1] = 0,

u[i] = max{B[i- 1] - A9[i]+ u[i - 1], 0}, where i = 2, . . ., n,

L[1]= A[1]+ t + B[1] - d[1],L[i] = A[1]+ A9

[j ]+ t + B[i]+ u[i] - d[i], where i = 2, . . . , n and j =2 . . ., i,

L[m]= max{L[i]}, where i=1,. . . , n and m=position of the latest job.If m=1, go to step 7; else, continue to step 3.

Step 3. Temporarily swap jobs in positions m and m-1 and re-calculate the

5Scheduling a two-machine ¯ owshop: two heuristics

Problem Data: t=10, t9 =10

Job 1 2 3 4 5 6 7A 18 20 20 21 19 15 22B 17 18 22 16 23 25 15d 100 80 98 50 61 61 81

Step 1. Due data range = 100 - 50 = 50Initial sequence (EDD order): 4 - 5 - 6 - 2 - 7 - 3- 1

Step 2: Job A A 9 B u d L

4 21 21 16 0 50 - 35 19 20 23 0 61 + 136 15 20 25 3 61 + 382 20 20 18 8 80 + 377 22 22 15 4 81 + 513 20 20 22 0 98 + 571 18 20 17 2 100 + 72

L[m]= + 72, where m is position 7.

Step 3. Swap Jobs 3 and 1. L[p]= + 77. Since L[p] ³ L[m], move Jobs 1 and 3 back totheir original positions.

Step 4. Move Job 6 to the ® rst position and slide Jobs 4 and 5 down one position inthe sequence. L [p]= + 67. Since L[p] £ L[m], the change is kept and m isposition 7.

Step 5. Since B[5]< B[6], swap Jobs 7 and 3. L[p]= + 65. Since L[p]< L[m], thechange is kept and m is still position 7. An updated tableau is given below:

Job A A9

B u d L

6 15 20 25 0 61 - 114 21 21 16 4 50 + 165 19 20 23 0 61 + 282 20 20 18 3 80 + 273 20 20 22 1 98 + 317 22 22 15 1 81 + 631 18 20 17 0 100 + 65

Because u[m]= 0, this step is complete. Go to Step 7.

Step 6. (Disregard due to Step 5)

Step 7. Sequence: 6 - 4 - 5 - 2 - 3- 7 - 1 Lmax = + 65

Table 1. Example problem using heuristic 1.

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tableau. Let L[p]= max{L[i]}, where p is the position of the latest job.If L[p] ³ L[m], swap jobs in positions m and m-1 back and go to step 4; elselet L[m]¬ L[p]and let m = p and repeat step 3.

Step 4. For i=2,. . ., m-1, if min{A[i]} ³ A[1], go to step 5. If min{A[i]}< A[1],temporarily move the job in position i to the ® rst position and shift allpassed jobs down one in the sequence. Re-calculate the tableau and ® ndL[p] as before. If L[p] £ L[m], make this move permanent. Let L[m]= L[p]and m = p, then go to step 5. Else, change the sequence back and removethe job in position i from consideration. Repeat step 4.

Step 5. If u[m]= 0 or m = 1, go to step 7. Else, for jobs i = m-1,. . ., 2, temporarilyswap jobs in positions i-1 and i if B[i-1]< B[i], or B[i-1]= B[i] andA

9[i-1]> A[i]. Calculate L[p] as before. If L[p]> L[m], swap the jobs back

to their original positions. Else, make the swap permanent, and letL[m]= L[p] and m = p. If u[m]= 0, go to step 7; else, continue to next i.(If the set of jobs i is exhausted, continue to step 6.)

Step 6. If at least one permanent swap was made for any i, repeat step 5. Else, goto step 7.

Step 7. The ® nal sequence is taken from the Job column in the tableau as readfrom the top to bottom, with the minimum Lmax = L[m].

Depending on the number of jobs, this heuristic may not be very e� cient to applywith hand calculations. However, the computer code for these seven steps is fairlystraightforward to generate. One could also use a computer spreadsheet to implementthis heuristic.

In step 1, an initial sequence of the jobs is established using EDD order orJohnson’s algorithm. The range of the due dates, which is simply found by subtractingthe earliest due date from the latest due date, is used to determine which sequencingmethod should be applied. A range of 22 time units was established as the value indeciding which method to use. It is somewhat inherent that as the due dates are spreadfurther apart, the EDD order will provide a good Lmax result for the job set in mostinstances. The value of 22 time units was determined by testing problems usingdiŒerent ranges of due dates for both of the sequencing methods. Two graphs tosupport this conclusion can be found in Figs 1 and 2. In these ® gures, Heuristic l-E isthe results of using Hl with EDD order as the initial sequence, while Heuristic l-J is theresults of using Hl with Johnson’s algorithm as the initial sequence. Each ® gure showsthat EDD order provides the better starting point when the due date range is largerthan 22 time units and Johnson’s algorithm when the range of due dates is smallerthan 22 time units. This outcome does assume that the ¯ owshop is starting in the nullstate. If this is not the case, the heuristic given in the next section should be considered.

Step 2 simply calculates the important values for the initial sequence, such asmachine A processing time including blocking, delay prior to processing on machineB, and the job lateness, which are represented by A 9 , u and L respectively. In step 3, acheck is made to determine if the latest job can be moved forward in the sequence toimprove the measure of performance. While this step is not always eŒective in makingan improvement, there do exist job sets where it will be bene® cial in ® nding a bettersolution.

When the system is started from the null state, deciding on the job to place in the® rst position of the sequence is of monumental importance. One can gain or lose

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7Scheduling a two-machine ¯ owshop: two heuristics

Figure 1. Test results for the decision rule in heuristic 1, step 1±7 jobs.

F igure 2. Test results for the decision rule in heuristic 1, step 1±10 jobs.

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several time units based on this choice. The key number used for making this decisionis the roundtrip travel time of the transporter, or t + t

9. All jobs sequenced after the

® rst position which have processing times on machine A that are less than thisroundtrip time, will suŒer a delay waiting for the transporter to return from deliveringthe previous job to machine B. Therefore, A will become blocked by this job. So forjobs located after the ® rst position, the actual time each will spend at A is themaximum of its processing time at this machine and the roundtrip travel time of thetransporter. This value is represented by A

9i.

When the system is starting in the null state, the transporter is sitting idle andwaiting for the ® rst job to ® nish processing on A. Thus, the ® rst job will never bedelayed waiting for transport and will consequently never cause any blocking. Step 4attempts to place a job with the shortest possible machine A processing time, wherethis time is less than the roundtrip transport time, in the ® rst position without causingthe initial solution to deteriorate. As stated above, if a job sequenced after the ® rstposition has a processing time at A which is less than the roundtrip travel time, itsactual time at this machine is equivalent to this roundtrip time. So the schedule couldlose an amount of time equal to this diŒerence.

For example, suppose the roundtrip travel time is 20 units and Jobs 1 and 2 havemachine A processing times of 15 and 22 units respectively. If Job 1 is placed ® rst inthe sequence, it will ® nish processing at machine A at time 15, while Job 2 will ® nishon this machine at time 37. But if Job 2 is placed ® rst, it will ® nish at time 22, while Job1 will ® nish at time 37. However, Job 1 must now wait for the transporter, so it willactually not be removed from machine A until time 42. Hence, the schedule lost 5 timeunits because the time spent at A for Job 1 was increased from 15 to 20. Also, no timeis lost on Job 2 regardless of its position because it has a processing time on A which isgreater than the roundtrip transport time.

An incrementa l improvement of the processing delay at machine B incurred by thelatest job is attempted in step 5. If the latest job has to wait in the buŒer for processingon machine B, a swap of adjacent jobs sequenced before it will be made in an attemptto decrease this delay to zero. Pairwise exchanges are made that `push’ the jobs withthe shortest machine B processing times toward the latest job. This will allow thedelay to be dispersed among the jobs located earlier in the sequence. However, no jobsare exchanged that will cause the Lmax value to increase. It can be noted that asimprovements are made during this step, the job with the maximum lateness couldchange. So the value taken by m during this step might also change. In Step 6, one isdirected to repeat Step 5 if an improvement was made or continue on to the next step.Step 7 simply states the sequence and Lmax solution generated by the heuristic.

4.3. Proposed Heuristic 2 ( H2)The second proposed heuristic uses an entirely diŒerent approach than the ® rst

and is stated below. An example using this heuristic is given in Table 2 for a problemcontaining seven jobs.

Step 1. Schedule all two-job combinations within the job set and, for each pair,calculate the Lmax solution for both possible sequences.

Step 2. For each pair, determine the sequence that provides the minimum Lmax

value. (Ties are broken arbitrarily.) In this sequence, give the job locatedin the ® rst position a `win’. The total number of wins for each job shouldbe recorded.

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Step 3. Order the complete job set according to the non-increasing number of`wins’ for each job and calculate Lmax for the resulting sequence. (Ties arebroken arbitrarily.) This solution is a local optimum for minimizing Lmax

based on two-job competition .

This heuristic is much more e� cient to use than the ® rst heuristic, as the computationsare very simple. If the ¯ owshop is starting in the null state, the A[i]value is used for theprocessing time on machine A for the job located in the ® rst position. However, whenthe heuristic is applied to a ¯ owshop in a busy state, the A

9[i]value should be used for

the job in the ® rst position of each pair in step 1 when calculating the correspondingschedules. This is necessary because the scheduled jobs will be placed behind anexisting job set already being processed in the ¯ owshop. Thus, the transporter will bedelivering a job to machine B when the ® rst job from the calculated sequence beginsprocessing.

5. ResultsTest problems were created in order to evaluate the eŒectiveness of the proposed

heuristics. In each problem, a set of values for each job i had to be generated for thefollowing ® ve variables: processing time on machine A (Ai), processing time onmachine B (Bi), due date (di), travel time from A to B (t), and travel time from B to A(t 9 ). Job sets of size 7, 10, and 20 were used which contain 112 problems in each.These 112 problems are divided into equal groups of 28 based on due date ranges of 0,25, 50 and 100. Each group is broken down into sets of 7 problems that have a speci® crange of values for the ® ve variables mentioned above. Table 3 shows these speci® c

9Scheduling a two-machine ¯ owshop: two heuristics

Problem Data: t=10, t9 =14

Job 1 2 3 4 5 6 7A 15 40 9 28 20 23 31B 36 29 19 16 20 35 26d 82 100 75 89 84 96 91

Steps 1 and 2: The results for the ® rst pair are as follows:

Pair: (1,2)Sequence: 1 - 2 Lmax = - 6

2 - 1 Lmax = + 33

Since the sequence 1- 2 results in the minimum Lmax , this is the preferred order for thistwo-job type schedule. Thus, Job 1 receives a `win’ .

Pair Win Pair Win Pair Win Job No. of wins

(1,2) 1 (1,3) 3 (1,4) 1 1 5(1,5) 1 (1,6) 1 (1,7) 1 2 1(2,3) 3 (2,4) 2 (2,5) 5 3 6(2,6) 6 (2,7) 7 (3,4) 3 4 0(3,5) 3 (3,6) 3 (3,7) 3 5 4(4,5) 5 (4,6) 6 (4,7) 7 6 3(5,6) 5 (5,7) 5 (6,7) 6 7 2

Step 3. The solution is given below by arranging the jobs in order of the non-increasingnumber of `wins’ .Sequence: 3 - 1 - 5 - 6 - 7 - 2 - 4 Lmax = + 118

Table 2. Example problem using heuristic 2.

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range values. Thus, each heuristic was tested on a total of 336 problems, which wereall randomly generated.

5.1. Flowshop starting in the null stateBoth of the proposed heuristics, along with minimum slack ordering, will now be

compared based on their solutions to a set of test problems. Slack time is the amountof time remaining before a job must begin processing if it is going to be completed ontime. It is calculated as follows:

slack i = di - A9i - t - Bi, for all i.

So, minimum slack ordering implies sequencing the jobs in order of non-decreasingslack times. This method (Sk) is being evaluated so that the heuristic solutions can becompared to an established sequencing rule. Results for the scenario of the ¯ owshopstarting in the null state are presented ® rst, followed by the case where the schedule ismade when the ¯ owshop is already in a busy state. Each solution method was codedusing the C programming language on an IBM compatible computer. In addition, asimple branch-and-bound code was developed to enumerate the test problemscontaining 7 and 10 jobs. The optimal solution obtained by this method could thenbe compared with the heuristic and minimum slack solutions. Enumeration onproblems with 20 jobs was not included.

As Table 4 and Figure 3 indicate, heuristic 1 produces better overall results thanheuristic 2 and minimum slack ordering when compared with the optimal solution.For problems containing 7 and 10 jobs, heuristic 1 found the optimal solution in 172

10 J. W . S tevens and D. D. Gemmill

Problem number

1 - 7 8 - 14 15 - 21 22 - 28

5 £ Ai £ 45 5 £ Ai £ 45 15 £ Ai £ 25 15 £ Ai £ 255 £ Bi £ 45 5 £ Bi £ 45 15 £ Bi £ 25 15 £ Bi £ 25

di: low rangea di: high rangeb di: low rangea di: high rangeb

10 £ t £ 15 t = 10 t = 10 10 £ t £ 1510 £ t 9 £ 15 t 9 = 10 t 9 = 10 10 £ t 9 £ 15

aAn arbitrary range such that many jobs are ® nished late.

bAn arbitrary range such that most or all jobs are ® nished early.

Table 3. Parameters used for the test problems.

For non-opt ional solns: meanNo. of No. optimal time to optimal (high value)

No. of Due date testjobs range problems H1 H2 Sk H1 H2 Sk

7 0 28 23 12 0 3.8 (8) 5.3 (13) 13.2 (34)7 25 28 18 10 3 7.0 (16) 7.4 (25) 16.4 (41)7 50 28 22 6 2 2.7 (5) 8.5 (24) 10.8 (31)7 100 28 21 8 5 9.1 (24) 8.5 (27) 13.8 (37)

10 0 28 25 17 2 2.0 (3) 5.2 (21) 13.0 (44)10 25 28 21 7 0 7.3 (14) 5.5 (14) 13.2 (46)10 50 28 20 6 4 6.5 (21) 5.5 (14) 12.6 (40)10 100 28 22 3 2 12.3 (23) 8.5 (27) 13.1 (48)

Table 4. Summary of test problem results for the ¯ owshop starting in the null state.

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out of 224 attempts, or about 77% of the time. In contrast, the optimal result was onlyobtained 69 times, or about 31% , with heuristic 2. The minimum slack sequence onlyfound the optimal in 18 of the test problems. In further analysing these results forheuristic 1, it did ® nd a few more optimal answers for problems with 10 jobs than forthose with 7, as well as for those with a due date range of 0 time units as opposed toproblems without equal due dates. Heuristic 2 does have one trend worth mentioning.As the due date range increases, the number of optimal solutions found decreases.

Also shown in Table 4 are general statistics regarding the non-optimal results. Fornon-optimal solutions obtained by heuristic 1, the mean distance from the optimumvaried from a low of 2.0 time units to a high of 12.3 time units. When comparing thesemean values, heuristics 1 and 2 always did much better than the minimum slackordering. However, heuristic 2 had a lower mean result in 4 of the 8 problem groups,including 3 of 4 groups containing 10 jobs. When comparing the highest value, orfarthest distance from the optimal, heuristics 1 and 2 again performed more favourablythan the minimum slack method. Heuristic 1 had an equal to or lower value thanheuristic 2 in 7 of the 8 job groups.

In comparing each of the solution methods to one another instead of to theoptimal, heuristic 1 clearly provides the better Lmax results. A summary of this data isgiven in Table 5. For the comparison of heuristics 1 and 2, results for heuristic 1 wereat least as good as those for heuristic 2 in 299 out of the 336 test problems. In regardsto job set size and range of due dates, heuristic 1 de® nitely gives the better Lmax valuesin each instance. Heuristic 2 appears to have its best results when the due date range is0 time units. For this case, it has a solution which is at least as good as that forheuristic 1 in approximately 55% of the 84 test problems.

When heuristics 1 and 2 are compared to minimum slack ordering, both havesuperior results. Heuristic 1 has an equal or better Lmax value in 323 of the 336

11Scheduling a two-machine ¯ owshop: two heuristics

Figure 3. Number of optimal solutions vs due date range in the null state.

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problems and heuristic 2 in 295 of the problems. The minimum slack method seems tohave about the same success against heuristic 1 regardless of the problem type, butdoes seem to have a couple of small trends in success against heuristic 2. Speci® cally,as the number of jobs in the set decreases, the minimum slack sequence obtains anincreasing number of better or equivalent Lmax solutions. This trend is also the case asthe due date range increases, where it’ s at least as good as heuristic 2 in 43 out of 84problems.

5.2. Flowshop in a busy stateA comparison will now be made between heuristic 2 and minimum slack ordering

for the ¯ owshop in a busy state. This means that the ¯ owshop is in operation when thejob set to be scheduled is ready to be processed. The test problems used to make thiscomparison are identical to those used in the previous section, as is the calculation ofslack time. Heuristic 1 is not included in the results presented here. When the¯ owshop is in a busy state, the third step of heuristic 1 serves no purpose sincenothing can be gained at machine A based on the job placed in the ® rst position.Though this heuristic might be useful if this step was eliminated, it was not tested aspart of this paper.

A comparison between heuristic 2 and minimum slack ordering with the optimalsolution was made ® rst. A summary of this data is given in Table 6 and shows thatheuristic 2 found more optimal Lmax values in the problems containing 7 and 10 jobs.It was able to ® nd the optimal result in 174 out of 224 problems, or about 78% of thetime, while the minimum slack sequence found 122 optimal solutions. It appears that

12 J. W . S tevens and D. D. Gemmill

Comparison: number of better solutions

H1 vs H2 H1 vs Sk H2 vs SkNo. of Due date No. of testjobs range problems H1 H2 Same H1 Sk Same H2 Sk Same

7 0 28 16 1 11 27 0 1 24 3 27 25 28 14 6 8 23 2 3 15 4 97 50 28 21 1 6 26 0 2 12 6 107 100 28 18 2 8 22 1 5 11 2 15

10 0 28 10 2 16 26 0 2 24 0 410 25 28 20 5 3 26 2 0 19 5 410 50 28 20 2 6 23 2 3 14 6 810 100 28 23 2 3 26 0 2 16 3 9

20 0 28 12 0 16 27 0 1 26 1 120 25 28 15 9 4 22 3 3 17 4 720 50 28 19 6 3 25 2 1 18 4 620 100 28 22 1 5 26 1 1 14 4 10

Totals:

7 112 69 10 33 98 3 11 62 14 3610 112 73 11 28 101 4 7 73 14 2520 112 68 16 28 100 6 6 75 13 24

0 84 38 3 43 80 0 4 74 3 725 84 49 20 15 71 7 6 51 13 2050 84 60 9 15 74 4 6 44 16 24

100 84 63 5 16 74 2 8 41 9 34

Table 5. Summary of heuristic comparisons for the ¯ owshop starting in the null state.

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heuristic 2 is more eŒective when the due date range is small and less eŒective whenthis range increases. At a range of 0 time units, heuristic 2 found the minimum Lmax

solution for all problems with a job set size of 7 and 10.Also shown in Table 6 are general statistics pertaining to the non-optimal

solutions. The mean number of time units above the minimum Lmax value wasmuch smaller for heuristic 2 than for minimum slack ordering in all groups of testproblems. It had a low of 0.0 units and a high of 9.3 units. These average values forheuristic 2 were lower in problems with 7 jobs than in those with 10 jobs, where thelargest average occurred when the due date range was 50 time units. The mean timevalues for minimum slack sequences are shown to be consistent for the diŒerentproblem types. For this non-optimal data, the highest value, or time distance, fromthe optimum is also given in this table. Again, heuristic 2 results are always less thanthose obtained using the minimum slack rule.

In comparison against one another instead of with the optimal, both methodscame up with the same result in 192 out of 336 problems, or about 57% of the time.A summary of this data is given in Table 7. Heuristic 2 found the better solution in 122of the remaining 144 problems. The minimum slack ordering seemed to do its bestwhen the number of jobs was 20 or when the due date range was 50. Heuristic 2seemed to do its best when the due date range is 0.

5.3. SummaryA total of 336 test problems were generated to evaluate the proposed heuristics.

The numerical results show that heuristic 1 is a preferred method for scheduling this¯ owshop when it starts in the null state. In 88% of the test problems, heuristic 1produced an Lmax value which was at least as good as that for the other two solutionmethods. The results also indicate that heuristic 2 is a good method for scheduling this¯ owshop when it is in a busy state. It was found that heuristic 2 obtained an Lmax

result equal to or less than that for the minimum slack sequence in 93% of the testproblems. Excellent solutions are generated when the due dates for the job set aretightly bound and near-optimal results in most other instances. Also, its computationa lrequirements are very simple.

6. ConclusionIn this paper, the focus of study was on the n=2=F=Lmax problem with travel times

13Scheduling a two-machine ¯ owshop: two heuristics

For non-optimalNumber solutions: mean time toOptimal optimal (high value)

No. of Due date No. of testjobs range problems H2 Sk H2 Sk

7 0 28 28 16 0.0 (0) 8.5 (18)7 25 28 21 15 3.1 (7) 11.1 (26)7 50 28 19 18 5.9 (15) 10.4 (21)7 100 28 18 14 4.7 (17) 9.9 (27)

10 0 28 28 16 0.0 (0) 10.3 (31)10 25 28 20 12 8.0 (28) 9.7 (45)10 50 28 22 18 9.3 (18) 12.0 (26)10 100 28 18 13 8.6 (23) 10.2 (29)

Table 6. Summary of test problem results for the ¯ owshop in a busy state.

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between machines and blocking. The key factor is that the travel time of thetransporter from the second machine to the ® rst, after a job has been delivered, issigni® cant. This, along with the fact that there is only one transporter given, creates asituation where the ® rst machine can be blocked by a job that has already ® nished itsprocessing. The literature review indicated that a polynomial time algorithm isunlikely to exist for the problem under consideration, since it is very similar toother problems which have been shown to be polynomial complete. Most of theresearch dealing with two machines and travel times have used the performancemeasure of makespan. The measure of performance of Lmax used here adds anothervariable to the problem, namely the varying due dates for each job. But more researchinvolving travel times is being done with the advances in ¯ exible manufacturingsystems.

Two heuristics were proposed as possible methods for ® nding good solutions tothe stated problem. The results presented clearly imply that heuristic 1 is the preferredmethod for solving the ¯ owshop problem when it starts in the null state and heuristic2 when the ¯ owshop is in a busy state. Good solutions, though not always optimal,were found in most instances for minimizing the maximum lateness.

There are many directions in which future research could be directed for thistype of scheduling problem. One obvious variation is to modify the measure ofperformance. A practical criteria would be to look at costs based on completing a jobearly or late. If a job is done early, it will incur an inventory cost until its due date.When a job is ® nished late, a penalty cost is imposed by a customer. Thus, one would

14 J. W . S tevens and D. D. Gemmill

Comparison: Number ofbetter solutions

H2 vs SkNo. of Due date No. of testjobs range problems H2 Sk Same

7 0 28 12 0 167 25 28 12 0 167 50 28 8 4 167 100 28 10 0 18

10 0 28 12 0 1610 25 28 13 1 1410 50 28 7 1 2010 100 28 13 2 13

20 0 28 10 0 1820 25 28 8 2 1820 50 28 8 8 1220 100 28 9 4 15

Totals:7 112 42 4 66

10 112 45 4 6320 112 35 14 63

0 84 34 0 5025 84 33 3 4850 84 23 13 48

100 84 32 6 46

Table 7. Summary of heuristic comparisons for the ¯ owshop in a busy state.

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evaluate the lateness of each individual job and obtain a result for the entire set basedon total costs. With a performance measure like this, the added consideration ofinserted idle time on each machine is necessary. Other performance measures thatmight be useful are minimizing earliness, average lateness, and machine utilization .

Other potential areas for future research involve the modi® cation of the problemitself. A larger number of machines could be used, either in series or parallel, as well asmore transporters. The buŒer preceding the second machine could have deterministicspace. Individual job attributes could also be changed so that some ready times arenonzero, processing times are sequence dependent, cancellation and preemption ispossible, or general randomness is employed. There is literally an in® nite number of¯ owshop scheduling problems that are possible, it really depends on the needs andpreferences of the particular researcher or practitioner.

ReferencesBAKER , K. R ., 1974, Introduction to Sequencing and Scheduling (New York: Wiley).CË ETIN KAYA, F . C., 1994, Lot streaming in a two-stage ¯ owshop with set-up, processing and

removal times separated. Journal of the Operational Research Society, 45, 1445±1455.CONWAY, R. W., MAXWELL, W. L., and M ILLER , L. W., 1967, Theory of Scheduling (Reading,

Mass: Addison-Wesley).GAREY, M. R., and JOHNSON , D . S., 1979, Computers and Intractability (San Francisco:

F reeman).JACKSON, J. R ., 1955, Scheduling a production line to minimize maximum tardiness. Research

Report U3, Management Sciences Research Project, UCLA.JOHN SON , S. M., 1954, Optimal-two- and three-stage production schedules with setup times

included. Naval Research Logistics Quarterly, 1, 61±68.MACCARTHY, B. L., and LIU, J., 1993, Addressing the gap in scheduling research: a review of

optimization and heuristic methods in production scheduling. International Journal ofProduction Research, 31, 59±79.

MAGGU , P. L., DAS, G ., and KUMAR , R., 1981, On equivalent-job for job-block in 2xnsequencing problem with transportation-times. Journal of the Operations ResearchSociety of Japan, 24, 136±146.

MAGGU , P. L., SINGHAL, M. L., MOHAMMAD , N ., and YADAV, S. K., 1982, On n-job, 2-machine¯ ow-shop scheduling problem with arbitrary time lags and transportation times of jobs.Journal of the Operations Research Society of Japan, 25, 219±227.

M ITTEN , L. G., 1959, A scheduling problem: an analytical solution based upon two machines, njobs, arbitrary start and stop lags, and common sequence. The Journal of IndustrialEngineering, 10, 131±135.

PANWALKA R, S. S., 1991, Scheduling of a two-machine ¯ owshop with travel time betweenmachines. Journal of the Operational Research Society, 42, 609±613.

PROUST, C., 1992, Using Johnson ’s algorithm for solving ¯ owshop scheduling problems.Proceedings of the Summer School on Scheduling Theory and its Applications,INRIA (Bonas, F rance).

SMITH , W. E., 1956, Various optimizers for single-state production. Naval Research LogisticsQuarterly, 3, (1).

STERN, H . I., and VITN ER , G., 1990, Scheduling parts in a combined production-transportationwork cell. Journal of the Operational Research Society, 41, 625±632.

SULE, D . R ., and HUANG , K . Y., 1983, Sequencing on two and three machines with setup,processing, and removal time separated. International Journal of Production Research,21, 723±732.

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