Permutation vs. non-permutation flow shop schedules

Operations Research Letters 10 (1991) 281-284 North-Holland

July 1991

Permutation vs. non-permutation flow shop schedules C h r i s N. P o t t s University of Southampton, Faculty of Mathematical Studies, Southampton, S09 5NH, UK

D a v i d B. S h m o y s * Cornell Universi(v, ORIE, Engineering and Theory Center, Ithaca, N Y 14853, USA

D a v i d P. W i l l i a m s o n * * MIT, Laboratory for Computer Science, 545 Technology Sq., Cambridge, MA 02139, USA Received June 1990 Revised July 1990

In studying the m-machine flow shop scheduling problem, it is c o m m o n practice to focus attention on permutation schedules. We show that for the problem of minimizing the m a x i m u m completion time, this assumption can be a costly one, by exhibiting a family of instances for which the value of the best permutation schedule is worse than that of the true optimal schedule by a factor of more than 2I V/re.

flow-shop scheduling; permutation schedules; approximation algorithms

In the flow shop scheduling problem, there are n jobs and m machines, and each job j, j = 1, 2 . . . . . n, consists of m operations where the i-th operation is to be performed, in order, by machine i for a specified time p,j. In other words, each job j is first processed by machine 1, then machine 2, and so on, until it completes its processing by machine m; the time at which job j completes its last operation shall be denoted Cj. The objective is to find a nonpreemptive schedule that minimizes C m a x = maxjC r the maximum completion time of any job. This optimum shall be denoted C*ax. The recent paper of Lawler, Lenstra, Rinnooy Kan

* This research was supported in part by N S F PYI Award CCR-89-96272 with matching support from UPS, IBM and Sun and by the Cornell Computation Optimization Project. ** This research was supported in part by a N S F graduate fellowship and Siemens Research Agreement of June 10, 1987 on Joint Research in Learning Systems.

and Shmoys [2] surveys many of the known results pertaining to this problem. Permutation schedules constitute an important subclass of schedules, where the order in which each machine processes the jobs is identical for all machines. Most research on flow shop scheduling problems has focused on permutation schedules, due to the relative combinatorial simplicity of schedules that can be specified simply by giving a permutation of the jobs. Unfortunately, this simplicity is bought at the price of drastically inferior schedules, and the purpose of this note is to study the worst-case behavior of the ratio of the maximum completion time of an optimal permutation schedule, denoted C*ax(~r), to the optimum value C*~x. If 0 ( J ) denotes this ratio for an instance J , we will prove that p is not bounded by any constant, by exhibiting a family of instances {J2,: n ~ N ) for which P(Jm) > l[v/m + ½], where the instance J,, involves m machines.

0167-6377/91/$03.50 © 1991 - Elsevier Science Publishers B.V. (North-Holland)

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The instance J 2 , is relatively easy to describe: there are n jobs, 2n machines, and Pu =

1 0

ifi=j+nori=n+l-j, otherwise.

When PA/= 0, this value should rather be interpreted as an arbitrarily small positive constant; in other words, each job must still be processed on each machine. It is easy to see that C*ax = 2; since each job requires 2 units of processing, any schedule must take at least that long, and if the jobs are processed in the order 1, 2 . . . . . n on machines 1, 2 . . . . . n, but then in the order n, n - 1 . . . . . 1 on machines n + 1, n + 2 . . . . . 2n, all jobs are completed at the end of the second time unit. To prove the main result of this note, we need only prove a lower bound on the length of the optimal permutation schedule. The fundamental insight into analyzing the length of permutation schedules for these instances is the following easy fact: Fact 1. For the instance J 2 , , Cm~x= n + 1 for the schedules given by either of the permutations 1 , 2 . . . . . n o r n , n - 1 . . . . . 1. Proof. The schedules are given in Figure 1.

[]

This simple fact will have important consequences. Consider an arbitrary permutation of the n jobs Jl, J2 . . . . . j , and let j,,, Ji2 ..... Ji~ be a subsequence of jobs such that J4
Time 0

I

2

3 ....n

n*l

Machine i FInI-I

morphic to J2s. F r o m this it follows that if there is an increasing (or decreasing) subsequence of length s in the permutation, then for this permutation schedule, CmaX> S + 1. A well-known result of Erd6s and Szekeres [1] states that in any sequence of k 2 + 1 distinct integers, there is either an increasing sequence of length k + 1, or else there is a decreasing sequence of length k + 1. Thus, this result can be combined with Fact 1 to prove an ~2(v~) bound on p(~¢,,). In order to prove a tight bound, we will need to use some of the ideas in the proof of the E r d 6 s Szekeres result in a slightly stronger setting. Theorem 1. For each instance J2n, n = 1, 2 . . . . .

>--- 12?

½l,

Proof. Let Jl, J2 . . . . . Jn denote a permutation of the jobs 1, 2 . . . . . n. We will show that if the jobs are scheduled in this order, then the value of Cmax for this schedule will be at least 2 ~ + ½, and since the value of Cmax is an integer, this implies the claimed result. The proof will isolate a particular subsequence of jobs that will, by themselves, imply this lower bound. Let l denote the length of the longest increasing subsequence in Jl, .]'2. . . . . Jn, and let l i denote the length of the longest increasing subsequence that starts with the element ji. Note that l, ~ {1 . . . . . l }. Consider the multiset

S = {li: i = 1 . . . . . n } ,

Time 0

I

2

3 ....n

n*l

Machine I I--A- l

F-q

nFi--]

n+2

+

and this bound is tight.

FTI n*l

July 1991

n

[]

77

n*l ~

n*2

2n

FInI'l (a)

I~]

2n

D

(b)

Fig. 1. (a) The p e r m u t a t i o n schedule 1, 2 . . . . . n; (b) the p e r m u t a t i o n s c h e d u l e n, n

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and let rnk denote the n u m b e r of times that k occurs in S. It is evident from the jobs of a longest increasing subsequence that m k > 0 for k = 1 . . . . . I. For k = l . . . . . l, we will show that the ( n o n e m p t y ) subsequence containing all jobs j, for which l, = k must complete processing no earlier than k + m k. If we let k * denote the value k {1, 2 . . . . . l } for which k + m~ is maximized, we will also show that

= 2v/~,andsor> 2~+½. Sincek*+mk. is an integer, we obtain the desired inequality. T o see that this b o u n d is tight, consider the schedule given by the p e r m u t a t i o n 1, 3, 2, 6, 5, 4, 10, 9, 8, 7 . . . . ; the pattern of increasingly longer blocks of decreasing sequences should be clear from this prefix. For n = 10, this schedule is given in Figure 2. Extrapolating from this example, we see that if

k* +m,.>_ [2¢ +½ l

½k(k-1)
By combining these two facts, we obtain the claimed lower bound. Let j,,, j,: . . . . . j,, denote the subsequence of jobs j, for which l, = k. N o t e that s = mk. If Ju and j,, are two jobs of this subsequence with u < v, it is easy to see that ju 1,., since j , can be placed at the head of the longest increasing sequence starting with i.. Thus, li, =/,~ = . . . . l,, implies that J}~' J,2. . . . . J,, forms a decreasing sequence. N o t e that j o b j,, must be processed for one time unit on machine n + 1 -Ji,, and after that J,2 must be processed for one time unit on machine n + 1 - J i : , and so forth, so that j o b j,, completes its processing on machine n no earlier than after s = m k units of time. But now consider the increasing subsequence of jobs of length k that starts with j); call this sequence j&, j&+ ,, .... j,, ..... . This sequence must now be processed on machines n + 1 . . . . . 2n. Once again, first j,, must be processed for one time unit on machine n +j,,, and this must be followed by j~.... on machine n +-Ji,+l' and so forth, so that j,~+~_, is completed on machine 2n at least k time units later. Thus, the subsequence requires in total at least k + m k time units to complete processing. To prove the second fact, suppose that k + m k < r, for each k = 1 . . . . . l. Then

then this schedule for ,-¢2, has Cmax = k + 1, which equals the b o u n d claimed in the theorem. []

½k(k + l),

The algorithm of ROck and Schmidt [3,4] proves that p ( J ) < [½m] for any instance J . It remains an interesting open question to determine whether there is a family of instances with an asymptotically larger ratio than O(~/m) or, from the opposite (and m o r e interesting) point of view, whether T1me

0

1

2

3

4

5

Machine 1 l-~

I-q

D 71

EEl

l-q l-q

,o771 ,, 1-TI ,° [-7]

l

~_. (k + rnk) <_ lr. "--1

,4

But note that

[-7]

l

E k=½l(l+l) k=l

D

/

and

E mk=n. k=l

,6 ,7

Thus,

r> n / l + ½(l+ 1). Applying elementary calculus, we see that the right-hand side achieves its m i n i m u m when l

19 2o

r 7] r 7] [-7]

Fig. 2. A good permutation schedule

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there is an algorithm that always delivers a permutation schedule with Cmax < O(~/-m)C~*ax.

References [1] P. Erd/Ss and A. Szekeres, "A combinatorial problem in geometry", Compositio Math. 2, 463-470 (1935). [2] E.L. Lawler, J.K. Lenstra, A.H.G. Rinnooy Kan, and D.B.

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Shmoys, "Sequencing and scheduling: algorithms and complexity', Technical Report BS-R8909, Centre for Mathematics and Computer Science, Amsterdam, 1989. [3] H. Ri~ck and G. Schmidt, "Machine aggregation heuristics in shop scheduling", Technical Report Bericht 82-11, Fachbereich 20 Mathematik, Technische Universit~it Berlin, 1982. [4] H. R&zk and G. Schrnidt, "Machine aggregation heuristics in shop-scheduling", Methods Oper. Res. 45, 303-314 (1983).