On-line machine scheduling with batch setups

On-line machine scheduling with batch setups Lele Zhang, Andrew Wirth Department of Mechanical Engineering The University of Melbourne, VIC 3010, Australia Abstract We study a class of scheduling problems with batch setups for the online-list and online-time paradigms. Jobs are to be scheduled in batches for processing. All obs in a batch start and complete together, and a constant setup is prior to each batch. The obective is to minimize the total completion time of all obs. We primarily consider the special cases of these problems with identical processing times, for which efficient on-line heuristics are proposed and their competitive performance is evaluated. eyword: on-line scheduling; batch setups. 1 Introduction This paper considers a class of on-line scheduling problems with batch setups, where n independent non-preemptive obs are to be processed on a single machine or one of m identical parallel machines. A machine can process at most one ob at a time. Each ob i must be assigned to a batch, which consists of a set of obs processed consecutively on a machine. A batch setup s is incurred at the start of each batch. The completion time of a ob i is the time when the last ob of the batch that includes ob i completes its processing, that is, the completion time of the batch. The obective is to minimize the total completion time of all the obs. Thus, our problem can be stated as on-line machine scheduling with sequence-independent batch setup times in the batch availability model, and also denoted as 1, P m s, F = 1 C i [1,, 3], where F represents the number of families. This sort of problem is motivated by various real life applications in manufacturing areas and storage systems. One application, mentioned in [4, 5], is the logging of tass in a storage system. Log data can be consecutively Corresponding author. Tel.: +61-3-8344 485; wirtha@unimelb.edu.au 1

written on diss. Each write-on can be considered as a batch of obs and a constant setup for dis-write occurs for each batch. Another example is from part-type production in flexible manufacturing systems [6]. All parttypes must be mounted on a pallet for processing. The obs on the same pallet form a batch and are completed at the same time. A standardized pallet setup time precedes each batch. Previous wor mainly considered the single machine scheduling problem. Coffman et al. [7] provided two main properties of optimal solutions for their batch-sizing problem to minimize the total flow time 1 s, F = 1 F i : (i) obs are processed in shortest processing time order, and (ii) batch sizes, the numbers of obs in batches, are in non-increasing order. Further, they proposed a dynamic programming algorithm which could optimally solve the problem in O(n log n) time. Webster and Baer [8] summarized the results in [7, 9, 10], and for the problem with identical processing times, they showed that, ignoring the integer requirement, the optimal number of batches η = 1 4 + np s 1 and the optimal batch size of the th batch is n η + s(η +1) p s p, where p and s are the common processing and setup times. An on-line variant of the problem in the presence of release dates while minimizing the total flow time 1 r i, s, F = 1 F i was discussed by Gfeller et al. [4]. They proposed a -competitive GREEDY algorithm as well as two lower bounds for the special case with identical processing times. In the on-line general case, they showed that any on-line algorithm is at best ( n ɛ)-competitive for any ɛ > 0, and any on-line algorithm without unnecessary idle time cannot be better than n-competitive. Besides, they introduced an O(n 5 ) dynamic programming algorithm for the off-line version of their problem with a fixed ob sequence in [5]. For the parallel machine problem, Cheng et al. [6] studied the off-line problem of batching and scheduling simultaneously available obs on identical parallel machines to minimize total completion time. They developed an O(mn (m+1) ) dynamic programming algorithm to solve the general problem. They also showed that the special case of identical processing times reduces to single machine scheduling problem, which can be solved optimally by previous algorithms [9, 11]. The comprehensive review papers [1,, 3, 8] provide general definitions and realistic applications for machine scheduling problems with setups for various machine configurations, performance measures and availabilities. In this paper, we consider the problems of single machine and identical parallel machine scheduling with batch setups for the on-line list and onlinetime paradigms. More precisely, in the online-list paradigm, all obs are available at time zero and are presented one by one in some sequence. Once a ob is presented, an on-line algorithm must assign it to some machine and a batch immediately, without any the information about any subsequent obs. In the online-time paradigm, a ob becomes available at its release time r i (also nown as its arrival time), and an on-line algorithm may schedule the

ob upon its arrival or delay the decision-maing until a later time. From the nature of scheduling with batch setups, we see that in the online-time setting obs processed in a batch must arrive before the start of that batch s processing. Furthermore, once that processing starts, no extra obs can be added to that batch. We evaluate on-line algorithms in terms of their competitive performance. Let I be a problem instance, A(I) be the obective function value by algorithm A for I and OP T (I) be the optimal off-line value. We say A is a c-competitive algorithm if A(I) OP T (I) c for all I. Furthermore, we say A has a competitive performance ratio of R A if R A = inf{c 1 : c, for all I}. The general notation is listed below. Notation: A I m n s M l B (l) A(I) OP T (I) an on-line algorithm; an instance of obs; the number of parallel machines; the number of obs; the constant setup requirement, which is incurred before the processing of each batch; the lth machine; the th batch on machine l; the start time of the setup for batch B (l) S (l) n (l) the number of obs in batch B (l) ; J i the ith presented or released ob; C i the completion time of J i, equal to the completion time of the batch B (l) such that i B (l) ; p i the processing time of J i ; r i the release time of J i ; superscript the corresponding optimal value. Table 1 summarizes the the bounds proven in this paper. Online-List Paradigm Add Table 1 here. Now we consider the problems in the online-list paradigm, where obs are presented one by one and an on-line algorithm has to schedule a ob to a machine and a batch immediately upon its presentation. It is clear that for identical processing times, it is unnecessary to have any idle times in this paradigm. So we shall assume that none of our algorithms for the problems with identical processing times allow idle time. ; 3

.1 Lower bounds for a single machine scheduling The following proposition provides a lower bound on the competitive performance ratio for the a single machine scheduling problems with identical processing times and arbitrary processing times, respectively. Proposition 1. No on-line algorithm is better than 1.049-competitive for the 1 online-list, p i = p, s, F = 1 C i problem. Proof. Two problem instances are constructed to be scheduled by an on-line algorithm A. If A assigns the first two obs in separate batches, then the third ob shows up. Otherwise, no more ob is presented. Without loss of generality, we assume that s < p. For the instance I with two obs, the A schedule is sj 1 J and so A(I) = (s + p), whereas an optimal off-line solution is sj 1 sj and then OP T (I) = 3(s + p). On the other hand, for the instance I with three obs, the A schedule may be either sj 1 sj sj 3 or sj 1 sj J 3 and the obective function value satisfies A(I ) min {6s + 6p, 5s + 7p} = 6s + 6p by the assumption that s < p. An optimal solution for I is sj 1 J sj 3 as s < p, and so OP T (I ) = 4s + 7p. Now we minimize the following expression { } s + 4p 6s + 6p max,, 3s + 3p 4s + 7p and find that R A 109 + 34 3 109 + 1 1.049, where the optimum is achieved when s p = 109 3 10. Hence the result follows. 4

. P m online-list, p i = p, s, F = 1 C i In this section, we consider scheduling obs on m identical parallel machines. First we define a positive integer λ as follows: λ = ηs p, where η is a constant and will be set to different values at various stages of the following discussion. Then, it immediately follows from the definition function that p(λ 1) < sη λp. (1) Now we introduce a simple heuristic for this case and then consider its competitive performance. Heuristic Uniform-Batch-Size (U BS): Set η =. Whenever a new ob is presented, assign it to the machine with the least number of obs. In case of a tie, choose the machine with a smaller index. Every λ obs assigned to the same machine form a batch. (An example of a UBS schedule is given in Figure 1. Add Figure 1 here. Proposition. R UBS = 3 problem. for the P m online-list, p i = p, s, F = 1 C i Proof. Before we consider the competitive performance of heuristic U BS, we remind ourselves of some properties of optimal solutions. We recall Lemma 3 of [6] which states that, for the equal processing time problem, there exists an optimal solution for which the difference in the numbers of obs between any two machines is not greater than 1. It is easy to see that, for any input instance, the number of obs processed on some machine in the UBS schedule may be assumed to be the same as that in an optimal solution, since all the machines are identical. In addition, because all obs have identical processing times, we also assume that the obs assigned to some machine in the UBS schedule are processed on the same machine in the optimal solution, and that their relative processing positions are also the same. In the remainder of the proof, we shall mainly consider the scenario of an arbitrary machine l for 1 l m, and similar arguments can be applied to other machines. We let a + 1 and d be the number of batches on machine l and the number of obs in the last batch, respectively. Obviously, the number of obs in a 5

batch can never exceed λ. In addition, the first a batches are the same size, and each of them contains exactly λ obs. Therefore, the number of the obs processed on machine l can be written as aλ + d, if a 0. Note that if a = 1, which implies that n < m, then the UBS schedule, which has each ob processed on different machines, is exactly the same as the optimal solution, and then the problem becomes trivial. Thus, without loss of generality, we may assume that a 0 for the rest of the discussion. Now we consider the obs completed in the first a batches and in the last batch, respectively. For batch B (l), 1 a if a 1: The completion time of B (l), the th batch processed on machine l for 1 a and a 1, is (s + λp). Thus the total completion time of the obs in batches B (l) 1,..., B(l) a is given by i B (l),1 a C i = aλ(a + 1)(s + λp). We see that the number of the obs in the first a batches is aλ, and the completion time of the ith ob in this set in an optimal solution is at least s+ip. It follows that the optimal total completion time of these obs satisfies i B (l) C,1 a i aλs + aλp(aλ + 1) = 1 (a λ p + aλp + aλs). Let 1 = 6 i B (l),1 a C i 4 i B (l),1 a C i. By (1) and η =, we have λp p < s λp. It follows that So we obtain 1 a λ(λp s) + aλ(s + p λp) + aλp > 0. For batch B (l) a+1 : i B (l) C,1 a i i B (l),1 a C i < 3. The completion time of batch B (l) a+1 is s(a + 1) + p(aλ + d) and so the total completion time of the obs in the batch is written as: C i = d[s(a + 1) + p(aλ + d)] = aλdp + ads + ds + d p. i B (l) a+1 Recall that we assume that the processing order in the optimal solution is the same as that in the UBS schedule, because all obs are identical. That is 6

to say, in the optimal off-line schedule, the obs in the batches B (l) 1,..., B(l) a precede the obs in B (l) a+1. Hence, the optimal completion time of the ith ob in B (l) a+1 cannot be smaller than s + p(aλ + i), and then the optimal total completion time of the obs in B (l) a+1 satisfies i B (l) a+1 C i ds + aλdp + dp(d + 1) = 1 (aλdp + ds + d p + dp). Again, we let = 6 i B (l) Ci 4 a+1 i B (l) C i. Since d λ, similar to a+1 the preceding case we have Hence we prove that ad(λp s) + d(s + p dp) + dp > 0. i B (l) a+1 i B (l) a+1 C i C i < 3. It is derived from the above cases that i B (l) i B (l) applicable to any machine, and so we draw the conclusion that n i=1 C i n < 3 i=1 C i. C i C i < 3. This inequality is Furthermore, we see that R UBS 3. Now we consider a lower bound of the competitive performance ratio for U BS. Again, we consider the situation of an arbitrary machine l. First we recall some properties of optimal solutions for the corresponding single machine scheduling problem from [8]: the optimal number of batches η = 1 4 + np s 1 and the size of the th batch is n η + s(η +1) p s p. We can rewrite the number of obs in the th batch as: n = n η s p. Next, we consider an instance for which s = p and n = m(g +G) for a large integer G such that G + G is divisible by 4. Then, by UBS, the parameter λ equals and (G +G) 4 batches are formed and processed on machine l. It follows that the total completion time of the obs on machine l is given by i M l C i = 3s 16 (G + G)(G + G + 4) = 3sG4 16 + O(G3 ). As for an optimal solution, by the optimal properties, G batches are formed and the number of obs in the th batch is G + 1. So the optimal total 7

completion time is obtained by Ci = M l = η n (s + s =1 n ) =1 [ η (G + 1 ) s + s =1 ] (G + 1 ) as η = G, =1 = s η η η (G + 5G + 3) (3G + 4) + =1 =1 =1 3 So we have = sg4 8 + O(G3 ). i M l C i i M l C i 3, as G. The proof is now complete. We note that UBS can be applied to the 1 online-list, p i = p, s, F = 1 C i problem as well and, besides, the results of Proposition is also valid for that problem. 3 Online-Time Paradigm In this section, we adopt the problem in the online-time scheduling environment. We recall the statement of the problems in the online-time paradigm that any ob in a batch must arrive before that batch s processing starts, and no extra obs can be added to that batch, once its processing starts. In this section, we shall discuss parallel machine scheduling first. We propose a new heuristic for the P m online-time, r i, p i = p, s, F = 1 C i and a lower bound of the competitive ratio for the case of m =. Furthermore, we consider the problem of a single machine scheduling, for which we introduce an on-line heuristic as well as a lower bound. 3.1 P m online-time, r i, p i = p, s, F = 1 C i 3.1.1 A New Heuristic We remind ourselves of the definition expression of the parameter λ, that is, λ = ηs p. Now we set η = 1 for the following heuristic. 8

Heuristic Sync (Sy): Do not schedule until all machines are idle and some obs are available. Let n t be the number of unscheduled obs at some scheduling time t. Start a new batch on each machine, and then assign min{λ, nt m } obs to each of machines 1,..., n t m n t m and assign min{λ, n t m } obs to each of the remaining machines. In case there are more than mλ obs, select the earliest released obs. Remars: Without loss of generality, we may assume that the first released ob arrives at time 0. Suppose that there is an instance I for which the first ob is available after time 0, that is, r 1 > 0. Then we can obtain another instance I which is derived by decreasing the release time of each ob in I by r 1. It is easy to see that the on-line and the optimum total completion times both decrease by r 1 n from I to I. Thus, Sy(I) OP T (I) < Sy(I ) OP T (I ) Sy, we may assume that r 1 = 0.. As we are interested in the worst performance of We notice that there may exist a scenario in which, at time t, a batch B 1 starts on machine 1 whilst there is no batch that starts on machines l 0,..., m for some l 0 with l 0 m. For convenience, we may add a dummy batch on each of those machines such that n (l) = 0 for l 0 l m. Now we give some preliminaries for the following discussion of Sy s performance. Classifications: Let Ī = {I 1, I,... } be a set of time intervals defined as follows: Machine 1 is never idle during any interval I q and all machines are idle throughout the time period between I q and I q+1. Let BI = {BI 1, BI,... } be a set of time intervals such that a batchinterval BI contains the processing period of B (1), the th batch on machine 1, that is, BI = [S (1), S (1) s + n (1) p], where S (1) and n (1) are the start time and the number of obs of batch B 1. We divide an interval I q into batch-groups in the following way. A batch-group G q in I q, = 1,,..., ends with a batch-interval BI in which the number of obs is smaller than mλ, or possibly ends with the last batch-interval of I q. Figure gives an example of Sy schedules. 9

Add Figure here. Proposition 3. 1 m R Sy for the P m online-time, r i, p i = p, s, F = 1 C i problem. Proof. Consider an arbitrary interval I q. For a batch-group G q we let a + 1 be the number of batches on machine 1 and let h be the index of its first batch. Also, we let N (l) be the total number of obs in the batch-group on machine l. If a 1, by the definition of batch-groups we see that each of batches 1,..., a on any machine must have λ obs. Hence, N (l) = aλ+n (l) h+a, where n (l) h+a is the number of obs in the last batch of Gq on machine l. Then, the total completion time of the obs in G q on machine l of the Sy schedule is i G q,m l C i = S (1) h N (l) a(a + 1) + λ(s + λp) + n (l) h+a [a(s + λp) + s + n(l) h+a p]. On the other hand, a lower bound for the optimal total completion time of the obs in G q satisfies Ci m r i + s i G q i G q l=1 N (l) + p m l=1 N (l) m l=1 ( N (l) m + 1)( m m l=1 + p m l=1 N (l) m l=1 ( N (l) + 1)(m m m N (l) m l=1 m N (l) m l=1 N (l) ) m m l=1 + m N (l) ). m We notice that the difference between N l 1 and N l for l 1, l with 1 l 1, l m, is at most 1. Also, because all the obs have the same processing time and the machines are all identical, we can rewrite the above lower bound as Ci = i G q m l=1 i G q,m l m l=1 i G q,m l C i [ m r min N (l) l=1 r i + sn (l) + sn (l) + + (l) (l) pn (N + 1) ] + 1), (l) (l) pn (N where r min = min r i for i G q. We let l be the difference between i G q,m l C i and i G q,m l C i for some machine l. By (1) and η = 1, we 10

have λp p < s λp. In addition to N (l) = aλ + n (l) h+a, we have l r min N (l) + sn (l) + pn (l) (N (l) n (l) h+a (as + aλp + s + n(l) h+a p) = r min N (l) + sn (l) S (1) h aλ(s + p λp) + > r min N (l) + sn (l) N (l) + aλp + pn(l) h+a S (1) h N (l). + 1) S (1) h N (l) aλ + (λp s)( a λ + an(l) h+a ) (s + λp)(a + 1) If = 1, then r min = S 1 h and thus l > r min N (l) + sn (l) > 0. Otherwise, from the definition of G q as well as heuristic Sy, we see that the earliest released ob in G q must arrive after the start time of B1 h 1. That is to say, r min > Sh 1 1 = S1 h s n1 h 1p. In this scenario, if h 3 or q, then we have r min > Sh 1 1 > s + p. Furthermore, we can obtain l > r min N (l) > r min N (l) + sn (l) S (1) h + sn (l) + (S (1) > (s + p)n (l) n (1) (l) h 1pN (s + p)n (l) λpn (l) N (l) h s n1 h 1 p)n S (1) h > 0, since n (1) h 1 λ and λp < s + p. Therefore, for the cases of = 1 q, h 3 and q, we obtain the following result: N Ci C i = i G q i G q m l > 0. l=1 Thus, for the rest of the proof we need only consider the case of q = 1 and h = for G 1, that is, the situation in which the first batch processed on machine 1 in G 1 is B1. For this case, we consider the first two batch-groups together. Again, we let a + 1 be the number of batches on machine 1 in G 1. Let u and v be the smallest superscripts of batches B(l 1) 1 and B (l ) a+ such that n (l 1) 1 < n (1) 1 and n (l ) a+ < n(1) a+ respectively. If n(l) 1 = n (1) 1 and/or n (l) for all l with 1 l m, then let u = m + 1 and/or v = m + 1. a+ = n(1) a+ We note that there are m possible cases with regard to the values of u and v. We shall analyze the case of n (1) 1 = n () 1 + 1 = = n (m) 1 + 1 and 11

n (1) a+ = = n(m 1) a+ = n (m) a+ + 1, that is, u = and v = m. Then, for the other cases, analogous arguments can lead to the same result as below. By the assumption that the first ob is released at time 0, the total completion time of the obs in the two batch-groups is given by m C i = i G 1 1 G1 = l=1 i G 1 1 G1,M l m l=1 [n (l) 1 C i aλ(a + 1)(s + λp) (s + n(l) 1 p) + aλ(s + n(1) 1 p) + + n (l) a+ (as + aλp + s + n(1) 1 p + n(l) a+ p)]. Let N 1 denote the total number of obs in the two batch-groups. By hypothesis, N 1 = amλ+mn 1 1 +mn1 a+ m. Then the optimal total completion time of these obs satisfies Ci sn 1 + p N 1 m ( N 1 m + 1)(N 1 m N 1 m ) i G 1 1 G1 + p N 1 m ( N 1 m + 1)(m N 1 + m N 1 m ) = m(aλ + n (1) 1 1 + n (1) a+ )[s + p (aλ + n(1) 1 + n (1) a+ )]. We let = i G 1 1 G1 C i i G 1 1 G1 C i. Since λp s and n (1) a+ 1, > amλ(s + λp as λp) + amn (1) a+ (λp s) + (n(1) 1 1)(sm + amλp + mpn (1) a+ ) + mpn (1) 1 mp + pn (1) a+ pn(1) amλ(a 1)(λp s) p(m 1)(n (1) 1 1) 0. 1 + aλp + as + s + mpn (1) 1 mp + p pn (1) 1 Consequently, we have shown that i I q C i < i I q Ci R Sy. for all l. Thus Next we provide a lower bound for R Sy. Consider an instance I consisting of m obs with r 1 = 0, r i = ɛ for i m and s < p. The Sy schedule is sj 1 sj (on machine 1) and [idle]sj i (on machine i 1) for 3 i m, where the idle period equals s + p. On the other hand, an optimal solution is sj 1 (on machine 1) and [ɛ]sj i (on machine i) for i m. Thus, Sy(I) = (s + p) + (m 1)(s + p), OP T (I) = (s + p) + (m 1)(ɛ + s + p) m(s + p) as ɛ 0. 1

It follows that R Sy Sy(I) OP T (I) 1 m. Now the proof of this proposition is complete. We notice that an obvious variant of heuristic Sy can be applied to the 1 r i, p i = p, s, F = 1 C i problem with a competitive ratio of. 3.1. A Lower Bound for P r i, p i = p, s, F = 1 C i Proposition 4. No on-line algorithm is better than the P r i, p i = p, s, F = 1 C i problem. + 6 -competitive for Proof. We construct instances based on the scheduling mechanism of an online algorithm A. The constructed instances all start with obs available at time 0 and all have s = p. We let X 1 and X be the start times of J 1 and J by A, respectively. Without loss of generality, we assume that X 1 X. Now we consider the following possible cases. 1. J 1 and J are assigned to the same machine. In this case, there are no further obs. Let the instance of this case be I. Then the total completion time satisfies A(I) (X 1 + s + p) and the optimum is OP T (I) = (s + p). Hence A(I) OP T (I) = X 1 + s + p s + p s + p s + p = 4, as s = p. 3. J 1 and J are scheduled to different machines. Depending on the value of X, either no further ob arrives or two obs arrive at time X + ɛ, where ɛ is a small positive number. Let I denote the instance consisting of two obs, and then we have A(I ) = X 1 + X + (s + p) whereas OP T (I ) = (s + p). Besides, let I be the instance of four obs. For I A can either schedule J 3 and J 4 on the same machine after J 1 or J, or put them on different machines. We remind ourselves that obs processed in a batch must arrive before the start of the batch processing. Thus, J 3, J 4 must be assigned to new batches for r 3 = r 4 = X + ɛ > X X 1. Also, we note that if obs 3 and 4 are placed on the same machine, the total completion time is smaller if they are processed within a batch, since s = p. Thus, A(I ) min{3(x 1 + s + p) + (s + p) + X + s + p, 3(X + s + p) + (s + p) + X 1 + s + p, (X 1 + s + p) + (X + s + p) + (s + p)} min{3x 1 + X + 6s + 8p, X 1 + X + 6s + 6p}, 13

by the assumption that X 1 X. An optimal solution of instance I is sj 1 J on M 1 and [X + ɛ]sj 3 J 4, where [X + ɛ] is an idle period for M. Thus, we have OP T (I ) = (s + p) + (X + s + p). The maximum of A(I ) and A(I ) OP T (I ) OP T (I ) { X1 + X + (s + p) max, (s + p) min is greater than or equal to { 3X1 + X + 6s + 8p, X 1 + X + 3s + 3p X + 4s + 8p X + s + 4p Since X 1 0 and s = p, we rewrite the above expression as: { } A(I ) max OP T (I ), A(I ) OP T (I ) { { X + 6p X + 0p max, min 6p X + 16p, X }} + 9p. X + 8p We minimize the maximum, and by some algebra we find that { } A(I ) max OP T (I ), A(I ) +, OP T (I ) 6 }}. where the minimum is achieved when X 1 = 0 and X = ( 4)p. From the discussion above, we see that R A min{ 4 3, + 6 } = 1.115) for any on-line algorithm A. Hence the result follows. + 6 ( 3. 1 online-time, r i, p i = p, s, F = 1 C i Now we turn our attention to the shop type of a single machine. We notice that the on-line scheduling environment of the case considered in this section is the same as the corresponding problem 1 online-time, r i, s, F = 1 F i in [4], except for the obective function. The two obectives of the total completion time and the total flow time are equivalent when all obs are available at the same time or for off-line scheduling problem. We cannot say which one is more appropriate, since one may be more realistic in some cases, and may be less in others. As an essential part of our discussion, we consider this 1 online-time, r i, p i = p, s, F = 1 C i problem in this section. In [4], the authors proposed a -competitive on-line heuristic for the problem, and also showed that no on-line algorithm could achieve better results. We shall prove that for 1 online-time, r i, s, F = 1 C i this lower bound on the competitive ratio for all on-line algorithms is 5+1, and further develop a simply implemented heuristic, which guarantees a competitive ratio bounded between 5 3 and 1 + 3 5. 14

3..1 A Lower Bound Proposition 5. No on-line algorithm is better than the 1 online-time, r i, p i = p, s, F = 1 C i problem. 5+1 -competitive for Proof. We consider the following scenario. The first ob is released at time 0 with p = ɛs, where ɛ is an arbitrary small number which we shall let tend to zero. Suppose that an on-line algorithm A allocates the machine to J 1 at time X 1. Depending on the value of X 1, either no further ob arrives or n 1 obs arrive at time X 1 + ɛ. In the latter case, as ɛ 0, A can at best assign the last n 1 obs in one batch immediately succeeding the first batch, whereas an off-line optimal solution may have all n obs processed in one batch starting at time X 1 +ɛ. Hence the on-line total completion time of the n obs is greater than or equal to n(x 1 + s) + (n 1)[s + (n 1)p], whilst the off-line optimum is not greater than n(x 1 + ɛ + s + np). A may choose the best value of X 1 to minimize max{ X 1+s+p s+p, n(x 1+s)+(n 1)[s+(n 1)p] n(x 1 +ɛ+s+np) }. Now we let ɛ 0, then p 0 and finally let n. It follows that { X1 + s + p max, n(x } 1 + s) + (n 1)[s + (n 1)p] max s + p { 1 + X 1 s, 1 + s X 1 + s n(x 1 + ɛ + s + np) }. Some algebra shows that the minimum is obtained when X 1 = s( 5 1) and equals 5+1. Therefore, we conclude that R A 5+1 for all on-line algorithms. 3.. A New Heuristic Now we introduce a new heuristic for the 1 r i, p i = p, s, F = 1 C i problem, which is a variant of heuristic Sy, and then evaluate its performance. Again we recall the definition function of λ, λ = ηs, and further we set η = 5 3 for the following heuristic. Heuristic Wait-Half-Setup (W HS): Keep the machine idle until time s. If both the machine and some unscheduled ob(s) are available, then assign the λ earliest released obs to the machine to form a new batch. If the number of available obs is smaller than λ, then schedule all of them to the machine in a new batch. Proposition 6. 5 3 R W HS 1 + problem. p 3 5 for the 1 r i, p i = p, s, F = 1 C i 15

Proof. Let c = 1 + 1 η. The proof of this proposition is analogous to that of Proposition 3, and thus we adopt its notation. We assume, without loss of generality, that the first ob is released at time 0. We say a batch is full if it consists of λ obs; otherwise, we say it is non-full. Then we recall that, in an arbitrary processing interval I q without idle time, batch-group G q contains a group of batches ending with a non-full batch or possibly with the last batch of I q. We let a + 1 be the number of batches in G q and let h be the index of the first batch. As we now consider the a single machine scheduling case, we omit the superscript of the symbols for batches. Thus, the number of obs in G q is given by N = aλ + n a+h, where n a+h is the number of obs in the last batch of G q. Furthermore, the total completion time of the obs in G q can be written as: C i = S h N + i G q aλ(a + 1)(s + λp) A lower bound for the optimum satisfies Ci r i + sn + pn (N + 1) i G q i G q where r min = min r i for i G q. + n a+h (as + s + pn ). () r min N + sn + pn (N + 1), (3) With regard to the values of h, and q, we consider the following cases. (A summary of the analysis is given in Table 1??.) We let l be the difference between c i G q Ci and i G q C i in Case l, as set out below. 1. h = 1 ( = 1, q = 1) Add Table 1 here. By the assumption that the first ob arrives at time 0, we have r min = 0 and S 1 = s in this case. By () and (3), the difference 1 satisfies 1 aλ s η λp p(1 + η) + + a η (λp η s ) } {{ } } {{ } δ (1) 1 δ (1) + n a+h a(λp η s) + s η s pn a+h(η 1) + η 16 p(1 + η) η } {{ } δ (1) 3.

Now we consider the values of δ (1) 1, δ(1) and δ (1) 3 respectively. If follows from (1) that δ (1) 0 and also δ (1) p(λ 1) 1 > η λp p(1 + η) + η = λp( 1 η 1 ) + + η p(η > 0, since η = 5 3. η ) Finally, we consider δ (1) 3. A batch may contain at most λ obs, and so n a+h λ. Thus, δ 3 must satisfy Consequently, we see that 1 > 0. δ (1) 3 > s η ηs + p η > 0.. h = 3, = and q = 1 In this case, batch-group G 1 1 contains two batches B 1 and B. By the definitions of W HS and G q, we see that B 1 is full whereas B is nonfull, that is, n 1 = λ, n λ 1, and also the earliest released ob in G 1 must arrive after the start time of B, that is, r min > S = 3s +λp. Thus, we rewrite () and (3) for this case as: i G q i G q C i = ( 5s + λp + n aλ(a + 1)(s + λp) p)n + + n a+h (as + s + pn ) ( 5s + λp p)n aλ(a + 1)(s + λp) + + n a+h (as + s + pn ), Ci ( 3s + λp)n + sn + pn (N + 1). Then we obtain 3. h, = 1 N [ (1 + 1 η )(3s ] + λp) s λp + p ] [ 3s > N η + λp η s p(λ 1) > sn η (3 + η η ) by (1), > 0. + 1 17

In the scenario of = 1 and h, B h is the first batch of I q, and the obs processed in or after B h cannot arrive before time S h and thus the smallest release date of G q satisfies r min S h > 3s. Then the following inequality bounds the value of 3. 4. h 4, 3 S h N (1 + 1 η ) S hn + sn + 1 > 0. When h 4 and 1, we observe that the batch immediately preceding of G q, B h 1, is a non-full batch and thus we have n h 1 λ 1 and r min > S h s pn h 1 > 5s. Also by ((1-)), the difference 4 for this case is bounded below by 5. h =, = and q = 1 4 N (r min + r min η S h + s ) + 1 > N (S h s pn h 1 + 5s η S h + s ) > sn η (5 η η ) > 0. From this case onwards, we shall consider the batch-groups G q 1 and G q together instead of Gq only. That is to say, in this case we consider G 1 1 and G1 consisting of the first a + h batches. The on-line total completion time is C i = (N 1 + N )( 3s + pn 1) i G 1 1 G1 + aλ(a + 1)(s + λp) + n a+h (as + s + pn ). Now we provide two lower bounds for the optimal total completion time of the obs in G 1 1 and G1. If the optimal solution starts all the obs in G 1 3s 1 at or after time, then a lower bound for the optimum is given by LB (5) 1 = 3s(N 1 + N ) + p(n 1 + N )(N 1 + N + 1). Otherwise, if the optimal solution starts processing some of the obs in G 1 3s 1 before time, then it must start another batch for the processing of the obs in G 1. Thus, another lower bound is LB (5) = s(n 1 + N ) + sn + p(n 1 + N )(N 1 + N + 1). 18

If LB (5) 1 LB (5), then the difference 5 between c i G 1 1 G1 C i and i G 1 C 1 G1 i satisfies 5 3s(N 1 + N ) η pn 1 = N 1 η + aλ + p(n 1 + N )(N 1 + N + 1) η aλ(as + s + λp) n a+h (as + s + n a+hp ) p(n + η + 1) + (pn 1 pηn 1 + 3s ) } {{ } δ (5) 1 3s η s λp + p + p η + a(λp η s) } {{ } δ (5) + n a+h 3s η s pn a+h(η 1) + η p(1 + η) η } {{ } δ (5) 3 + p(n 1 + N ) +a( λp η s). We consider the signs of δ (5) 1, δ(5) and δ (5) 3. Since n a+h λ and N 1 = n h 1 λ 1, by (1) we see that Thus, we obtain 5 > 0. δ (5) 1 > s(3 + η η ) > 0, δ (5) > s η (3 η η ) > 0, δ (5) 3 > s η (3 η η ) > 0. Otherwise, if LB (5) 1 > LB (5), then we have 6. h = 3, = and q = 5 s(1 + η)(n N 1 ) η > N 1(δ (5) 1 sη s) η > sn 1( η ) η > 0. + 5 19

As for Case 5, we consider the obs in G 1 and G. For q =, there must be machine idle time preceding G 1. Also, by W HS we see that the earliest released ob in G 1 arrives at S the start time of B and the earliest released ob in G must arrive after time S. We let r min be the smallest release time of the obs in G 1 and G, and then we have r min S > 3s + pn 1. Thus we can write the on-line total completion time and a lower bound for the optimum as: aλ(s + λp)(a + 1) C i = (S + s + pn 1 )(N 1 + N ) + i G 1 G i G 1 G + n a+h (as + s + pn ), Ci (r min + s)(n 1 + N ) + p(n 1 + N )(N 1 + N + 1). Since r min S > 3s + pn 1, the difference 6 satisfies 6 (N 1 + N )(r min S + r min s ) + 5 > 0. η 7. h = 3, = 3 and q = 1 In this case, G 1 and G1 3 consist of B and B 3,..., B a+3, respectively. From the values of h, and q, we see that G 1 1 and G1 both contain a non-full batch, namely, N 1 = n 1 λ 1 and N = n λ 1. This implies that the obs in the batch-groups G 1 and G1 3 must arrive after time s 3s and + pn, respectively. Following observations similar to those in Case 5, we can write the on-line total completion time of the obs in G 1 and G1 3 and two lower bounds for the optimum as: C i = (N 1 + N )( 5s + pn aλ(a + 1)(s + λp) + pn 1 ) + i G 1 G1 3 i G 1 G1 3 C i min + n a+h (as + s + pn ), (N 1 + N )( 5s + pn ) + p(n 1 + N )(N 1 + N + 1), } {{ } LB (7) 1 (N 1 + N )( 3s + pn ) + sn + p(n 1 + N )(N 1 + N + 1), } {{ } 0 LB (7)

where LB (7) 1 and LB (7) are defined similar to those in Case 5. We note that the term pn (N 1 + N ) of LB (7) comes from the processing requirements of the obs in G 1 1. Since we may assume that obs are processed in order of non-decreasing release dates in an optimal solution, the obs in G 1 and G1 3 must be processed after those in G1 1. An argument analogous to that in Case 5 leads to the same result that the difference 7 between c i G 1 G1 C 3 i and i G 1 C G1 i is greater 3 than 0. We conclude from the above arguments that the inequality c i G q C Gq i +1 i G q C Gq i > 0 is valid for all h, q and. Hence we obtain c n +1 i=1 C i n i=1 C i > 0. Now we consider the following instance I to determine a lower bound for the competitive ratio R W HS. The first ob of I is available at time 0 and n 1 obs arrives at time s + ɛ. We let the positive number ɛ 0 and then the identical processing time p 0. The on-line and the optimal off-line obective function values are given below: W HS(I) = 3s (n 1)(5s + np) + p +, n(3s + ɛ + np) OP T (I) =. Let ɛ 0, then p 0 and finally n, and it follows that Hence the result follows. W HS(I) OP T (I) 5 3. 4 Conclusions and Future Wor In this paper we discuss four problems of scheduling machines with a common batch setup in the batch availability model. For both the online-list and the online-time scenarios, we introduce lower bounds and new on-line heuristics to each case with equal processing requirements. Furthermore, we establish the lower and upper bounds on the competitive performance ratio for each heuristic. Table summarizes the the bounds proven in this paper. Future wor can extend the above discussion in various ways. With tighter lower bounds on the optimal obective function values or better defined properties of the optimal (off-line) solutions, can the competitive ratios of the proposed heuristics be improved? Also, given the general case with arbitrary processing times, is there any on-line algorithm which can guarantee r-competitive for any r? Moreover, issues with other obectives lie the minimization of total weighted flow time or the total weighted completion time remain to be approached in on-line scheduling environments. 1

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