S. ABERS Vohra [3] then gave an algorithm that is.986-competitive, for all m 70. Karger, Phillips and Torng [] generalized the algorithm and proved a

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1 BETTER BOUNDS FOR ONINE SCHEDUING SUSANNE ABERS y Abstract. We study a classical problem in online scheduling. A sequence of jobs must be scheduled on m identical parallel machines. As each job arrives, its processing time is known. The goal is to minimize the makespan. Bartal, Fiat, Karlo and Vohra [3] gave a deterministic online algorithm that is.986-competitive. Karger, Phillips and Torng [] generalized the algorithm and proved an upper bound of.945. The best lower bound currently known on the competitive ratio that can be achieved by deterministic online algorithms it equal to.837. In this paper we present an improved deterministic online scheduling algorithm that is.93-competitive, for all m. The algorithm is based on a new scheduling strategy, i.e., it is not a generalization of the approach by Bartal et al. Also, the algorithm has a simple structure. Furthermore, we develop a better lower bound. We prove that, for general m, no deterministic online scheduling algorithm can be better than.85-competitive. Key words. makespan minimization, online algorithm, competitive analysis AMS subject classications. 68Q0, 68Q5, 90B35. Introduction. We study a classical problem in online scheduling. A sequence of jobs must be scheduled on m identical parallel machines. Whenever a job arrives, its processing time is known in advance, and the job must be scheduled immediately on one of the machines, without knowledge of any future jobs. Preemption of jobs is not allowed. The goal is to minimize the makespan, i.e., the completion time of the last job that nishes. Algorithms for this scheduling problem are used in multiprocessor scheduling. Moreover, the problem is important because it is the root of many problem variants where, for instance, preemption is allowed, precedence constraints exist among jobs, or machines run at dierent speeds. The problem was rst investigated by Graham [0]. In fact, Graham also studied the oine version of the problem, when all jobs are known in advance. The problem of computing an optimal oine schedule for a given job sequence in NP-hard [9]. Graham gave a fast scheduling heuristic that achieves a good approximation ratio. He developed the well-known ist algorithm that takes the given jobs one by one and always schedules them on the least loaded machine. Clearly, ist is also an online algorithm. Following [3], we call a deterministic online scheduling algorithm A c-competitive if, for all job sequences = J ; J ; : : : ; J n, A() c OP T (); where A() is the makespan of the schedule produced by A and OP T () is the makespan of an optimal schedule for. Graham's ist algorithm is ( m )-competitive. Galambos and Woeginger presented an algorithm that is ( m m)-competitive, where m > 0, but m tends to 0 as m goes to innity. It was unknown for a long time whether there is an algorithm that achieves a competitive ratio of c, c <, for general m. Bartal, Fiat, Karlo and A preliminary version of this paper was presented at the 9th Annual ACM Symposium on Theory of Computing (STOC), 997. y Max-Planck-Institut fur Informatik, Im Stadtwald, D{663 Saarbrucken, Germany. albers@mpi-sb.mpg.de. Work supported in part by the EU ESPRIT TR Project N. 044 (ACOM-IT).

2 S. ABERS Vohra [3] then gave an algorithm that is.986-competitive, for all m 70. Karger, Phillips and Torng [] generalized the algorithm and proved a competitive ratio of.945, for all m. This has been the best upper bound known so far for general m. For the special case m = 4, Chen, van Vliet and Woeginger [6] developed an algorithm that is.733-competitive. With respect to lower bounds, Faigle, Kern and Turan [7] showed that no deterministic online algorithm can have a competitive ratio smaller than ( m ) for m = and m = 3. Thus, for these values of m, ist is optimal. Faigle et al. [7] also proved that no deterministic online algorithm can be better than.707-competitive, for any m 4. The best lower bound known so far for general m is due to Bartal, Karlo and Rabani [4] who showed that no deterministic online algorithm can have a competitive ratio smaller than.837, for m For more work on related online scheduling problems see, for instance, [,, 5,, 4]. In this paper we present an improved deterministic online algorithm for the scheduling problem dened above. The algorithm is.93-competitive, for all m. Our algorithm is based on a new scheduling strategy, i.e., it is not a generalization of the approach by Bartal et al. [3]. Moreover, the algorithm has a simple structure. At any time, the algorithm maintains a set S of b m c machines with a low load and a set S of d m e machines with a high load. Every job is either scheduled on the least loaded machine in S or on the least loaded machine in S. The decision, which of the two machines to choose, depends on the ratio of the load on machines in S to the load on machines in S. A description of the algorithm is given in Section. A detailed analysis follows in Section 3. We also develop a better lower bound for online scheduling. In Section 4 we show that if a deterministic online scheduling algorithm is c-competitive for all m 80, then c :85.. The new scheduling algorithm. For the description of the algorithm we need some denitions. et the load of a machine be the sum of the processing times of the jobs already assigned to it. At any time, the algorithm maintains a list of the machines sorted in non-decreasing order by current load. et Mi t denote the machine with the i-th smallest load, i m, after exactly t jobs have been scheduled. In particular, M t is the machine with the smallest load and Mm t is the machine with the largest load. We denote by li t the load of machine Mi t, i m. Note that the load lm t of the most loaded machine is always equal to the current makespan. As previous algorithms [3, ], our new scheduling strategy tries to prevent schedules in which the load on all machines is about the same. If all machines have the same load, with all previous jobs being very small, an adversary can present an additional large job and force a competitive ratio of ( m ). This is the worst-case scenario for ist. Our new algorithm, called M, always tries to maintain k machines with a low load and m k machine with a high load, where k = b m c. The goal is to always have a schedule in which the total load l on the k lightly loaded machines is at most times the total load h on the m k heavily loaded machines, for some to be specied later. A schedule satisfying l h is always prepared to handle a large incoming job and can easily maintain a competitive ratio of c, where c is.93. Algorithm M always schedules a new job J t with processing time p t on the least loaded machine as long as l h is satised after the assignment. Note that during this assignment, the load l on the lightly loaded machines does not necessarily increase by p t because the least loaded machine might become one of the machines Mi t, k < i m. If an assignment of J t to the least loaded machine results in l > h, then M considers scheduling J t on the machine with the (k + )-st smallest load.

3 BETTER BOUNDS FOR ONINE SCHEDUING 3 However, if this assignment increases the makespan and the new makespan exceeds c ( l + h )=m, then J t is nally scheduled on the least loaded machine, ignoring the violation of l h. Note that l + h is the sum of the processing times of all jobs that have arrived so far, and thus ( l + h )=m is a lower bound on the optimum makespan. Algorithm M: Set c = :93, k = b m (c )k j= c and j = 0:9m. Set =. (c )(m k) Every new job J t is scheduled as follows. et l be the sum of the loads on machines M; t : : : ; Mk t if J t is scheduled on the least loaded machine. Similarly, let h be the sum of the loads on machines M t ; : : : ; M t k+ m if J t is scheduled on the least loaded machine. et t m be the makespan, i.e. the load of the most loaded machine, if J t is scheduled on the machine with the (k + )-st smallest load. Recall that lm t is the makespan before the assignment of J t. Schedule J t on the least loaded machine if one of the following conditions holds. (a) l h (b) t m > lt m and t m > c l+ h m Otherwise schedule J t on the machine with the (k + )-st smallest load. Theorem.. Algorithm M is.93-competitive, for all m. Before analyzing the algorithm in the next section, we discuss the choice of. First observe that 0 < <, for m. The inequality 0 < holds because c > = and k > j; thus (c )k j= > 0. Inequality < holds because (c )k (c )(m k) and j= > 0. In fact, for even m, = (c ) j=m 0:686 c and, for odd m, tends to this value as m goes to innity. Always setting = 0:686 in the algorithm M asymptotically results in the same competitive ratio of.93. Choosing = (c )k j= has two advantages. () We can prove a competitiveness of (c )(m k).93 for even small m. () In the analysis we can do symbolic calculations where a xed = 0:686 would require numeric calculations. 3. Analysis of the algorithm. We present a detailed proof of Theorem.. The analysis presented by Graham [0] for the ist algorithm, combined with the observation that algorithm M only schedules a job on the machine with the (k + )-st smallest load if the resulting makespan does not exceed.93 times the optimum makespan, shows that M is c-competitive, where c = maxf( m ); :93g, for all m. This gives the desired bound for small m. For m 8, the following analysis applies. Consider an arbitrary job sequence = J ; J ; : : : ; J n. et p t be the processing time of J t, t n. We will show that M schedules every job J t, t n, such that M( t ) :93 OP T ( t ); where M( t ) is the makespan of the schedule produced by M on the subsequence t = J ; J ; : : : ; J t and OP T ( t ) is the makespan of an optimal schedule for t. 3.. The basic analysis. Suppose that M has already scheduled the rst t jobs and that a competitive ratio of c = :93 was maintained at all times. et = tx s= is the sum of the loads on all machines after J t is assigned. p s :

4 4 S. ABERS If the makespan does not change during the assignment of J t, then by induction hypothesis there is nothing to show. Also, if the makespan changes but is bounded from above by c m, then we are done because m is a lower bound on the optimum makespan for t. Thus we concentrate on the case that during the assignment of J t, the makespan increases and exceeds c m. Condition (b) in algorithm M implies that J t is scheduled on the least loaded machine. et l = l t be the load of the least loaded machine immediately before J t is assigned. First we consider the case that l (c ) m = 0:93 m. We have If p t m, then M( t ) = l + p t (c ) m + p t: M( t ) (c ) m + m c m c OP T ( t): If p t = ( + ) m, for some positive, then M( t ) = l + p t (c ) m + ( + ) m c ( + ) m = c p t c max st p s c OP T ( t ): Here we use the fact that max st p s is also lower bound on the optimum makespan. In the remainder of this proof we will study the situation that the load on the least loaded machine is greater than (c ) m, i.e., l = (c + ) m for some positive. Since l cannot be greater than m, we have 0 < c = 0:077. Note that all machines must have a load of at least (c + ) m. Since J t is assigned to the least loaded machine and the makespan after the assignment is greater than c m, we have p t > c m l ( ) m ( + ) m. Our goal is to show that the sequence t = J ; J ; : : : ; J t contains m jobs, each with a processing time of at least ( + ) m. Then there are m + jobs with a processing time of at least ( + ) m, two of which must be scheduled on the same machine in an optimal schedule. Thus If p t m, then OP T ( t ) ( + ) m : M( t ) = l + p t (c + ) m + m c( + ) m c OP T ( t): If p t = ( + ) m for some positive, then M( t ) = (c + ) m + p t (c + + ) m c maxf( + ); ( + )g m c OP T ( t ): In each case, Theorem. is proved. It remains to show that the sequence t = J ; J ; : : : ; J t does contain m jobs, each with a processing time of at least ( + ) m.

5 BETTER BOUNDS FOR ONINE SCHEDUING Identifying large jobs. We have to analyze jobs in the sequence t. et time s, s t, denote the point of time immediately after J s is scheduled. (Time 0 is the point of time before any jobs are scheduled.) For any time s, s t, let s be the total load on the m machines, i.e., s = sx r= Note that t =. Definition 3.. At any time s, s t, the schedule constructed by M is called steady if the total load on the k lightly loaded machines M s ; : : : ; Mk s is at most times the total load on the m k heavily loaded machines M s ; : : : ; M s k+ m. In the following, when refering to machines M s; : : : ; M s m, we will often drop s when the meaning is clear. emma 3.. At time t, i.e. immediately before J t is scheduled, M's schedule is not steady. Proof. Immediately before J t is scheduled, the total load on the machines M ; : : : ; M k is at least p r : l = k(c + ) m > k(c ) m = ((c )k j ) m + j = (c )(m k) m + j m m : If M's schedule was steady, then the total load on machines M k+ ; : : : ; M m would be at least l. Thus the total load before the assignment of J t would be at least t ( + ) l > k(c ) m + (m k)(c ) m + j m : Here we used the facts l > k(c ) m and l > (c )(m k) m + j m. Thus t > (c ) + j m = + (c ) + j m > because = (c )k j= (c )(m k) (c ) j=m 7 c 0 0: > 0. We have a contradiction. and hence (c ) + j m > 0: Analyzing load. Definition 3.3. At any time s, s t, a machine is called full if its load is at least (c + ) m. Recall that at time t, all machines have a load of at least (c + ) m and, thus, are full. For i = ; : : : ; m, let t i be the most recent time when exactly i machines were full. Note that t < t < : : : < t m = t : Of particular interest to us will be the time t m bjc when exactly m bjc machines were full. et t 0, t m bjc t 0 < t, be the most recent time when M's schedule

6 6 S. ABERS was steady. If M's schedule was not steady during the time interval [t m bjc ; t ], then let t 0 = t m bjc. et f be the number of machines that are full at time t 0. Our goal is to show that at time t 0, the total load on the non-full machines M ; : : : ; M m f in M's schedule is at most (c :5)(m f) m. We will show this using the following two lemmas. et X = (c ) c Y = (c ) c j m : emma 3.4. If at time t 0, the total load on the non-full machines M ; : : : ; M m f in M's schedule were greater than (c :5)(m f) m, then the total load at time t would satisfy > X + Y ( c m ) (bjc+). The proof of emma 3.4 is presented in the Appendix. emma 3.5. X + Y ( c m ) (bjc+) Proof. We have ( c m ) (bjc+) ( c m ) j e cj=m : The rst inequality follows because bjc + j. Thus, X + Y ( c m ) (bjc+) (c ) c + ( (c ) c j m) ecj=m = ( ) + ( 0:93 :930:45) e 0:9:93 : :93 :93 Evaluating the last expression gives that it is at least. We summarize the results of emmas 3.4 and 3.5. emma 3.6. At time t 0, the total load on the non-full machines M ; : : : ; M m f is at most (c :5)(m f) m Tracing the assignment of large jobs. We now identify jobs with a processing time of at least ( + ) m. emma 3.7. During the time interval (t m k ; t 0 ], f m + k jobs, each of size at least ( + ) m, are scheduled. Proof. At time t m k, m k machines are full. At time t 0, f machines are full, where f m bjc. Consider the f m + k steps in (t m k ; t 0 ] at which the number of full machines increases. Since at least m k machines are full, the number of full machines can only increase if a job is scheduled on the least loaded machine. By emma 3.6, at time t 0, the total load on the m f least loaded machines is at most (c :5)(m f) m. This implies that at time t0, the load on the least loaded machine is at most (c :5) m. Thus, at any of the f m + k steps in (t m k; t 0 ] at which the number of full machines increases, the load on the least loaded machine is at most (c :5) m. Hence jobs of size at least (c + ) m (c :5) m = ( + ) m are introduced. emma 3.8. At time t m k, each of the machines M ; : : : ; M f m+k+ has a load of at most (c :5) m. The total load on the machines M f m+k+; : : : ; M k is at most (c :5)(m f) m. Proof. The machine with the (f m + k + )-st smallest load at time t m k becomes the least loaded machine no later than time t 0, when f machines are full. Thus, if at time t m k, machine M f m+k+ (or any machine M i with i f m + k) had a load greater than (c :5) m, then the load of the least loaded machine at

7 BETTER BOUNDS FOR ONINE SCHEDUING 7 time t 0 would be greater than (c :5) m. emma 3.6 implies that this is impossible. Similarly, if at time t m k, machines M f m+k+ ; : : : ; M k had a total load of at least (c :5)(m f) m, then the total load on M ; : : : ; M m f at time t 0 would be at least (c :5)(m f) m. Again, emma 3.6 gives the desired statement. emma 3.9. During the time interval (t 0 ; t ], m f jobs of size at least ( +) m are scheduled. Proof. By the denition of t 0, M's schedule is not steady during [t 0 +; t ]. et s [t 0 + ; t ] be any of the m f time steps in [t 0 + ; t ] at which the number of full machines has just increased. If the load on the least loaded machine is at most (c :5) m when J s is scheduled, then p s (c + ) m (c :5) m = ( + ) m. Suppose that immediately before the assignment of J s, the least loaded machine has a load greater than (c :5) m. et ls be the load of machine M k+ k+ and suppose l s = (c ++) k+ m for some non-negative. By the denition of t0, at least m bjc machines are full at any time in [t 0 ; t ]. Thus, s (m bjc)(c + ) m + bjc(c :5) m + (m k) m (c ) bjc m + (m bjc) m + (c ) j m + (m j) m + : The second inequality follows because m k m. Since M's schedule is not steady, M would prefer to schedule J s on machine M k+ but cannot because Hence, l s k+ + p s > c( s + p s )=m: p s ( c m s l s k+ )=( c m ) c m s l s k+ c(c j m ) m (c ) m + c( j m ) m m + c m m : The load l s = (c ++) k+ m cannot be greater than (3 c ) m since otherwise we would have, at time t, m k machines each with a load greater than (3 c ) m and k machines each with a load of at least (c + ) m, resulting in a total load greater than. Thus, 4 c and p s ((c ) cj m ) m + (c cj m ) m + ( c )( c ) m = ((c ) (c ) cj m ) m + ( cj m ) m 0:567 m + 0:44 m ( + ) m for all 0:. Recall that our is at most c = 0:077. We now consider the time t m k bjc when exactly m k bjc machines are full. et t 00 be the earliest point of time in the interval [t m k bjc ; t m k ] at which the machine with the (k + )-st smallest load has a load greater than (c :5) m. emma 3.0. During the time interval (t 00 ; t m k ], every job is scheduled on the least loaded machine. Proof. We rst show that at any time s [t 00 ; t m k ], M's schedule is steady. emma 3.8 implies that at time s the total load on the lightly loaded machines

8 8 S. ABERS M ; : : : ; M k is at most l = k(c :5) m. By the denition of t00, the total load on the heavily loaded machines M k+ ; : : : ; M m at time s is at least h = (m k bjc)(c ) m + bjc(c :5) m = (m k)(c ) m bjc We show that at time s, the total load on the lightly loaded machines is at most times the load on the heavily loaded machines. This holds if l h, i.e., if which is equivalent to m : k(c :5) m k(c ) j= ((m (c )(m k) k)(c ) m bjc m ); (c )(kbjc + (m k)j) j bjc (c )k(m k): This in turn holds if jm j k(m k): (c ) The left-hand side is at most 0:45m, and the right-hand side is 4 m for even m at and least 0:46m for odd m 9. Thus, at time s, M's schedule is steady. Now consider job J s+ scheduled at time s +. et l s be the load on the least loaded machine at time s. We have l s (c :5) m. et p be a processing time such that l s + p = (c :5) m. The property stated in emma 3.8 must also hold at time s because the load on the lightly loaded machines M ; : : : ; M k can only be smaller. Thus, if J s+ has a processing time of at most p, scheduling J s+ on the least loaded machine results in a total load of at most k(c :5) m on machines M ; : : : ; M k. Since the total load on machines M k+ ; : : : ; M m is at least (m k)(c ) m bjc m, the calculations of the preceding paragraph show that M's schedule must be steady after the assignment. Suppose that J s+ has a processing time p s+ > p and that scheduling J s+ on the least loaded machine results in a load of k(c :5) m + m on machines M ; : : : ; M k, for some > 0. This implies that at time s +, the load on any of the machines M k+ ; : : : ; M k+bjc must be at least (c :5 + ) m. With the above denitions of l and h, we conclude that after the assignment of J s+ to the least loaded machine, the total load on M ; : : : ; M k is at most l + m and the total load on M k+; : : : ; M m is at least h + bjc m. Since, for m 8, we have bjc and, M's schedule must be steady. emma 3.. At time t 00, the load on machine M k+ is at most (c :5) m. Proof. If t 00 > t m k bjc, then the lemma follows from the denition of t 00. We show that t 00 cannot be equal to t m k bjc. Recall that f m bjc. Thus, emma 3.8 implies that at time t m k, machine M k bjc+ has a load of at most (c :5) m. If t 00 = t m k bjc, then there are bjc steps in (t 00 ; t m k ] at which the number of full machines increases. By emma 3.0, at all these steps, the jobs are assigned to the least loaded machine. Thus at time t 00, the load of machine M k+ cannot be greater than the load of machine M k bjc+ at time t m k. This means that M k+ has a load of at most (c :5) m at time t00, contradicting the choice of t 00. emma 3.. During time interval (0; t m k ], m k jobs of size at least ( + ) m are scheduled.

9 BETTER BOUNDS FOR ONINE SCHEDUING 9 Proof. et i be the number of machines that are full at time t 00. Consider the i steps in (0; t 00 ] at which the number of full machines increases. At any of these steps, before the assignment of the job, the load on M and M k+ is at most (c :5) m each, see emma 3.. Thus jobs of size at least (c + ) m (c :5) m ( + ) m must be scheduled. At the m k i steps in (t 00 ; t m k ] at which the number of full machines increases, jobs are scheduled on the least loaded machine (emma 3.0). The least loaded machine has a load of at most (c :5) m and we conclude again that jobs of size at least ( + ) m must be scheduled. emma 3.7 as well as emmas 3.9 and 3. imply the following statement. emma 3.3. During time interval (0; t ], m jobs of size at least ( + ) m are scheduled. By the discussion immediately preceding Section 3., the proof of Theorem. is complete. 4. The lower bound. We develop an improved lower bound for deterministic scheduling algorithms. Theorem 4.. et A be a deterministic online scheduling algorithm. If A is c-competitive for all m 80, then c :85. Proof. We will construct a job sequence such that A() :85 OP T (). The job sequence consists of several rounds. We assume that m is a multiple of 40. Round : m jobs with a processing time of w = 0:0. Round : m jobs with a processing time of x = 0:06. Round 3: Subround 3.: 9 m jobs with a processing time of y 0 = 0:8. Subround 3.: m jobs with a processing time of y 0 = 0:4. Round 4: Subround 4.: m jobs with a processing time of z = 0:5. Subround 4.: m jobs with a processing time of z 4 = y = 0:6. 3 Subround 4.3: m jobs with a processing time of z 40 3 = y = 0:78. 3 Subround 4.4: m jobs with a processing time of z 40 4 = 0:84. Subround 4.5: m + jobs with a processing time of z 0 5 =. Note that in the fourth round, m + jobs have to be scheduled. In the following, when analyzing the various subrounds, we will often compare the makespan produced by an online algorithm A in a subround to the optimum makespan at the end of the subround. It is clear that the optimum makespan during the subround can only be smaller. Analysis of Round : Clearly, in order to maintain :85-competitiveness, online algorithm A must schedule the m jobs in Round on dierent machines. Analysis of Round : Algorithm A must schedule the m jobs in Round on dierent machines. Otherwise, A's makespan would be at least w + x = 0:3. Since the optimum makespan during the round is always at most w + x = 0:07 and 0:3 > 0:07 :857, A would not be.85-competitive. At the end of the second round, A has a load of l = w + x = 0:07 on each of its machines. Analysis of Round 3 : At the end of Subround 3., the optimum makespan is at most x + y = 0:34. On each of 9 m machines, OPT schedules an x-job and 0 a y -job. The remaining m machines have an x-job and 0 jobs of size w. If A 0 does not schedule the jobs in Subround 3. on dierent machines, then its makespan is at least w + x + y = 0:634 > :853(x + y ). The optimum makespan after

10 0 S. ABERS Subround 3. is y + x = 0:40. In an optimal schedule, m machines have a y 0 - job, m machines have a y -job and two x-jobs. The remaining machines have a y -job and at most three w-jobs. Online algorithm A must schedule the jobs of Subround 3. on dierent machines and these machines may not contain any y -job since otherwise A's makespan is at least w+x+y +y = 0:75 > :87(y +x). At the end of Round 3, the least loaded machine in A's schedule has a load of l 3 = w + x + y = 0:35. Analysis of Round 4 : Subround 4.: After Subround 4., the optimum makespan is y + y = 0:68. In an optimal schedule, m machines contain a y 0 and a y. m 9 machines contain a z, two w and two x. m machines contain two y 0. Algorithm A must schedule all z -jobs on dierent machines. Otherwise its makespan would be at least l 3 + z = :35 > :98(y + y ). Subround 4. : At the end of the subround, the optimum makespan is y + z = 0:78. In OPT's schedule, m machines have a y and a z. m machines have a 0 y and a y. m machines have two y 5, three w and three x. m machines have a 4 z and some of them have two additional w and x. Algorithm A must schedule each z -job on a machine not containing any z or z. Otherwise its makespan would be at least l 3 + z + z = :45, which is greater than :856(y + z ). Subround 4.3 : The optimum makespan after the subround is 3y = 0:846. In an optimal schedule m machines have a y, a z and an x. m machines have a z 4, two 3 x and four w. m machines have a z m machines three y 0. m machines two 40 y. As before, A may not schedule any z 3 -job on a machine containing a z, a z or a z 3 because this would result in a makespan of at least l 3 +z +z 3 = :57 > :855(3y ). Subround 4.4 : The optimum makespan is y + z = 0:9. In OPT's schedule, all the z-jobs are scheduled on dierent machines. m machines having a z 0 also contain a y. ( )m machines containing a z 0 also have a y, an x and up to three w. The m machines having a z 4 also have a y. Machines having a z 3 also have three x. Machines having a z 4 also have an x. At this point, OPT is left with m machines 0 on which it has to schedule m jobs with a processing time of x and m jobs with a 4 4 processing time of y. This can be done by scheduling (a) m machines with ten x 40 and one y each and (b) 3 m machines with three y 40. As usual, A may not schedule a z 4 -job on a machine having already any z-job; otherwise its makespan is at least l 3 + z + z 4 = :69 = :88(y + z ). Subround 4.5 : The online algorithm A must schedule one of the z 5 -jobs on a machine already containing another z-job, because a total of m + jobs have to be scheduled in Round 4. This gives a makespan of at least w + x + y + z + z 5 = :85. We will show that OPT can schedule all the jobs with a makespan of if m 80. An optimal schedule is as follows. m machines have a z 0 5. m machines have two 4 3 z. m machines have a z , two w and two x. m machines have a z 40 3 and a y. m machines have a z 5, one y, two w and one x. m machines have a z 0 and a 9 y. m machines have three y 40, two w and two x. OPT has m machines left on 40 which it has to schedule one z 5 and m jobs of size x. This can be done if at least 5 two machines are left, i.e. if m 80. OPT can use one machine for the z 5 -job and the remaining machines for the x-jobs. 5. Open problems. An interesting problem is to formulate and analyze a generalization of the algorithm M that, at any time, is allowed to schedule a new job on any of the m machines. In such an algorithm, the ratio of the load on the i-th smallest machine to the load on the (i + )-st smallest machine has to be bounded by some i, i m. The problem is to specify i 's and a proper scheduling rule that is able to maintain these values. A rst step in this direction is to maintain three

11 BETTER BOUNDS FOR ONINE SCHEDUING set S, S and S 3 of m=3 machines with a low, medium and high load, respectively. More generally, with respect to the scheduling problem studied here, a fundamental open problem is to develop randomized online algorithms that achieve a competitive ratio smaller than the deterministic lower bound, for all m. Appendix. We prove emma 3.4. For convenience, we state the lemma again. emma 3.4. If at time t 0, the total load on the non-full machines M ; : : : ; M m f in M's schedule were greater than (c :5)(m f) m, then the total load at time t would satisfy > X + Y ( c m ) (bjc+). Proof. In order to prove the lemma, we have to keep track of the load on the m machines during the entire time interval [t 0 ; t]. For i = f; : : : ; m +, let Z i = X + Y ( c m ) (i m+bjc) : We will show by induction on i that for i = f; : : : ; m, (5.) ti ti > Z i ; where is a non-negative potential that we will dene in a moment. Using the inequality tm tm > Z m, we will then prove > Z m+. We rst explain the purpose of the potential. We want to show that during the time interval (t 0 ; t ], every time another machine becomes full, a job J with a large processing time p must be scheduled. Since, by the denition of t 0, M's schedule is not steady in (t 0 ; t ], M would prefer to assign J to machine M k+. However, M schedules J on the least loaded machine, causing another machine to become full. This implies that an assignment of J to machine M k+ results in an increased makespan that exceeds c times the average load on the machines, i.e., J's processing time p must be large. In some cases, when M k+ has a high load, we will not be able to argue that J's processing time is greater than a certain value. In these cases we will pay some \missing processing time" out of the potential. This way we can ensure that J's amortized processing time is greater than the desired value. J's amortized processing time is the actual processing time plus the change in potential. Formally, the potential is dened as follows. At time t 0, we color some of the load in M's schedule. More precisely, on each of the machines M ; : : : ; M k+m f we color the load that is above level (c ) m. We can imagine that we draw a horizontal line at level (c ) m across M's schedule and color the load on machines M ; : : : ; M k+m f that is above this line. Note that this way, a job might be partially colored. During time interval (t 0 ; t], the colored load is updated as follows.. Whenever M schedules a job that causes one more machine to become full, we choose the least loaded machine with colored load among M k+ ; : : : ; M m and remove the color from its load.. Whenever a job is assigned to a machine with colored load, we color that job. 3. After the nal job J t is scheduled, the color is removed from all machines. At any time, let I = fijm i has colored loadg and let c i, i I, be the amount of the load that is colored on M i. Dene = X ii c i : During the interval (t 0 ; t ], the following invariants hold. (I) Whenever M schedules a job that causes one more machine to become full, there is a machine in fm k+ ; : : : ; M m g with colored load.

12 S. ABERS (I) If a machine has colored load, then all its load above level (c ) m is colored. (I3) At any time, if machine M k+ has load (c + ) m for some positive, then c i m for all i I with i k +. (I4) At any time, there exists a machine among M ; : : : ; M k with colored load at least m. Invariant (I) holds because at time t 0 there are m f machines in fm k+ ; : : : ; M m g with colored load, exactly m f more jobs are scheduled in (t 0 ; t ] that cause a machine to become full and every time this happens, by update rule, the number of machines in fm k+ ; : : : ; M m g with colored load is reduced by exactly. Invariant (I) follows from update rule. (I3) is immediate from (I). Invariant (I4) holds because initially, at time t 0, we color load on the k lightly loaded machines that are full and these machines remain in the set of lightly loaded machines during (t 0 ; t ]. Base of the induction. In order to prove inequality (5.) for i = f, we have to evaluate t 0, the total load on the m machines at time t 0. We will show that (5.) t 0 t 0 > Z f : This implies that inequality (5.) holds for i = f because between time t 0 and time t f the number of full machines remains the same and whenever the load on the m machines increases by p, the potential increases by at most p, see update rule. If M's schedule is not steady at time t 0, i.e., t 0 = t m bjc and f = m bjc, then (5.3) Z f = X + Y = (c ) j m (c ) bjc m : By assumption, at time t 0, the load on the non-full machines is greater than (c :5)(m f) m = (c :5)bjc m = (c )bjc m bjc m. The load on the full machines is at least (c )f m + t 0 = (c )(m bjc) m + t0. We obtain (5.4) t 0 t 0 > (c ) bjc m : Inequalities (5.3) and (5.4) give the desired bound. We study the case that M's schedule is steady at time t 0. The total load on the non-full machines is greater than (c :5)(m f) m. The load on each machine M m f + ; : : : ; M k is at least (c + ) m. Thus the total load l on the k lightly loaded machines M ; : : : ; M k is l > (c + )(k m + f) m + (c :5)(m f) m = (c )k m (m f) m + (k m + f) m (c )k m j m + (f m + bjc) m + (k m + f) m : Note that the load (k m + f) m will go into the potential. Since the schedule is steady, the total load h on the heavily loaded machines M k+ ; : : : ; M m is at least times the above expression. Neglecting the term (k m + f) m, we obtain h > (c )(m k) m + (f m + bjc) m : The load (c )(m k) m can ll the machines M k+; : : : ; M m up to a level of (c ) m. Of the additional load (f m + bjc) m, at least a fraction of f k m k is

13 BETTER BOUNDS FOR ONINE SCHEDUING 3 located on machines M k+m f + ; : : : ; M m and does not go into the potential. Thus, We have to show that (5.5) t 0 t 0 = l + h t 0 > (c ) j m + (f m + bjc)( + f k = X + Y + (f m + bjc)( + m k ) m : f k f k m k ) m (f m + bjc)( + m k ) m Y (( c m ) (f m+bjc) ) holds for every f fm bjc; : : : ; mg. This proves inequality (5.). We dene functions g(x) = (x m + bjc)( + x k m k ) m h(x) = Y (( c m ) (x m+bjc) ): The function g(x) is a polynomial of degree, and h(x) is an exponentially increasing function. Obviously, g(m bjc) = h(m bjc) = 0. We will show that (5.6) g 0 (m bjc) > h 0 (m bjc) and (5.7) g(y) > h(y) for some y > m: This implies that g(x) > h(x) must hold for all x (m bjc; m]. (If g(z) h(z) were true for some z (m bjc; m], then g(x) < h(x) for all x > z.) Recall that, as mentioned in the proof of emma 3., 7. Also, evaluating Y 0 with its actual parameters gives Y 0:99. We have g 0 (m bjc) = ( + 0 ( + ( = 0:8 m : m bjc k m k ) m 7 m k bjc m k ) m j m ) m The last inequality holds because bjc j and m k m=. Also, h 0 (m bjc) = Y ln(( c m ) ) 0:99 m ln(( c m ) m ) < 0:79 m. The last inequality holds because ln(( c m ) m ) is decreasing in m and evaluates to less than.63 for m 4. This shows (5.6). For the proof of (5.7), let y = m + j bjc. Then g(y) 0:9m( + 0) 7 m 0:35. Also, h(y) = Y (( c m ) 0:9m ) 0:99(( c m ) 0:9m ). The last expression is decreasing in m and less than 0:34 for all m 4. The proof of inequality (5.5) is complete. Induction step. We show that if inequality (5.) holds for i, then it also holds for i. et s i be the earliest point of time when exactly i machines are full. We have t i < s i t i. For all s [t i ; s i ], (5.8) s s > Z i :

14 4 S. ABERS This is because of the induction hypothesis and the fact that if the load on the m machines increases by p between time t i and time s i, then the potential increases by at most p. et si be the total load on the m machines at time s i and let l = l si be k+ the load on machine M k+ at time s i. Suppose l = (c + ) m for some > 0. The job J si that causes the i-th machine to become full is scheduled on the least loaded machine. Since M's schedule is not steady at time s i, M would prefer to schedule J si on machine M k+. Since this is not possible, condition (b) in algorithm M implies that the processing time p si of J si must satisfy which is equivalent to l + p si > c( s i + p si ) ; m p si > ( c m s i l)=( c m ): Consider the change in potential during the assignment of J si. Update rule and invariants (I){(I3) imply that the potential drops by at least m. p si > ( c m s i l)=( c m ) + m ( c m (Z i + si ) l)=( c m ) + m ( c m Z i + c m m (c + ) m )=( c m ) + m = ( c m Z i (c ) m )=( c m ) = cy m ( c m ) (i m+bjc) : The second inequality follows because of inequality (5.8). The third inequality holds because at time s i, there is at least one machine in fm k+ ; : : : ; M m g with colored load, i.e., si m. Thus, si si si si + p si > Z i + cy m ( c m ) (i m+bjc) = X + Y ( c m ) (i m+bjc) = Z i : The induction step is complete because during time interval (s i ; t i ] the inequality is maintained. We nally have to prove (5.9) > Z m+ : Our inductive proof shows t t > Z m. Job J t is scheduled on the least loaded machine and by assumption l + p t > c(t +pt) m, where l = (c + ) m is the load of the least loaded machine at time t, i.e., immediately before J t is scheduled. Recall that at time t we remove the color from the load in M's schedule. Invariant (I4) implies that the potential at time t must decrease by at least m. Calculations identical to that in the inductive step show inequality (5.9).

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