Theoretical Computer Science. Optimal algorithms for online scheduling with bounded rearrangement at the end

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1 Theoretical Computer Science 4 (0) Content lit available at SciVere ScienceDirect Theoretical Computer Science journal homepage: Optimal algorithm for online cheduling with bounded rearrangement at the end Xin Chen a, Yan Lan b, Attila Benko c, György Dóa c, Xin Han a, a Software School, Dalian Univerity of Technology, China b Dalian Neuoft Intitute of Information, China c Department of Mathematic, Univerity of Pannonia, Vezprém, Hungary a r t i c l e i n f o a b t r a c t Article hitory: Received 30 May 0 Received in revied form 5 July 0 Accepted 9 July 0 Communicated by D.-Z. Du Keyword: Online cheduling Bounded rearrangement In thi paper, we conider an online non-preemptive cheduling problem on two related machine, where at mot K job are allowed to be rearranged, but only after all job have been revealed and (temporarily) cheduled. We minimize the makepan, and we call the problem a Online cheduling with bounded rearrangement at the end (BRE), which i a emi-online problem. Job arrive one by one over lit. After all the job have been arrived and cheduled, we are informed that the input equence i over; then at mot K already cheduled job can be reaigned. With repect to the wort cae ratio, we cloe the gap between the lower bound and upper bound, improving the previou reult a well. Epecially, for the lower bound, (i) for an improved lower bound + i obtained, which i better than (+) ++ bound + (Liu et al. (009) [9]); (ii) for 5 <, an improved lower (+) i obtained, which i better than (Liu et al. (009) [9]). For the upper + ++ i obtained, which i optimal and bound, (i) for and K =, a new upper bound + + better than the one + bound in Liu et al. (009) [9]; (ii) for + 5 < and K =, an upper + i propoed, which i optimal and better than the previou one in Liu + et al. (009) [9]; (iii) for < + 5 alo optimal and better than the previou one min{ + + and K =, an upper bound (+) i obtained, which i ++, (+) } in Liu et al. (009) [9]. + 0 Elevier B.V. All right reerved.. Introduction In thi paper, we conider an online non-preemptive cheduling problem on two related machine with bounded rearrangement to minimize the completion time, called Online cheduling with bounded rearrangement at the end, BRE for hort, which i a emi-online problem. Job arrive one by one over lit, i.e., after the incoming job ha been aigned, the new job arrive. After all job have arrived, we are informed that there i no further job; then at mot K already cheduled job can be reaigned, where K 0 i a fixed integer. When K = 0 and =, thi problem degenerate into one of the mot fundamental cheduling problem on two machine, aigning job online to identical parallel machine to minimize the completion time [6]; the fundamental (offline) cheduling problem i trongly NP-hard, if there are m machine, and m i not part of the input. [5]. Correponding author. Tel.: addree: cx.dlut@gmail.com (X. Chen), lanyan@neuoft.edu.cn (Y. Lan), benko.attila@almo.vein.hu (A. Benko), doagy@almo.vein.hu (G. Dóa), hanxin.mail@gmail.com (X. Han) /$ ee front matter 0 Elevier B.V. All right reerved. doi:0.06/j.tc

2 670 X. Chen et al. / Theoretical Computer Science 4 (0) Table Lower bound. [, + 5 ) [ + 5, ) [, ) Previou reult Our reult (+) ++ (+) ++ (+) ++ + (+) Table Upper bound. [, + 5 ) [ + 5, ) [, ) Previou reult min{ +, (+) }, K = + +, K =, K = + (+) Our reult, K =, K = +, K = Fig.. The thick curve are for our upper bound which are equal to the lower bound of the problem, i.e., our upper bound are optimal. Related model: Our problem i related to three online model, conidering the poibility of reaigning ome job from ome machine to another. The following online model have been invetigated in the lat year: (i) cheduling with bounded migration [0]; (ii) cheduling with a buffer [7,,8,4,,]; (iii) cheduling with the poible rearrangement of any K job at any time when a new job come, (without knowledge that the equence i ended or not), denoted by BR for hort [3]. Tan and Yu [] defined three further imilar problem where the rearrangement can be done only after all job are revealed and cheduled: (i), the lat job of any machine can be rearranged to the other machine, (ii), the lat K job of the equence can be rearranged, (iii) any K job can be rearranged. In thi paper we will deal with the third problem, what we denote a BRE. Problem BR eem to be more flexible than BRE, ince in the latter cae the rearrangement can be done only once, at the end of the equence. But it i worthy to note an advantage of the latter condition comparing to the former one: in cae of BRE, when we do the rearrangement, we are already informed that there i no further job, while in cae of BR the rearrangement mut be done in uch a way, that we cannot know whether the equence i to be continued, or not. Thu, at thi moment we cannot tate that BR i really more flexible than BRE. Our contribution: In thi paper, we ue competitive ratio to evaluate online algorithm, which i one of the tandard meaure. If an online algorithm alway achieve a olution within a factor ρ of the offline optimum, we ay the online algorithm i ρ-competitive. Our reult are ummarized in Table and (refer to Fig. ).. Preliminarie Scheduling on two related machine Input: Given two machine M, M with peed and repectively, and a et of job J = {j,..., j n } aociated with proceing time p : J R +, Output: Schedule J on M and M uch that the maximal completion time of M and M i minimized. If all the job are known in advance, then we ay the problem i offline. If job are revealed incrementally, i.e., one by one, once the current job i given we have to immediately chedule or aign it and the aignment cannot be changed in the future, then thi verion of the problem i called online. Rearrangement: After all job have been aigned to the machine, we are informed that the equence i over, then at mot K 0 already cheduled job can be reaigned to other machine, where K i a contant. Then any algorithm conit of two main part, the cheduling phae, and then (after being informed that the equence i over), the reaignment phae. In the problem of online cheduling with rearrangement on two related machine, if K = 0, the problem i totally online, if K = n, where n i the number of job in the input, then the problem i offline. In thi paper, we mainly tudy the problem with K < n, which i between online and offline verion.

3 X. Chen et al. / Theoretical Computer Science 4 (0) Given an online algorithm A, if for any input J we have A(J) ρ OPT(J), where A(J) and OPT(J) are the cot by online algorithm A and an optimal algorithm repectively, then we ay online algorithm A i ρ-competitive. On the other ide, if there i an input J for any determinitic online algorithm A, we have A(J) ρ OPT(J), then we ay ρ i the lower bound of the problem. Furthermore ome online algorithm A if ρ = ρ, we ay algorithm A i optimal. In the following, we denote {j, j,..., j t } a J t for t. Let L i t be the load of machine M i after dealing with job j t for i in the cheduling phae. If time t i clear from the context or t i the current time, we ue L i to replace L i t. Function ρ() i defined a the competitive ratio of our online algorithm, for hort, we ue ρ. 3. Lower bound In thi ection, we give new lower bound for problem BRE. And the analyi i imilar with the one for problem BR in [3]. Lemma. For any K, we have the lower bound for problem BRE a below: (i) for + 5 no online algorithm ha it competitive ratio trictly le than (+) + ; (ii) for 5 no online algorithm ha it competitive ratio trictly le than ++ + ; (iii) for > no online algorithm ha it competitive ratio trictly le than. + + Proof. Let ϵ > 0 be a ufficiently mall number uch that /ϵ i integer. Let t be /ϵ. The firt t job are mall job, each one ha a proceing time exactly ϵ. Cae : + 5. The following lower bound wa firt given in [9] for problem BR. Our proof i impler than the one in [9]. For the ake of completion, the detail of the proof are given below. After the t job have been aigned on machine, if L t + or ++ L t +, we are informed that job j ++ t i the lat job. After the lat job j t i given, at mot K job can be reaigned, where K i a contant, we have L t + Kϵ or ++ L t + at mot + ϵ. So the competitive ratio i at leat + L L t (+) t min + ϵ, + ϵ Kϵ( + ) ++, + ϵ( + ) a ϵ goe to zero the lower bound (+) i implied. Ele we conider the next cenario: L t < + + +, + + L t < Kϵ. On the other hand, the optimal value i Then we are informed that job j t+ ha a proceing time and i the lat job. Then the optimal value OPT(J t+ ) =. If j t+ i aigned on M after the poible rearrangement, then L t+ + L t Kϵ Kϵ ele M accept j t+ then the completion time on M L t+ + L t Kϵ ( + ) + + Kϵ, Kϵ = ( + ) + + Kϵ. In both cae the lower bound (+) i implied a ϵ approache to zero. ++ Cae : + 5, aume L t, then the lat job j + t+ ha a proceing time p(j t+ ) =. Then OPT(J t+ ) =. The completion time of any online algorithm i min L +, L + t t Kϵ min, + L t Kϵ min +, + L t Kϵ = + L t Kϵ + Kϵ. In thi cae, when ϵ approache to zero, the competitive ratio approache to +. Ele if L t < +, then L t +. Next two job j t+ and j t+ with p(j t+ ) = + L t and p(j t+) = p(j t+ ) arrive and we are informed that job j t+ i the lat job. Note that p(j t+ ) p(j t+ ). The optimal value OPT(J t+ ) i equal to p(j t+ ) = + L t by the following aignment: j t+ M and J t+ \ {j t+ } M. If j t+ or j j+ i aigned on M then the completion time i at leat L t + p(j t+) Kϵ = L t Kϵ.

4 67 X. Chen et al. / Theoretical Computer Science 4 (0) Ele both j t+ and j t+ are aigned on M, then L t+ L + t p(j t+) + p(j t+ ) Kϵ = ( + ) + L t Kϵ = 3 + L t Kϵ The completion time L t+ / approache to zero. 3 L t Kϵ (by ). L t Kϵ. In both cae, the competitive ratio approache to f () = when ϵ + Cae 3: >, aume L t +, then job j t+ with p(j t+ ) = arrive and it i the lat job. Then OPT(J t+ ) =. The completion time of any online algorithm i min L t +, L t + Kϵ min, + Kϵ = + Kϵ (by > ). + + In thi cae, when ϵ approache to zero, the competitive ratio approache to + +. Ele if L t <, then + L t. Next two job j + t+ and j t+ with p(j t+ ) = ( + )L t and p(j t+ ) = p(j t+ ) arrive and the input end. Note that p(j t+ ) > p(j t+ ) for >. The optimal value OPT(J t+ ) = ( + )L t by the following aignment: j t+ M and J t+ \ {j t+ } M. If j t+ or j j+ i aigned on M, then the completion time i at leat L t + p(j t+) Kϵ = ( + )L t Kϵ. Ele both j t+ and j t+ are aigned to M, then L t+ L t + p(j t+) + p(j t+ ) Kϵ = ( + )L t Kϵ. The completion time L t+ / ( + )L t zero. + Kϵ. In both cae, the competitive ratio approache to when ϵ approache to + 4. Upper bound In thi ection, we give two optimal online algorithm for and repectively. In the optimal algorithm for, we allow to reaign only one job, i.e., K =. In the optimal algorithm for, we allow to reaign at mot two job, i.e., K =. We firt give ome definition and ome ueful obervation. Let t = l max t, P t +, which i a natural lower bound for the problem, where lt denote the ize of the longet job o far (among the firt t job) and P t denote the total ize o far. If t = l t we ay that the lower bound i determined by the longet ize (o far), otherwie we ay that the lower bound i determined by the total ize (o far). Obervation. P t ( + ) t. Given a function ρ() >, conider a chedule jut after aigning job j t, if L t > ρ() t then we ay that M i overloaded ele underloaded; if L t > ρ() t then we ay that M i overloaded, ele underloaded. If both machine are underloaded at time t, we ay the chedule i underloaded. Note that both machine cannot be overloaded at the ame time. Alo note that in cae the chedule i underloaded, then naturally the competitive ratio i not violated (at leat at the current point of the running). But in the oppoite cae, if the chedule i overloaded, then the competitive ratio till can be valid, if the optimum value i bigger than the lower bound. Lemma. If M i overloaded then L > ρ() + ρ() L and if M i overloaded then L > ρ() + ρ() L. Proof. If M i overloaded then L > ρ(), then uing Obervation, we have L ( + ) L ( + ρ()). Hence L > ρ() + ρ() L. If M i overloaded then L > ρ() and L ( + ) L ( + ρ()). Hence L > ρ() + ρ() L.

5 X. Chen et al. / Theoretical Computer Science 4 (0) An optimal algorithm for We give an online algorithm with a competitive ratio ρ() = + for all which algorithm ue only one arrangement, + i.e., K =, and which i optimal for all. The idea are a follow: before the input end, we keep the low machine underloaded all the time; when the input end, if neceary, we find an appreciate job in M and migrate it to M. Algorithm LC (Larget Change). Let job j with ize p be the incoming job. Then update the lower bound.. Aign j to M, if L + p ρ(), ele j M. 3. If the input doe not end, goto Step. 4. Ele if M i overloaded then migrate a job from M to M uch that the final makepan decreae a much a poible by the migration, (if there exit uch job). Theorem 3. Algorithm LC i ρ = + -competitive for any. + Proof. Let ρ = +. Let L + i be the load on M i when the input end for i before the migration. Let L i be the load on M i after the migration for i. It i trivial to ee if max{l, L /} ρ, thi lemma hold. On the other hand, according to the algorithm, if L ρ then max{l, L /} ρ. Hence we conider the cae: max{l, L /} > ρ. In thi cae, before the migration M i overloaded. Aume L = ρ + x with ome x > 0. Then ( + ) L ( + ) L = + x = x. () + + Then we have two lemma. Lemma 4. If max{l, L /} > ρ then there i no job of ize in [x, x + ] on M jut before the migration. Proof. Aume there i a job j with ize p [x, x + ]. Then from () we get L = L p ρ and L = L + p + = ρ, which contradict to the fact + max{l, L /} > ρ. Hence thi lemma hold. () + + = Lemma 5. If max{l, L /} > ρ then jut before the migration the total ize of all the job with ize in (0, x) on M i le than + + x. + + Proof. Let X be the total ize of all the job with ize in (0, x) on M. Aume thi lemma doe not hold, i.e, X > + +x. + + Let t be the time when the input end. Let j r with a ize p (0, x) be the lat job aigned on M, i.e., job j r arrived at time r t. Since X > 0, j r i well-defined, i.e. there exit uch job. According to the algorithm, if job j r i aigned on M, then M would be overloaded. So we have r L + r p > ρ r ρ P + ρ X + L +. + Since L r L x by () and p < x, we have X < L + p + + r L r ρ = ρ + < ρ + x + + ρ x ( + ) = x = + + L + + r ρ p + + x, which caue a contradiction. The aumption i not true and thi lemma hold. By (), Lemma 4 and 5, the total ize of all the job with ize larger than + x i larger than + ρ + x x ( + ) = ( + ) + = = + +. (3)

6 674 X. Chen et al. / Theoretical Computer Science 4 (0) Oberve that in an optimal chedule at leat one job of ize greater than + x i aigned on M or all the job with ize greater than + x are on M, hence we have OPT min + x, + + = + min x,. (4) ( + ) If OPT + (+) = (+)+ (+) max{l, L /} L P = (+), then ince P/( + ), we have (+) ( + ) = + ( + ) + ( + ) ρ OPT. Ele OPT + x, then by () and ince ρ >, we have max{l, L /} L = ρ + x = ρ + x ρ( + x) ρ OPT. Hence thi theorem hold. 4.. An optimal algorithm for We propoe an algorithm which rearrange at mot two job at the end, and it i ρ()-competitive, refer to the following table. The competitive ratio matche the lower bound of the problem, thu the algorithm i optimal. We need two technical ratio b() and c(), which are defined in the following table. We firt give ome lemma which will be ueful in analyzing our algorithm, then propoe and analyze the algorithm. ρ() b() c() I := < + 5 I := + 5 (+) We call a job big, if it ize i bigger than b() ele call it mall. Note that at any point of the execution, it i well defined whether a job i big or mall. If a job i mall at ome time, it never become big. But a big job can become mall at ome time later, ince the value of lower bound can increae. The next Obervation can be checked eaily for both cae regarding I or I. Obervation. For, we have b() = (ρ() )( + ) > ( + ). 4 Lemma 6. Any time, the number of big job i at three. Proof. It follow from that the total ize of all job i P ( + ), while the total ize of four big job would be more than 4b() > ( + ) by Obervation. Obervation 3. For, we have ρ() + = ( + c())b(). To make eay checking the validity of the next obervation we give here a table about the ued expreion. Then the obervation can be checked eaily by ome imple calculation. ρ() ( + )(3 ρ()) (+) I ++ I + (+)( +) ρ() ρ() + + ρ() + ρ() ( + ) 3 c() + Obervation 4. For [, + 5 ), we have ρ ( + ), c() = ρ + ρ > 0. Obervation 5. For, we have ( + )(3 ρ()) ρ(). Proof. For I, the olution of equation = are approximately a follow = , =.3473 and =.53, thu hold if I. Obervation 6. For, we have ρ() ρ() ρ() + ρ(). Proof. For I, to check + 3, one need to validate + 4 3, i.e = + 0 which hold in the conidered interval I. Obervation 7. For, we have ρ() + ρ() > c().

7 X. Chen et al. / Theoretical Computer Science 4 (0) The following table i for the next obervation. ρ() (+) I ++ I + +c() c() ρ() + ρ() ρ() + ρ() Obervation 8. For [ + 5, ], we have c() +c() = +. Obervation 9. For, we have ρ() +c() c() ( + ), or in equivalent form c() ρ() + ρ(). Obervation 0. For, we have ρ() + ρ() +. Proof. To check 3 + we need to ee that 4 ( + ) 3 + = hold, which i the ame 3 + a + which hold if I. Obervation. For, we have b() c()( + b()). Lemma 7. Aume M i i overloaded, for i =,. If M j i overloaded after a job migrate from M i to M j, where j = 3 i, then the job mut be a big job. Proof. Note that the total ize of proceing time i at mot ( + ). We have L > ρ() if M i overloaded, and L > ρ() if M i overloaded. Let j be the job migrated from an overloaded M i to M j, where j = 3 i. After migration, M j become overloaded. Then the ize of job j mut be bigger than ρ() + ρ() ( + ) = ( + ) (ρ() ) = b(), where the lat inequality hold from Obervation. Hence job j i big. Lemma 8. Suppoe that a big job i aigned on M (among other job or alone). Then M cannot be overloaded. Proof. If a big job i aigned to the low machine, then L > b(). We have L ( + ) L < ( + b()), by Obervation and 6, L /L + b() b() Thu M cannot be overloaded by Lemma. = ρ() ρ() ρ() + ρ() An online algorithm and it analyi In our algorithm, there are two phae, the cheduling phae and the reaignment phae. In the cheduling phae,we try to keep the next two propertie through the whole execution, which alo give u a help in the reaignment phae. Let l denote the total load of the mall job aigned on M, without taking into account the big job, we call it a the retricted load. P: there are two mall job on M with ize p and p uch that (L + p + p ) c()(l p p ), where p and p can be zero. It mean that after moving at mot two mall job from M to M, in the modified chedule L c() l hold. P: (L p) c() (l + p), where p i the ize of the lat job aigned on M (p = 0 if there i no job aigned on M ). Scheduling phae of algorithm SMF (low machine firt). Let job j with ize p be the incoming job. Then update the lower bound and alo l ince ome big job may become mall.. Job j i big: if (L + p) ρ() then j M ; otherwie j M. 3. Job j i mall: if L c() l then aign job j to M otherwie to M. 4. Update L, L, and l. If the input end, goto the reaignment phae, ele goto Step. Lemma 9. If there are three big job in the input, at leat one of the three job i aigned on M. Proof. Let j i, j k, j l be the three big job, which have ize larger than b()ρ. If one of the three job i aigned on M, then we are done. Without lo of generality, aume job j i and j k are aigned on M and job j l arrive later than the two other job. We prove that job j l mut be aigned to M at Step of the cheduling phae. According to the definition of the big job, after j l arrive, L p(j i )+p(j k ) > b(). If job j l i aigned on M, L < (+) b() = (+) ( + ) (ρ() ) = ( + )(3 ρ()) ρ() by Obervation and 5. Lemma 0. For, if propertie P and P keep holding during the whole execution of the chedule phae, SMF i ρ()-competitive after the reaignment phae. Proof. Suppoe that the tatement doe not hold. Since machine M and M cannot be overloaded at the ame time, we have the following two cae.

8 676 X. Chen et al. / Theoretical Computer Science 4 (0) Reaignment phae of algorithm SMF (low machine firt). If M i overloaded then move the lat job from M to M.. Ele if M i overloaded then {there are at mot two big job on M } (a) If there exit a job j uch that migrating j from M to M both machine are underloaded, then migrate job j from M to M. (b) Ele i. if there i at mot one big job on M, then move two mall job from M to M to enure L c() l in the modified chedule, ii. ele if there are two big job on M, A. if L > c() l, then move the lat mall job from M to M {to enure L c() l }, and B. move the maller big job from M to M. Cae A: M i overloaded at the end of the cheduling phae. Let t be the current time. In thi cae we perform Step of the reaignment phae. We prove that after the reaignment, both machine are underloaded. Let j k be the lat job aigned in the cheduling phae to M, where k t. If j k wa aigned on M a a big job, then M cannot be overloaded by the algorithm. Hence job j k wa aigned on M a a mall job. In the reaignment tep, job j k i migrated from M to M. After the migration, if M i overloaded then job j k would have been big by Lemma 7, which contradict with the fact that job j k i mall. Hence after the reaignment, M i till underloaded. Aume M i till overloaded after the reaignment. Then L k > ρ() t. Note that job j k wa aigned on M at Step 3 in the cheduling phae. Thu l L k k c() the job we get contradiction a follow: P L + k L > ρ() k t + ρ() t = ρ() + t ( + ) t P, c() c() > ρ() t. Then for the total ize of all c() where we ued Obervation 9. Thu, the aumption that M i overloaded doe not hold, i.e., both machine are underloaded after the reaignment tep in thi cae. Cae B: M i overloaded at the end of the cheduling phae. By Lemma 8, there i no big job on M after the cheduling phae. Let n b be the number of big job on the input. By Lemma 6 we have n b 3. Since M i overloaded, by Lemma 8 and 9, all the big job are on M, thu n b. If the execution pae through Step (a) of the reaignment phae, then both machine are underloaded. Next we conider the cae where the execution pae through Step (b) of the reaignment phae, there are the following three ubcae. Subcae : n b = 0. By property P we can reaign two mall job from M to M to make ure that L c()l. We claim that after the migration M i underloaded. By Lemma 7, after migrating the firt mall job from M to M, M i till underloaded. Since the execution doe not top at Step (a) in the reaignment, after migrating the firt mall item M i till overloaded. Again by Lemma 7, after the econd migration M i till underloaded. Moreover after the moving L l c() hold by the guarantee of property P. Since there i no big job, thi i the ame a L L c(). If thi chedule would be M overloaded, then L /L > ρ() ρ() by Lemma. However by Obervation 7, + ρ() + ρ(), which caue a contradiction. c() Hence after the reaignment M i underloaded. Subcae : n b =. We reaign two mall job from M to M to make ure that L c()l by property P. By the above argument, after the migration M i underloaded. Suppoe that the chedule i M -overloaded after the migration. We have L > ρ(). Since the ize of the big job i at mot, we have l > ρ() = (ρ ), and L c()l > c()(ρ ). Then by Obervation 9 the total ize i L + L > c()(ρ ) + ρ = (c()(ρ ) + ρ) ρ(ρ ) + ρ + ρ ρ = ( + ), + ρ where the lat inequality hold from Obervation 0, i.e. which caue contradiction. Hence M i underloaded after the reaignment. Subcae 3: n b =. Let the two big job be j x and j y, which are aigned on M, and both are bigger than b(). Aume job j x i not larger than j y, i.e., p(j x ) p(j y ). In thi cae, we reaign the lat job j k from M to M to get L c()l by property P, where L i the load of M jut after the firt migration and l i the retricted load of M jut after the firt migration, then reaign job j x from M to M. After migrating job j x on M, by Lemma 8 M i underloaded. If M i underloaded too, then we are done. Otherwie, aume that M i overloaded. Next we prove thi i not poible.

9 X. Chen et al. / Theoretical Computer Science 4 (0) If M i overloaded, then L + p(j x) > ρ. Becaue the total ize i P ( + ), we have l + p(j y) < ( + ) L p(j x) < ( + ) ρ. (5) For [, + 5 ), by Obervation 4 we have ρ < L + p(j x) c()l + p(j y) = (c() )l + l + p(j y) < ( + ρ) + (c() )l, (ρ ) < ρ + ρ l, thu we have l > ( + ρ), which contradict with (5). For [ + 5, ], i.e., > +, we are going to prove that after the rearrangement the load of M i at mot ρ()opt, where OPT i the value by an optimal olution. In thi cae, an optimal algorithm chedule two big job together on one machine or eparately on two machine, then we have a new lower bound, i.e., OPT min p(j x ), p(j x) + p(j y ) = p(j x ). Since L c()l, we have L L + l + c() c() ( + ) p(j x) p(j y ) c() ( + ) p(j x) c(). + c() + c() Uing thi inequality, if p(j x ) then by Obervation 8 L c() + c() ( ) = + = (ρ() ) (ρ() )p(j x) then L + p(j x) ρ()p(j x ) ρ()opt. Ele p(j x ) < then L + p(j x) ( + ) p(j x) + Hence M i underloaded. + p(j x ) = ( + ) + + p(j + x ) ( + ) = p(j ( + ) x) < = ρ(). Lemma. Propertie P and P keep holding through the whole execution in the cheduling phae of the algorithm. Proof. We ue induction approach to prove thi lemma. It i not difficult to ee jut after the firt job of the equence arrive, P and P hold. Aume propertie P and P hold at time t. Let job j t be the next job, with ize p. Regarding property P, if the current job j t i aigned to M by the algorithm, then property P till hold, ince the retricted load of M, i.e. the value of l can only increae. Thu conider the cae j t i aigned to M next. If j t i aigned on M a a big job, i.e., p > b(), then M i underloaded by the algorithm, i.e., L ρ(), where i the lower bound at time t. Thu we have L p < (ρ() b()) c()b() < c()p c()(l + p), ince ρ() b() c()b() hold by Obervation 3. Ele job j t i aigned a a mall job, i.e., it happen at Step 3 of cheduling phae. We have L p c()l c()(l + p). Hence P hold at time t. Let u conider the validity of property P at time t. (We uppoe again that propertie P and P hold at time t.) Firt we note that if the next job j t i mall and aigned to M by the algorithm (in Step 3), then property P trivially hold, ince reaigning only j t to the low machine, the inequality L + j t c() (l j t ) i atified by the algorithmic rule. Thu uppoe in the following that j t i big, or it i mall and aigned to M. Let S t be the et of all the mall job on M at time t. (Thu j t / S t follow by the previou note.) If S t = then property P trivially hold without any rearrangement ince l t = 0. Thu let u uppoe that S t i not empty. Let job j r S t be that job which i aigned to M at the latet time, where r t, i.e. j r i the lat job what i ever aigned to M and it i mall at moment t. Then, ince S t i not empty, and j t / S t, follow that r < t. It mean that all job which are aigned to M after time r are big job when they come, and they remain big until time t. Since any time at mot two big job can be aigned to M, at moment r there are at mot two big job on M. If there i no big job on M at time r, then the value of l r cannot change until time t, that i, l t = l r. If there i one big job on M at time r which become mall until time t, let thi job be denoted by j x. Then l t = l + r p(j x) hold. Finally, if there are two big job on M at time r which become mall until time t, let thee job be denoted by j x and j y, then l t = l r +p(j x)+p(j y ).

10 678 X. Chen et al. / Theoretical Computer Science 4 (0) Furthermore, by the definition of job j r, the inequality hold l t L r, ince all the job aigned on M after time r are big at time t. Now we ditinguih two cae. Cae. Job j r wa aigned on M a a big job. Thi mean that if j r i aigned to M, then M would become to be overloaded, i.e., L r + p(j r) > ρ r. By Lemma and Obervation 9, we have L + r p(j r) ρ > + ρ c(). L r Since L r = L r + p(j r), by (6) we have L t + p(j r) L r + p(j r) > c()l r = c()(l r p(j r)) c()(l t p(j r)). Cae. Job j r wa aigned on M a a mall job. Then, a we have een, there can be three cae, according to that how many big job are at moment r on the fat machine which job become to be mall till moment t. Cae.. If there i not uch job, then l t = l r. Then property P hold trivially, ince after time r the retricted load of the fat machine doe not change, and the load of the low machine can only increae, and at time r hold the next inequality, L + r p(j r) > c()l r, ince j r i aigned to the fat machine. Thu by reaigning j r to the low machine at moment t property P hold. Cae.. There i one big job on M, we denote it by j x, which become to be mall until time t. Then l t = l + r p(j x). Similarly to the previou cae, by reaigning j r and j x to the low machine, L t + p(j r ) + p(j x ) > c() l t p(j r) p(j x ) hold. Cae.3. There are two job on M, j x and j y, which job become to be mall until time t. Then l t = l + r p(j x) + p(j y ). Since job j x and j y wa big at time r, min{p(j x ), p(j y )} > b() r. Hence l r ( + ) r p(j x ) p(j y ) < ( + b()) r, and b() r < L t + p(j x) + p(j y ). By Obervation, we have b() c()( + b()). Thu we have L t + p(j x) + p(j y ) > c()(l t p(j x) p(j y )). Hence thi lemma hold. Remark: in thi paper we cloe the gap between the lower bound and upper bound for all cae except for K = and < <. So the open quetion i to cloe the gap for the cae K = and < <. (6) Acknowledgement The third author wa partially upported by Project TAMOP-4../B-0/ The lat author wa partially upported by the Fundamental Reearch Fund for the Central Univeritie. Reference [] G. Dóa, L. Eptein, Online cheduling with a buffer on related machine, J. Comb. Optim. 0 () (00) [] G. Dóa, L. Eptein, Preemptive online cheduling with reordering, in: ESA, 009, pp [3] G. Dóa, Y. Wang, X. Han, H. Guo, Online cheduling with rearrangement on two related machine, Theoret. Comput. Sci. 4 (8 0) (0) [4] M. Englert, D. Özmen, M. Wetermann, The power of reordering for online minimum makepan cheduling, in: Proc. 48th Symp. Foundation of Computer Science, FOCS, 008, [5] M.R. Garey, D.S. Johnon, Strong np-completene reult: motivation, example and implication, J. ACM 5 (978) [6] R.L. Graham, E.L. Lawler, J.K. Lentra, A.H.G. Rinnony Kan, Optimization and approximation in determinitic equencing and cheduling: a urvey, Ann. Dicrete Math. 5 (979) [7] H. Kellerer, V. Kotov, M.G. Speranza, Z. Tuza, Semi on-line algorithm for the partition problem, Oper. Re. Lett. (5) (997) [8] S. Li, Y. Zhou, G. Sun, G. Chen, Study on parallel machine cheduling problem with buffer, in: Proc. of the nd International Multiympoium on Computer and Computational Science, IMSCCS 007, 007, pp [9] M. Liu, Y. Xu, C. Chu, F. Zheng, Online cheduling on two uniform machine to minimize the makepan, Theoret. Comput. Sci. 40 ( 3) (009) [0] N. Sivadaan, P. Sander, M. Skutella, Online cheduling with bounded migration, Math. Oper. Re. 34 () (009) [] Z. Tan, S. Yu, Online cheduling with reaignment, Oper. Re. Lett. 36 () (008) [] G. Zhang, A imple emi on-line algorithm for p//c max with a buffer, Inform. Proce. Lett. 6 (997)