Approxmatng the optmal compettve rato for an ancent onlne schedulng problem Ln Chen 1 Desh Ye 1 Guochuan Zhang 1 1 College of Computer Scence, Zhejang Unversty, Hangzhou, 310027, Chna chenln198662@zju.edu.cn, zgc@zju.edu.cn Abstract We consder the classcal onlne schedulng problem P C max n whch jobs are released over lst and provde a nearly optmal onlne algorthm. More precsely, an onlne algorthm whose compettve rato s at most (1 + ɛ) tmes that of an optmal onlne algorthm could be acheved n polynomal tme, where m, the number of machnes, s a part of the nput. It substantally mproves upon the prevous results by almost closng the gap between the currently best nown lower bound of 1.88 [21] and the best nown upper bound of 1.92 [15]. It has been nown by follore that an onlne problem could be vewed as a game between an adversary and the onlne player. Our approach extensvely explores such a structure and bulds up a completely new framewor to show that, for the onlne over lst schedulng problem, gven any ɛ > 0, there exsts a unform threshold K whch s polynomal n m such that f the compettve rato of an onlne algorthm s ρ 2, then there exsts a lst of at most K jobs to enforce the onlne algorthm to acheve a compettve rato of at least ρ O(ɛ). Our approach s substantally dfferent from that of [19], n whch an approxmaton scheme for onlne over tme schedulng problems s gven, where the number of machnes s fxed. Our method could also be extended to several related onlne over lst schedulng models. Keywords: Compettve analyss; Onlne schedulng; Dynamc programmng. 1 Introducton Very recently Günther et al. [19] come up wth a nce noton called Compettve rato approxmaton scheme for onlne problems. Formally speang, t s a seres of onlne algorthms {A ɛ : ɛ > 0}, where A ɛ has a compettve rato at most (1 + ɛ) tmes the optmal compettve rato. Naturally, a compettve rato approxmaton scheme could be seen as an onlne verson of the PTAS (polynomal tme approxmaton scheme) for the offlne problems. Usng such a noton, they provde nearly optmal onlne algorthms for several onlne schedulng problems where jobs arrve over tme, ncludng Qm r j, (pmtn) w j c j as well as P m r j C max, where m s the number of machnes. The algorthm runs n polynomal tme when m s fxed. That s a great dea for desgnng nearly optmal onlne algorthms, that motvates us to revst the classcal onlne problems whch stll have a gap between upper and lower bounds. However, the technque of Günther et al. [19] heavly reles on the structure of the optmal soluton for the over tme schedulng problem, through whch they can focus on jobs released durng a tme wndow of a constant length. It thus seems hard to generalze to other onlne models. Clearly, the frst onlne schedulng problem whch should be revsted s P C max, a fundamental problem n whch jobs are released over lst. Ths ancent schedulng model admts a smple 1
algorthm called LS (lst schedulng) [18]. Its compettve rato s 2 1/m that acheves the best possble for m = 2, 3 [14]. Nevertheless, better algorthms exst for m = 4, 5, 6, 7, see [10] [16] [22] for upper and lower bounds for onlne schedulng problems where m tang these specfed values. Many more attentons are pad to the general case where m s arbtrary. There s a long lst of mprovements on upper and lower bounds, see [1] [7] [20] for mprovements on compettve algorthms, and [1] [8] [17] for mprovements on lower bounds. Among them the currently best nown 1+ln 2 2 1.9201 [15], whle the best nown lower bound s 1.88 [21]. We refer upper bound s 1 + the readers to [23] for a nce survey on ths topc. Although the gap between the upper and lower bounds are relatvely small, t leaves a great challenge to close t. In ths paper we tacle ths classcal problem by provdng a compettve rato approxmaton scheme. The runnng tme s polynomal n the nput sze. More precsely, the tme complexty related to m s O(m Λ ) where Λ = 2 O(1/ɛ2 log 2 (1/ɛ)). It s thus polynomal even when the number of machnes s a part of the nput. To smplfy the noton, throughout ths paper we use compettve scheme nstead of compettve rato approxmaton scheme. General Ideas We try to gve a full pcture of our technques. Gven any ɛ > 0, at any tme t s possble to choose a proper value (called a scalng factor) and scale all the jobs released so far such that there are only a constant number of dfferent nds of jobs. We then represent the jobs scheduled on each machne by a tuple (called a trmmed-state) n whch the number of each nd of jobs remans unchanged. Composng the trmmed-states of all machnes forms a trmmed-scenaro and the number of dfferent trmmed-scenaros we need to consder s a polynomal n m, subject to the scalng factors. Gven a trmmed-scenaro, we can compute the correspondng approxmaton rato (comparng wth the optmal schedule), whch s called an nstant approxmaton rato. Specfcally, f the schedule arrves at a trmmed-scenaro when the adversary stops, then the compettve rato equals to the nstant approxmaton rato of ths trmmed-scenaro. Formal defntons wll be gven n the next secton. Note that the nstant approxmaton rato of every trmmed-scenaro could be determned (up to an error of O(ɛ)) regardless of the scalng factor. To understand our approach easly we consder the onlne schedulng problem as a game. Each tme the adversary and the scheduler tae a move, alternatvely,.e., the adversary releases a job and the onlne scheduler then assgns the job to a machne. It transfers the current trmmed-scenaro nto a new one. Suppose the adversary wns the game by leadng t nto a certan trmmed-scenaro wth an nstant approxmaton rato ρ, forcng the compettve rato to be at least ρ. The ey observaton s that f he has a wnnng strategy, he would have a wnnng strategy of tang only a polynomal number (n m) of moves snce the game tself conssts of only a polynomal number of dstnct trmmed-scenaros. A rgorous proof for such an observaton reles on formulatng the game nto a layered graph and assocatng the schedulng of any onlne algorthm wth a path n t. Gven the observaton, the onlne problem ass f the adversary has a wnnng strategy of C = poly(m) moves, startng from a trmmed-scenaro where there s no job. Such a problem could be solved va dynamc programmng, whch decomposes t nto a seres of subproblems that as whether the adversary has a wnnng strategy of C < C moves, startng from an arbtrary trmmed-scenaro. Varous extensons could be bult upon ths framewor. Indeed, compettve schemes could be 2
acheved for Rm C max and Rm Cp where p 1 s some constant and C s the completon tme of machne. The runnng tmes of these schemes are polynomal when m s a constant. In addton to compettve schemes, t s nterestng to as f we can acheve an optmal onlne algorthm. We consder the sem-onlne model P p j q C max, where all job processng tmes are bounded. We are able to desgn an optmal onlne algorthm runnng n (mq) O(mq) tme. It s exponental n both m and q. Recall that the compettve rato of lst schedulng for P C max s 2 1/m. Throughout the paper we focus on onlne algorthms whose compettve rato s no more than 2. We assume that m 2. 2 Structurng Instances To tacle the onlne schedulng problem, smlarly as the offlne case we want to well structure the nput nstance subject to an arbtrarly small loss. However, n the onlne settng we are not aware of the whole nput. The nstance needs scalng n a dynamc way. Gven any 0 < ɛ 1/4, we may assume that all the jobs released have a processng tme of (1 + ɛ) j for some nteger j 0. Let c 0 be the smallest nteger such that (1 + ɛ) c 0 1/ɛ. Let ω be the smallest nteger such that (1 + ɛ) ω 3. Let SC = {(1 + ɛ) jω j 0, j N}. Consder the schedule of n (n 1) jobs by any onlne algorthm. Let p max = max j {p j }. Then LB = max{ n j=1 p j/m, p max } s a trval lower bound on the maespan. We choose T LB SC such that T LB LB < T LB (1 + ɛ) ω, and defne job j as a small job f p j T LB (1 + ɛ) c 0, and a bg job otherwse. T LB s called the scalng factor of ths schedule. Let L s h be the load (total processng tme) of small jobs on machne h. An (ω + c 0 + 1)- tuple st h = (η h c 0, η h c 0 +1,, ηh ω) s used to represent the jobs scheduled on machne h, where η h ( c 0 + 1 ω) s the number of bg jobs wth processng tme T LB (1 + ɛ) on machne h, and η h c 0 = L h s /(T LB (1 + ɛ) c 0 ). We call such a tuple as a state (of machne h). The frst coordnate of a state mght be fractonal, whle the other coordnates are ntegers. The load of a state s defned as LD(st h ) = ω = c 0 (1 + ɛ) η 4LB. Composng the states of all machnes forms a scenaro ψ = (st 1, st 2,, st m ). Thus, any schedule could be represented by (T LB, ψ) where T LB SC s the scalng factor of the schedule. Specfcally, f the adversary stops now, then the compettve rato of such a schedule s approxmately (up to an error of O(ɛ)): ρ(ψ) = C max (ψ)/op T (ψ) where C max (ψ) = max j LD(st j ), and OP T (ψ) s the maespan of an optmal soluton for the offlne schedulng problem n whch jobs of ψ are taen as an nput (here small jobs are allowed to splt). We defne LD(ψ) = h LD(st h) and P max (ψ) the largest processng tme (dvded by T LB ) of jobs n ψ (P max (ψ) = (1 + ɛ) c 0 f there s no bg job n ψ). Obvously, OP T (ψ) LB = max{ld(ψ)/m, P max (ψ)} 1. The above rato s regardless of the scalng factor and s called an nstant approxmaton rato. We can use a slghtly dfferent (ω + c 0 + 1)-tuple τ = (ν c0, ν c0 +1,, ν ω ) to approxmate a state, where each coordnate s an nteger. It s called a trmmed-state. Specfcally, τ s called a smulatng-state of st h f ν = η h for c 0 < ω and η h c 0 ν c0 η h c 0 + 2. 3
We defne LD(τ) = ω = c 0 ν (1 + ɛ) and restrct our attenton on trmmed-states whose load s no more than 4LB + 2(1 + ɛ) c 0. There are at most Λ 2 O(1/ɛ2 log 2 (1/ɛ)) such nds of trmmedstates (called feasble trmmed-states). We sort these trmmed-states arbtrarly as τ 1,, τ Λ, and defne a Λ-tuple φ = (ξ 1, ξ 2,, ξ Λ ) to approxmate scenaros, where ξ = m and 0 ξ m s the number of machnes whose correspondng trmmed-state s τ. Indeed, φ s called a trmmedscenaro and specfcally, t s called a smulatng-scenaro of ψ = (st 1, st 2,, st m ) f there s a one to one correspondence between the m states (.e., st 1 to st m ) and the m trmmed-states of φ such that each trmmed-state s the smulatng-state of ts correspondng state. Recall that n ψ, jobs are scaled wth T LB, thus 1 max{1/mld(ψ), P max (ψ)} < (1 + ɛ) ω. We may restrct our attentons to trmmed-scenaros satsfyng 1 max{1/mld(φ), P max (φ)} < (1 + ɛ) ω + 2(1 + ɛ) c 0, where smlarly we defne LD(φ) = j ξ jld(τ j ), and P max (φ) the largest processng tme of jobs n φ. Trmmed-scenaros satsfyng the prevous nequalty are called feasble trmmed-scenaros. Notce that there are Γ (m + 1) Λ dfferent nds of feasble trmmed-scenaros. we sort them as φ 1,, φ Γ. As an excepton, we plug n two addtonal trmmed-scenaros φ 0 and φ Γ+1, where φ 0 represents the ntal trmmed-scenaro n whch there are no jobs, and φ Γ+1 represents any nfeasble trmmed-scenaro. Let Φ be the set of these trmmed-scenaros. We defne ρ(φ) = C max (φ)/op T (φ) as the nstant approxmaton rato of a feasble trmmed-scenaro φ, n whch C max (φ) = max j {LD(τ j ) : ξ j > 0}, and OP T (φ) s the maespan of the optmum soluton for the offlne schedulng problem n whch jobs of φ are taen as an nput and every job (ncludng small jobs) should be scheduled ntegrally. As an excepton, we defne ρ(φ 0 ) = 1 and ρ(φ Γ+1 ) =. Furthermore, notce that except for φ Γ+1, C max (φ) 4(1 + ɛ) ω + 2(1 + ɛ) c 0 20, whch s a constant. Thus we can dvde the nterval [1, 20] equally nto 19/ɛ subntervals and let = {1, 1 + ɛ,, 1 + ɛ 19/ɛ}. We round up the nstant approxmaton rato of each φ to ts nearest value n. For smplcty, we stll denote the rounded value as ρ(φ). Lemma 1 If φ s a smulatng-scenaro of ψ, then ρ(ψ) O(ɛ) ρ(φ) ρ(ψ) + O(ɛ). Proof. It can be easly seen that OP T (ψ) OP T (φ) OP T (ψ) + 3(1 + ɛ) c 0. Meanwhle C max (ψ) C max (φ) C max (ψ) + 2(1 + ɛ) c 0. Note that OP T (ψ) 1 and the lemma follows drectly. 2 Consder the schedulng of n jobs by any onlne algorthm. represented by a lst as The whole procedure could be (T LB (1), ψ(1)) (T LB (2), ψ(2)) (T LB (n), ψ(n)), where ψ() s the scenaro when there are jobs, and T LB () s the correspondng scalng factor. Here ψ() changes to ψ( + 1) by addng a new job p +1, and the reader may refer to Appendx A to see how the coordnates of a scenaro change when a new job s added. Let µ 0 be the smallest nteger such that (1+ɛ) µ 0 4(1+ɛ) ω+c 0+1 and R = {0, (1+ɛ) c 0,, (1+ ɛ) µ 0/ω +ω 1 }. We prove that, f a scenaro ψ changes to ψ by addng some job p n, then there 4
exsts some job p n R such that φ changes to φ by addng p n, and furthermore, φ and φ are the smulatng-scenaros of ψ and ψ, respectvely. Ths suffces to approxmate the above scenaro sequence by the followng sequence φ 0 φ(1) φ(2) φ(n), where φ() s the smulatng-scenaro of ψ(), and φ 0 s the ntal scenaro where there s no job. We brefly argue why t s ths case. Suppose T LB s the scalng factor of ψ. Accordng to the onlne algorthm, p n s put on machne h where st h = (η c0,, η ω ). Let τ = (ν c0,, ν ω ) be ts smulatng state n φ. If p n /T LB < (1 + ɛ) c 0 and η c0 + p n /T LB ν c0, then φ s stll a smulatng scenaro of ψ and we may set p n = 0. Else f ν c0 < η c0 + p n /T LB ν c0 + 1, we may set p n = (1 + ɛ) c 0. For the upper bound on the processng tme, suppose p n /T LB s so large that the prevous load of each machne (whch s no more than 4LB 4(1 + ɛ) ω ) becomes no more than (1 + ɛ) c 0 p n /T LB. It then maes no dfference by releasng an even larger job. A rgorous proof nvolves a complete analyss of how the coordnates of a trmmed-scenaro change by addng a job belongng to R (see Appendx B), and a case by case analyss of each possble changes between ψ and ψ (see Appendx C). 3 Constructng a Transformaton Graph We construct a graph G that contans all the possble sequences of the form φ 0 φ(1) φ(2) φ(n). Ths s called a transformaton graph. For ease of our followng analyss, some of the feasble trmmed-scenaros should be deleted. Recall that 1 max{1/mld(φ), P max (φ)} < (1 + ɛ) ω + 2(1 + ɛ) c 0 s satsfed for any feasble trmmed-scenaro φ, and t may happen that two trmmed-scenaros are essentally the same. Indeed, f (1 + ɛ) ω max{1/mld(φ), P max (φ)} < (1 + ɛ) ω + 2(1 + ɛ) c 0, then by dvdng (1 + ɛ) ω from the processng tmes of each job n φ we can derve another trmmed-scenaro φ satsfyng 1 max{1/mld(φ ), P max (φ )} < 1 + 2(1 + ɛ) c0 ω, whch s also feasble. If φ s a smulatng-scenaro of ψ, then φ s called a shfted smulatng-scenaro of ψ. It s easy to verfy that the nstant approxmaton rato of a shfted smulatng scenaro s also smlar to that of the correspondng scenaro (see Appendx D). In ths case φ s deleted and we only eep φ. Let Φ Φ be the set of remanng trmmed-scenaros. We can prove that, for any real schedule represented as ψ(1) ψ(2) ψ(n), we can fnd φ 0 φ(1) φ(2) φ(n) such that φ() Φ s ether a smulatng-scenaro or a shfted smulatng-scenaro of ψ(). The reader can refer to Appendx D for a rgorous proof. Recall that when a trmmed-scenaro changes to another, the adversary only releases a job belongng to R. Let ζ = R and α 1,, α ζ be all the dstnct processng tmes n R. We show how G s constructed. We frst construct two dsjont vertex sets S 0 and A 0. For every φ Φ, there s a vertex s 0 S 0. For each s 0, there are ζ vertces of A 0 ncdent to t, namely a 0 for 1 j ζ. The node a 0 represents the release of a job of processng tme α j to the trmmed-scenaro φ. Thus, S 0 A 0 along wth the edges forms a bpartte graph. Let S 1 = {s 1 s0 S 0} be a copy of S 0. By schedulng a job of α j, f φ could be changed to φ, then there s an edge between a 0 and s1. We go on to buld up the graph by creatng an arbtrary number of copes of S 0 and A 0, namely S 1, S 2, and A 1, A 2, such that S h = {s h s0 S 0}, 5
A h = {a h a0 A 0}. Furthermore, there s an edge between s h and a h f and only f there s an edge between s 0 and a0, and an edge between ah and sh+1 f and only f there s an edge between a 0 and s1. The nfnte graph we construct above s the transformaton graph G. We let G n be the subgraph of G nduced by the vertex set ( n =0 S ) ( =0 n 1 A ). 4 Best Response Dynamcs Recall that We can vew onlne schedulng as a game between the scheduler and the adversary. Accordng to our prevous analyss, we can focus on trmmed-scenaros and assume that the adversary always releases a job wth processng tme belongng to R. By schedulng a job released by the adversary, the current trmmed-scenaro changes nto another one. We can consder the nstant approxmaton rato as the utlty of the adversary who tres to maxmze t by leadng the schedulng nto a (trmmed) scenaro. After releasng n jobs, f he s satsfed wth the current nstant approxmaton rato, then he stops and the game s called an n- stage game. Otherwse he goes on to release more jobs. The scheduler, however, tres to mnmze the compettve rato by leadng the game nto trmmed-scenaros wth small nstant approxmaton ratos. Consder any n-stage game and defne ρ n (s n ) = ρ(φ ). It mples that f the game arrves at φ eventually, then the utlty of the adversary s ρ(φ ). Notce that the adversary could release a job of processng tme 0, thus n-stage games nclude -stage games for < n. Consder a n 1. If the current trmmed-scenaro s φ and the adversary releases a job wth processng tme α j, then all the possble schedules by addng ths job to dfferent machnes could be represented by ) = {s n : sn s ncdent to an 1 }. The scheduler tres to mnmze the compettve rato, and he nows that t s the last job, thus he would choose the one wth the least nstant approxmaton rato. Thus we defne ρ n (a n 1 ) = mn{ρ n (s n ) : sn N(an 1 )}. N(a n 1 Knowng ths beforehand, the adversary chooses to release a job whch maxmzes ρ n (a n 1 Let N(s n 1 ) = {a n 1 : a n 1 s ncdent to s n 1 } and thus we defne ρ n (s n 1 ) = max{ρ n (a n 1 ) : a n 1 Iteratvely applyng the above argument, we can defne ρ n (a h 1 ρ n (s h 1 j N(s n 1 )}. ) = mn{ρ n (s n ) : sh N(ah 1 )}, ) = max{ρ n (a h 1 ) : a h 1 j N(s h 1 )}. ). The value ρ n (s h ) means that, f the current trmmed-scenaro s φ, then the largest utlty the adversary could acheve by releasng n h jobs s ρ n (s h ). Notce that we start from the empty schedule s 0 0, thus ρ n(s 0 0 ) s the largest utlty the adversary could acheve by releasng n jobs. 6
4.1 Boundng the number of stages The computaton of the utlty of the adversary reles on the number of jobs released, however, theoretcally the adversary could release as many jobs as he wants. In ths secton, we prove the followng theorem. Theorem 1 There exsts some nteger n 0 O((m + 1) Λ /ɛ), such that ρ n (s 0 ) = ρ n 0 (s 0 ) for any φ Φ and n n 0. To prove t, we start wth the followng smple lemmas. Lemma 2 For any 1 h n, ρ n (s h ) ρ n(s h 1 ). Proof. The proof s obvous by notcng that the adversary could release a job wth processng tme 0. 2 Lemma 3 For any 0 h n and Γ + 1, ρ n (s h ). Proof. The lemma clearly holds for h = n. Suppose the lemma holds for some h 1, we prove that the lemma s also true for h 1. Recall that ρ n (a h 1 ) = mn {ρ n (s n ) : sh N(ah 1 )}. We prove that ρ n (a h 1 ). To ths end, we only need to show that, we can always put α j to a certan machne so that φ s not transformed nto φ Γ+1. We apply lst schedulng when α j s released. Suppose by schedulng α j n ths way, φ s transformed nto φ Γ+1, then α j = (1 + ɛ) µ for 1 µ ω and LB = max{1/mld(φ) + α j /m, P max (φ), α j } < (1 + ɛ) ω + 2(1 + ɛ) c 0. Furthermore, suppose α j s put to a machne whose trmmed-state s τ. Then LD(τ) + α j 4(1 + ɛ) ω + 2(1 + ɛ) c 0. Now t follows drectly that LD(τ) > 3(1+ɛ) ω. Notce that we put α j to the machne wth the least load. Before α j s released, the load of every machne n φ s larger than 3(1 + ɛ) ω, whch contradcts the fact that φ s a feasble trmmed-scenaro. Therefore, applyng lst schedulng, φ can always transform to another feasble trmmedscenaro, whch ensures that ρ n (a h 1 ). Thus ρ n (s h 1 ) = max j {ρ n (a h 1 ) : a h 1 N(s h 1 )}. Lemma 4 If there exsts a number n N such that ρ n+1 (s 0 ) = ρ n(s 0 ), then for any nteger h 0, ρ n+h (s 0 ) = ρ n(s 0 ). Proof. We prove the lemma by nducton. Suppose t holds for h. We consder h + 1. Obvously ρ n+h (s n+h ) = ρ n+h+1 (s n+h+1 ) = ρ(φ ). Accordng to the computng rule, ρ n+h+1 (a n+h ) = mn {ρ n+h+1 (s n+h+1 ) : s n+h+1 N(a n+h )}, 2 Recall that s n+h+1 N(a n+h 1 ρ n+h (a n+h 1 N(a n+h ). Hence, ρ n+h+1 (a n+h ) = mn{ρ n+h (s n+h ) f and only f s 1 N(a0 ) = ρ n+h (a n+h 1 ). ) : s n+h N(a n+h 1 )}. ), and thus t s also equvalent to sn+h 7
Usng analogous arguments, we can show that ρ n+h+1 (s n+h ) = ρ n+h (s n+h 1 ). Iteratvely applyng the above procedure, we can fnally show that ρ n+h+1 (s 1 ) = ρ n+h(s 0 ). Smlarly, ρ n+h(s 1 ) = ρ n+h 1 (s 0 ). Accordng to the nducton hypothess, we now ρ n+h (s 1 ) = ρ n+h 1(s 0 ) = ρ n(s 0 ), and ρ n+h+1(s 1 ) = ρ n+h (s 0 ) = ρ n(s 0 ). Meanwhle ρ n+h (a 0 ) = mn{ρ n+h (s 1 ) : s1 N(a0 )} = mn{ρ n (s 0 ) : s1 N(a0 )}, ρ n+h+1 (a 0 ) = mn{ρ n+h+1 (s 1 ) : s1 N(a0 )} = mn{ρ n (s 0 ) : s1 N(a0 )}. Thus t mmedately follows that ρ n+h (a 0 ) = ρ n+h+1(a 0 ). Furthermore, ρ n+h+1 (s 0 ) = max{ρ n+h+1 (a 0 ) : a 0 N(s 0 )} j = max{ρ n+h (a 0 ) : a 0 N(s 0 )} = ρ n+h (s 0 ). j The lemma holds for h + 1. 2 Now we arrve at the proof of Theorem 1. Defne Z(n) = φ Φ \{φ Γ+1 } ρ n(s 0 ) as the potental functon. Accordng to the prevous lemmas, Z(n + 1) Z(n), and f Z(n 0 + 1) = Z(n 0 ), then Z(n) = Z(n 0 ) for any n n 0. Furthermore, f Z(n+1) > Z(n), then Z(n+1) Z(n) ɛ. Suppose Z(n + 1) > Z(n), then t follows drectly that Z(n + 1) > Z(n) > > Z(1). Recall that Z(1) 0 and Z(n + 1) 20( Φ 1) O((m + 1) Λ ), thus n + 1 O((m + 1) Λ /ɛ). Furthermore, t can be easly verfed that f Z(n + 1) = Z(n), then ρ n+1 (s 0 ) = ρ n(s 0 ) for any φ Φ. Thus, by settng n 0 = O((m + 1) Λ /ɛ), Theorem 1 follows. Let n 0 be the smallest nteger satsfyng Theorem 1. Let ρ = ρ n0 (s 0 0 ), and ρ(s0 ) = ρ n 0 (s 0 ). Now t s not dffcult to see that, the optmal onlne algorthm for P C max has a compettve rato around ρ. A rgorous proof of such an observaton depends on the followng two facts. 1. Gven any onlne algorthm, there exsts a lst of at most n 0 jobs such that by schedulng them, ts compettve rato exceeds ρ O(ɛ). 2. There exsts an onlne algorthm whose compettve rato s at most ρ + O(ɛ). The frst fact could be proved va G n0, where ρ = ρ n0 (s 0 0 ) ensures that n 0 jobs are enough to acheve the lower bound. The readers may refer to Appendx E.1 for detals. The second observaton could be proved va G n0 +1, where ρ n0 +1(s 0 ) = ρ n 0 +1(s 1 ) = ρ(s0 ) for every φ. Each tme a job s released, the scheduler may assume that he s at the vertex s 0 where ρ n0 +1(s 0 ) ρ, and fnd a feasble schedule by leadng the game nto s 1 where ρ n 0 +1(s 0 ) = ρ n 0 +1(s 1 ) ρ. After schedulng the job he may stll assume that he s at s 0. The readers may refer to Appendx E.2 for detals. Usng the framewor we derve, compettve schemes could be constructed for a varety of onlne schedulng problems, ncludng Rm C max and Rm Cp for constant p. Addtonally, f we restrct that the processng tme of each job s bounded by q, then an optmal onlne algorthm for P p j q C max could be derved (n (mq) O(mq) tme). The readers may refer to Appendx F for detals. 8
5 Concludng Remars We provde a new framewor for the onlne over lst schedulng problems. We remar that, through such a framewor, nearly optmal algorthms could also be derved for other onlne problems, ncludng the -server problem (despte that the runnng tme s rather huge, whch s exponental). As nearly optmal algorthms could be derved for varous onlne problems, t becomes a very nterestng and challengng problem to consder the hardness of dervng optmal onlne algorthms. Is there some complexty doman such that fndng an optmal onlne algorthm s hard n some sense? For example, gven a constant ρ, consder the problem of determnng whether there exsts an onlne algorthm for P C max whose compettve rato s at most ρ. Could t be answered n tme f(m, ρ) for any gven functon f? We expect the frst exctng results along ths lne, that would open the onlne area at a new stage. References [1] S. Albers. Better bounds for onlne schedulng. SIAM Journal on Computng, 29:459 473, 1999. [2] S. Albers and M. Hellwg. Sem-onlne schedulng revsted. Theoretcal Computer Scence, 443:1 9, 2012. [3] J. Aspnes, Y. Azar, A. Fat, S. Plotn, and O. Waarts. On-lne routng of vrtual crcuts wth applcatons to load balancng and machne schedulng. Journal of the ACM, 44(3):486 504, 1997. [4] A. Avdor, Y. Azar, and J. Sgall. Ancent and new algorthms for load balancng n the l p norm. Algorthmca, 29:422 441, 2001. [5] B. Awerbuch, Y. Azar, E. Grove, M. Kao, P. Krshnan, and J. Vtter. Load balancng n the l p norm. In Proceedngs of the 36th Annual Symposum on Foundatons of Computer Scence (FOCS), pages 383 391, 1995. [6] Y. Azar, J. Naor, and R. Rom. The compettveness of on-lne assgnments. In Proceedngs of the 3rdAnnual ACM-SIAM Symposum on Dscrete Algorthms (SODA), pages 203 210, 1992. [7] Y. Bartal, A. Fat, H. Karloff, and R. Vohra. New algorthms for an ancent schedulng problem. Journal of Computer and System Scences, 51:359 366, 1995. [8] Y. Bartal, H. Karloff, and Raban Y. A better lower bound for on-lne schedulng. Informaton Processng Letters, 50:113 116, 1994. [9] P. Berman, M. Charar, and M. Karpns. On-lne load balancng for related machnes. Journal of Algorthms, 35:108 121, 2000. [10] B. Chen, A. van Vlet, and G.J. Woegnger. New lower and upper bounds for on-lne schedulng. Operatons Research Letters, 16:221 230, 1994. 9
[11] T. Cheng, H. Kellerer, and V. Kotov. Sem-on-lne multprocessor schedulng wth gven total processng tme. Theoretcal computer scence, 337(1):134 146, 2005. [12] Y. Cho and S. Sahn. Bounds for lst schedules on unform processors. SIAM Journal on Computng, 9(1):91 103, 1980. [13] T. Ebenlendr and J. Sgall. A lower bound on determnstc onlne algorthms for schedulng on related machnes wthout preempton. In Proceedngs of the 9th Worshop on Approxmaton and Onlne Algorthms (WAOA), pages 102 108, 2012. [14] U. Fagle, W. Kern, and G. Turán. On the performane of onlne algorthms for partton problems. Acta Cybernet, 9:107 119, 1989. [15] R. Flescher and M. Wahl. On-lne schedulng revsted. Journal of Schedulng, 3:343 353, 2000. [16] G. Galambos and G. Woegnger. An on-lne schedulng heurstc wth better worst case rato than Graham s lst schedulng. SIAM Journal on Computng, 22:349 355, 1993. [17] T. Gormley, N. Rengold, E. Torng, and J. Westbroo. Generatng adversares for requestanswer games. In Proceedngs of the 11th Annual ACM-SIAM symposum on Dscrete algorthms (SODA), pages 564 565, 2000. [18] R. L. Graham. Bounds for certan multprocessng anomales. Bell System Techncal Journal, 45:1563 1581, 1966. [19] E. Günther, O. Maurer, N. Megow, and A. Wese. A new approach to onlne schedulng: Approxmatng the optmal compettve rato. In Proceedngs of the 24th Annual ACM-SIAM Symposum on Dscrete Algorthms (SODA), 2013. To appear. [20] D. Karger, S.J. Phllps, and E. Torng. A better algorthm for an ancent schedulng problem. In Proceedngs of the 5thAnnual ACM-SIAM Symposum on Dscrete Algorthms (SODA), pages 132 140, 1994. [21] J. F. Rudn III. Improved Bound for the Onlne Schedulng Problem. PhD thess, The Unversty of Texas at Dallas, 2001. [22] J.F. Rudn III and R. Chandrasearan. Improved bound for the onlne schedulng problem. SIAM Journal on Computng, 32:717 735, 2003. [23] J. Sgall. On-lne schedulng. In Amos Fat and Gerhard J. Woegnger, edtors, Onlne Algorthms: The State of the Art, pages 196 231. Sprnger, 1998. 10
A Addng a new job to a scenaro Before we show how a scenaro s changed by addng a new job, we frst show how a scenaro s changed when we scale ts jobs usng a new factor T SC and T > T LB. A.1 Re-computaton of a scenaro Let (T LB, ψ) be a real schedule at any tme where ψ = (st 1, st 2,, st m ). If we choose T > T LB to scale jobs, then a bg job prevously may become a small job (.e., no greater than T (1 + ɛ) c 0 ). Suppose T = T LB (1 + ɛ) ω, then a job wth processng tme T LB (1 + ɛ) j s denoted as T (1 + ɛ) j ω now, hence a state st = (η c0,, η ω ) of ψ becomes ŝt = (ˆη c0,, ˆη ω ) where ˆη = η +ω for > c 0 (we let η = 0 for > ω), and ˆη c0 = ω c0 = c 0 T LB (1 + ɛ) η T (1 + ɛ) c 0 = ω c0 = c 0 (1 + ɛ) η (1 + ɛ) ω c. 0 The above computaton could be vewed as shftng the state leftwards by ω bts, and we defne a functon f to represent t such that f (st) = ŝt. Smlarly the scenaro ψ changes to ˆψ = (f (st 1,, f (st m )) and we denote f (ψ) = ˆψ. A.2 Addng a new job Agan, let (T LB, ψ) be a real schedule at any tme where ψ = (st 1, st 2,, st m ). Suppose a new job p n s released and scheduled on machne h where st h = (η c0, η c0 +1,, η ω ), and furthermore, ψ changes to ψ. We determne the coordnates of ψ n the followng. Consder p n. If p n T LB (1+ɛ) ω then we defne the addton st h +p n /T LB = st h n the followng way where st h = ( η c0,, η ω ). If p n /T = (1 + ɛ) µ for c 0 + 1 µ ω, then η µ = η µ + 1 and η j = η j for j µ. If p n /T (1 + ɛ) c 0, then η c0 = η c0 + p n /(T LB (1 + ɛ) c 0 ) and η j = η j for j c 0. Let ψ = (st 1,, st h 1, st h, st h+1,, st m ) be a temporal result. If ψ s feasble, whch mples that max{ld( ψ)/m, P max ( ψ)} [1, (1 + ɛ) ω ), then ψ = ψ. Otherwse ψ s nfeasble and there are two possbltes. Case 1. max{1/mld( ψ), P max ( ψ)} (1 + ɛ) ω. It s not dffcult to verfy that max{1/mld( ψ), P max ( ψ)} < (1 + ɛ) 2ω, thus f 1 ( ψ) s feasble and we wrte ψ = f 1 ( ψ). Case 2. 1 max{1/mld( ψ), P max ( ψ)} < (1 + ɛ) ω whle LD( st h ) > 4(1 + ɛ) ω,.e., st h s an nfeasble state. In ths case the compettve rato of the onlne algorthm becomes larger than 2. Thus job p n s never added to st h f t s scheduled accordng to an onlne algorthm wth compettve rato no greater than 2. Otherwse, (1 + ɛ) ω p n /T LB < (1 + ɛ) (+1)ω for some 1. It s easy to verfy that, by addng p n to the schedule, the scalng factor becomes T LB (1+ɛ) ω. Thus ψ = (st 1,, st m) where st j = f (st j ) for j h, and st h = f (st h ) + p n /(T LB (1 + ɛ) ω ). 11
B Addng a new job to a trmmed-scenaro Notce that a trmmed-scenaro could also be vewed as a scenaro, thus addng a new job to t could be vewed as addng a new job to a scenaro, and then roundng up the coordnates of the resulted scenaro to ntegers. Specfcally, we restrct the processng tme of the job added s ether 0 or (1 + ɛ) µ for µ c 0. We wll show later that t s possble to put an upper bound on the processng tmes. B.1 Re-computaton of a trmmed-scenaro To re-compute a trmmed-scenaro φ, we tae φ as a scenaro wth scalng factor T LB = 1. Suppose we want to use a new factor (1 + ɛ) ω to scale jobs, then each trmmed-state of φ, say τ, s recomputed as f 1 (τ). Notce that ts frst coordnate may be fractonal, we round t up and let g 1 (τ) = f 1 (τ) where v for a vector means we round each coordnate v of v to v. We defne g teratvely as g (τ) = g 1 (g 1 (τ)). Notce that f τ s feasble (.e., LD(τ) 4(1 + ɛ) ω + 2(1 + ɛ) c 0 ), then g (τ) s feasble for any 1. Thus, we defne g (φ) = φ = (ξ 1, ξ 2,, ξ Λ ) where ξ j = h:g (τ h )=τ j ξ h. Specfcally, f {h : g (τ h ) = τ j } =, then ξ j = 0. We have the followng lemma. Lemma 5 For any nteger 0, feasble state st h and feasble trmmed-state τ, the followng holds: (1 + ɛ) ω LD(f (st h )) = LD(st h ), LD(τ) (1 + ɛ) ω LD(g (τ)) LD(τ) + The proof s smple through nducton. B.2 Addng a new job (1 + ɛ) c0+ω LD(τ) + 2(1 + ɛ) c0+ω. =1 Suppose the feasble trmmed-scenaro φ becomes φ by addng a new job p n = (1 + ɛ) µ, and furthermore, the job s added to a machne whose trmmed-state s τ j. We show how the coordnates of φ s determned. There are two possbltes. Case 1. If c 0 µ ω, then by addng a new job p n = (1 + ɛ) µ to a feasble trmmed-state τ j, we smply tae τ j as a state and compute τ j = τ j + p n accordng to the rule of addng a job to states. Consder the m trmmed-states of φ, we replace τ j wth τ j whle eepng others ntact. By dong so a temporal trmmed-scenaro φ s generated and we compute LB( φ) = max{1/mld(φ) + p n /m, P max (φ), p n }. There are three possbltes. Case 1.1 LB( φ) < (1+ɛ) ω +2(1+ɛ) c 0 and LD( τ j ) < 4(1+ɛ) ω +2(1+ɛ) c 0. Then τ j s a feasble trmmed-state and suppose τ j = τ j. Then φ = φ,.e., φ = (ξ 1, ξ 2,, ξ Λ ) where ξ j = ξ j 1, ξ j = ξ j + 1 and ξ l = ξ l for l j, j. Case 1.2 LB( φ) < (1 + ɛ) ω and LD( τ j ) 4(1 + ɛ) ω + 2(1 + ɛ) c 0. Then τ j s nfeasble and φ = φ Γ+1. 12
Case 1.3 LB( φ) (1 + ɛ) ω. It can be easly verfed that LB( φ) < (1 + ɛ) 2ω. Notce that g 1 ( τ j ) s always feasble, thus φ = g 1 ( φ),.e., for each trmmed-state τ of φ, we compute g 1 (τ). Snce g 1 (τ) s always feasble, they made up of a feasble trmmed-scenaro φ. Remar. There mght be ntersecton between Case 1 and Case 3. Indeed, f (1 + ɛ) ω LB( φ) < (1 + ɛ) ω + 2(1 + ɛ) c 0, and τ s feasble, then by addng p n the trmmed-scenaro φ changes nto φ = φ accordng to Case 1 and g 1 (φ ) accordng to Case 3. Here both φ and g 1 (φ ) are feasble trmmed-scenaros. Ths s the only case that φ + p n may yeld two dfferent solutons. In the next secton we wll remove φ f both φ and g 1 (φ) are feasble. By dong so φ+p n yelds a unque soluton, but currently we just eep both of them so that Theorem 2 could be proved. Case 2. If (1 + ɛ) ω µ < (1 + ɛ) (+1)ω then agan we tae τ j as a state and compute τ j = g (τ j ) + p n /(1 + ɛ) ω. We re-compute φ as g (φ) = (ˆξ 1, ˆξ 2,, ˆξ Λ ). Then we replace one trmmed-state g (τ j ) wth τ j and ths generates φ. It s easy to verfy that φ s feasble. Remar 2. Notce that the number of possble processng tmes of job p n could be nfnte, however, we show that t s possble to further restrct t to be some constant. Let p n = (1 + ɛ) µ. Let µ 0 be the smallest nteger such that (1 + ɛ) µ 0 4(1 + ɛ) ω+c0+1. If µ = ω + l wth µ 0 /ω and 0 l ω 1, then φ s re-computed as g (φ). Notce that for any feasble trmmed-state τ, LD(τ) 4(1+ɛ) ω +2(1+ɛ) c 0 < 4(1+ɛ) ω+1, thus LD(g (τ)) (1+ɛ) c 0, whch mples that g (τ) = (0, 0,, 0) f τ = (0, 0,, 0) and g (τ) = (1, 0, 0,, 0) otherwse. Thus, g (φ) = g µ0 /ω (φ). The above analyss shows that by addng a job wth processng tme p n = (1 + ɛ) ω+l for µ 0 /ω and 0 l ω 1 to any feasble trmmed-scenaro φ s equvalent to addng a job wth processng tme p n = (1 + ɛ) µ 0/ω ω+l to φ. Thus, when addng a job to a trmmed-scenaro, we may restrct that p n R = {0, (1 + ɛ) c 0,, (1 + ɛ) µ 0/ω +ω 1 }. C Smulatng transformatons between scenaros The whole secton s devoted to prove the followng theorem. Theorem 2 Let φ be the smulatng-scenaro of a feasble scenaro ψ. If accordng to some onlne algorthm (T, ψ) changes to (T, ψ) by addng a job p n 0, then φ could be transformed to φ ( φ φ 0, φ Γ ) by addng a job p n R = {0, (1 + ɛ) c 0,, (1 + ɛ) µ 0/ω +ω 1 } such that φ s a smulatng-scenaro of ψ. Let τ θ(h) n φ be the smulatng-state of st h n ψ. Before we gve the proof, we frst present a lemma that would be used later. Lemma 6 Let φ be a smulatng-scenaro of ψ. For any 1, f f (st h ) = (η c 0, η c 0 +1,, η ω) and g (τ θ(h) ) = (ν c 0, ν c 0 +1,, ν ω), then ν = η for > c 0 and η c 0 ν c 0 η c 0 + 2. Proof. Let st h = (η c0, η c0 +1,, η ω ) and τ θ(h) = (ν c0, ν c0 +1,, ν ω ). We frst prove the lemma for = 1. 13
It s easy to verfy that ν = η for > c 0. Furthermore, ω c0 ν c = c 0 = 0 (1 + ɛ) ν (1 + ɛ) ω c 0 ω c0 = c 0 +1 (1 + ɛ) η + (1 + ɛ) c 0 (η c0 + 2) (1 + ɛ) ω c + 1 0 η c 0 + 1 + 2(1 + ɛ) ω < η c 0 + 2 Thus the lemma holds for = 1. If the lemma holds for = 0, then t also holds for = 0 + 1. The proof s the same. 2 Now we come to the proof of Theorem 2. Proof. Let ψ = (st 1, st 2,, st m ) and φ = (ξ 1, ξ 2,, ξ Λ ). Recall that τ θ() s the smulatng-state of st n φ. Notce that LD(st ) LD(τ θ() ) LD(st ) + 2(1 + ɛ) c 0, t follows that 1/mLD(ψ) 1/mLD(φ) 1/mLD(ψ)+2(1+ɛ) c 0. Meanwhle, P max (ψ) = P max (φ) as long as ψ (0, 0,, 0). Suppose job n s assgned to machne h n the real schedule. Let st h = (η c0,, η ω ) and τ θ(h) = (ν c0,, ν ω ). Recall that η c0 ν c0 η c0 + 2 and η = ν for > c 0. There are two possbltes. Case 1. p n /T (1 + ɛ) ω. Let st h = st h + p n /T = (η c 0,, η ω). We defne p n n the followng way. If p n /T = (1 + ɛ) µ for c 0 + 1 µ ω, then p n = p n /T. If p n /T (1 + ɛ) c 0, η c 0 ν c0, then p n = 0. η c 0 > ν c0, then p n = (1 + ɛ) c 0. Let τ θ(h) + p n = (ν c 0,, ν ω), then ν c 0 = ν c0 + p n/((1 + ɛ) c 0 ), n both cases η c 0 ν c 0 η c 0 + 2. By addng p n to ψ, the scalng factor may or may not be changed. If T = T, the state of machne h n ψ s st h. We consder τ ζ(h) + p n. Snce LD(τ ζ(h) + p n) LD(st h ) LD(τ ζ(h)) LD(st h ) 2(1+ɛ) c 0, and st h s a feasble state, τ ζ(h) +p n s also a feasble trmmed state. Meanwhle max{1/mld(ψ ), P max (ψ )} < (1 + ɛ) ω, thus by addng p n to φ, the scalng factor of the trmmed-scenaro s also not updated, whch mples that the trmmed-state of machne h n φ s τ ζ(h) + p n. It can be easly verfed that n ths case, φ s the smulatng-scenaro of ψ. Otherwse T > T and the state of machne h s f 1 (st h ) n ψ. We compute LB = max{1/mld(φ)+ p n/m, P max (φ), p n}. Snce LB = max{1/mld(ψ) + p n /m, P max (ψ), p n } > (1 + ω) ω, t follows drectly that LB > (1 + ω) ω. Meanwhle LB < (1 + ω) 2ω, thus the trmmed-state of machne h n φ s g 1 (τ ζ(h) + p n). 14
We compare st h and τ ζ(h) +p n = (ν c 0,, ν ω). Obvously η l = ν l for l c 0 and η c 0 ν c 0 η c 0 + 2. Accordng to Lemma 6, g 1 (τ ζ(h) + p n) s a smulatng-state of f 1 (st h ), whch mples that φ s a smulatng-scenaro of ψ. Remar. Recall that when (1 + ɛ) ω LB < (1 + ɛ) ω + 2(1 + ɛ) c 0, φ + p n may yeld two solutons ˆφ and g 1 ( ˆφ), as we have clamed. Our above dscusson chooses ˆφ f the scalng factor of the real schedule does not change, and chooses g 1 ( ˆφ) when the the scalng factor of the real schedule changes. Case 2. For some 1, (1 + ɛ) ω p n /T < (1 + ɛ) (+1)ω. Then we defne p n = p n /T at frst. Let f (st h ) = (η c 0,, η ω), g (τ ζ(h) ) = (ν c 0,, ν ω), then accordng to Lemma 6 we have η = ν for c 0 < ω and η c 0 ν c 0 η c 0 + 2. Then t follows drectly that g (τ ζ(h) ) + p n s a smulatng-state of f (st h ) + p n. Thus, by addng p n, φ s a smulatng-scenaro of ψ. Furthermore, f p n > (1 + ɛ) µ 0/ω +ω 1, then suppose p n = (1 + ɛ) ω+l for some µ 0 /ω and 0 l ω 1. Due to our prevous analyss, p n could be replaced by a job wth processng tme p n = (1 + ɛ) µ 0/ω +l. The trmmed-scenaro φ stll transforms nto φ by addng p n. 2 D Deleton of equvalent trmmed-scenaros Recall that the addton φ + p n may yeld two solutons, φ and g 1 (φ ) where both of them are feasble. To mae the result unque, φ s deleted from Φ f g 1 (φ ) s feasble and Φ s the set of the remanng trmmed-scenaros. We have the followng smple lemma. Lemma 7 If φ and g 1 (φ) are both feasble trmmed-scenaros, then ρ(φ) ρ(g 1 (φ)) O(ɛ). Wth fewer trmmed-scenaros, Theorem 2 may not hold, however, we have the followng lemma. Lemma 8 Suppose by releasng job n wth p n R and schedulng t onto a certan machne, the feasble trmmed-scenaro φ changes to ˆφ. Furthermore, g 1 (φ) s also feasble. Then there exsts p n R such that by schedulng t on the same machne, g 1 (φ) changes to φ and furthermore, ether φ = ˆφ or φ = g 1 ( ˆφ). Proof. Suppose job n s scheduled onto a machne of trmmed-state τ = (ν c0,, ν ω ) n φ, then we put p n onto a machne of trmmed-state g 1 (τ) = (ν c 0,, ν ω) n g 1 (φ). If p n = 0 then obvously we can choose p n = 0. Otherwse let p n = (1 + ɛ) µ and there are three possbltes. Case 1. µ ω c 0. ω c0 If by addng p n, the scalng factor of φ does not change, then we compare ν c 0 = ω c0 = c (1+ɛ) ν +(1+ɛ) µ 0 ν (1+ɛ) ω c 0 c 0 + 1. If ν c 0 = y, then p n = 0. Otherwse y = ν c 0 + 1, wth y = = c 0 (1+ɛ) ν (1+ɛ) ω c 0 then p n = (1 + ɛ) c 0. It can be easly verfed that g 1 (τ) + p n = g 1 (τ + p n ) and g 1 ( ˆφ) = g 1 (φ) + p n. Otherwse by addng p n the scalng factor of φ ncreases, then we defne p n n the same way and t can be easly verfed that ˆφ = g 1 (φ) + p n. Case 2. ω c 0 < µ 2ω. 15
In ths case we defne p n = (1 + ω) µ ω and the proof s smlar to the prevous case. Notce that n both case 1 and case 2, p n (1 + ω) ω. As LD(g 1 (τ)) 4 + 2(1 + ɛ) c 0 ω, LD(g 1 (τ)) + p n 4(1 + ɛ) ω, thus we can add p n to g 1 (τ) drectly (wthout changng the scalng factor). Furthermore, max{1/m[ld(g 1 (φ)) + p n], P max (g 1 (g 1 (φ))), p n} (1 + ɛ) ω, thus by addng p n to g 1 (φ), the scalng factor does not change, thus n both cases, φ = g 1 (φ) + p n. Case 3. µ > 2ω. Suppose µ = ω + l wth 2 and 0 l ω 1. Then p n = (1 + ɛ) µ ω. Accordng to the defnton of g, g (φ) = g 1 (g 1 (φ)), thus φ = ˆφ. 2 Combnng Theorem 2 and Lemma 8, we have the followng theorem. Theorem 3 Let φ Φ be the smulatng-scenaro or shfted smulatng-scenaro of a feasble scenaro ψ. If accordng to some onlne algorthm (T, ψ) changes to (T, ψ) by addng a job p n 0, then φ could be transformed to φ Φ ( φ φ 0, φ Γ ) by addng a job p n R = {0, (1+ɛ) c 0,, (1+ ɛ) µ 0/ω +ω 1 } such that φ s a smulatng-scenaro or shfted smulatng-scenaro of ψ. E The nearly optmal strateges for the adversary and the scheduler E.1 The nearly optmal strategy for the adversary We prove n ths subsecton that, by releasng at most n 0 jobs, the adversary can ensure that there s no onlne algorthm whose compettve rato s less than ρ O(ɛ). We play the part of the adversary. Consder G n0. Notce that ρ = ρ n0 (s 0 0 ) = max j{ρ n0 (a 0 0,j ) : a0 0,j N(s0 0 )}, thus there exsts some j 0 such that a 0 0,j 0 N(s 0 0 ) and ρ n 0 (a 0 0,j 0 ) = ρ. We release a job wth processng tme α j0. Suppose due to any onlne algorthm whose compettve rato s no greater than 2, ths job s scheduled onto a certan machne so that the scenaro becomes ψ, then accordng to Theorem 2 and the constructon of the graph, there exsts some s 1 ncdent to a 0 0,j 0 such that ether φ s a smulatng-scenaro of ψ, or φ s a shfted smulatngscenaro of ψ. As ρ n0 (a 0 0,j 0 ) = mn {ρ n0 (s 1 ) : s1 N(a0 0,j 0 )}, t follows drectly that ρ n0 (s 1 ) ρ. If ρ(φ ) = ρ n0 (s 1 ) ρ, then we stop and t can be easly seen that the nstant approxmaton rato of ψ s at least ρ O(ɛ) (by Lemma 1). Otherwse we go on to release jobs. Suppose after releasng h 1 jobs the current scenaro s ψ and φ s ts smulatng-scenaro or shfted smulatng-scenaro, furthermore, ρ n0 (s h 1 ) ρ. As ρ ρ n0 (s h 1 ) = max j {ρ n0 (a h 1 ) : a 0 N(sh 1 )}, thus there exsts some j 0 such that a h 1 0 N(s h 1 ) and ρ n0 (a 0 0 ) ρ. We release the h-th job wth processng tme α j0. Agan suppose ths job s scheduled onto a certan machne so that the scenaro becomes ψ, then there exsts some s h ncdent to ah 1 0 such that φ s ether a smulatng-scenaro or a shfted smulatng-scenaro of ψ. As ρ n0 (a h 1 0 ) = mn {ρ n0 (s h ) : sh N(a0 0 )}, t follows drectly that ρ n0 (s h ) ρ. If ρ(φ ) = ρ n0 (s h ) ρ, then we stop and t can be easly seen that the nstant approxmaton rato of ψ s at least ρ O(ɛ). Otherwse we go on to release jobs. Snce ρ(φ ) = ρ n0 (s n 0 ), we stop after releasng at most n 0 jobs. 16
E.2 The nearly optmal onlne algorthm We play the part of the scheduler. Notce that ρ n0 +1(a 0 ) = mn{ρ n0 +1(s 1 ) : s1 N(a0 )} = mn{ρ(s 1 ) : s1 N(a0 )}, ρ(s 0 ) = ρ n0 +1(s 0 ) = max{ρ n0 +1(a 0 ) : a 0 N(s 0 )}. j Suppose the current scenaro s ψ wth scalng factor T. Let φ Φ be ts smulatng-scenaro or shfted smulatng-scenaro, and furthermore, ρ(s 0 ) ρ. Let p n be the next job the adversary releases. We apply lazy schedulng frst,.e., f by schedulng p n onto any machne, ψ changes to ψ (the scalng factor does not change) whle φ s stll a smulatng-scenaro or shfted smulatng-scenaro of ψ, we always schedule p n onto ths machne. Otherwse, Accordng to Theorem 2 and Lemma 8, p n(h) could be constructed such that f ψ changes to ψ by addng p n to machne h, then φ changes to φ by addng p n to the same machne such that φ s a smulatng-scenaro or shfted smulatng-scenaro of ψ. Notce that the processng tme of p n(h) may also depend on the machne h. We show that, f p n could not be scheduled due to lazy schedulng, then p n(h) = p n for every h. To see why, we chec the proofs of Theorem 2 and Lemma 8. We observe that, f p n(h) (1+ɛ) c 0+1 for some h, then p n(h) = p n for every h (the processng tme p n(h) only depends on p n /T ). Otherwse, t mght be possble that p n(h 1 ) = 0 for some h 1 whle p n(h 2 ) = (1 + ɛ) c 0 for another h 2. However, f ths s the case then p n should be scheduled on machne h 1 accordng to lazy schedulng, whch s a contradcton. Thus, p n(h) = (1 + ɛ) c 0 for every h. Now we decde accordng to G n0 +1 whch machne p n should be put onto. As p n R, let α j0 = p n, then we consder ρ n0 +1(a 0 0 ) = mn {ρ n0 +1(s 1 ) : s1 N(a0 0 )}. Recall that ρ(s 0 ) ρ accordng to the hypothess, then ρ n0 +1(a 0 0 ) ρ, whch mples that there exsts some s 1 0 ncdent to a 0 such that ρ n0 +1(s 1 0 ) = ρ(s 0 0 ) ρ. Thus, we can schedule p n to a certan machne, say, machne h 0, so that φ transforms to φ 0. And thus n the real schedule we schedule p n onto machne h 0. Let ψ be the current scenaro, then φ 0 s ts smulatng-scenaro or shfted smulatng-scenaro wth ρ(s 0 0 ) ρ. Thus, we can always carry on the above procedure. Snce the nstant approxmaton rato of each smulatng-scenaro or shfted smulatng-scenaro s no greater than ρ, the nstant approxmaton rato of the correspondng scenaro s also no greater than ρ + O(ɛ). F Extensons We show n ths secton that our method could be extended to provde approxmaton schemes for varous problems. Specfcally, we consder Rm C max, Rm h Cp h for some constant p 1 (and as a consequence Qm C max, Qm h Cp h and P m h Cp h could also be solved). We menton that, f we restrct that the number of machnes m s a constant (as n the case Rm C max and Rm h Cp h ), then our method could be smplfed. We also consder the sem-onlne model P p j q C max where the processng tme of each job released s at most q. In ths case an optmal algorthm could be derved n (mq) O(mq) tme. Notce that our prevous dscussons focus on fndng nearly optmal onlne algorthms, however, for 17
onlne problems, we do not now much about optmal algorthms. Only the specal cases P 2 C max and P 3 C max are nown to admt optmal algorthms. Unle the correspondng offlne problems whch always admt exact algorthms (sometmes wth exponental runnng tmes), we do not now whether there exsts such an algorthm for onlne problems. Consder the followng problem, does there exst an algorthm whch determnes whether there exsts an onlne algorthm for P C max whose compettve rato s no greater than ρ. We do not now whch complexty class ths problem belongs to. An exact algorthm, even wth runnng tme exponental n the nput sze, would be of great nterest. Related wor. For the objectve of mnmzng the maespan on related and unrelated machnes, the best nown results are n table 1. There s a huge gap between the upper bound and lower bound except for the specal case Q 2 C max. However, the standard technque for Q 2 C max becomes extremely complcated and can hardly be extended for 3 or more machnes. For the objectve of ( h Cp h )1/p,.e., the L p norm, not much s nown. See table 1 for an overvew. We further menton that when p = 2, Lst Schedulng s of compettve rato 4/3 [4]. Table 1: Lower and upper bounds on the compettve rato for determnstc problems lower bounds upper bounds Q C max 2.564 [13] 5.828 [9] Q2 C max (2s + 1)/(s + 1) for s 1.61803 (2s + 1)/(s + 1) for s 1.61803, 1 + 1/s for s 1.61803 [12] 1 + 1/s for s 1.61803 [12] R C max Ω(log m) [6] O(log m) [3] P ( h Cp h )1/p 2 Θ(ln p/p) [4] R ( h Cp h )1/p O(p) [5] Much of the prevous wor s drected for sem-onlne models of schedulng problems where part of the future nformaton s nown beforehand, and most of them assume that the total processng tme of jobs (nstead of the largest job) s nown. For such a model, the best nown upper bound s 1.6 [11] and the best nown lower bound s 1.585 [2]. F.1 Rm C max In ths case, we can restrct beforehand that the processng tme of each job, say, j, on machne h (1 h m) s p jh {(1 + ɛ) : 0, N}. There s a nave algorthm Al 0 that puts every job on the machne wth the least processng tme, and t can be easly seen that the compettve rato of ths algorthm s m. Snce m s a constant, t s a constant compettve rato onlne algorthm, and thus we may restrct on the algorthms whose compettve rato s no greater than m. Gven any real schedule, we may frst compute the maespan of the schedule by applyng Al 0 on the nstance and let t be Al 0 (C max ), then we defne LB = Al 0 (C max )/m and fnd a scalng factor T SC such that T LB < T (1 + ɛ) ω. Smlarly as we do n the prevous sectons, we can then defne a state for each machne of the real schedule wth respect to T and then a scenaro by combnng the m states. Snce OP T mt (1 + ɛ) ω, f the real schedule s produced by an onlne algorthm whose compettve rato s no greater than m, then the load of each machne s bounded by m 2 T (1 + ɛ) ω, and ths allows us to bound the number of dfferent feasble states by 18
some constant, and the number of all dfferent feasble scenaros s also bounded by a constant (dependng on m and 1/ɛ). We can then defne trmmed-states and trmmed-scenaros n the same way as before. Specfcally, a trmmed-state s combned of m trmmed-states drectly (t s much smpler snce the number of machnes s a constant). Agan, a feasble trmmed-state s a trmmed-state whose load could be slghtly larger than m 2 T (1 + ɛ) ω (to nclude two addtonal small jobs), and a feasble trmmed-scenaro s a trmmed-scenaro such that every trmmed-state s feasble. Transformatons between scenaros and trmmed-scenaros are exactly the same as before and we can also construct a graph to characterze the transformatons between trmmed-scenaros, and use t to approxmately characterze the transformaton between scenaros. All the subsequent arguments are the same. F.2 Rm h Cp h when p 1 s a constant Here C h denotes the load of machne h. Agan we can restrct beforehand that the processng tme of each job, say, j, on machne h (1 h m) s p jh {(1 + ɛ) : 0, N}. Consder the nave algorthm Al 0 that puts every job on the machne wth the least processng tme and let C h (Al 0 ) be the load of machne h due to ths algorthm. Snce x p m h=1 s a convex functon, we now drectly that OP T m( C h(al 0 ) ) p m h=1 C h(al 0 ) p m and thus the compettve rato of Al 0 s also m and agan we may restrct on the algorthms whose compettve rato s no greater than m. Gven any real schedule, we may frst compute the objectve functon of the schedule by applyng Al 0 on the nstance and let t be Al 0 ( h Cp h ), then we defne LB = [Al 0( h Cp h )/m]1/p and fnd a scalng factor T SC such that T LB < T (1 + ɛ) ω. Consder any schedule produced by an onlne algorthm whose compettve rato s no greater than m, then ts objectve value should be bounded by mal 0 ( h Cp h ), whch mples that the load of each machne n ths schedule s bounded by [mal 0 ( h Cp h )]1/p = m 2/p LB. Agan usng the fact that m s a constant, we can then defne a state for each machne of the real schedule wth respect to T and then a scenaro by combnng the m states. Trmmed-states and trmmed-scenaros are defned smlarly, all the subsequent arguments are the same as the prevous subsecton. Remar. Our method, however, could not be extended n a drect way to solve the more general model Rm h f(c h) f the functon f fals to satsfy the property that f(a)/f(b) = f(a)/f(b) for any > 0. Ths s because we neglect the scalng factor when we construct the graph G and compute the nstant approxmaton rato for each trmmed-scenaro. Indeed, the nstant approxmaton rato s not dependent on the scalng factor for all the objectve functons (.e., C max and h Cp h ) we consder before, however, f such a property s not satsfed, then the nstant approxmaton rato depends on the scalng factor and our method fals. m F.3 P p j q C max We show n ths subsecton that, the sem-onlne schedulng problem P p j q C max n whch the largest job s bounded by some nteger ζ (the value q s nown beforehand), admts an exact onlne algorthm. Agan we use the prevous framewor to solve ths problem. The ey observaton s that, n 19