A Better Algorithm For an Ancient Scheduling Problem. David R. Karger Steven J. Phillips Eric Torng. Department of Computer Science

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1 A Better Algorith For an Ancient Scheduling Proble David R. Karger Steven J. Phillips Eric Torng Departent of Coputer Science Stanford University Stanford, CA Abstract One of the oldest and siplest variants of ultiprocessor scheduling is the on-line scheduling proble studied by Graha in 966. In this proble, the obs arrive on-line and ust be scheduled non-preeptively on identical achines so as to iniize the akespan. The size of a ob is known on arrival. Graha proved that the List Processing Algorith which assigns each ob to the currently least loaded achine has copetitive ratio ( =). Recently algoriths with saller copetitive ratios than List Processing have been discovered, culinating in Bartal, Fiat, Karlo, and Vohra's construction of an algorith with copetitive ratio bounded away fro. Their algorith has a copetitive ratio of at ost ( =7) :986 for all ; hence for > 7, their algorith is provably better than List Processing. We present a ore natural algorith that outperfors List Processing for any 6 and has a copetitive ratio of at ost :945 for all, which is signicantly closer to the best known lower bound of :837 for the proble. We show that our analysis of the algorith is alost tight by presenting a lower bound of :9378 on the algorith's copetitive ratio for large. Introduction Scheduling n obs on achines is one of the ost widely studied probles in coputer science. One of its earliest and siplest variants is the on-line scheduling proble introduced by Graha [5] in 966. The achines are identical, and the n nonpreeptable, single task obs are all independent. Job i has size J i >. The obs arrive one by one, and each ob ust be iediately and irrevocably scheduled without knowledge of later obs. The size of a ob is known on arrival, and the obs are executed only after the scheduling is copleted. The goal is to iniize the akespan the copletion tie of the last ob to nish. This proble is also referred to as load-balancing. Supported by NSF Grant CCR-957, NSF Young Investigator Award CCR , and grants fro Mitsubishi Corporation and OTL. For exaple, the \achines" can be counication channels, and the \obs" can be requests for counication bandwidth. When a custoer requests bandwidth, a channel ust be chosen on which the required bandwidth is iediately and peranently ade available to the custoer. Under this interpretation, the goal is to iniize the axiu load on any channel. Because of the online nature of the proble, the perforance of a scheduler is easured by its copetitive ratio. For a ob sequence, let A() denote the akespan of algorith A's schedule, and let OPT() denote the iniu akespan of all -achine schedules for. The copetitive ratio of A is dened by def A() C A = sup OPT() ; where the supreu is over all nonepty ob sequences. The natural question is, how sall can C A be? Graha showed that the List Processing Algorith (List for brevity) which assigns each ob to the least loaded achine has copetitive ratio exactly. Starting in 99, several algoriths were developed [4, 6] with better copetitive ratios than List, and recently Bartal, Fiat, Karlo, and Vohra [] gave an algorith with a copetitive ratio of :986. Faigle, Kern, 7 and Turan [3] proved lower bounds on the achievable deterinistic copetitive ratio of for = and 3 and + p for 4 (so List is optial for = and 3). Bartal, Karlo, and Rabani [] iproved this lower bound for 4 to + p + which becoes :837 for large enough. This paper presents a deterinistic algorith, ALG, that is ore natural than the algorith of Bartal et al. The algorith ALG uses a paraeter that aects its behavior; the best choice of depends on. For 6, there is an such that ALG outperfors List; furtherore, ALG :945 has a copetitive ratio of :945 for all. Figure shows how ALG (under the best choice of ) copares to List, Bartal et al's algorith, and the asyptotic lower bound. We show that our analysis of ALG is alost tight by presenting a lower bound of :9378 on the

2 Karger, Phillips, Torng copetitive ratio List nuber of achines (log scale) BFKV Algorith New Algorith Lower Bound for Large Figure : Coparison of our algorith to List, Bartal et al's algorith, and the asyptotic lower bound copetitive ratio of ALG (for each choice of ) for large. Denition of ALG In order to do better than List, we ust see what List does wrong in its worst case, a sequence of ( ) obs of size followed by one ob of size. List assigns the large ob to a achine that already has sall obs while the optial schedule assigns the large ob to its own achine. The proble is that List keeps the schedule so at that when the large ob arrives it has to go on a relatively tall achine. The algorith ALG strives to aintain an ibalance in the processor loads so that a large ob can always be assigned to a relatively short achine. However, it cannot create too large an ibalance because this would iediately iply a poor copetitive ratio. follows: when ob t arrives, it is placed on the tallest achine M t k such that ht k + J t A t k. The paraeter deterines the degree of ibalance that ALG tries to aintain. Note that with the above denition of A t, any very tall ob will be placed on the shortest achine. The algorith of Bartal et al is a discrete version of ALG where a ob can only be placed on achine M t (the sallest achine) or achine M t ; :445. The reader ay wonder why we use the average height of the shorter achines before adding the new ob rather than the average height of the shorter achines after adding the new ob (i.e. coparing h t k + J t to A t+ k where k is the rank of achine M t k with ob t at tie t + ). Furtherore, why not copare the new height of the achine to the average height of all the achines (i.e. copare h t k +J t to A t+ )? Both of these are better approxiations to OPT(), so they would see to be better choices. However, neither algorith is better than ( )-copetitive. In particular, when these algoriths schedule a long sequence of equal-sized obs, they distribute alost all of the obs evenly on slightly ore than half the achines while the rest of the achines get only a sall nuber of obs. An adversary can then insert approxiately large obs to create a copletely at schedule, after which a single large ob akes the akespan ties the optial akespan. In the next two sections we prove the following theore. Theore.. For each 6, there exists soe such that ALG outperfors List, and for :945, ALG is -copetitive for all. h k n - o k - k o (a) Our Algorith h k n (b) List Figure : Scheduling any sall obs, then one big ob Dene the height (or load) of a achine to be the su of the sizes of all obs currently assigned to it. Definition.. Tie t is the tie ust before the t th ob is scheduled. At tie t: Mi t is the (i + ) st shortest achine (e.g. M t is the shortest achine), h t i is the height of Mi t ( i < ), At i is the average height of the i shortest achines, i, and A t =. Note that A t is the average height of all achines at tie t and is thus a lower bound on OPT. Definition.. The algorith ALG works as 3 Upper Bound: Fixed Fix 6, and x, <. This section shows how to copute an upper bound on the copetitive ratio of ALG on achines, by induction on the length of the ob sequence. Let be a ob sequence of length n, noralized so that the total size of all obs in is (and thus A n+ = ). Let n be the rst n obs in, and assue inductively that ALG ( n ) OP T ( n ). The rest of this section is devoted to showing that ALG () OP T (). The following lower bounds on OPT() will be used. Fact 3.. The following quantities are lower bounds on OPT():. ties the total size of all obs in (= ).. The largest ob in ( J n ). 3. Twice the size of the ( + ) st largest ob in.

3 A Better Algorith For an Ancient Scheduling Proble 3 If ALG () 6= h n +J n, then the denition of ALG and the inductive assuption easily iply ALG () OP T (). If ALG () or ALG () J n, then Fact 3. (part () or () respectively) iplies ALG () OPT(). Therefore, for the rest of this section, assue (3.) (3.) (3.3) ALG () = h n + J n ; ALG () ; ALG () J n : We prove that for 6, the ( + ) st largest ob in has size at least (3.4) ALG () (for a suitable choice of ), which cobined with Fact 3. part (3) proves ALG () OP T (). Definition 3.. Dene by = h n. Dene =. Dene b =. Dene a = b. Definition 3.. A achine is tall if it has height at least ; otherwise it is short. Let s be the current nuber of short achines (so A t s is the average height of the short achines at tie t). A ob is large if it has size b. Definition 3.3. An elevating ob raises a achine fro being short to being tall, and we say that an elevating ob elevates the achine it is placed on. Each achine is tall by tie n, so there are elevating obs in. We will prove that the last ob is large and that all the elevating obs are large. Lea 3. below shows that a large ob has size at least ALG (), as required by (3.4) above. The sequence is divided into two phases. In the rst phase, A t s < a, and each elevating ob in this phase ust go on the sallest achine and is therefore large. A t s increases onotonically with t until we reach the second phase (called the kickstart) when A t s a. The kickstart is analyzed in reverse: starting with the last elevating ob in and oving backwards in tie, we show that each elevating ob is large. Eventually a tie t is reached where so uch processing tie is taken by the obs in the tall achines and the later elevating obs that we ust have A t s < a. Thus the rst phase of the proof applies to all preceding ties. 3. Large obs Lea 3.. satises <. Proof. The fact that > follows fro = h n A n < A n+ =. To see that, we have fro (3.) (3.) and (3.3) above that h n + J n J n and h n + J n. The rst inequality gives (3.5) J n hn : Substituting this upper bound for J n into the second inequality gives h n ( + ) ; which siplies to = h n. Lea 3.. b ALG () Proof. Equation (3.5) above gives ALG () = h n + J n h n ( + ) = ( ). The lea follows fro the denition b =. ALG (). Thus a large ob has size at least Lea 3.3. If p :9, then ob n is large. Proof. Job n raises a achine fro height to at least =, so J n +. Job n is large whenever +. Setting =, the worst case, this siplies to p :9. 3. The First Phase This section proves that can be broken into two phases, such that A t s < a during the rst phase, A t s a in the second phase, and all elevating obs in the rst phase are large. Lea 3.4. If ob t is an elevating ob for achine M t k, then At k a. Proof. Suppose ob t goes on achine Mk t. The rule for ALG iplies that A t k ht k + J t. Since ob t is an elevating ob, we have h t k + J t. By transitivity, A t k, so At k > a. Lea 3.5. If ob t is an elevating ob and A t s < a, then ob t goes on the shortest achine M t and is large. Proof. For all < i < s, A t i < At s < a. Therefore, by Lea 3.4, elevating ob t cannot go on any short achine except the shortest, achine M. t Since M t is the shortest achine, h t A t s < a. Job t is an elevating ob, so h t + J t which iplies J t > a = b. Therefore, ob t is large. Lea 3.6. If A t+ s Proof. < a then A t s At+ s < a. Case : ob t is not elevating. The nuber of short achines is the sae at ties t and t +. The only dierence between tie t and tie t + is the absence of ob t at tie t. Therefore, A t s A t+ s.

4 4 Karger, Phillips, Torng Case : ob t is elevating. There is one ore short achine at tie t than at tie t +. We show that ob t ust be placed on the shortest achine at tie t. This iplies A t s A t+ s, since the average height of the short achines is increased by the reoval fro consideration of the shortest such achine. Assue that elevating ob t is placed on short achine Mk t, with k >. Now At k = At+ k (since for < k, h t = ht+ ). But A t+ k A t+ s < a, and A t k a by Lea 3.4, giving a contradiction. Therefore, ob t ust go on achine M. t Corollary 3.. If A t s < a, then A u s A t s < a for all u < t. If A t s a, then A u s a for all u > t. Corollary 3.. Each elevating ob that arrives during the rst phase is large. 3.3 The Kickstart This section shows that all elevating obs that arrive during Phase are large. Consider the elevating obs in reverse order of arrival. Let e s be the s th elevating ob fro the last to arrive; say it arrives at tie t s. Given lower bounds on the heights e ; : : :; e s, we derive a lower bound on e s (note s is the nuber of short achines P at tie t s ). We continue until the lower bound on s e is so large that the fact that A n+ = forces s < a, proving that tie t s ust be in Phase. If each lower bound on e, < s, exceeds b, then we have proven that all elevating obs that arrive during Phase are large. Lea 3.7. Given,, and lower bounds for e e s, and assuing that s a, a lower bound for e s for ALG can be coputed by solving the following linear progra. Miniize e s subect to (3.6) (3.7) (3.8) (3.9) + s = e + + ax( e s; ); s < s a = k= ; Proof. At any tie t n, the size of the obs that arrive before t and the size of the obs that arrive after t ust su to. The su of the obs that arrive before tie t s is at least, while the P su of the s obs that arrive after tie t s is at least = e + J n, and J n + (fro Lea 3.3). This iplies equation (3.6). Refer to gure 3. The s equations in (3.7) follow fro the fact that e s cannot go on any of the tall achines (because it is an elevating ob). Refer to gure 4. Equation (3.8) is part of the hypothesis of the lea, while equation (3.9) is the denition of. h k ts h k ts tall achines - s - ε a =.5 - ε - elevating obs short achines last ob - < - Figure 3: Equation (3.6) Figure 4: Equation (3.7) k s length of all obs = elevating ob e s Corollary 3.3. For any,, and, let e ; : : :; e i ; : : :; e s be successive solutions to the linear progra of Lea 3.7. If e is large for < s and there is no feasible solution to the linear progra for e s, then s Phase is large. < a and each elevating ob that arrives during Proof. The fact that the linear progra of Lea 3.7 for e s is infeasible iplies that the assued conditions at tie t s cannot hold which eans s < a. Therefore, Phase ust last at least until tie t s. We have calculated the iniu values of all later elevating obs, and, by assuption, they are large. Therefore, all elevating obs that arrive during Phase ust be large. Note that Corollary 3.3 depends on. We reove this dependence by showing that the worst value of for our analysis is =. Lea 3.8. Assue that :. If the conditions of Corollary 3.3 hold (so the kickstart is successful) for =, then they hold for all,. Proof. Let a superscript on a variable denote the value of that variable with equal to the superscript (e.g. b is the value of b when = ). Lea 3.8 is proven by showing that if e e k > b and the linear progra for e k+ is infeasible, then for soe k, we

5 A Better Algorith For an Ancient Scheduling Proble 5 have e e > b and the linear progra for e + is infeasible. The full proof is included in the appendix. Thus to deterine if ALG is -copetitive on achines, we need to solve the linear progras described in Lea 3.7 and Corollary 3.3 only for the case =. Corollary 3.4. If the conditions of Corollary 3.3 hold for and (and = ), so that ALG is -copetitive for achines, then ALG is - copetitive for achines for. Proof. This follows fro the fact that the constraints of Lea 3.7 are tighter for larger. Lea 3.9. When 6 3, there exists an less than and at ost :943 such that ALG is -copetitive on achines. Proof. We have written code to solve the linear progras of Lea 3.7 for 6 3 and s. It uses binary search to nd the sallest (within a tolerance of :) such that the conditions of Corollary 3.3 are satised, and hence ALG () OPT(). In all cases, < and : Upper Bound: Large The linear progras of Lea 3.7 only apply to a specic value of, and it is ipossible to solve the for all. To give a bound for all large, we discretize the linear progra of Lea 3.7, eectively regarding groups of achines as having the sae height. Let x be a constant such that > x(x ). Let b = d x e, and let i = b i bc for all variables i. Unless is a ultiple of b, we change the constraint on h in equation (3.7) to the weaker constraint. We siilarly change the constraint on. Furtherore, for b, we change the constraint on h in equation (3.7) to the weaker constraint h =. We weaken constraint (3.6) by discarding the ters + (our lower bound for J n ) and e for in s 's block (s b < (s +)b). This changes the linear progra of Lea 3.7 to the following linear progra: Miniize e s subect to (4.) (4.) (4.) (4.3) (4.4) =(s +)b + (s + )ba + ax( b e s; ); s b e = (s + )b < b = ; b < a (s +)b b = b b k= k ; s + < Note that we have grouped all the achines into x blocks of b achines with one last block of the tallest achines of size less than or equal to b (this follows fro the assuption that > x(x )). In particular, the linear progras for e i and e are identical when i =. Using this fact to siplify the equations, is easily eliinated fro equation (4.). Lastly is eliinated fro equations (4.) and (4.) by pessiistically assuing that there are b achines in the last block (this is an upper bound). The linear progra now siplies to the following one: Miniize e s subect to x x =s + s + (s + )a + ax(ats e s; ) x = s + a = s + < x k= = e k ; s + < x This linear progra gives a valid bound for any > x(x ). Therefore, we now have a technique to prove that for soe, ALG is -copetitive for >. Lea 4.. When :945 and > 3, ALG is -copetitive. Proof. We have written code that solves the linear progra above, and for = :945 and x = 5, (and hence for 5 3 = 995), A tk s eventually falls below a and all elevating obs that arrive after t k are large. Again, by Corollary 3., the rest of the elevating obs are large. By Lea 3.3, the last ob is large as well which eans that there are + large obs. Therefore, in all cases ALG () OPT() and the lea follows. This copletes the proof of our ain result. We have actually shown that for >, ALG is - copetitive for :943 (by running the code once for x = and = :943). We expect that ALG is -copetitive for :943 for all, but we lack the coputing resources to verify :943 for all up to. 5 Lower Bounds We now prove lower bounds on the perforance of ALG for both sall and large. We rst prove for = 4 and = 5 that for no is ALG better than -copetitive. We then show that our analysis of ALG is alost tight by describing a sequence of obs

6 6 Karger, Phillips, Torng such that ALG :9378 is not :9378-copetitive on that sequence. Siilar sequences can be constructed to prove that ALG is not :9378-copetitive for < :9378, and these will appear in the ournal version of this paper. Clearly, if > :9378, then for large enough, ALG is no better than :9378-copetitive on a long sequence of equal sized obs. 5. Lower Bound for Sall Lea 5.. For no is ALG better than ( )- copetitive for = 4 or = 5. Proof. First consider the case for = 4. For 5, ALG 7 is clearly worse than :75-copetitive on a long sequence of equal sized obs. For < 5, it 7 is not hard to verify that ALG ust act exactly like List on the sequence ; ; ; ; ; ; ; ; 4 giving ALG a akespan of 7 while the optial akespan is easily seen to be 4. Thus, for = 4, C ALG :75. The proof is alost identical for = 5. For 5, 7 ALG is clearly worse than :8-copetitive on a long sequence of equal sized obs. For < 5, it is not 7 hard to verify that ALG ust act exactly like List on the sequence ; ; ; ; ; ; ; ; ; ; 5; 5; 5; 5; 5; giving ALG a akespan of 8 while the optial akespan is easily seen to be. Thus, for = 5, C ALG :8. 6 Lower Bound for Large 6. Sketch The lower bound ob sequence, which deonstrates that ALG :9378 is not :9378-copetitive for large, is extracted fro the upper bound proof, which shows that a sequence that is bad for ALG rst causes it to create a at schedule and then hits it with one nal ob of size. The ethod of constructing a at schedule closely follows the kickstart. The sequence akes the average achine height, as in the upper bound. The ain part of (Phases and 3) forces ALG to create a schedule with ( ) tall achines (with height ) and short achines (with height a = = ). The nal part of (Phase 4) then forces ALG to follow the kickstart by repeatedly setting the height of the next ob to be ust large enough to prevent its going on any of the tall achines. The paraeter is set so that the rst obs in Phase 4 have height so that two can be scheduled o-line on one achine. Finally, the last ob of size follows. The one coplication is that we need to ensure that can be scheduled with a akespan of ust. To do this, the rst part of (Phase ) consists of a large nuber of extreely sall obs which the o-line algorith can use as ller. The rest of this section is organized as follows. Sections 6. through 6.5 give details of phases through 4. Section 6.6 shows how OPT can pack the obs so that all achines have the sae height. Throughout, we assue that is large so that O(=) ters can be ignored. 6. Phase Phase consists of a sequence of innitesial obs which bring the average achine height to c, where c = :38 (the reason for this value is given in Section 6.3). - ε.5- ε c - Figure 5: Schedule at end of Phase The algorith places the innitesial obs so that the achine heights obey the relation h = A. This gives a recurrence for h, whose solution is h = h + where the binoial coecient r r(r ):::(r k+) k r = when k =. Then ; k is dened by r k = for real r and positive integer k, and k! h + i A = = h + : i i< Since A = c, we have h + = c=. Lastly, applying Stirling's approxiation gives (6.5) (6.6) h = c A = c + O + O 6.3 Phase Phase consists of obs which are placed on the current shortest achine and bring that achine's height to exactly a = =. The value of c (the average achine height after Phase ) is deterined to ensure

7 A Better Algorith For an Ancient Scheduling Proble 7 - ε - ε.5-ε c.5- ε c - (- λ) - λ Figure 6: Schedule at end of Phase Figure 7: Schedule at end of Phase 3 that the algorith does indeed place each ob in Phase on the achine with current sallest height. The value of c is bounded by two constraints. First, a Phase ob ust not be placed on a achine whose height is already a. Let A and h be the values of A and h respectively at the end of Phase. Since h increases with, the tightest constraint on c is when the last Phase ob is being inserted, and the constraint involves adding a ob of size a h and requiring that it be too tall to go on a height a achine: a + (a h ) > A = a( ) + A ( )A ; Substituting the approxiations (6.5) and (6.6) above gives (for large ): c < a( ) ( + ) :38; using = :39 and = :6 (deterined in Section 6.5). The second constraint on c is that a Phase ob ust not be placed on a achine whose height is not yet a, but that is not the current sallest achine. Soe straightforward calculations show that this is a weaker constraint on c. Hence we take c = : Phase 3 Phase 3 consists of ( ) obs which will be placed on the current shortest achine and bring that achine's height to exactly. For this to work, we need to check that the algorith will not place a Phase 3 ob on any achine except the sallest at the tie. The sallest ob of Phase 3 has size c > :54 (using = :6, deterined below), so clearly no Phase 3 ob will be placed on any noninial-height achine of height a or less. The average of the achine heights at any tie in Phase 3 is at ost ( )( )+a :74, while adding any Phase 3 ob to a achine of height would raise it to height at least + :54 = :44 > :74. Hence the algorith places each Phase 3 ob on the sallest achine at the tie. 6.5 Phase 4 Phase 4 consists of obs where each ob size is inial subect to the constraint that the algorith places the ob on the sallest achine at the tie. The paraeter is chosen so that the rst ob of Phase 4 has size =, breaking the kickstart. Thus satises: + = > A = (( )( ) + a); which gives as a function of. The constraint that the algorith places the ob on the sallest achine at the tie gives a lower bound for J, which, after soe calculations, is: J > = + (3= a) ; so we set J = = + (3= a) (plus a tiny perturbation). Lastly, the fact that the set of all obs has total size deterines a value for. Setting the su of all the ob sizes to be and solving for gives = :6 and = : Packing To prove that OPT =, we need to show that the obs can be assigned to achines so that each achine has height. This is done as follows: to ake roo for the ob of size (which gets its own achine), we

8 8 Karger, Phillips, Torng place the six sallest Phase obs on achines, 3 per achine. We place the six sallest Phase 4 obs on 3 achines, per achine. Each of the reaining Phase 4 obs gets its own achine. The reaining Phase obs are paired with Phase 3 obs the largest Phase ob with the sallest Phase 3 ob, the second largest Phase ob with the second sallest Phase 3 ob, and so on. Finally the innitesial Phase obs bring are used as ller to ake each achine have height. To check that this assignent does not ake any achine too tall, observe the following facts: The sallest Phase obs have size a c < =3. The sallest Phase 4 obs have size =. The sallest Phase 3 ob has size c, while the largest Phase ob has size = for a total of less than. Furtherore, since the schedule after Phase is convex, if the sallest Phase 3 ob and the largest Phase add to less than, then so do the second sallest Phase 3 ob and the second largest Phase ob, and so on. 7 Extensions We have been able to iprove our algorith slightly by elding it with the algorith that copares to the average height of all achines. Definition 7.. The algorith ALG ; works as follows: when a ob J arrives, it is placed on the tallest achine k such that h k + J (( )A k + A + ). For a udicious choice of, ALG ; outperfors ALG. Unfortunately we can currently only prove a sall iproveent of about : in the copetitive ratio for large. One unfortunate property of ALG is that its worst case perforance ratio occurs on a set of identically sized obs. One ight desire an algorith that perfors well on such a ob set, even at the cost of increasing the copetitive ratio. We have shown such a tradeo for ALG, naely that if :5 < <, then there is an > such that ALG is ( )-copetitive for large. The details will be in the full paper. Clearly, for saller, ALG achieves a better akespan when all the obs are the sae size. A related observation is that any algorith which does well in the worst case cannot be optial in the average case it is easy to show that for any < there is a such that any -copetitive algorith, given a long sequence of identical obs, ust construct a schedule whose akespan exceeds ( + )OPT. This leaves three interesting open probles. First, to close the gap between the upper and lower bounds for C(ALG ). Second, is ALG ( )-copetitive for :5? And third, fro the analysis of bad ob sequences for ALG in Section 5, can one derive a better lower bound for all algoriths, or deterine how an algorith could iprove on C(ALG )? 8 Acknowledgeents We thank Alan Hu for contributing the exaple deonstrating that ALG cannot outperfor List when = 4. References [] Y. Bartal, A. Fiat, H. Karlo, and R. Vohra. New algoriths for an ancient scheduling proble. In Proc. 4nd ACM Syp. on Theory of Coputing, pages 5{ 58, 99. [] Y. Bartal, H. Karlo, and Y. Rabani. Private Counication. [3] U. Faigle, W. Kern, and G. Turan. On the perforance of on-line algoriths for particular probles. Acta Cybernetica, 9:7{9, 989. [4] G. Galabos and G. Woeginger. An on-line scheduling heuristic with better worst case ratio than Graha's list scheduling. SIAM J. Coput., :349{355, 993. [5] R. Graha. Bounds for certain ultiprocessing anoalies. Bell Syste Technical Journal, 45:563{ 58, 966. [6] P. Narayanan and R. Chandrasekaran. Optial online algoriths for scheduling. Manuscript, University of Texas at Dallas, 99. Appendix We present here the full proof of Lea 3.8. Proof. Let. Let a superscript on a variable denote the value of that variable with equal to the superscript (e.g. b is the value of b when = ). We show that if e e k > b and the linear progra for e k+ is infeasible, then for soe k, e e > b and the linear progra for e + is infeasible. Let =. We have b = = + while b =. Meanwhile, a = = + while a =. This gives us the following two dierences: (b) = b b = and (a) = a a =. We prove that for i k, if the linear progra for e i is feasible, then (e i ) = e i e i (b), so e i is large if e i is large. If the linear progra for e i is feasible for i k, then the fact that the linear progra for e k+ is infeasible iplies that the linear progra for e is k+ also infeasible (since the constraints are all tighter in the latter progra). For the reainder of this proof, considering the linear progra for elevating ob e s, s k, arriving

9 A Better Algorith For an Ancient Scheduling Proble 9 at tie t s. We reserve the superscript on all variables (including h and A) to denote the value of and oit the, t s. Assue that e e s are indeed (b) larger than their counterparts for =. We show that e s is as well by showing that if we set e s = e s + (b), and iniize h i subect to (3.7), then h i h i for each of the tall achines M i at this tie. Let (A i ) = A i A i and let (h i ) = h i h i for each tall achine M i. Using this notation, we want to prove that (h i ) for i s. Proving this will be easy for i close to s because, in these cases, h i = which iplies (h i) =. The only proble we ight have arises when h i nally rises above. Let be the inial index where A e s. In this case, fro Equation (3.7), we get (h ) (A ) (b). If we can prove that (h ) (A ), then a siple induction proves that for l, (A l ) (A ) which iplies (h l ) (A ) >. Thus we need ( )(A ) which eans we need (A ). For :5, this eans we ( ) need (A ) :7 while for :5 :, this eans we need (A ) :79. If s, then (A ) = (a)s+s + = s For :5, this iplies (A ) :7 which, fro the arguent above, iplies that for l, (h l ). Siilarly, if 4s, (A ) For :5 :, this eans (A ) :84 :79 which, fro the arguent above, iplies that for l, (h l ). Thus we siply need to prove that for :5 that h i does not rise above for i < s and that for :5 :, that h i does not rise above for i < 4s. For the reainder of this proof, =. Let be the rst point where A e s. Clearly, this is the rst point where h rises above. Substituting s( )+( s)( ) for A and siplifying leads to 4+ s. For, 4+ while for : :5, This copletes the proof of the lea.