«Û +(2 )Û, the total charge of the EH-pair is at most «Û +(2 )Û +(1+ )Û ¼, and thus the charging ratio is at most

Transcription

1 ÁÑÔÖÓÚ ÇÒÐÒ ÐÓÖØÑ ÓÖ Ù«Ö ÅÒÑÒØ Ò ÉÓË ËÛØ ÅÖ ÖÓ ÏÓ ÂÛÓÖ ÂÖ ËÐÐ Ý ÌÓÑ ÌÝ Ý ØÖØ We consider the following buffer management problem arising in QoS networks: packets with specified weights and deadlines arrive at a network switch and need to be forwarded so that the total value of forwarded packets is maimized. If a packet is not forwarded before its deadline, it is lost and brings no profit. The main result of the paper is an online competitive algorithm the first deterministic algorithm for this problem with competitive ratio below 2. We also study the -ÙÒÓÖÑ case, in which all packets have the same span (the difference between the deadline and the arrival time), for which we give an algorithm with ratio 5 Ô Our algorithm achieves the same ratio in a more general scenario when all packets are similarly ordered. Our last two results concern the 2-uniform case, for which we give an algorithm with ratio 1377 and a matching lower bound. ½ ÁÒØÖÓÙØÓÒ One of the issues arising in IP-based QoS networks is how to manage the packet flows at a router level. In particular, in case of overloading, when the total incoming traffic eceeds the buffer size, the buffer management policy needs to determine which packets should be dropped by the router. Kesselman Ø Ðº [6] postulate that the packet drop policies can be modeled as combinatorial optimization problems. Of the two models proposed in [6], the one relevant to this work is called Ù«Ö ÑÒÑÒØ ÛØ ÓÙÒ ÐÝ, and is defined as follows: Packets arrive at a network switch. Each packet is characterized by a positive weight and a deadline before which it must be transmitted. Packets can only be transmitted at integer time steps. If the deadline of a packet is reached while it is still being buffered, the packet is lost. The goal is to maimize the weighted number of forwarded packets. It is easy to see that this buffer management problem is equivalent to the online version of the following unit job scheduling problem. We are given a set of unit-length jobs, with each job specified by a triple (Ö Û ), where Ö and are integral release times and deadlines, and Û is a non-negative real weight. One job can be processed at each integer time. We use the term ÛØ ØÖÓÙÔÙØ or Ò for the total weight of the jobs completed by their deadline. The goal is to compute a schedule that maimizes the weighted throughput. ÔÖØÑÒØ Ó ÓÑÔÙØÖ ËÒ ÍÒÚÖ ØÝ Ó ÐÓÖÒ ÊÚÖ ¾¾½º ËÙÔÔÓÖØ Ý ÆË ÖÒØ Ê¹ ¼ Ò Ê¹¼¾¼º ÑÖ ÛÓØ ºÙÖºÙ Ý ÅØÑØÐ ÁÒ ØØÙØ Ë Ê ØÒ ¾ ¹½½ ÈÖ ½ Þ ÊÔÙк ÈÖØÐÐÝ ÙÔÔÓÖØ Ý ÁÒ ØØÙØ ÓÖ ÌÓÖØÐ ÓÑÔÙØÖ ËÒ ÈÖÙ ÔÖÓØ ÄƼ¼¼ Ó ÅËÅÌ Êµ Ò ÖÒØ ½¼½¼½ Ó Î Êº ÐÐ ØÝÑغ ºÞ 1

2 In the online version of the problem jobs arrive in time and the algorithm needs to schedule one of the pending jobs without knowledge of the future. An online algorithm is ʹÓÑÔØØÚ if its gain on any instance is at least 1Ê times the optimal (offline) gain on this instance. The ÓÑÔØØÚ ÖØÓ of is the infimum of such values Ê. It is common in the literature to view the behavior of an online algorithm as a game between and an ÚÖ ÖÝ, who issues the jobs in Á and schedules them in order to maimize the ratio between his gain and and the gain of. È Ø ÛÓÖº A simple greedy algorithm that always schedules the heaviest available job is 2- competitive, and this is the best previous bound for deterministic algorithms for this problem. A lower bound of 1618 was shown in [1, 4, 5]. Some restrictions on instances of the problem have been studied in the literature [6, 1, 4]. Let the span of a job be the difference between its deadline and the release time. In ¹ÙÒÓÖÑ Ò ØÒ the span of each job is equal eactly. In ¹ÓÙÒ instances, the span of each job is at most. The lower bound of 1618 in [1, 4, 5] applies even to 2-bounded instances. A matching upper bound for the 2-bounded case was presented in [6]. The algorithms for the 2-uniform instances were studied by Zhu Ø Ðº [1], who established a lower bound of ½ ¾ (Ô 3 + 1) 1366 and an upper bound of Ô This bound is tight for memoryless algorithms [2], that is, algorithms which base their decisions only on the weights of pending jobs and are invariant under scaling of weights. Finally, the first deterministic algorithms with competitive ratio lower than 2 for the -bounded instances appear in [2]. These ratios, however, depend on, and approach 2 as increases. A randomized 1582-competitive algorithm for the general case was given in [2]. For 2-bounded instances, there is a 125-competitive algorithm [2] and this is optimal [4]. For the 2-uniform case the currently best lower bound for randomized algorithms is 1172 [2]. ÇÙÖ Ö ÙÐØ º We present several deterministic online algorithms for the buffer management problem. Our main result is a deterministic 1939-competitive algorithm for the general case. This is the first deterministic algorithm for this problem with ratio better Ô than 2. For the -uniform case, we give an algorithm with ratio , independent of. In fact, this ratio holds in a much more general scenario when all jobs are similarly ordered, that is, when Ö Ö implies for all jobs. Finally, we completely solve the 2-uniform case: we give an algorithm with competitive ratio 1377 and a matching lower bound. Note that this ratio is strictly in between the previous lower and upper bounds from [1]. ¾ ÌÖÑÒÓÐÓÝ Ò ÆÓØØÓÒ A ÙÐ Ë specifies which jobs are eecuted, and for each eecuted job it specifies an integral time Ø, Ö Ø, when it is scheduled; only one job can be scheduled at any Ø. The ØÖÓÙÔÙØ or Ò of a schedule Ë is the total weight of the jobs eecuted in Ë. If is a scheduling algorithm, by Ò (Á) we denote the gain of the schedule computed by on Á. The optimal gain on Á is denoted by ÓÔØ(Á). A job is ÔÒÒ in schedule Ë at time Ø if Ö Ø and has not been scheduled before Ø. Note that all jobs released at time Ø are considered pending. An instance is ¹ÓÙÒ if Ö for all jobs. Similarly, an instance is ¹ÙÒÓÖÑ if Ö = for all. The difference Ö is called the ÔÒ of a job. An instance is called ÑÐÖÐÝ ÓÖÖ when the release times and deadlines are similarly ordered, that is if Ö Ö implies for any two jobs and. Given two jobs, we say that ÓÑÒØ if either (i), or (ii) = and Û Û, or (iii) =, Û = Û and. ((iii) only ensures that ties are broken in some arbitrary but 2

3 consistent way.) Given a non-empty set of jobs Â, the ÓÑÒÒØ job in  is the one that dominates all other jobs in Â; it is always uniquely defined as dominates is a linear order. A schedule Ë is called ÒÓÒÐ ÖРعÐÒ if for any jobs scheduled in Ë at time Ø and scheduled later in Ë, either is released strictly after time Ø, or dominates. I.e., at any time Ø, the job scheduled by Ë is the dominant pending job in Ë. Any schedule can be easily converted into a canonical earliest-deadline schedule by rearranging its jobs. Thus we may assume that offline schedules are canonical earliest-deadline ¹ÓÑÔØØÚ ÐÓÖØÑ We start with some intuitions that should be helpful in understanding the algorithm and its analysis. The greedy algorithm that always eecutes the heaviest job (H-job) is not better than 2-competitive. An alternative idea is to eecute the earliest deadline job at each step. This algorithm is not competitive at all, as possibly many jobs of very small weight are eecuted even if there are heavy jobs pending with only slightly larger deadlines. A natural refinement of this approach is to focus on sufficiently heavy jobs, of weight at least «times the maimal weight, and chose the earliest deadline job among those (an E-job). As it turns out, this algorithm is also not better than 2-competitive. The general idea of our new algorithm is to alternate H-jobs and E-jobs. Although this simple algorithm, as stated, has ratio no better than 2, with several seemingly minor modifications, we can achieve competitive ratio smaller than 2. ÐÓÖØÑ ÒÐ We use parameters «= and ½½ = and a boolean variable, ½½ initially set to Ð, that stores information about the previous step. At a given time step Ø, update the set of pending jobs (remove jobs with deadline Ø and add jobs released at Ø). If there are no pending jobs, go to the net time step. Otherwise, let be the heaviest pending job (breaking ties in favor of dominant jobs) and the dominant job among the pending jobs with weight at least «Û. Schedule either or according to the following procedure: = Ð ØÒ schedule = ØÒ set ØÖÙ Ð set Ð = Ø + 1 and Û Û ØÒ schedule Ð schedule A job scheduled while = Ð is called an Ç¹Ó if =, or an ¹Ó otherwise. A job scheduled while =ØÖÙ is called an ͹Ó. A job scheduled in the last case is called an À¹Ó. The letters stand for Çbvious, arly, Írgent, and Àeaviest. Variable is ØÖÙ iff the previous job was an E-job. Thus, in terms of the labels, the algorithm proceeds as follows: If an O-job is available, we eecute it. Otherwise, we eecute an E-job, and in the net step either a U-job (if available) or an H-job. Note that the condition = guarantees that there is a pending job if is ØÖÙ: if = Ø then = by the definition of dominance; so if an E-job is eecuted at time Ø, Ø and will be pending in the net step. ÌÓÖÑ º½ ÒÐ ¹ÓÑÔØØÚ ØÖÑÒ Ø ÐÓÖØÑ ÓÖ ÙÒØ¹Ó ÙÐÒº 3

4 ÈÖÓÓ The proof is by a charging scheme. Fi an arbitrary (offline) schedule Ú. For each job eecuted in Ú, we partition its weight Û into several Ö, and assign each charge to a job eecuted by ÒÐ. If the total charged to each job of ÒÐ were at most ÊÛ, the Ê-competitiveness of ÒÐ would follow by summation over all jobs. Our charging scheme does not always meet this simple condition. Instead, we divide the jobs of ÒÐ into disjoint groups, where each group is either a single O-job, or an EH-pair (an E-job followed by an H-job), or an EU-pair (an E-job followed by a U-job). This is possible by the discussion of types of jobs before the theorem. For each group we prove that its ÖÒ ÖØÓ is at most Ê, where the charging ratio is defined as the total charge to this group divided by the total weight of the group. This implies that ÒÐ is Ê-competitive by summation over all groups. ÖÒ Ñº Let be the job eecuted at time Ø in Ú. Denote by and, respectively, the job eecuted by ÒÐ and the heaviest pending job at time Ø. (If there are no pending jobs, introduce two dummy jobs of weight 0. This does not change the algorithm.) Then is charged to ÒÐ s jobs, according to the following rules. (EB) If is eecuted by ÒÐ before time Ø, then charge (1 )Û to and the remaining Û (1 )Û to. (EF) Else, if is eecuted by ÒÐ after time Ø, then charge Û to and (1 )Û to. (NF) Else, if is an H-job, Û Û, and Ú eecutes after time Ø, then charge Û to and (1 )Û to the job scheduled by ÒÐ at time Ø + 1. (U) Else, charge Û to. (Note that this case includes the case =.) We label all charges as EB, EF, NF, U, according to which case above applies. We also distinguish upward, forward, and backward charges, defined in the obvious way. Thus, for eample, in case (EB), the charge of Û (1 )Û to is a backward EB-charge. The letters in the labels refer to whether was eecuted by ÒÐ, and to the charge direction: ecuted-ackward, ecuted- orward, Æon-eecuted-orward, Ípward. We start with several simple observations that will be used later in the proof, sometimes without eplicit reference. Ç ÖÚØÓÒ ½ Consider the eecution of ÒÐ. At time Ø, let be the heaviest job and the dominant job of weight at least «Û. Let be any pending job. Then (1.1) Û Û. (1.2) If dominates then Û «Û. ÈÖÓÓ Inequality (1.1) is trivial, by the definition of. Inequality (1.2) follows from the fact that dominates all pending jobs with weight at least «Û. Ç ÖÚØÓÒ ¾ Suppose that at time Ø ÒÐ schedules a job that receives a forward NFcharge from the job Ð scheduled at time Ø 1 in Ú. Then Ð is pending at time Ø. ÈÖÓÓ Denote by the heaviest pending job of ÒÐ at time Ø 1. By the definition of NF-charges, Ð =, Ð is pending at time Ø 1, is eecuted as an H-job, Ø + 1, and Û Ð Û. Therefore Ð Ø + 1, since otherwise ÒÐ would eecute Ð (or some other job) as a U-job at step Ø 1. Thus Ð is also pending at step Ø, as claimed. Ç ÖÚØÓÒ Suppose that at time Ø ÒÐ schedules a job of type O or E, and is the heaviest pending job at time Ø. Then (3.1) The upward charge to is at most Û. (3.2) If Ú eecutes after time Ø then the upward charge is at most «Û. (3.3) The forward NF-charge to is at most (1 )Û. ÈÖÓÓ Denote by the job eecuted at time Ø in Ú. If is scheduled by ÒÐ before time Ø, then the upward charge is (1 )Û «Û and claims (3.1) and (3.2) hold. Otherwise is 4

5 pending at time Ø. Then the upward charge is at most Û Û and (3.1) follows. If Ú eecutes after time Ø, then, since Ú is canonical, job must dominate, and (3.2) follows from (1.2). (3.3) follows by Observation 2 and (1.1). Ç ÖÚØÓÒ Suppose that at time Ø ÒÐ eecutes an H-job that is eecuted after time Ø in Ú. Then the upward charge to is at most Û. ÈÖÓÓ Let be the job eecuted at time Ø in Ú. In case (EB), the upward charge to is at most (1 )Û Û. In all other cases, is pending at time Ø, so Û Û. In cases (EF) and (NF), the charge is Û Û. In case (U) the charge is Û, but since (NF) did not apply, this case can occur only if Û Û. This completes the proofs of all observations. Now we eamine the charges to all groups in our partition of ÒÐ s jobs: single O-jobs, EH-pairs, and EU-pairs. Ç¹Ó º Let = be an O-job eecuted at time Ø. The forward NF-charge is at most (1 )Û. If gets a backward EB-charge then gets no forward EF-charge and the upward charge is at most «Û ; the total charging ratio is at most 2 + «Ê. If does not get a backward EB-charge then the forward EF-charge is at most (1 )Û and the upward charge is at most Û ; the charging ratio is at most 3 2 Ê. ¹Ó º Before considering EH-pairs and EU-pairs, we estimate separately the charges to E-jobs. Suppose that at time Ø ÒÐ eecutes an E-job, and the heaviest job is. We claim that is charged at most «Û + (2 )Û. We have two cases. Case 1: gets no backward EB-charge. Then, in the worst case, gets an upward charge of Û, a forward NF-charge of (1 )Û, and a forward EF-charge of (1 )Û. Using (2 )Û = 2«Û and «Û Û, the total charge is at most (2 )Û + (1 )Û «Û + (2 )Û as claimed. Case 2: gets a backward EB-charge. Then there is no forward EF-charge and the upward charge is at most «Û by (3.2). Let Ð be the job scheduled at time Ø 1 in Ú. If there is an NF-charge, it is generated by Ð and then Ð is pending for ÒÐ at time Ø by Observation 2. If there is a forward NF-charge and Ð, then Ð is pending at the time when Ú schedules. (Note that in this case job Ð cannot be eecuted by ÒÐ, because charges EB, EF did not apply to Ð.) Consequently, receives at most a backward EB-charge Û (1 )Û Ð, a forward NF-charge (1 )Û Ð, and an upward charge «Û. The total is at most «Û + Û «Û +(2 )Û as claimed. Otherwise, there is no forward NF-charge, or there is a forward NF-charge and Ð. In the second case, Ð is pending at Ø, and Ð dominates, so the forward NF-charge is at most (1 )Û Ð (1 )Û. With the backward EB-charge of Û and upward charge of at most «Û, the total is at most «Û + (2 )Û as claimed. À¹ÔÖ º Let be the E-job scheduled at time Ø, the heaviest pending job at time Ø, and ¼ the H-job at time Ø + 1. By the algorithm, =. Note that, since ÒÐ did not eecute as a O-job at time Ø, is still pending after the E-step and Û ¼ Û. We now estimate the charge to ¼. There is no forward NF-charge, as the previous step is not an H-step. If there is a backward EB-charge, the additional upward charge is at most Û ¼ and the total is at most (1 + )Û ¼. If there is no EB-charge, the sum of the upward charge and a forward EF-charge is at most Û ¼ + (1 )Û ¼ (1 + )Û ¼. With the charge to of at most «Û +(2 )Û, the total charge of the EH-pair is at most «Û +(2 )Û +(1+ )Û ¼, and thus the charging ratio is at most 2 + «Û + (2 1)Û ¼ Û + Û ¼ 2 + «Û + (2 1)Û ¼ «Û + Û ¼ «+ 2 1 «+ 1 = Ê

6 The first step follows from Û «Û. The net epression is decreasing in Û ¼ as 2 1 1, so the maimum is at Û ¼ = Û. ͹ÔÖ º As in the previous case, let, and denote the E-job scheduled at time Ø and the heaviest pending job at time Ø. By and ¼ we denote the scheduled U-job and the heaviest pending job at time Ø + 1. As in the case of EH-pairs, = and Û ¼ Û. Job gets no backward EB-charge, since it epires, and no forward NF-charge, since the previous step is not an H-step. The upward charge is at most Û ¼, the forward EF-charge is at most (1 )Û. With the charge to of at most «Û + (2 )Û, the total charge of the EU-pair is at most «Û + (2 )Û + Û ¼ + (1 )Û, so the charging ratio is at most 2 + «Û + Û ¼ Û Û + Û 2 + «Û + (1 )Û ¼ «Û + Û ¼ 2 + «+ 1 «+ = Ê In the first step, we apply bounds Û «Û and Û Û ¼. The net epression is decreasing in Û ¼, so the maimum is at Û ¼ = Û. Summarizing, we now have proved that the charging ratio to all job groups is at most Ê, and the Ê-competitiveness of ÒÐ follows. ¾ Ô ½¼µ¹ÓÑÔØØÚ ÐÓÖØÑ ÓÖ ËÑÐÖÐÝ ÇÖÖ ÂÓ We now consider the case when the jobs are similarly ordered, that is Ö Ö implies for all jobs. Note that, in particular, this covers the case when all jobs on input have the same span. ÐÓÖØÑ ËÑÐ We use one parameter «= Ô and a boolean variable, initially set to Ð. At a given time step Ø, update the set of pending jobs (remove jobs with deadline Ø and add jobs released at Ø). If there are no pending jobs, go to the net time step. Otherwise, let be the heaviest pending job (breaking ties in favor of dominant jobs) and the dominant job among the pending jobs with weight at least «Û. Schedule either or according to the following procedure: = Ð ØÒ schedule = Ø + 1 ØÒ set ØÖÙ Ð schedule ; set Ð A job scheduled while =Ð is called an Ç¹Ó if = or = Ø+1, and an ¹Ó otherwise. A job scheduled while =ØÖÙ is called an À¹Ó. The intuition behind the algorithm is very similar to ÒÐ, and, in fact, it is simpler. It schedules O-jobs, if available, and if not, it schedules an E-job followed by an H-job; note that the condition = guarantees that there will be a pending job if is ØÖÙ. ÌÓÖÑ º½ ËÑÐ 5 ÙÐÒ ÓÖ ÑÐÖÐÝ ÓÖÖ Ò ØÒ º Ô ¹ÓÑÔØØÚ ØÖÑÒ Ø ÐÓÖØÑ ÓÖ ÙÒØ¹Ó ÈÖÓÓ The proof is by a charging scheme. The steps of the algorithm can be divided into single O-steps, and EH-pairs. This is because each E-job must be followed by an H-job, and each H-job 6

7 must be preceded by an E-job. Similarly as for ÒÐ, we give a charging scheme and show that the charging ratio to each group is at most Ê. ÖÒ Ñº Let be the job eecuted at time Ø in Ú. Denote by and, respectively, the job eecuted by ËÑÐ and the heaviest pending Ô job at time Ø. (Without loss of generality, we can assume that such jobs eist.) Let = Then is charged to ËÑÐ s jobs, according to the following rules. (EB) If is eecuted by ËÑÐ at or before time Ø, then charge Û to. (NF) Else, if Ø+1 and is an H-job, then charge Û to and (1 )Û to the job scheduled by ËÑÐ at time Ø + 1. (U) Else, charge Û to. We start with some general observations. If Ú schedules ¼ before then ¼ : if is available at the time of scheduling ¼ then this follows from the definition of canonical schedules, and otherwise from the assumption of similar ordering. Ç ÖÚØÓÒ ½ Consider the eecution of ËÑÐ. At time Ø, let be the heaviest job and the dominant job of weight at least «Û. Let be any pending job. Then (1.1) Û Û. (1.2) If dominates then Û «Û. Ç ÖÚØÓÒ ¾ Suppose that at time Ø ËÑÐ schedules a job (of type O or E), and is the heaviest pending job at time Ø. Then (2.1) The upward charge to is at most Û. (2.2) If Ú eecutes after time Ø then the upward charge is at most «Û. (2.3) The forward NF-charge to is at most (1 )Û. (2.4) If Ú eecutes after time Ø then forward NF-charge is at most (1 )Û. (2.5) If Ú does not eecute after time Ø then the charge to is at most ÊÛ. The proofs of (1.1)-(2.3) are the same as for ÒÐ. For (2.4), let Ð be the job eecuted at time Ø 1 in Ú; since is scheduled later we have Ð. By the definition of NF-charges, Ð is pending at time Ø. This, by the choice of, implies that Û Ð Û and (2.4) follows. In (2.5), does not get a backward EB-charge, the upward charge is at most Û and the forward NF-charge is at most (1 )Û. The total is at most (2 )Û (2 )Û «= ÊÛ. Ç¹Ó º Let be an O-job eecuted at time Ø, and let be the heaviest pending job. By the algorithm, either = or = Ø + 1 (i.e., epires). If gets no backward charge, the charging ratio is at most Ê by (2.5). If gets a backward EB-charge then Ø + 1 and thus =. The upward charge is at most «Û = «Û and the forward NF-charge is at most (1 )Û. So the charging ratio is at most 2 + «Ê. À¹ÔÖ º Let be the E-job scheduled at time Ø, be the heaviest pending job at time Ø, and ¼ be the H-job at time Ø + 1. Let and ¼ be the jobs eecuted at times Ø and Ø + 1 in Ú, respectively. By the algorithm, we have =, which implies as otherwise is chosen as. Furthermore, Ø + 2 and Û ¼ Û. Case 1: Ú schedules after time Ø. Then ¼ dominates or ¼ =, as Ú schedules ¼ at Ø + 1 when is pending. Also, dominates. Thus the U-charges to and ¼ are each at most «Û. In addition, could receive an NF-charge of at most (1 )Û and the EB-charges are Û + Û ¼. The total charge to the EH-pair is at most (2 )Û + 2«Û + Û ¼, and the charging ratio is at most 1 + (1 )Û + 2«Û Û + Û ¼ 1 + (1 )Û + 2«Û (1 )«+ 2«1 + = Ê Û + Û «+ 1 7

8 The first inequality follows from Û ¼ Û, and the net inequality holds because its left-hand side is decreasing in Û «Û (note that 1 2«). Case 2: Ú does not schedule after time Ø. We estimate the charges to and ¼ separately. The charging ratio of is at most Ê by (2.5). We claim that the upward charge to ¼ is at most Û ¼. If ¼ is scheduled before Ø, then there is no charge. If ¼ Ø+2, then this follows by the NF-charge definition. Otherwise ¼ and thus Ö ¼ Ö Ø and ¼ is pending at Ø. Thus Û ¼ Û, as otherwise ¼ would be chosen as at time Ø (note that = ¼ by the case condition.) So the upward charge is at most Û «Û ¼ Û ¼ and the claim is proven. The total charge to ¼, including a possible EB-charge, is at most (1+ )Û ¼ and the charging ratio is at most Ê. ¾ = ÊÛ ¼ ¾¹ÍÒÓÖÑ ÁÒ ØÒ In this section we consider 2-uniform instances, where each job satisfies = Ö +2. Let É 1377 be the largest root of É + É ¾ 4É + 1 = 0. First, we prove that no online algorithm for this problem can be better than É-competitive. Net, we show that this lower bound is in fact tight. º½ ÄÓÛÖ ÓÙÒ In this sub-section we prove that no deterministic online algorithm for 2-uniform instances can have a competitive ratio smaller than É Recall that É is defined as the largest root of É + É ¾ 4É + 1 = 0. Fi some 0 2É 2. We define a sequence Ψ, = 12, as follows. For = 1, Ψ ½ = É 1. Inductively, for 1, let Ψ ½ = É(2 É)Ψ + 3É ¾ 5É + 1 É(2 É Ψ ) = (2 É)Ψ (É 1) ¾ 2 É Ψ where the second equality follows from É + É ¾ 4É + 1 = 0. ÄÑÑ º½ ÓÖ ÐÐ Û Ú 1 É Ψ É 1º ÙÖØÖÑÓÖ Ø ÕÙÒ Ψ ÓÒÚÖ ØÓ 1 ɺ ÈÖÓÓ Substituting Ψ = Þ + 1 É, we get Þ ½ = ¾ÉµÞ ½ Þ. If 0 Þ 2É 2, then 0 Þ ½ ¾É ¾É Þ. Thus 0 Þ 2É 2 for all, lim ½ Þ = 0, and the lemma follows. ¾ ÌÓÖÑ º¾ ÌÖ ÒÓ ØÖÑÒ Ø ÓÒÐÒ ÐÓÖØÑ ÓÖ Ø 2¹ÙÒÓÖÑ ÛØ ÓÑÔØØÚ ÖØÓ ÑÐÐÖ ØÒ Éº ÈÖÓÓ The proof is by contradiction. Assume that there eists a (É )-competitive algorithm, for some 0. We develop an adversary strategy that will force s ratio to be bigger than É. Throughout this proof, to simplify notation, we will identify jobs by their weight. Thus, when we say job Ü, we mean the job with weight Ü. Such job will be always either uniquely defined by the contet, or there will be two such identical jobs, in which case one of them can be chosen arbitrarily. At each step Ø, we distinguish ÓÐ ÔÒÒ Ó, that is, those that were released at time Ø 1 but not eecuted, from the newly released jobs. Without loss of generality, we can assume that there is always (ecept for the last step) eactly one old pending job. (All old pending jobs 8

9 ecept the heaviest one can be ignored. The situation with no old pending jobs can be simulated by pretending that we have an old pending job with weight 0.) By a ÓÒ ÙÖØÓÒ we mean a pair Ü Ý, where Ü, Ý denote the old pending jobs of and the adversary, respectively. Let Ψ be the sequence from Lemma 5.1. For 1 define = 1 Ψ É 1 and = 2 É Ψ (É 1) ¾ É Note that by the first part of Lemma 5.1 we have 1 for all. We now describe the adversary strategy. Initially, we issue two jobs of some arbitrary weight Ü ½ 0. Both and the adversary eecute one job Ü ½ and we enter the configuration Ü ½ Ü ½. Suppose now that after 1 iterations, the current configuration is Ü Ü. The adversary now follows the following steps: issue one job Ü (1) eecutes Ü eecute Ü, Ü and halt Ð ( eecutes Ü ) at the net time step issue Ü (2) eecutes Ü eecute Ü, Ü, Ü, and halt Ð ( eecutes Ü ) (3) at the net time step issue two jobs Ü ½ = Ü eecute Ü, Ü, Ü If eecutes first Ü and then Ü, then after step (3) it will eecute one job Ü ½ = Ü, and the new configuration will be Ü ½ Ü ½, and the game continues. The adversary strategy is illustrated in Figure 1, where we assume that Ü = 1, and to simplify notation we write = and =. Jobs are represented by line segments, with dark rectangles representing if and when the job is eecuted by, while the lightly shaded rectangles represent job eecutions by the adversary. If the game completes 1 iterations, then define Ò and Ú to be the gain of and the adversary in these 1 iterations (that is, ending in configuration Ü Ü.) We claim that for any, either the adversary wins the game before iteration, or else in iteration we have (É )Ò Ú Ψ Ü (1) The proof of (1) is by induction on. For = 1, (É )Ò ½ Ú ½ (É )Ü ½ Ü ½ = Ψ ½ Ü ½, as claimed. Suppose that (1) holds after iteration 1, that is, when the configuration is Ü Ü. If eecutes Ü in step (1), then, denoting by the sequence of jobs up to this step, using the inductive assumption, and substituting the formula for, we have (É )Ò () Ú() = (É )Ò Ú + (É ) Ü (Ü + Ü ) [Ψ 1 + (É 1) ]Ü 0 contradicting the (É )-competitiveness of. If eecutes Ü and then Ü in step (2), then, again, denoting by ¼ the sequence of jobs up to this step, using the inductive assumption, and substituting the formulas for, we have (É )Ò ( ¼ ) Ú( ¼ ) = (É )Ò Ú + (É )(Ü + Ü ) (Ü + Ü + Ü ) [Ψ + É 1 + (É 1) ]Ü 0 9

10 adversary move (no new jobs) adversary move a algorithm and adversary move a algorithm and adversary move algorithm move a adversary move 1100 a b algorithm and adversary move 1100 a b 1100 net iteration a b b algorithm and adversary move a b b adversary move 1100 algorithm move a b Figure 1: The adversary strategy. and this again contradicts the (É )-competitiveness of. The only other possibility is that will eecute first Ü, then Ü, and then it will have no choice but eecute Ü. In the new configuration Ü ½ Ü ½ we will have (É )Ò ½ Ú ½ (É )Ò Ú + (É )(Ü + Ü + Ü ) ( Ü + 2 Ü ) [Ψ + É(1 + + ) ( + 2 )]Ü = Ü Ψ ½ = Ü ½ Ψ ½ completing the proof of (1). From (1) and from Lemma 5.1, we obtain that for large enough we will have (É )Ò Ú Ψ Ü (1 É+ )Ü. But this contradicts the (É )-competitiveness of, since, denoting by ± the sequence of all jobs issued up to this time (including the pending jobs Ü ), we have ¾ (É )Ò (±) Ú(±) = (É )Ò Ú + (É )Ü Ü (1 É + )Ü + (É )Ü Ü = 0 º¾ ÍÔÔÖ ÓÙÒ In this section we present an online algorithm for 2-uniform jobs with competitive ratio É. Given that the 2-uniform case seems to be the most elementary case of unit job scheduling (without being trivial), our algorithm (and its analysis) is surprisingly difficult. Recall, however, that any algorithm for 2-uniform instances whose competitive ratio is better than Ô 2 cannot be memoryless (see [2]). In other words, in addition to the pending jobs, it needs to use some information about the jobs that were already eecuted or that epired. Further, when the adversary uses the strategy from the lower bound proof in the previous section, any online algorithm needs to behave in an essentially unique way in order to be É-competitive. Our algorithm was in fact designed to match 10

11 this optimal strategy, and then etended (by interpolation) to other adversarial strategies. Thus we suspect that the compleity of the algorithm is inherent in the problem and cannot be avoided. We start with some intuitions. Let be our online algorithm. Suppose that at a certain time Ø we have one old pending job Þ, and two new pending jobs with. In some cases, the decision which job to eecute is easy. If Þ, can ignore Þ and eecute in the current step. If Þ, can ignore and eecute Þ in the current step. If Þ, faces a dilemma: it needs to decide whether to eecute Þ or. In general, the choice will be made based on the ratio Þ. If Þ eceeds a certain threshold (possibly dependent on the past computation), we eecute Þ, otherwise we eecute. We can handle all three cases by using a parameter, say, and making the decision according to the following procedure: ÈÖÓÙÖ ËËÈ If Þ + (1 ) schedule Þ, otherwise schedule. To derive an online algorithm, say, we eamine the adversary strategy in the lower bound proof. Consider the limit case, when ½, and let = lim ½ = É(É 1) and = lim ½ = É(É 1) ¾. Assume that in the previous step two jobs Þ were issued, so that the current configuration is Þ Þ. If the adversary now issues a single job, then needs to do the following: if Þ, eecute Þ, and if Þ, then eecute. (The tie for = Þ can be broken either way.) Thus in this case we need to apply ËËÈ «with the threshold «= 1 = (É 1)É. Then, suppose that in the first step eecuted Þ, so that in the net step is pending. If the adversary now issues a single job, then (assuming ) must to do the following: if Þ, eecute, and if Þ, then eecute. Thus in this case we need to apply ËËÈ with the threshold = = É 1. Suppose that we eecute. In the lower-bound strategy, the adversary would now issue two jobs in the net step. But what happens if he issues a single job, say? Routine calculations show that, in order to be É-competitive, needs to use yet a different parameter in ËËÈ. This parameter is not uniquely determined, but it must be at least = (3 2É)(2 É), which is greater than É 1. Further, it turns out that the same value can be used on subsequent single-job requests. Our algorithm is derived from the above analysis: on a sequence of single-job requests in a row, use ËËÈ with parameter «in the first step, then in the second step, and in all subsequent steps. In general, of course, two jobs can be issued at each step (or more, but only two heaviest jobs need to be considered.) We think of an algorithm as a function of several arguments. The values of this function on the boundary are determined from the optimal adversary strategy, as eplained above. The remaining values are obtained through interpolation. We now give a formal description of our algorithm. We define constants «= É 1 É 027 = É = 3 2É 2 É 039 We also use two functions () = min 1 ««Æ( ) = «+ (1 )[ + ( )()] where 0 1 and «. Note that for the parameters, within their ranges, we have 0 () 1, «Æ( ). Function Æ also satisfies Æ(1 ) = «, Æ(0 ) for any, and Æ(0 «) =, Æ(0 ) =. ÐÓÖØÑ ËÛغ Fi a time step Ø. Let Ù Ú be the two jobs released at time Ø 1 (we assume that Ù Ú) and (where ) be the jobs released at time Ø. Let Þ ¾ Ù Ú be the old pending 11

12 job. Let also be the parameter of ËËÈ used at time Ø 1 (initially = «). Then ËÛØ chooses the job to eecute as follows: If Þ = Ù run ËËÈ with = Æ( Ú Ù ), If Þ = Ú run ËËÈ with = «. ÌÓÖÑ º ÐÓÖØÑ ËÛØ É¹ÓÑÔØØÚ ÓÖ Ø 2¹ÙÒÓÖÑ ÛÖ É 1377 Ø ÐÖ Ø ÖÓÓØ Ó É + É ¾ 4É + 1 = 0º The proof of the theorem is by tedious case analysis of an appropriate potential function, and is included in the appendi. ÓÒÐÙ ÓÒ We established the first upper bound better than 2 on the competitiveness of deterministic scheduling unit jobs to maimize weighted throughput. There is still a wide gap between our upper bound of 1939 and the best known lower bound of. Closing or substantially reducing this gap is a challenging open problem. We point out that our algorithm ÒÐ is not memoryless, as it uses one bit of information about the previous step. Whether it is possible to reduce the ratio of 2 with a memoryless algorithm remains an open problem. Another intriguing open problem is to determine the competitiveness of the case when the jobs are similarly ordered, or when they all have the same span. We gave an algorithm for this version with competitive ratio The lower bound of applies to similarly ordered jobs. As for the uniform span case, to the best to our knowledge, the best lower bound is 1377 given in this paper for the 2-uniform case. ÊÖÒ [1] N. Andelman, Y. Mansour, and A. Zhu. Competitive queueing policies in QoS switches. In ÈÖÓº ½Ø ËÝÑÔº ÓÒ ÖØ ÐÓÖØÑ Ëǵ, pages ACM/SIAM, [2] Y. Bartal, F. Y. L. Chin, M. Chrobak, S. P. Y. Fung, W. Jawor, R. Lavi, J. Sgall, and T. Tichý. Online competitive algorithms for maimizing weighted throughput of unit jobs. In ÈÖÓº ¾½ Ø ËÝÑÔº ÓÒ ÌÓÖØÐ ÔØ Ó ÓÑÔÙØÖ ËÒ ËÌ˵, to appear. [3] S. Blake, D. Black, M. Carlson, E. Davies, Z. Wang, and W. Weiss. An architecture for differentiated services. Internet RFC [4] F. Y. L. Chin and S. P. Y. Fung. Online scheduling for partial job values: Does timesharing or randomization help? ÐÓÖØÑ, pages , [5] B. Hajek. On the competitiveness of online scheduling of unit-length packets with hard deadlines in slotted time. In ÓÒÖÒ Ò ÁÒÓÖÑØÓÒ ËÒ Ò ËÝ ØÑ, pages , [6] A. Kesselman, Z. Lotker, Y. Mansour, B. Patt-Shamir, B. Schieber, and M. Sviridenko. Buffer overflow management in QoS switches. In ÈÖÓº Ö ËÝÑÔº ÌÓÖÝ Ó ÓÑÔÙØÒ ËÌǵ, pages ACM,

13 Ò ÐÓÖØÑ ÓÖ Ø ¾¹ÍÒÓÖÑ Recall that the algorithm uses the following constants «= É 1 É 027 = É = 3 2É 2 É 039 where É is the largest root of É + É ¾ 4É + 1 = 0. We also use the following two functions () = min 1 ««Æ( ) = «+ (1 )( + ( )()) where 0 1. Note that these functions satisfy the following properties: () 0 for any Æ : [01] [«] [«] Æ(1 ) = «for any Æ(0 ) for any ÐÓÖØÑ ËÛغ Fi a time step Ø. Let Ù Ú be the two jobs released at time Ø 1 (we assume that Ù Ú) and (where ) be the jobs released at time Ø. Let Þ ¾ Ù Ú be the old pending job. Let also be the parameter of ËËÈ used time Ø 1 (initially = «). Then ËÛØ chooses the job to eecute as follows: If Þ = Ù run ËËÈ with = Æ( Ú Ù ), If Þ = Ú run ËËÈ with = «. ÒÐÝ º We use a potential function argument to show that ËÛØ is É-competitive. The proof for the 2-uniform cases is based on a ÔÓØÒØÐ ÙÒØÓÒ argument. We define below a potential function Φ that maps all possible configurations into real numbers. (In general, a configuration at a given step may encode all information about the computation up to this step, both for the algorithm and the adversary. Here it is sufficient only to remember the jobs issued in last two steps.) Intuitively, the potential represents ¼ savings at a given step. At each time step, an online algorithm and the adversary eecute a job. The proof is based on the following lemma which can be proven by a simple summation over all steps. ÄÑÑ º½ ÄØ Ò ÓÒÐÒ ÐÓÖØÑ ÓÖ ÙÐÒ ÙÒØ Ó º ÄØ Φ ÔÓØÒØÐ ÙÒØÓÒ ØØ 0 ÓÒ ÓÒ ÙÖØÓÒ ÛØ ÒÓ ÔÒÒ Ó Ò Ø ØÔ Ø Ê Ò Ú + Φ (2) ÛÖ Φ ÖÔÖ ÒØ Ø Ò Ó Ø ÔÓØÒØÐ Ò Ò Ú ÖÔÖ ÒØ ³ Ò Ø ÚÖ ÖÝ Ò Ò Ø ØÔº ÌÒ Ê¹ÓÑÔØØÚº Each configuration is defined by the parameter of ËËÈ used in the previous step, the jobs released in the previous step, and the pending jobs Ü Ý ¾ Ù Ú of the algorithm and the adversary respectively. The configuration is thus described by a five-tuple Ù Ú Ü Ý. To simplify the 13

14 notation we write Φ ÜÝ (Ù Ú ) instead of Φ(Ù Ú Ü Ý ). We define Φ ÙÙ (Ù Ú ) = Ú¾ Ù (1 É ()) + Ù() Φ ÙÚ (Ù Ú ) = Ú(É 1 + ()) + Ù( () 1É) Φ ÚÙ (Ù Ú ) = Ù ÉÚ Φ ÚÚ (Ù Ú ) = Ú(1 É) where = 2 É ¾ = «(É 1) = É ¾ + 3É 6. Note that 0 Recall that by and we denote the jobs released at this step and that we assume that. By Lemma A.1 it is sufficient to prove that Φ ÜÝ (Ù Ú ) + É Ò ËÛØ Ú + Φ Ü¼ Ý ¼( ¼ ) (3) where ¼ are the parameters of ËËÈ used in this and the net steps, and Ü Ý Ü ¼ Ý ¼ are the pending jobs of the algorithm and the adversary in this and the net steps respectively. We start by introducing several lemmas and facts that let us avoid a number of technical issues later in the proof. ÄÑÑ º¾ Φ ÜÝ (Ù Ú ) Ø Ø ÓÐÐÓÛÒ ÔÖÓÔÖØ ÓÖ ÒÝ Ù Ú ÛÖ Ú Ùµ (1) Φ ÙÙ (Ù Ú ) Φ ÙÚ (Ù Ú ) (2) Φ ÚÙ (Ù Ú ) Φ ÚÚ (Ù Ú ) (3) Φ ÚÙ (Ù Ú ) Ù = Φ ÚÚ (Ù Ú ) Ú ÈÖÓÓ (1) We have to show that Ú ¾ Ù (1 É ()) + Ù() Ú(É 1 + ()) + Ù( () 1É) Note that the epression on the left-hand side is minimized for Ú = Ù, and the epression on the right-hand side is maimized for Ú = Ù. We may thus assume that Ú = Ù and reduce the inequality to 1 É É 1 1É which holds as + 1É = 2É 2. (2) Note that Φ ÚÙ (Ù Ú ) = Ù ÉÚ Ú ÉÚ = Φ ÚÚ (Ù Ú ). (3) We have Φ ÚÙ (Ù Ú ) Ù = Ù ÉÚ Ù = ÉÚ +(Ú Ú) = (1 É)Ú Ú = Φ ÚÙ (Ú Ú ) Ú. ¾ ÄÑÑ º ÄØ É Ø ÖØ Ø ÖÓÓØ Ó É + É ¾ 4É + 1 = 0º Ì ÓÐÐÓÛÒ ÒÕÙÐØ ÓÐ For the purpose of this analysis, it is also convenient to define Ð = () 7É ¾ 6É 5 0 (4) 7É ¾ + 14É 6 0 (5) É ¾ + 3É 6 0 (6) 9É ¾ + 32É 27 0 (7) ( Ð) = «+ (1 )( + ( )Ð) = Æ( ) Note that satisfies the following properties: (1 Ð) = «for any Ð (8) (0 Ð) for any Ð 0 (9) 14

15 Solutions of the following linear programs will be used to obtain bounds on certain epressions. In these programs denotes a constant such that «. ÄÔ1 : ÄÔ2 : ÄÔ3 : ma (1 É) É ma (1 É)( + ) ma + s.t. s.t. s.t. + (1 ) 1 + (1 ) 1 + (1 ) (É 1 + ( )) = (1 ( ) 1É) = solution: = 0 solution: = 1 solution: = 0 = 1 = 1 = 1 We are now ready to prove that (3) holds. Let ËËÈ be the algorithm eecuted in the previous step. We distinguish the following cases. Case 1: Suppose that Ù was eecuted in the previous step. Then Ú is pending for ËÛØ and the algorithm applies ËËÈ «. Case 1a: If Ú «+ (1 inequalities «) then ËËÈ «schedules Ú, and we have to verify the following four Ù + Φ ( «) Φ ÚÙ (Ù Ú ) + ÉÚ ma + Φ ( «) Ú + Φ ( «) Φ ÚÚ (Ù Ú ) + ÉÚ ma + Φ ( «) Note that the first and the third inequalities are equivalent by Lemma A.2. Also, the last inequality implies the second one by the same lemma. Without the loss of generality we may assume that Ù = 1. After substitution, the third and the fourth inequalities reduce to (1 É)Ú + ÉÚ Ú + ¾ (1 É) (10) (1 É)Ú + ÉÚ + (É 1 ) É (11) The right-hand sides of inequality (10) is maimized for = 0 and the inequality holds trivially. After rearranging, and using the fact that 1 = Ù Ú «+ (1 «), inequality (11) reduces to 0 (1 É) which holds. Case 1b: If Ú «+ (1 «) then ËËÈ «schedules, and we need to verify the following inequalities Φ ÚÙ (Ù Ú ) + É Ù + Φ ( «) ma + Φ ( «) Φ ÚÚ (Ù Ú ) + É ma Ú + Φ ( «) + Φ ( «) 15

16 Note that the first and the third inequalities are equivalent by Lemma A.2. Also, the last inequality implies the second one by the same lemma. We may assume that Ù = 1, and the remaining inequalities reduce to (1 É)Ú + É Ú + ( É) (12) (1 É)Ú + É + (1 É) (13) After using the case condition, the inequality (12) reduces to 0 «and holds. The inequality (13) reduces to Ú + which holds by the case condition. Case 2: Now we assume that Ù was not eecuted in the previous step. So Ù is now the pending job for ËÛØ, and the algorithm applies ËËÈ Ðµ, where = Ú Ù. Case 2a: If Ù ( Ð) + (1 ( Ð)) then the algorithm eecutes and we need to show the following four inequalities Φ ÙÙ (Ù Ú ) + É Ù + Φ ( ( Ð)) ma + Φ ( ( Ð)) Φ ÙÚ (Ù Ú ) + É ma Ú + Φ ( ( Ð)) + Φ ( ( Ð)) Once again, note that the last inequality implies the second one by Lemma A.2. We assume that Ù = 1, and the remaining inequalities reduce to Ú ¾ (1 É Ð) + Ð + É 1 + É (14) Ú(É 1 + Ð) Ð 1É + É Ú + É (15) Ú(É 1 + Ð) Ð 1É + É + (1 É) (16) ÈÖÓÓ Ó ÒÕÙÐØÝ (16). We start by regrouping the terms to reduce (16) to the following inequality Ú ¾ (1 É Ð) + Ð 1 + (1 É) É We now face the following problem: we wish to maimize the right-hand side of this inequality, subject to the case condition and conditions 0,. For now, the term (Ú Ð) from the case condition is treated like a constant, say, where «. It is easy to see now, that the above problem can be formulated as a linear program with variables. In fact, it is equivalent to the program ÄÔ1. We conclude that it is enough to verify the inequality for = 1(Ú Ð) = 0, as this is the solution of ÄÔ1. Further, the left-hand side of (16) is minimized for Ú = 0 and Ð = 1, so overall this inequality reduces to 2(É 1) + 1É which holds. ÈÖÓÓ Ó ÒÕÙÐØÝ (14). Once again, we start by regrouping the terms to reduce (14) to Ú(É 1 + Ð) Ð 1É Ú + (1 É) É As in the proof of (16) we notice that it is enough to verify the inequality for = 1(Ú Ð) = 0, as this is the solution of ÄÔ1. The inequality now reduces to [«Ú + (1 Ú)( + ( )Ð)][ Ú ¾ (É 1 + Ð) + Ð 1] + É

17 Let Ä(Ú Ð) denote the epression on the left-hand side of this inequality. The function Ä is a polynomial of third degree in Ú. The coefficient of Ú is (É 1 + Ð)( «+ ( )Ð) 0, so to prove that the inequality holds it is enough to show that for any Ð ¾ [01], Ä(0 Ð) 0, Ä(1 Ð) 0, and that the second derivative ¾ Ä Ú ¾ (Ú Ð) 0. Since Ä(0 Ð) = [ + ( )Ð][Ð 1] + É 1 is quadratic in Ð, Ä(00) = 0, and the coefficient of Ð ¾ is ( ) 0, it is sufficient to show that Ä ¼Ðµ (0) 0 to prove that Ä(0 Ð) 0. Indeed, Ä ¼Ðµ (0) = ( ) +, and the inequality Ð Ð ( ) + 0 reduces to 7É ¾ 6É 5 0 which holds. Since, Ä(1 Ð) = 0, it remains to verify that the second derivative ¾ Ä Ú ¾ (Ú Ð) 0. Since, ¾ Ä (Ú Ð) = (É 1 + Ð)[4Ú( «+ ( )Ð) 2(«Ú + (1 Ú)( + ( )Ð))] Ú¾ it is enough to verify that «Ú + (1 Ú)( + ( )Ð) 2Ú( «+ ( )Ð) 0 Since this is a linear inequality and the coefficient of Ú is 3(«( )Ð) 0, it is enough to verify it for Ú = 1. The inequality now reduces to 3«2( + ( )Ð) 0, and it is sufficient to show that it holds for Ð = 1 as the coefficient of Ð is 2( ) 0 and the inequality is linear in Ð. The inequality now reduces to É ¾ + 3É 6 0 and holds. ÈÖÓÓ Ó ÒÕÙÐØÝ (15). We start by regrouping the terms to obtain reduce (15) to Ú(É 1 + Ð) Ð 1É (1 É) + (1 É) Similarly to the proof of (16) we notice that it is enough to verify the inequality for = = 1, as this is the solution of ÄÔ2. It is now enough to prove (Ú Ð)(Ú(É 2 + Ð) Ð 1É) + É 1 0 The coefficient of Ð ¾ is (1 Ú)( )(Ú 1) 0 so it is enough to verify that the inequality holds for Ð = 01. For Ð = 0 the inequality reduces to («Ú + Ú)( Ú(2 É + ) 1É) + É 1 0 (17) This is a quadratic inequality in Ú. The coefficient of Ú ¾ is ( «)(2 É+) 0 so it is enough to verify that the derivative with respect to Ú is negative for Ú = 1. Let denote the epression on the left-hand side of (17). Then ¼ (Ú) = («)( Ú(2 É + ) 1É) («Ú + (1 Ú))(2 É + ) The inequality ¼ (1) 0 can be reduced to 4 3É, so it holds. For Ð = 1 the inequality reduces to («Ú + (1 Ú) )( Ú(2 É) 1É) + É

18 This again is a quadratic inequality in Ú with coefficient of Ú ¾ equal ( «)(2 É) 0. So to show that it holds it is enough to verify that the derivative in Ú is negative for Ú = 1. The derivative of the epression on the left-hand side is («)( Ú(2 É) 1É) + («Ú + (1 Ú) )(É 2) and for Ú = 1 it reduces to 7É ¾ 14É which holds. Case 2b: Suppose that Ù ( Ð) + (1 ( Ð)). ËËÈ Ðµ eecutes Ù. We need to verify the following inequalities: Ù + Φ ( ( Ð)) Φ ÙÙ (Ù Ú ) + ÉÙ ma (18) + Φ ( ( Ð)) Ú + Φ ( ( Ð)) Φ ÙÚ (Ù Ú ) + ÉÙ ma (19) + Φ ( ( Ð)) Note that the last inequality implies the second one by Lemma A.2. We may assume that Ù = 1, and the remaining inequalities reduce to Ú ¾ (1 É Ð) + Ð + É 1 + ¾ (1 É ((Ú Ð))) + ((Ú Ð)) (20) Ú(É 1 + Ð) Ð 1É + É Ú + ¾ (1 É ((Ú Ð))) + ((Ú Ð)) (21) Ú(É 1 + Ð) Ð 1É + É + (É 1 + ((Ú Ð))) + ( ((Ú Ð)) 1É) (22) ÈÖÓÓ Ó ÒÕÙÐØÝ (20). It is enough to show that (1 Ú)(É + Ð 1) ((Ú Ð)). By the case condition 1(Ú Ð) so it is enough to show (Ú Ð)(Ú 1)(É + Ð 1) ((Ú Ð)) We distinguish two cases. If (Ú Ð) then ((Ú Ð)) = 1, (1 Ú)( «+ ( )Ð) «and it is enough to show ( «)(É + Ð 1) ( «+ ( )Ð) This inequality is linear in Ð so it is enough to verify it for Ð = 01. Actually it is enough to verify it for Ð = 0 as ( «) ( ) 0. For Ð = 0 the inequality reduces to (É 1) which holds. If (Ú Ð) then ((Ú Ð)) = ((Ú Ð) «)( «), and we need to show ( «)(Ú Ð)(1 Ú)(É + Ð 1) ((Ú Ð) «) Note that (Ú Ð) «= (1 Ú)( «+ ( )Ð), so it is enough to show ( «)(Ú Ð)(É + Ð 1) ( «+ ( )Ð) (23) This is a linear inequality in Ú. So it is enough to verify it for Ú = 01. Actually, it is enough to verify it for Ú = 1, as the coefficient of Ú is ( «)(É+Ð 1)(«( )Ð) 0. The inequality 18

19 reduces to ( «)«(É + Ð 1) («+ ( )Ð) 0. This inequality is linear in Ð and the coefficient of Ð is [( «)«( )] 0, so it is enough to verify it for Ð = 0. The inequality reduces now to «(É 1) 0 which holds. ÈÖÓÓ Ó ÒÕÙÐØÝ (21). It is enough to show that (Ú 1)(É 2 + Ð) ((Ú Ð)) We distinguish two cases. If (Ú Ð), then ((Ú Ð)) = 1, and (1 Ú)( «+( )Ð) «. The inequality reduces to ( )(2 + É Ð) ( «+ ( )Ð) 0 This is a linear inequality in Ð. The coefficient of Ð is ( «) ( ) 0, so it is enough to verify the inequality for Ð = 1. The inequality now reduces to É 1, which holds. If (Ú Ð), then ((Ú Ð)) = ((Ú Ð) «)( «), and the inequality reduces to (Ú Ð)(2 + É Ð)( «) ( «+ ( )Ð) It is now enough to prove that 2 + É Ð É + Ð 1 for any Ð ¾ [01], to show that this inequality follows from (23). For Ð = 0 the inequality reduces to 3 + 2É, which holds. For Ð = 1 the inequality reduces to 3 2É +, which also holds. ÈÖÓÓ Ó ÒÕÙÐØÝ (22). We first observe that the problem of maimizing the right-hand side of (22) subject to case conditions can be formulated as the linear program ÄÔ3. It is therefore enough to verify the inequality for = 1(Ú Ð) = 0. As in the proof of (20) we distinguish two cases. If (Ú Ð) the inequality reduces to (Ú Ð)(Ú(É 1 + Ð) Ð 1É + É) 3 2É This is a quadratic inequality in Ú. The coefficient of Ú ¾ is (É 1 + Ð)(«( )Ð) 0 so we need to verify that it holds for Ú = 01. For Ú = 0 the inequality reduces to [ +( )Ð]( Ð 1É+É) 3 2É. This is a quadratic inequality in Ð. For Ð = 0 it reduces to ( 1É + É) 3 2É which holds. For Ð = 1 it reduces to (2 É) 3 2É which also holds. The coefficient of Ð ¾ is ( ) 0 so the inequality holds. For Ú = 1 the inequality reduces to «3 2É which holds. If (Ú Ð) the inequality reduces to (Ú Ð)(Ú(É 1 + Ð) Ð 1É + É) 1 1É ((Ú Ð) «)( «) Since the coefficient of Ú ¾ is (É 1 + Ð)(«( )Ð) 0 it is enough to verify the inequality for Ú = 0 and Ú = 1. For Ú = 1 the inequality reduces to «1 1É which holds. For Ú = 0 the inequality reduces to [ + ( )Ð]( Ð 1É + É) 1 1É ( «+ ( )Ð)( «) This is a quadratic inequality in Ð. For Ð = 0 it reduces to (É 1É) 1 1É which holds. For Ð = 1 it reduces to (2 É) 1 1É ( «)( «), which in turn reduces to 9É ¾ + 32É 27 0, which holds. The coefficient of Ð ¾ is ( )( ) 0, so the inequality holds. This completes the proof of inequality (3). The É-competitiveness of ËÛØ follows now from Lemma A.1. 19