Tight lower bounds for semi-online scheduling on two uniform machines with known optimum

CEJOR https://doi.org/10.1007/s10100-018-0536-9 ORIGINAL PAPER Tight lower bounds for semi-online scheduling on two uniform machines with known optimum György Dósa 1 Armin Fügenschuh 2 Zhiyi Tan 3 Zsolt Tuza,5 Krzysztof Węsek 6,7 Springer-Verlag GmbH Germany, part of Springer Nature 2018 Abstract This problem is about scheduling a number of jobs on two uniform machines with given speeds 1 and s 1, so that the overall finishing time, i.e., the makespan, is earliest possible. We consider a semi-online variant (introduced for equal speeds) by Azar and Regev, where the jobs are arriving one after the other, while the scheduling algorithm knows the optimum value of the corresponding offline problem. One can B Armin Fügenschuh fuegenschuh@b-tu.de György Dósa dosagy@almos.vein.hu Zhiyi Tan tanzy@zju.edu.cn Zsolt Tuza tuza@dcs.uni-pannon.hu Krzysztof Węsek wesekk@hsu-hh.de; wesekk@mini.pw.edu.pl 1 Department of Mathematics, University of Pannonia, Veszprém, Hungary 2 Brandenburgische Technische Universität Cottbus-Senftenberg, Platz der Deutschen Einheit 1, D-0306 Cottbus, Germany 3 Department of Mathematics, Zhejiang University, Hangzhou 310027, People s Republic of China Department of Computer Science and Systems Technology, University of Pannonia, Veszprém, Hungary 5 Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences, Budapest, Hungary 6 Helmut Schmidt University/University of the Federal Armed Forces Hamburg, Holstenhofweg 85, 2203 Hamburg, Germany 7 Faculty of Mathematics and Information Science, Warsaw University of Technology, ul. Koszykowa 75, 00-662 Warszawa, Poland

G. Dósa et al. ask how close any possible algorithm could get to the optimum value, that is, to give a lower bound on the competitive ratio: the supremum over ratios between the value of the solution given by the algorithm and the optimal offline solution. We contribute to this question by constructing tight lower bounds for all values of s in the intervals [ 1+ 21, 3+ 73 8 ] [1.3956, 1.3] and [ 5 3, + 133 9 ] [ 5 3, 1.7258], except a very narrow interval. Keywords Semi-online scheduling Makespan minimization Machine scheduling Lower bounds 1 Introduction We deal with the following scheduling problem: Given are two uniform machines, denoted by M1 and M2, that are both capable of processing incoming jobs. They only differ in the processing speed. We assume that machine M1 is working at some unit speed 1, and machine M2 is s times faster, with s 1. Hence when machine M1 processes a job of length L, then machine M2 can handle this job in L/s time. There is a number of incoming jobs of various lengths. The task is to assign the jobs to the two machines. It is desired to finish all incoming jobs as early as possible, that is, to minimize the makespan. In the offline variant of this problem, all jobs to be assigned are fully known in advance. If nothing is known about the jobs beforehand, we are faced with an online problem. We deal here with a semi-online problem, that means, the jobs are still not known individually, but we assume to have some further overall knowledge. In particular, we assume that the value of the solution to the corresponding offline problem, which we denote by OPT, is known in advance. Assume there is an algorithm A that solves the semi-online variant of the problem. This algorithm receives, besides the value of OPT, one job after the other in an unknown order and must now immediately decide to which of the two machines this job should be assigned (later changes are not possible). This algorithm A finally arrives at a makespan value M not smaller than OPT. Still, M can be compared to OPT by considering the OPT M ratio. Of course, it is preferred to have an algorithm where this ratio is close to 1 for any given input data. Thus the question arises, how close to 1 can we get? Is it possible to construct an algorithm that reaches this value, or is there a theoretical lower bound well above 1, that no algorithm will ever undercut, no matter how hard it tries? The paradigm of revealing the instance of a problem in parts, and the decision has to be made as the part is revealed, is naturally motivated by many real-world applications. As it was mentioned before, if no information is given in advance, then we call such a problem online, and if some partial information about the instance is known beforehand, then the scheme is called semi-online. The most common way of measuring the quality of an online or semi-online algorithm uses the notion of the mentioned competitive ratio. Assume that we are dealing with a minimization problem and the (offline) optimum value for an instance I is equal to OPT. Formally, an online algorithm is said to be r-competitive if for any instance I the value of the result of the algorithm A satisfies A OPT r, and the competitive ratio of an algorithm is defined as

Tight lower bounds... the infimum of such r. The question is: what is the best possible ratio for our (online or semi-online) problem? Formally, we would ask for the optimal competitive ratio, that is the infimum over all numbers r for which there exists an r-competitive algorithm. An algorithm is optimal if its competitive ratio matches the lower bound (i.e., value for which no algorithm can have a better competitive ratio). 1.1 Survey of the literature The online and various semi-online variations of the problem where a set of jobs has to be scheduled on m machines (with m N, m 2) and the objective to minimize the maximal completion time have been studied for decades. We assume (except where noted otherwise) that any job has to be done on one machine, and jobs arrive in a list one after another (list-online scheme). For more information we refer to the survey of Tan and Zhang (2013). We will first deal with the basic case of identical machines (and thus the speed of all machines is the same) in the pure online scheme. A classic work by Graham (1966) is probably the first step in this direction. It gives a (2 m 1 )-competitive algorithm, by using a heuristic of assigning each task to the least loaded machine. It was proved by Faigle et al. (1989) that for m equal to 2 or 3 this algorithm is in fact optimal. In the case of arbitrary m, many papers appeared in order to decrease the gaps between lower and upper bounds. For the lower bound, Gormley et al. (2000) showed that no online algorithm can have competitive ratio better than 1.8536 generally (when m can be arbitrarily large). For the upper bound, Albers (1999) proposed a 1.9230-competitive algorithm for any m, and if m tends to infinity, then work of Fleischer and Wahl (2000) gives an algorithm with competitive ratio tending to 1.9201. In the more general case of machines with arbitrary speeds, the best general bounds are the following. Berman et al. (2000) provided a 5.828-competitive algorithm, and Ebenlendr and Sgall (2015) proved a lower bound of 2.5650. If m = 2 (and the speeds are 1 and s), then the greedy strategy of Graham is again useful in every step we choose the machine which will finish the current job as soon as possible. Epstein et al. (2001) showed that such an algorithm is optimal and has competitive s ratio min{1 + 1+s, 1 + 1 s }. We will later see that in semi-online variants the situation is much more complex. How much can we gain if some information is known in advance? Here we are interested in the semi-online scheme when only the optimal offline value is known (OPT version, for short), although we wish to mention a strong relation with another semi-online version of the described scheduling problem, in which only the sum of jobs is known (SUM version) Kellerer et al. (1997), Angelelli et al. (2007, 2008), Ng et al. (2009), Dósa et al. (2011), Lee and Lim (2013), Kellerer et al. (2015). Namely, for any given m uniform (possibly non-identical) machines the optimal competitive ratio for the OPT version is at most the competitive ratio of the SUM version (see Dósa et al. 2011; for equal speeds it was first implicitly stated by Cheng et al. 2005). Azar and Regev (2001) were the first to investigate the OPT version for identical machines (under the name of bin stretching), although the observation about the relation with the SUM version implies that the first upper bound of 3 for the case of two

G. Dósa et al. identical machines follows from the work of Kellerer et al. (1997) (now it is known to be optimal). Azar and Regev (2001), Kellerer and Kotov (2013), Gabay et al. (2015) and Böhm et al. (2015) continued the progress for the case of more than two identical machines. Then, Böhm et al. (2017a) deals with an arbitrary number of bins and gives an algorithm with a stretching factor 2 3, and Böhm et al. (2017b) deals with the special case where the number of bins is three. It gives an algorithm with a stretching factor 11 8 = 1.3750 (this is the current champion for this problem), and also proves that 15 11 = 1.3636 is a lower bound for the problem. These listed publications also indicate that it is very hard to give tight result for such a problem. Since we are interested in the OPT version with non-identical speeds on two uniform machines, we will state the previous results in terms of s (recall that the speeds are 1 and s). Epstein (2003) was first to investigate this problem. She proved the following bounds for the optimal competitive ratio r (s): r (s) ] [ 3s+1 3s, 2s+2 2s+1 [ s ( 3 + 65 20 for s [1, q E 1.123] ] r 2s+2 (s) ), 2s+1 for s [ q E, 1+ 65 8 1.1328 ] r (s) = 2s+2 2s+1 for s [ 1+ 65 8, 1+ 17 1.2808 ] r r (s) = s for s [ 1+ 17, 1+ 3 2 1.3660 ] (s) : [ ] r (s) 2s+1 2s, s for s [ 1+ 3 2, 2 1.12 ] [ ] r (s) 2s+1 2s, s+2 s+1 for s [ 2, 1+ 5 2 1.6180 ] [ ] r (s) s+1 2, s+2 s+1 for s [ 1+ 5 2, 3 1.7321 ] r (s) = s+2 s+1 for s 3 where q E is the solution of 36x 135x 3 + 5x 2 + 60x + 10 = 0. It means that her bounds are proven optimal on some intervals, where the upper and lower bounds coincide. It will turn out, however, that some of her bounds are also optimal, where she finds a lower bound (but her upper bound can be decreased), or she finds an upper bound (but her lower bound can be increased). On two intervals the question about the tight bound remained open after the work of Epstein. These are [ 1, 1+ 17 1.2808 ] and [ 1+ 3 2 1.3660, 3 ] ; we shall call them left interval and right interval, respectively. Ng et al. (2009) considered only the right interval, and presented algorithms giving the following upper bounds: 2s+1 2s for s [ 1+ 3 2, 1+ 21 1.3956 ] 6s+6 r s+5 for s [ 1+ 21, 1+ 13 3 1.5352 ] (s) 12s+10 9s+7 for s [ 1+ 13 3, 5+ 21 12 1.710 ] 2s+3 s+3 for s [ 5+ 21 12, 3 ]

Tight lower bounds... and provided the following lower bounds: 3s+5 2s+ for s [ 2, 21 3 1.5275 ] 3s+3 3s+1 for s [ 21 3, 5+ 193 12 1.57 ] s+2 r 2s+3 for s [ 5+ 193 12, 7+ 15 12 1.5868 ] (s) 5s+2 s+1 for s [ 7+ 15 19, 9+ 193 1 1.6352 ] 7s+ 7s for s [ 9+ 193 1, 5 3 1.6667] 7s+ s+5 for s [ 5 3, 5+ 73 8 1.6930 ] Hence from the work of Ng et al. it turned out that Epstein s lower bound is in fact tight on the very left end of the right interval. Finally, Dósa et al. (2011) considered the left interval, and solved the problem there almost completely, as they provided the following bounds: [ 8s+5 5+ 205 5s+5 for s 18, 1+ 31 r (s) 2s+2 2s+1 for s r (s) ] 6 1.096 [ 1+ 31 6, 1+ 17 1.2808 { 3s+1 3s for s [ 1, q D 1.0710 ] 7s+6 s+6 for s [ q D, 1+ 15 12 1.0868 ] where q D is the unique root of equation 3s 2 (9s 2 s 5) = (3s + 1)(5s + 5 6s 2 ). It means that Epstein s lower bound is tight on the left end of the left interval, and her upper bound is tight on the right end of the left interval, and the problem about the optimal value remained open in the middle for a small interval. For a visual summary (with our contribution included), see Figs. 1, 2, 3 and. Whenever the dotted line (that represents an upper bound) is on an unbroken line (that represents a lower bound), the optimal competitive ratio is known. In our work we revisit the right interval. As a consequence of all previous results, before this publication the optimality question was unsettled on the interval ( 1+ 21, 3) (1.3956, 1.7321). ] 1.2 Our contribution We deal with the semi-online two uniform machines scheduling problem with a known opt condition, This problem is studied on two parts of the right interval, [ 1+ 21, 3+ 73 8 ] [1.3956, 1.3] and [ 5 3, + 133 9 ] [ 5 3, 1.7258],forwhichwegive new lower bound constructions. (These numbers are solutions of certain equations, and will be introduced formally in the following section.) We apply an adversary strategy, that is, depending on the current assignment of a given job the adversary defines the next job that makes life complicated for the algorithm. We show that the input can

G. Dósa et al. Fig. 1 Our new lower bound in comparison with existing lower and upper bounds from Epstein (2003), Ng et al. (2009), Dósa et al. (2011, 2018) Fig. 2 Zooming into the left part of Fig. 1 always be continued in such a malicious way that any kind of algorithm will at some step exceed the lower bound on the makespan. Marry, and you will regret it; don t marry, you will also regret it; marry or don t marry, you will regret it either way says the Danish philosopher Søren Kierkegaard. He describes a decision situation from which two different possible choices will lead into the future. But no matter how the person decides, the continuation will not lead to a happy end. Both can be seen as unhappy situations (from the person s point of view who has to make the decision). This in our scheduling setting will correspond to

Tight lower bounds... Fig. 3 Zooming into the right part of Fig. 1 Fig. Zooming into the middle part of Fig. 3, [q 3, q 5 ] is the narrow interval the five final cases that are described in Sect. 3.1. The scheduler (the algorithm) can make a decision, similar to the person deciding for or against a marriage. No matter how the algorithm decides, it will lead to an unhappy situation, since it is possible to generate further jobs, based on its decision, in such a way that the competitive ratio of the algorithm is relatively high. It will not be obvious at first sight that the algorithm is always trapped in a situation that leads to an unhappy ending; it evolves over several

G. Dósa et al. rounds of further jobs that are determined in what we call intermediate cases.atmost eight jobs need to be generated to close the trap. The moves of the algorithm and the generation of jobs can be seen as a two-player game, such as chess. One player is the algorithm, and the other player is the adversary who constructs malicious jobs. What our result then shows is that the player of the algorithm is checkmate after at most eight moves. Our purpose was not to give a difficult construction; however, we were not able to get the tight bounds with fewer than eight jobs. Let us have a look at Fig. 1 between q 1 and q 7. The old lower bounds are denoted by solid blue and green lines, and the (old) upper bound is dotted green line. Our new lower bound is red, it is on the green dotted line, i.e., the bound is proven to be optimal. It means that together with the upper bound result of Ng et al. (2009), we obtain that our new lower bound is tight (that is, the algorithms presented by Ng et al. are optimal) for any value of s in one of these intervals: [ 1+ 21, 3+ 73 8 ] [1.3956, 1.3], [ 5 3, 13+ 129 30 ] [ 5 3, 1.693], and [ 31+ 8305 72, 5+ 21 12 ] [1.6963, 1.7103]. Between 5 3 and 3 there remain two intervals where the question for an optimal algorithm remains open: the first interval (we call it the narrow interval) between approximately 1.693 and 1.6963, see Fig., and the second interval between approximately 1.7103 and 3. In our recent paper Dósa et al. (2018), we proved that our new lower bound is also optimal in the left part of the second interval. Furthermore, it turns out that Epstein s previous lower bound is optimal in the right side of the second interval, see Fig. 3. Therefore, together with the result proven in Dósa et al. (2018), we completely solve the case of any speed in [ 5 3, 3] except the narrow interval, where our new lower bound is very close to the upper bound of Ng et al. (2009) 2 Preliminaries and notations 2.1 Definitions For the sake of simplicity let us assume without loss of generality that OPT = 1. (This can be done by normalization: If OPT differs from being unit, all values are divided by the value of OPT.) We denote the prescribed competitive ratio by r. Theload of a machine means the sum of lengths of jobs being (so far) assigned there, during the scheduling procedure. Let q 1 := Let q 2 := 21+1 1.3956, which is the positive solution of 2s+1 2s = s+5 6s+6. 73+3 8 1.30, which is the positive solution of s+5 6s+6 = 5s+2 s+1. Let q 3 := 13+ 129 30 1.693, which is the positive solution of 12s+10 9s+7 = 18s+16 16s+7. Let q := 30+7 186 7 approx1.6955, which is the positive solution of 18s+16 16s+7 = 3s+10 8s+7. Let q 5 := 31+ 8305 72 1.6963, which is the positive solution of 3s+10 8s+7 = 12s+10 9s+7. Let q 6 := 5+ 21 12 1.7103, which is the positive solution of 12s+10 9s+7 = 2s+3 s+3. Let q 7 := + 133 9 1.7258, which is the positive solution of 12s+10 9s+7 = s+1 2. We note that in this paper we do not consider speeds between q 2 and 5 3 1.6667. Now we define the ratio of our new lower bound. It differs on several intervals, where

Tight lower bounds... the speed is called small, smaller regular, bigger regular, smaller medium or bigger medium. For short, we call s regular if s is smaller regular or bigger regular; and we call s medium if s is smaller medium or bigger medium. Note that the interval where s is medium is the narrow interval. Let r(s) = r 1 (s) := 6s+6 s+5, if q 1 s q 2 1.30 i.e. s is small r 2 (s) := 12s+10 9s+7, if 5 3 s q 3 1.693 i.e. s is smaller regular r 3 (s) := 18s+16 16s+7, if q 3 s q 1.6955 i.e. s is smaller medium r (s) := 8s+7 3s+10, if q s q 5 1.6963 i.e. s is bigger medium r 2 (s) := 12s+10 9s+7, if q 5 s q 7 1.7258 i.e. s is bigger regular As we will show in the very end (cf. Theorem 2), this function will be our lower bound on the optimal competitive ratio. As abbreviation, we will often write r for r(s). Now we define the so-called safe sets. A safe set is a time interval on some of the machines, and it is safe in the sense that if the load of the machine is in this interval, this enables a smart algorithm to finish the schedule by not violating the desired competitive ratio. It means that no matter what kind of jobs will be given after this point by the adversary, the competitive ratio of the algorithm will not be bigger than it was prescribed before the jobs started to come. In other words, from the point of view of a lower bound construction (the adversary), we must avoid that the algorithm can assign the current job in a way that the increased load of some machine will be inside a safe set. Such safe sets were already applied in the literature, for example in Angelelli et al. (2008), Cheng et al. (2005), Dósa et al. (2011), and also in our recent paper Dósa et al. (2018), considering the same model that is treated here, but from the algorithmic side (yielding upper bound results). The term safe sets was coined in Dósa et al. (2011) (c.f. the figure on page 69 of Dósa et al. 2011). Safe sets S 1 and S 3 are defined on the second machine, while safe sets S 2 and S are defined on the first machine. We introduce notations for the top, bottom, and length of the safe sets. Thus let S i =[B i, T i ] and D i = T i B i, for any 1 i, see Fig. 5. The boundaries of safe sets S i are defined as follows. We will show in Sect. 2.2 that these intervals are well-defined, in particular, that D i > 0 for all 1 i. 1. B 1 = s + 1 r, and T 1 = rs, thus D 1 = (s + 1)(r 1), 2. B 2 = s + 1 sr, and T 2 = r, thus D 2 = (s + 1)(r 1), Fig. 5 Safe sets. The horizontal axis represents time

G. Dósa et al. 3. B 3 = 2s 2r rs + 2, and T 3 = s(r 1), thus D 3 = 2r 3s + 2rs 2,. B = s 2r 3rs + 3, and T = r 1, thus D = (3r )(s + 1). In fact, the boundaries are determined by the following system of equations. T 1 = rs (1) T 2 = r (2) T 1 + B 2 = T 2 + B 1 = 1 + s (3) T 3 = T 1 s () T = T 2 1 (5) B 3 = B 1 D 1 = B 1 (T 1 B 1 ) (6) B = B 2 D 3 = B 2 (T 3 B 3 ) (7) We introduce some notation as abbreviations that are used in the sequel: c = 1 D 2 (8) b = B 3 B (9) and let a = T B 3 (10) d = b c B, (11) where the latter two are later only used in the case of s 5 3.Leteand f be determined by the following system of equations: f + e = d (12) f e = a + c (13) By easy calculations we get the following expressions in terms of r and s: c = rs r + s + 2, b = 2rs 2s 1, a = rs + 3r 2s 3, d = 6rs + 3r 7s 6, e = 1 2 r 3s + 3rs 5 2, f = 5 2 r s + 3rs 7 2.

Tight lower bounds... 2.2 Properties In this subsection, we show that the safe sets are well defined in the sense that on a machine these sets follow and do not intersect each other. Moreover, we prove several properties of the boundaries of the safe sets and the expressions introduced above. The (rather technical) proof of Lemma 1 can be found in Appendix. Let s 0 = 633+23 5 1.686. Lemma 1 (i) If s is small, regular or medium, (ii) If s is small, (iii) If s is regular or medium, 0 < B < T < B 2 < T 2 s. (1) 0 c B 3 < T 3 < B 1 < T 1. (15) b 0. (16) c b + c B B 3 2c. (17) B 3 + B = T. (18) B 2 + B 1 + c. (19) (iv) If s is smaller regular or smaller medium, (v) If s s 0 and s is smaller regular, B 2 B 3 + B s 1. (20) T 3B B 2. (21) d max{e, f } a + c 0. (22) d B. (23) a c. (2) T b + 2c c. (25) c + B 2 2d b. (26) c + T 2d b. (27) (vi) If s s 0 and s is smaller regular, or s is smaller medium, (vii) If s is bigger medium, c + B 2 d B 3. (28) B = a + c. (29) b 8c + 7a B 3. (30)

G. Dósa et al. (viii) If s is bigger regular, e a + c. (31) c + T 2e b. (32) 3 Final cases and intermediate cases The general idea to show that no semi-online algorithm knowing s and OPT can ever be better than the ratio r(s) is to construct a malicious sequence of jobs, forcing any algorithm to schedule them in such a way that they have a makespan of at least r.we construct this sequence iteratively, depending on the previous assignment choices of the algorithm. This leads to a number of cases that need to be considered separately. Some cases are final in this sense (similar to an endgame in chess): When entering the case, the algorithm is trapped, because we can then construct one or a few more jobs, which make the algorithm overshoot the desired makespan. That is, either the load on M1 is no less than T 2 = r, or the load on M2 is no less than T 1 = rs. The other cases are intermediate, since it can be viewed as a step to the final cases. We will denote by L 1 and L 2 the current load of machine M1 and M2, respectively. Moreover let L 1 and L 2 denote the increased load of that machine if the current job is assigned there. No more than two jobs arrive during each case. Therefore, we represent jobs with their size if there is no confusion. The following observation, whose proof is trivial, will be used frequently during the definition of final and intermediate cases. Observation 1 Given a set of jobs with total size 1 + s, if there exists a subset of jobs with total size 1 or total size s, then OPT = 1. 3.1 Final cases We will introduce below five final cases. The first four are valid for any considered s (small, regular or medium), while the last one is only valid when s is small. For each situation an adversary strategy will be described. Case F1 L 2 T 3 and L 1 + L 2 1. The next and last two jobs are B = s and C = 1 (L 1 + L 2 ) 0. Since L 1 + L 2 + B + C = 1 + s and B = s, OPT= 1 by Observation 1. IfB is assigned to M2, we get L 2 T 3 +s = T 1 by (), thus we are done. Otherwise B is assigned to M1. Then L 1 s T 2 by (1), and we are done again. Case F2 L 1 = B 2, L 2 B 3, and there exists a subset of already assigned jobs with total size c. The next and last jobs are B = D 2 > 0 and C = 1 + s (L 1 + L 2 + D 2 ). Since L 1 + L 2 + B + C = 1 + s and B + c = 1by(8), we get OPT = 1by Observation 1. Note that C = (T 2 + B 1 ) (B 2 + L 2 ) D 2 (T 2 + B 1 ) (B 2 + B 3 ) D 2 = B 1 B 3 = D 1 = T 1 B 1 = T 2 B 2 = D 2 > 0

Tight lower bounds... by (3) and (6). If any of B and C is assigned to M1, the lower bound holds since L 1 B 2 + D 2 = T 2. Otherwise both jobs go to M2 and we are done again as L 2 = 1 + s B 2 = T 1 by (3). Case F3 L 1 = T and L 2 = 0. The next and last two jobs are B = 1 and C = B 1. Since B = 1 and L 1 + L 2 + B + C = T + 1 + B 1 = T 2 + B 1 = 1 + s by (5) and (3), we have OPT = 1 by Observation 1. By(3) and (1), C = B 1 = 1 + s T 2 1. If any of B or C is assigned to M1, then L 1 T + 1 = T 2 by (5). Thus the lower bound holds. Otherwise both go to M2 and L 2 = 1 + s T > 1 + s B 2 = T 1 by (1) and (3), and we are done again. Case F T L 1 B 2 and L 1 + L 2 s 1. The next and last two jobs are B = 1 and C = 1 + s (L 1 + L 2 + B) 1 + s s = 1. Since B = 1 and L 1 + L 2 + B + C = 1 + s, wehave OPT = 1byObservation1. IfanyofB and C is assigned to M1, then L 1 T + 1 = T 2 by (5), thus the lower bound holds. Otherwise both go to M2 and L 2 = 1 + s L 1 1 + s B 2 = T 1 by (3), and we are done again. Case F5 (s is small) L 1 = B 2 and L 2 = B c. The next and last jobs are B = 1 (B c) and C = s B 2. Since L 2 +B = 1 and L 1 + L 2 + B + C = 1 + s,wehaveopt= 1byObservation1.By(8) and (17), B = 1 (B c) = D 2 + c (B c) = D 2 + 2c B D 2, and by (1), C = s B 2 T 2 B 2 = D 2.IfanyofB or C is assigned to M1, we are done: L 1 B 2 + D 2 = T 2. Otherwise both B and C are assigned to M2, and we are done again by (3): L 2 = 1 + s B 2 = T 1. 3.2 Intermediate cases We will introduce seven intermediate cases. The first one is valid for any considered s (small, regular or medium), while the others are only valid for special speeds s, as given below. Case I1 L 1 B 2, L 2 B 3, L 2 L 1 b, and there exists a subset of already assigned jobs with total size c. The next job is A = B 2 L 1 0. If A is assigned to M1, then L 1 = L 1+ A = B 2 and L 2 B 3, and we are in Case F2. If A is assigned to M2, then L 2 = L 2 + B 2 L 1 b + B 2 = B 3 B + B 2 = B 3 + D 3 = T 3

G. Dósa et al. by (9) and (7), and L 1 + L 2 = L 1 + L 2 + B 2 L 1 = L 2 + B 2 B 3 + B 2 = T 3 D 3 + B 2 = T 3 + B T 3 + B 2 = T 1 s + B 2 = 1 by (7), (1), () and (3), so we are in Case F1. Case I2 (s is small) L 1 = 0 and L 2 = B c. Let the next job be A = B 2. Suppose A is assigned to M2. Then L 2 =B c + B 2 = B 3 b c + B 2 B 3 B + B 2 = B 3 + D 3 =T 3 holds by (9), (17) and (7), and L 1 + L 2 = B 2 + B c 1 holds by (19). Thus Case F1 holds for the new loads L 1, L 2, and hence we are done. Otherwise A is assigned to M1. At this moment L 1 = B 2 and L 2 = B c. Then we are in Case F5. Case I3 (s is regular or medium) L 1 = c and L 2 = 0 (which implies that there exists a subset of already assigned jobs with total size c). Let the next job be A = T c. This is nonnegative by (25). Suppose A is assigned to M1. At this time L 1 = T and L 2 = 0. Case F3 holds, we are done. Now suppose A goes to M2. Then L 1 = c and L 2 = T c. Note that L 1 = c = 1 D 2 = T 2 T D 2 T 2 D 2 = B 2 by (8), (5) and (1), L 2 = T c = B 3 +a c B 3 by (10) and (2), and L 2 L 1 = T 2c b by (25). Hence we are in Case I1, and we are done. Case I (s is regular or medium) L 1 B, L 2 = L 1 a, and there exists a subset of already assigned jobs with total size c. Let the next job be A = T L 1. This is nonnegative since A = T L 1 T B 0by(1). Suppose A is assigned to M1. At this time L 1 = T, L 2 = L 1 a and L 1 + L 2 = T + L 1 a T + B a = B 3 + B s 1 by (10) and (20). We are in Case F. Now suppose A goes to M2. Then L 1 = L 1 B B 2 by (1), L 2 = L 2 + T L 1 = T a = B 3 by (10), and L 2 L 1 = L 2 + T 2L 1 = T a L 1 = B 3 L 1 B 3 B = b by (10) and (9). Hence we are in Case I1, and we are done. Case I5 (s is smaller regular or smaller medium) L 1 = d and L 2 = c (which implies that there exists a subset of already assigned jobs with total size c). Subcase 1 Assume 5 3 s s 0 1.686. Let the next job be A = T d. This is nonnegative by (23) and (1). If A is assigned to M1, then L 1 = T and L 1 + L 2 = T + c = T + 1 D 2 = T 2 D 2 = B 2 s 1

Tight lower bounds... by (8), (5) and (20). We meet the prerequisites of Case F, and we are done. Thus suppose A goes to M2. Then L 1 = d B < B 2 by (23) and (1), and L 2 = c + T d = c + a + B 3 d B 3 by (10) and (22). And by (27), L 2 L 1 = c + T 2d b. Hence we are in Case I1, and we are done. Subcase 2 Assume s 0 s q 3 or s is smaller medium. Let the next job be A = B 2 d. Note that A > 0by(23) and (1). If A is assigned to M1, then the new load on this machine is L 1 = B 2, and L 2 = L 2 = c B 3 by (15), thus we are in Case F2, and we are done. Thus suppose A goes to M2. Then L 1 = d B < B 2 by (23) and (1), L 2 = c + B 2 d B 3 by (28), and L 2 L 1 = c + B 2 2d b by (26). Hence we are in Case I1, and we are done. Case I6 (s is bigger regular) L 1 = e and L 2 = c. Let the next job be A = T e. Note that A B e d e 0by(1), (23) and (22). Suppose A is assigned to M1. Then L 1 = T and L 1 + L 2 = T + c = T + 1 D 2 = B 2 s 1 by (8), (5) and (20), thus Case F holds, and so we are done. Otherwise, A goes to M2. Then L 1 = e d B < B 2 by (22), (23) and (1), L 2 = c + T e = c + a + B 3 e B 3 by (10) and (31), and L 2 L 1 = c + T 2e b by (32). Hence we are in Case I1, and we are done. Case I7 (s is regular or medium) L 1 = B, and L 2 = b B, and there exists a subset of already assigned jobs with total size c. Let the next job be A = 2B. Suppose A is assigned to M1. Atthistime L 1 = 3B and L 2 = b B. Note that T L 1 = 3B B 2 by (21). Using (9) and (20), we get L 1 + L 2 = 3B + (b B ) = B 3 + B s 1 Hence we are in Case F, and we are done. We conclude that A is assigned to M2. Then L 1 = B < B 2 by (1), L 2 = b + B = B 3 by (9), and L 2 L 1 = b. Hence we are in Case I1, and we are done. Proof of lower bounds Consider an arbitrary algorithm to solve the semi-online scheduling problem, where the values s and OPT are known. In the following construction, we take the point of view of an adversary, and try to make the algorithm s life as hard as possible. More formally, we will show that the algorithm will provide a schedule having competitive ratio of at least r(s). Although the whole family of jobs has optimum 1, the adversary still has enough freedom to force the algorithm to an assignment where it ends up with a load L 1 on machine M1 with L 1 T 2 = r or a load L 2 on machine M2 with L 2 T 1 = rs.

G. Dósa et al..1 Construction when s is small The adversary decides that the first job is J 1 = B c. This job has a non-negative size by (17). Suppose J 1 goes to M2, then Case I2 is satisfied, and we are done. We conclude J 1 goes to M1. The second job is J 2 = c, where c 0by(15). Suppose J 2 goes to M2. Note that L 1 B B 2 by (15) and (1), and L 2 = c B 3 by (17). Moreover, L 2 L 1 = 2c B B 3 B = b by (17) and (9). We are in Case I1, and we are done. We conclude J 2 goes to M1. At this moment the loads are L 1 = B and L 2 = 0. The third job is J 3 = B 3. Suppose J 3 goes to M1. Since L 1 = B + B 3 = T holds by (18) and L 2 = 0, we are in Case F3, and thus we are done. We conclude J 3 goes to M2. At this moment the loads are L 1 = B B 2 by (1), L 2 = B 3, and L 2 L 1 = B 3 B = b by (9). We are in Case I1, and we are done. We remark that the sequence of jobs can be drawn as a decision tree, with the first job at its root node, and all other jobs at the subsequent nodes (see Fig. 6). A left branch means that the job at a node is assigned to machine M1, and a right branch means that it is assigned to machine M2. Note that this tree has in total a depth of six jobs, counting also those jobs that are added in the intermediate and final steps and are not shown in Fig. 6..2 Construction when s is regular or medium First, the adversary chooses the job J 1 = c, where c 0by(15). If J 1 goes to M1, then Case I3 is satisfied, and we are done. We conclude that J 1 goes to M2. We divide the further construction into two main cases, depending on the value of s. Case 1 s is smaller regular, smaller medium, or bigger regular. Subcase 1.1 s is smaller regular or smaller medium. The second job is J 2 = d.ifj 2 goes to M1, then Case I5 is satisfied, and we are done. Thus we conclude that J 2 goes to M2. At this point L 1 = 0 and L 2 = d + c = b B by (11). We continue the construction after Subcase 1.2. Subcase 1.2 s is bigger regular. The second job is J 21 = e, where e a + c 0by(22) and (31). If J 21 goes to M1, then the assumption of Case I6 is satisfied, and we are done. We conclude that J 21 goes to M2. The next job is J 22 = f, where f = a + c + e 0by(13), (22) and (31). Suppose J 22 goes to M1. Then L 1 = f and L 2 = J 1 + J 21 = c + e. By(22), (15) and (23), L 1 = f d B.By(13), L 1 L 2 = f c e = a. We are altogether in Case I, thus we are done. We conclude that J 22 goes to M2. Atthis point L 1 = 0 and L 2 = c + e + f = c + d = b B by (12) and (11). Now we join the treatments of these subcases, Subcase 1.1 and Subcase 1.2, and finish the construction. In both subcases now L 1 = 0 and L 2 = b B. Then comes J 3 = B. Suppose J 3 goes to M1. Then L 1 = B and L 2 =

Tight lower bounds... b B, so we are in Case I7, and we are done. We conclude that J 3 goes to M2, and at this moment L 1 = 0, L 2 = b = B 3 B B 3, we are in Case I1. Case 2 s is bigger medium. The second job is J 21 = c + a. IfJ 21 goes to M1, then the assumptions c + a = L 1 = L 2 + a B of Case I are satisfied because of (22) and (29), and we are done. We conclude that J 21 goes to M2. Then comes J 22 = 2c+2a. Suppose J 22 goes to M1. Then L 1 = 2c + 2a and L 2 = J 1 + J 21 = 2c + a. Thus by repeating the previous arguments, we are again in Case I, and we are done. We conclude that J 22 goes to M2. At this point L 1 = 0 and L 2 = c+3a. Then comes J 23 = c + a. Suppose J 23 goes to M1. Then L 1 = c + a and L 2 = c + 3a. Thus by applying (29), Case I is applicable again, and we are done. We conclude that J 23 goes to M2. At this moment L 1 = 0, L 2 = 8c+7a. By (30) we know that b 8c + 7a B 3, we are in Case I1, and we are done. Again, it is possible to sketch the above assignment steps in a decision tree, as explained at the end of Sect..1 (Fig. 6). The depth of this tree depends on the value of s. For smaller regular and smaller medium s, we have a depth of seven jobs, and for bigger medium and bigger regular s, we have a depth of eight jobs. 5 Main Theorem The following theorem summarizes the work done in the preceding sections. Theorem 2 The function r(s) (defined in Sect. 2.1) is a lower bound on the optimal competitive ratio for the two uniform machine semi-online scheduling problem with known optimal offline objective function value. Together with the algorithms from Ng et al. (2009), we then obtain: Fig. 6 Decision trees of the construction

G. Dósa et al. Corollary 1 The lower bound given by r(s) is tight for [q 1, q 2 ] [1.3956, 1.3], moreover for [ 5 3, q 3] [1.6667, 1.693] and [q 5, q 6 ] [1.6963, 1.7103]. 6 Conclusions and outlook Starting with the work of Epstein (2003) on this semi-online two uniform machines scheduling problem with known optimum, researchers have continued to close the gap between lower and upper bounds. As one can deduce from Fig. 1, this goal has been achieved for some large portions of the line [1, ). We contributed to this ultimate goal by giving new lower bounds and thus showing that some already existing algorithms (of Ng et al. 2009) are in fact best possible, so our bounds are tight. Our new results give insight into the difficulty of the problem: Why is it so hard to give the tight competitive ratio for this model? In part, an answer lies in the fact that not a single algebraic function can describe the tight lower bound. From what is known by now, at least six different piecewise-defined algebraic functions are necessary. And still, the question of the optimal competitive ratio is open on certain parts of the right interval, namely on (q 2, 5 3 ), (q 3, q 5 ), and (q 6, 3). For the latter, we proved in Dósa et al. (2018) that our presented lower bound between q 6 and q 7 and the old lower bound of Epstein (2003) between q 7 and 3 are in fact tight. Acknowledgements Gyorgy Dosa acknowledges the financial support of Szechenyi 2020 under the EFOP- 3.6.1-16-2016-00015. György Dósa s and Zsolt Tuza s work was jointly funded by the National Research, Development and Innovation Office NKFIH under the grant SNN 116095. Zhiyi Tan s work was supported by the National Natural Science Foundation of China (11671356, 1127132, 1171286). Krzysztof Węsek s work was partially supported by the European Union in the framework of European Social Fund through the Warsaw University of Technology Development Programme, realized by Center for Advanced Studies. Furthermore, Węsek s work was conducted as a guest researcher at the Helmut Schmidt University. Armin Fügenschuh s work was mostly carried out while being affiliated with the Helmut Schmidt University. Fügenschuh acknowledges the support by the German Research Association (DFG), grant number FU 860/1-1. Last but not least, we are grateful to the two anonymous referees for their various helpful comments on our manuscript. Appendix: Proof of Lemma 1 The technical proofs of the Lemmas can be easily carried out using a computer algebra system. We prepared a Maple worksheet that is available upon request from the authors. First, we present some lower and upper bounds on r(s). Lemma 2 (i) If s is small, regular or medium, r(s) r 2 (s). (33) r(s) 11s + 9 9s + 5 9s + 7 9s + 1. (3) { 3s + 1 r(s) <min 2s + 1, s + 3 } < s. (35) 3s + 2

Tight lower bounds... r(s) < s + 2 s + 1 2s + 1 s + 1. (36) 63s + 53 r(s) 5s + 1 s + 1 2. (37) r(s) > 3 > 3s + 2 2s + 2. (38) r(s) 2s + 1. 2s (39) (ii) If s is small, (iii) If s is regular or medium, (iv) If s is smaller regular or smaller medium, r(s) = r 1 (s) 5s + 2 s + 1 2 s. (0) r(s) 11s + 8 8s + 6 5s + 6 s +. (1) r(s) 6s + 5 6s + 1 2 s. (2) r(s) 3s + 5 2s + 5s + 3s +. (3) r(s) (v) If s s 0 and s is smaller regular, r(s) 18s + 16 16s + 7. () 17s + 1 15s + 6. (5) (vi) If s s 0 and s is smaller regular, or s is smaller medium, r(s) 7s + 7 7s + 2. (6) (vii) If s is bigger medium, r(s) s + 13 3s. (7) (vii) If s is bigger regular, r(s) s + 3 6s 3. (8) Proof Note that s > 0 in all cases. Hence in all inequalities below, we never multiply with negative numbers, thus the sense of the inequalities does not change.

G. Dósa et al. (i) Equation (33) The inequality r 1 (s) r 2 (s) is equivalent to (12s + 10)(s + 5) (6s+6)(9s+7) = 6s 2 +s+8 0, which holds for 0 < s 1+ 13 3 1.5352. Hence it holds for all s q 2. The inequality r 3 (s) r 2 (s) is equivalent to (12s+10)(16s+7) (18s+16)(9s+7) = 30s 2 26s 2 0, which holds for s 13+ 129 30 = q 3. The inequality r (s) r 2 (s) is equivalent to (12s+10)(3s+10) (8s+7)(9s+7) = 36s 2 +31s+51 0, which holds for 0 < s 31+ 8305 72 = q 5. Hence, r(s) r 2 (s) by the definition of r(s). Equation (3) The inequality r 2 (s) 11s+9 9s+5 is equivalent to (12s +10)(9s + 5) (11s + 9)(9s + 7) = 9s 2 8s 13 0. The inequality 11s+9 9s+5 9s+7 9s+1 is equivalent to (9s +1)(11s +9) (9s +5)(9s +7) = 18s 2 16s 26 0. Both inequalities hold for 0 < s + 133 9 = q 7. Equation (35) The inequality r 2 (s) < 2s+1 3s+1 is equivalent to (12s + 10)(2s + 1) (3s + 1)(9s + 7) = 3s 2 + 2s + 3 < 0, which holds for s > 1+ 10 3 1.387. Hence it holds for all s q 1. The inequality r 2 (s) < 3s+2 s+3 is equivalent to (12s+10)(3s+2) (s+3)(9s+7) = 1 s < 0, which holds for all s 1. The inequality 2s+1 3s+1 < s is equivalent to (3s +1) s(2s +1) = 2s 2 + 2s + 1 < 0, which holds for s > 1+ 3 2 1.3660. Hence it holds for all s q 1. In combination with (33) we thus obtain (35). Equation (36) The inequality r 2 (s) < s+2 s+1 is equivalent to (12s + 10)(s + 1) (s +2)(9s +7) = 3s 2 3s < 0, which holds for 0 < s < 3+ 57 6 1.7583. Hence it holds for all s q 7. The inequality s+2 s+1 2s+1 s+1 holds for any s 1. In combination with (33) we thus obtain (36). Equation (37) The inequality r 1 (s) 63s+53 5s+1 is equivalent to (5s+1)(6s+ 6) (63s + 53)(s + 5) = 18s 2 11s 19 0, which holds for s 11+ 189 36 1.3773. Hence it holds for all s q 1. The inequality r 2 (s) 63s+53 5s+1 is equivalent to (12s + 10)(5s + 1) (63s + 53)(9s + 7) = 27s 2 + 2s + 30 0, which holds for 0 < s + 133 9 = q 7.The inequality r 3 (s) 63s+53 5s+1 is equivalent to (18s + 16)(5s + 1) (63s + 53)(16s + 7) = 198s 2 + 169s + 285 0, which holds for 0 < s 169+ 25281 396 1.70016. Hence it holds for all q 3 s q. The inequality r (s) 63s+53 5s+1 is equivalent to (8s +7)(5s +1) (63s +53)(3s +10) = 171s 2 16s 23 0, which holds for s 73 6882 171 1.69311. Hence it holds for all q s q 5. By the definition of r(s),wehaver(s) 63s+53 5s+1. The inequality 63s+53 5s+1 s+1 2 is equivalent to 2(63s + 53) (s + 1)(5s + 1) = 5s 2 + 0s + 65 0, which holds for 0 < s + 133 9 = q 7. Equation (38) Since 63s+53 5s+1 > 3 for any s > 1, and 3 > 2s+2 3s+2 for any 0 < s < 2, (38) is obtained in combination with (37). Equation (39) The inequality r 1 (s) 2s+1 2s is equivalent to 2s(6s + 6) (2s +1)(s +5) = s 2 2s 5 0, which holds for s 1+ 21 = q 1.The

Tight lower bounds... inequality 63s+53 5s+1 2s+1 2s is equivalent to 2s(63s + 53) (2s + 1)(5s + 1) = 36s 2 21s 1 0, which holds for s > 7+ 705 2 1.39799. Hence it holds for all s 5 3. By the definition of r(s) and (37), (39) can be proved. (ii) Equation (0) The inequality r 1 (s) 5s+2 s+1 is equivalent to (s+1)(6s+6) (5s +2)(s +5) = s 2 3s 0, which holds for 0 < s 3+ 73 8 = q 2. The inequality 5s+2 s+1 2 s is equivalent to s(5s + 2) 2(s + 1) = 5s2 6s 2 0, which holds for 0 < s 3+ 19 5 1.7178. Hence it holds for all s q 2. Equation (0) is thus proved. (iii) Equation (1) The inequality 63s+53 5s+1 11s+8 8s+6 is equivalent to (8s+6)(63s+ 53) (11s + 8)(5s + 1) = 9s 2 9s 10 0, which holds for s 5 3. The inequality 11s+8 8s+6 s+ 5s+6 is equivalent to (11s + 8)(s + ) (8s + 6)(5s + 6) = 2s + s 2 0, which holds for s 1+ 17 1.28078. Hence it holds for all s 5 3. In combination with (37) we thus obtain (1). Equation (2) The inequality r 2 (s) 6s+5 6s+1 is equivalent to (6s + 1)(12s + 10) (6s + 5)(9s + 7) = 18s 2 15s 25 0, which holds for s 5 3.The inequality 63s+53 5s+1 6s+5 6s+1 is equivalent to (6s+1)(63s+53) (6s+5)(5s+ 1) = 108s 2 90s 152 0, which holds for s 15+ 209 36 1.6705. Hence it holds for all q 3 s q 5. By the definition of r(s) and (37), r(s) 6s+5 6s+1 equivalent to s(6s + 5) 2(6s + 1) = 6s 2 7s 2 0, which holds for any s is regular or medium. The inequality 6s+5 6s+1 2 s is for s 7+ 97 12 1.007. Hence it holds for all s 5 3, and thus (2) is obtained. Equation (3) The inequality r 2 (s) 2s+ 3s+5 is equivalent to (12s + 10)(2s + ) (3s + 5)(9s + 7) = 3s 2 + 2s + 5 0, which holds for s 5 3.The inequality 2s+ 3s+5 5s+ 3s+ is equivalent to (3s+)(3s+5) (2s+)(5s+) = s s 2 0, which holds for s 1+ 17 2 1.56155. Hence it holds for all s 5 3. In combination with (33) we thus obtain (3). (iv) Equation () The inequality r 2 (s) 18s+16 16s+7 is equivalent to (12s + 10)(16s + 7) (18s + 16)(9s + 7) = 30s 2 26s 2 0, which holds for 0 < s 13+ 129 30 = q 3. r 3 (s) 18s+16 16s+7 trivially holds when q 3 s q. (v) Equation (5) The inequality r 2 (s) 17s+1 15s+6 is equivalent to (12s + 10)(15s + 6) (17s + 1)(9s + 7) = 27s 2 23s 38 0, which holds for 0 < s 23+ 633 5 = s 0. (vi) Equation (6) The inequality 63s+53 5s+1 7s+2 7s+7 is equivalent to (63s+53)(7s+ 2) (5s + 1)(7s + 7) = 126s 2 105s 181 0, which holds for s 35+ 11361 8 1.68557. Hence it holds for all s s 0. Combining it with (37), we have r(s) 7s+7 7s+2.

G. Dósa et al. (vii) Equation (7) The inequality r (s) 13 3s s+ is equivalent to (13 3s)(8s + 7) (s + )(3s + 10) = 36s 2 + 31s + 51 0, which holds for 0 < s 31+ 8305 72 = q 5. Hence, r(s) 13 3s s+. (viii) Equation (8) The inequality r 2 (s) s+3 6s 3 is equivalent to (6s 3)(12s + 10) (s + 3)(9s + 7) = 36s 2 31s 51 0, which holds for s 31+ 8305 72 = q 5. Hence, r 2 (s) s+3 6s 3. Now we are in a position to prove Lemma 1. Proof of Lemma 1 (i) Equation (1) B = s 2r 3rs+ 3 > 0 is equivalent to r(s) < 3s+2 s+3, which is valid by (35). B T = (s 2r 3rs + 3) (r 1) = s 3r 3rs + < 0, since r(s) > 3 by (38). T B 2 = (r 1) (s + 1 sr) = sr + r s 2 < 0, since r(s) < s+2 s+1 by (36). B 2 T 2 = (s + 1 sr) r = s + 1 sr r < 0asr(s) >1. T 2 = r s is valid by (35). Equation (1) is thus proved. Equation (15) Since c = 1 D 2 = 1 (T 2 B 2 ) = B 2 +(1 T 2 ) = B 2 T by (8) and (5), we have c > 0by(1). c B 3 = ( rs r + s + 2) (2s 2r rs + 2) = r s < 0, since r(s) <s by (35). B 3 T 3 = (2s 2r rs+2) (sr s) = 2rs 2r +3s +2 < 0, since r(s) > 3s+2 2s+2 by (38). T 3 B 1 = (sr s) (s + 1 r) = sr + r 2s 1 < 0, since r(s) < 2s+1 s+1 by (36). B 1 T 1 = (s + 1 r) sr = s + 1 r sr < 0as r(s) >1. (15) is thus proved. Equation (16)By(39), r(s) 2s+1 2s. Hence, b = 2rs 2s 1 0. Equation (16) is thus proved. (ii) Equation (17) The first inequality of (17) is due to (16). (b + c) B = (2rs 2s 1 rs r +s +2) (s 2r 3rs+3) = rs+r 5s 2 0, since r(s) < 5s+2 s+1 by (0). By (9) and (16), B B 3. B 3 2c = (2s 2r rs + 2) ( 2rs 2r + 2s + ) = rs 2 0, since r(s) 2 s by (0). Equation (17) is thus proved. Equation (18) Since r(s) = 6s+6 s+5 when s is small, B 3 + B = rs r + 6s + 5 = r 1 = T, which is (18). Equation (19) (B 2 + B ) (1 + c) = (s + 1 rs + s 2r 3rs + 3) (1 rs r + s + 2) = 3rs r + s + 1 0, since r(s) > 3 > s+1 3s+1 by (38). Equation (19) is thus proved. (iii) Equation (20) B 2 (B 3 + B ) = (s + 1 rs) (2s 2r rs + 2 + s 2r 3rs + 3) = 3rs + r 5s 0, since r(s) 5s+ 3s+ by (3). (B 3 + B ) (s 1) = (2s 2r rs+ 2 + s 2r 3rs+ 3) (s 1) = rs r + 5s + 6 0, since r(s) s+ 5s+6 by (1). Equation (20) isthus proved. Equation (21) T 3B = (r 1) ( 9rs 6r + 12s + 9) = 9rs + 7r 12s 10 0, since r(s) r 2 (s) = 12s+10 9s+7 by (33). 3B B 2 = ( 9rs 6r + 12s + 9) (s + 1 rs) = 8rs 6r + 11s + 8 0, since by (1). Equation (21) is thus proved. 6s+5 0, since r(s) 6s+1 by (2). a + c = (rs + 3r 2s 3) + ( rs r + s + 2) = 2r s 1 0, since r(s) 11s+8 8s+6 Equation (22) e = 2 1 r 3s + 3rs 2 5

Tight lower bounds... r(s) s+1 2 by (37). By the definition of f and d, wehave f a + c 0, d e and d f. Equation (22) is thus proved. Equation (23) d B = (6rs + 3r 7s 6) (s 2r 3rs + 3) = 9rs + 5r 11s 9 0, since r(s) 11s+9 9s+5 by (3). Equation (23)isthus proved. Equation (2) a c = (rs + 3r 2s 3) ( rs r + s + 2) = 2rs + r 3s 5 0, since r(s) 2s+ 3s+5 by (3). Equation (2) isthus proved. Equation (25) T (b+2c) = (r 1) (2rs 2s 1 2rs 2r +2s +) = 3r 0, since r(s) > 3 by (38). Together with (15) and (16), (25) is thus proved. (iv) Equation (26)From()wehaver(s) 18s+16 16s+7. Therefore, c + B 2 2d b = ( rs r +s+2)+(s+1 rs) (12rs+6r 1s 12) (2rs 2s 1) = 16rs 7r + 18s + 16 0. Equation (26) is thus proved. (v) Equation (27) From(5) we have that r(s) = r 2 (s) 17s+1 15s+6. Therefore, c + T 2d b = ( rs r + s + 2) + (r 1) (12rs + 6r 1s 12) (2rs 2s 1) = 15rs 6r + 17s + 1 0. Equation (27) is thus proved. (vi) Equation (28)From(6)wehavethatr(s) 7s+7 7s+2. Therefore, c + B 2 d B 3 = ( rs r+s+2)+(s+1 rs) (6rs+3r 7s 6) (2s 2r rs+2) = 7rs 2r + 7s + 7 0. Equation (28) is thus proved. (vii) Equation (29) Since r(s) = 3s+10 8s+7 when s is bigger medium, B = s 2r 3rs + 3 = 8r s = a + c. Equation (29) is thus proved. Equation (30)From(7) we know r(s) 13 3s s+. Therefore, b (8c+7a) = (2rs 2s 1) ( 8rs 8r + 8s + 16 + 7rs + 21r 1s 21) = 3rs 13r +s + 0. Since r(s) = r (s) = 3s+10 8s+7 8s+7 15 for any s 5 3, 8c + 7a B 3 = ( 8rs 8r + 8s + 16 + 7rs + 21r 1s 21) (2s 2r rs + 2) = 15r 8s 7 0. Equation (30) is thus proved. (viii) Equation (31)From(8)wehavethatr 2 (s) s+3 6s 3. Therefore, e (a+c) = ( 1 2 r 3s+3rs 5 2 ) (rs+3r 2s 3 rs r+s+2) = 3rs 3 2 r 2s 3 2 0. (31) is thus proved. Equation (32) c + T 2e b = ( rs r + s + 2) + (r 1) (r 6s + 6rs 5) (2rs 2s 1) = 9rs r + 9s + 7 0, since r(s) 9s+7 9s+1 by (3). Equation (32) is thus proved. References Albers S (1999) Better bounds for online scheduling. SIAM Journal of Computing 29:59 73 Angelelli E, Speranza MG, Tuza Z (2007) Semi online scheduling on three processors with known sum of the tasks. J Sched 10:263 269 Angelelli E, Speranza MG, Tuza Z (2008) Semi-online scheduling on two uniform processors. Theoret Comput Sci 393:211 219 Azar Y, Regev O (2001) Online bin-stretching. Theoret Comput Sci 268(1):17 1 Berman P, Charikar M, Karpinski M (2000) On-line load balancing for related machines. Journal of Algorithms 35:108 121

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