Improved bounds for speed scaling in devices obeying the cube-root rule

: Speed scaling is a power management technology that involves dynamically changing the speed of a processor. This technology gives rise to dual-objective scheduling problems, where the operating system both wants to conserve energy and optimize some Quality of Service (QoS) measure of the resulting schedule. In the most investigated speed scaling problem in the literature, the QoS constraint is deadline feasibility, and the objective is to minimize the energy used. The standard assumption is that the processor power is of the form s α where s is the processor speed, and α > 1 is some constant; α ≈ 3 for CMOS based processors. In this paper we introduce and analyze a natural class of speed scaling algorithms that we call qOA. The algorithm qOA sets the speed of the processor to be q times the speed that the optimal ofﬂine algorithm would run the jobs in the current state. When α = 3, we show that qOA is 6.7-competitive, improving upon the previous best guarantee of 27 achieved by the algorithm Optimal Available (OA). We also give almost matching upper and lower bounds for qOA for general α . Finally, we give the ﬁrst non-trivial lower bound, namely e α − 1 / α , on the competitive ratio of a general deterministic online algorithm for this problem.


Introduction
Current processors produced by Intel and AMD allow the speed of the processor to be changed dynamically.Intel's SpeedStep and AMD's PowerNOW technologies allow the operating system to dynamically change the speed of such a processor to conserve energy.In this setting, the operating system must not only have a job selection policy to determine which job to run, but also a speed scaling policy to determine the speed at which the job will be run.Almost all theoretical studies we know of assume a processor power function of the form P(s) = s α , where s is the speed and α > 1 is some constant.Energy consumption is power integrated over time.The operating system is faced with a dual objective optimization problem as it both wants to conserve energy, and optimize some Quality of Service (QoS) measure of the resulting schedule.
The first theoretical study of speed scaling algorithms was in the seminal paper [16] by Yao, Demers, and Shenker.In the problem introduced in [16] the QoS objective was deadline feasibility, and the objective was to minimize the energy used.To date, this is the most investigated speed scaling problem in the literature [2,6,3,9,11,12,14,16,17].In this problem, each job i has a release time r i when it arrives in the system, a work requirement w i , and a deadline d i by which the job must be finished.The deadlines might come from the application, or might arise from the system imposing a worst-case quality-of-service metric, such as maximum response time or maximum slow-down.Since the speed can be made arbitrarily high, every job can always be completed by its deadline, and hence without loss of generality the job selection policy can be assumed to be Earliest Deadline First (EDF), as it produces a deadline feasible schedule whenever one exists.Thus the (only) issue here is to determine the processor speed at each time, i. e., find an online speed scaling policy, to minimize energy.

The story to date
In their seminal work, Yao, Demers, and Shenker [16] showed that the optimal offline schedule can be efficiently computed by a greedy algorithm YDS.They also proposed two natural online speed scaling algorithms, Average Rate (AVR) and Optimal Available (OA).Conceptually, AVR is oblivious in that it runs each job in the way that would be optimal if there were no other jobs in the system.That is, AVR processes each job i at the constant speed w i /(d i − r i ) throughout interval [r i , d i ], and the speed of the processor is just the sum of the processing speeds of the jobs.The algorithm OA maintains the invariant that the speed at each time is optimal given the current state, and under the assumption that no more jobs will arrive in the future.In particular, let w(x) denote the amount of unfinished work that has deadline within x time units from the current time.Then the current speed of OA is max x w(x)/x, this is precisely the speed that the offline optimum algorithm [16] would set in this state.Another online algorithm BKP is proposed in [6].BKP Table 1: Results on the competitive ratio for energy minimization with deadline feasibility.
Table 1 summarizes the results in the literature related to the competitive ratio of online algorithms for this problem.The competitive ratio of AVR is at most 2 α−1 α α .This was first shown in [16], and a simpler potential function based analysis was given in [3].This result is almost tight.In particular, the competitive ratio of AVR is least (2 − δ ) α−1 α α , where δ is a function of α that approaches zero as α approaches infinity [3].The competitive ratio of OA is exactly α α [6], where the upper bound is proved using an amortized local competitiveness argument.Thus the competitive ratio of AVR is strictly inferior to that of OA.The competitive ratio of BKP is at most 2(α/(α − 1)) α e α [6], which is about 2e α+1 for large α.This bound on the competitive ratio of BKP is better than that of OA only for α ≥ 5. On the other hand, the known lower bounds for general algorithms are rather weak and are based on instances consisting of just two jobs.In particular, [5] show a lower bound of (4/3) α /2 for any deterministic algorithm.If one tries to find the worst 3, 4, . . .job instances, the calculations get messy quickly.
The most interesting value of α is certainly 3 as in current CMOS based processors the dynamic power is approximately the cube of the speed (this is commonly called the cube-root rule) [8].It seems likely that α would be in the range [2,3] for most conceivable devices.The best known guarantee for α in this range is α α achieved by OA, which evaluates to 4 for α = 2 and 27 for α = 3.

Our contributions
In Section 3 we introduce and analyze a natural class of speed scaling algorithms, that we call qOA.The algorithm qOA sets the speed of the processor to be q ≥ 1 times the speed that the optimal offline algorithm would run the jobs in the current state, or equivalently q times the speed that the algorithm OA would run in the current state.In the worst-case instances for OA the rate that work arrives increases with time.Intuitively, the mistake that the algorithm OA makes in these instances is that it runs too slowly initially as it doesn't anticipate future arrivals.So the motivation of the definition of qOA is to avoid making the same mistake as OA, and to run faster in anticipation of further work arriving in the future.
We show, using an amortized local competitiveness analysis, that if q is set to 2 − 1/α, then the competitive ratio of qOA is which is at most 4 α /(2e 1/2 α 1/4 ).This bound is approximately 3.38 when α = 2, and 11.52 when α = 3. Setting q = 2 − 1/α is not necessarily the optimum value of q for our analysis (although it isn't too far off).For general α, it is not clear how to obtain the optimum choice of q for our analysis since this involves solving a system of high degree algebraic inequalities.For the case of α = 3 and that of α = 2, we can explicitly determine the the choice of q that gives best bound on the competitive ratios using our analysis.We show that qOA is at worst 2.4-competitive when α = 2, and at worst 6.7-competitive when α = 3.There are two main technical ideas in the competitive analysis of qOA.The first is the introduction of a new potential function which is quite different from the one used in the analysis of OA in [6] and the potential function used to analyze AVR in [3].The second idea is to use a convexity based argument in the analysis, instead of Young's inequality.The analysis in [7], and almost all of the amortized local competitiveness analyses in the speed scaling literature, rely critically on the Young's inequality.However, in the current setting, Young's inequality gives a bound that is too weak to be useful when analyzing qOA.Instead we observe that certain expressions that arise in the analysis are convex, which allows us to reduce the analysis of the general case down to just two extreme cases.
In Section 4 we give the first non-trivial lower bound on the competitive ratio for a general deterministic algorithm.We show no deterministic algorithm can have a competitive ratio less than e α−1 /α.Our lower bound is almost optimal since BKP achieves a ratio of about 2e α+1 , in particular, the base of the power, e, is the best possible.For α = 3, this raises the best known lower bound a modest amount, from 1.2 to 2.4.
Given the general lower bound of e α−1 /α, and that BKP achieves a ratio with a base of the power e, a natural question is whether there is some choice of the parameter q for which the competitive ratio of qOA varies with e as the base of the power.Somewhat surprisingly, we show that this is not the case and the base of the power cannot be improved beyond 4.In particular, in Section 5 we show that the competitive ratio of qOA can not be better than For large α this is about 4 α−1 /(αe).Note that this lower bound essentially matches our upper bound of 4 α 2e 1/2 α 1/4 for qOA.
Our results are summarized in the last two rows of Table 1.In particular we improve the competitive ratio in the case that the cube-root rule holds from [1.2, 27] to [2.4, 6.7] and in the case that α = 2 from [1. 1,4]

Other related results
There are now enough speed scaling papers in the literature that it is not practical to survey all such papers here.We limit ourselves to those papers most related to the results presented here.Surveys of the speed scaling literature include [1,10].
A naive implementation of the offline optimum algorithm for deadline feasibility YDS [16] runs in time O(n 3 ).Faster implementations for discrete and continuous speeds can be found in [11,13,14].[2] considered the problem of finding energy-efficient deadline-feasible schedules on multiprocessors.[2] showed that the offline problem is NP-hard, and gave O(1)-approximation algorithms.[2] also gave online algorithms that are O(1)-competitive when job deadlines occur in the same order as their release times.[4] investigated speed scaling for deadline feasibility in devices with a regenerative energy source such as a solar cell.

Formal problem statement
A problem instance consists of n jobs.Job i has a release time r i , a deadline d i > r i , and work w i > 0. In the online version of the problem, the scheduler learns about a job only at its release time; at this time, the scheduler also learns the work and the deadline of the job.We assume that time is continuous.A schedule specifies for each time a job to be run and a speed at which to run the job.The speed is the amount of work performed on the job per unit time.A job with work w run at a constant speed s thus takes w/s units of time to complete.More generally, the work done on a job during a time period is the integral over that time period of the speed at which the job is run.A job i is completed by d i if work at least w i is done on it during [r i , d i ].A schedule is feasible if every job is completed by its deadline.Note that the times at which work is performed on job i do not have to be contiguous, that is, preemption is allowed.If the processor is running at speed s, then the power is P(s) = s α for some constant α > 1.The energy used during a time period is the integral of the power over that time period.Our objective is to minimize the total energy subject to completing all jobs by their respective deadlines.An algorithm A is said to be c-competitive if for any instance, the energy usage by A is at most c times that of the optimal schedule.If S is a schedule then we use E S to denote the energy used by that schedule.If A is an algorithm, and we are considering a fixed instance, then we use E A to denote the energy used by the schedule produced by A on the instance.
We essentially prove Theorems 3.1, 3.2 and 3.3 in parallel; the proofs differ only at the end.We use an amortized local competitiveness analysis, and use a potential function Φ.Although our presentation here should be self-contained, for further background information on amortized local competitiveness arguments see [15].In this setting, the units of Φ will be energy, and thus, the derivative of Φ with respect to time will be power.Intuitively, Φ is a bank/battery of energy that qOA has saved (over some optimum solution) from the past, that it can use in the future if it lags behind the optimum.
Before defining the potential function Φ, we need to introduce some notations.We always denote the current time as t 0 .Since all of our quantities are defined with respect to the current time, we will drop t 0 for notational ease (unless there is cause for confusion).Let s a and s o be the current speed of qOA and the optimal algorithm OPT respectively.For any t 0 ≤ t ≤ t , let w a (t ,t ) denote the total amount of unfinished work for qOA at t 0 that has a deadline during (t ,t ].Define w o (t ,t ) similarly for OPT.Using this notation, recall that qOA runs at speed where q ≥ 1 will be some fixed constant depending on α.Let d(t ,t ) = max{0, w a (t ,t ) − w o (t ,t )} denote the excess unfinished work that qOA has relative to OPT among the already released jobs with deadlines in the range (t ,t ].We define a sequence of critical times t 0 < t 1 < t 2 < • • • < t h iteratively as follows: Let t 1 be the latest time such that d(t 0 ,t 1 )/(t 1 − t 0 ) is maximized.Clearly, t 1 is no more than the latest deadline of any job released thus far.If t i is earlier than the latest deadline, let t i+1 > t i be the latest time, not later than the latest deadline, that maximizes d(t i ,t i+1 )/(t i+1 − t i ).
We will refer to the intervals [t i ,t i+1 ] as critical intervals.We use g i to denote d(t i ,t i+1 )/(t i+1 − t i ), which is the density of the excess work with deadline in (t i ,t i+1 ].We note that g 0 , g 1 , . . ., g h−1 is a non-negative strictly decreasing sequence.To see this, suppose for the sake of contradiction that this does not hold, and let i be smallest index such that g i ≥ g i−1 .Then this implies that contradicting the choice of t i in our iterative procedure.Finally, we note that the quantities t i and g i depend on the current time t 0 and might change over time.
We define the potential function Φ as where β is some constant which we will optimize later.We first note some simple observations about Φ, t i and g i .
Observation 3.4.Φ is zero before any jobs are released, and after all jobs are completed.Proof.This directly follows as each g i = 0 by definition.
Observation 3.5.Job arrivals do not increase Φ, or change the definition of critical times.Also, job completions by either qOA or OPT do not change Φ.
Proof.Upon a job arrival, the work of both online and offline increases exactly by the same amount, and hence the excess work d(t ,t ) does not change for any t and t .For job completions, we note that d(t ,t ) and Φ are both a continuous function of the unfinished work, and the unfinished work on a job continuously decreases to 0 as it completes.
The critical times thus only change due to qOA or OPT working on the jobs.However, as we show next, this does not cause any discontinuous change in Φ.
Observation 3.6.The instantaneous change in critical times does not (abruptly) change the value of Φ.
Proof.There are three ways the critical times can change.The observations above imply that the potential function does not change due to any discrete events such as arrivals, job completions, or changes in critical intervals.Then, in order to establish that qOA is c-competitive with respect to energy, it is sufficient to show the following running condition at all times when there is no discrete change as discussed above: The fact that the running condition holds for all times establishes c-competitiveness follows by integrating the running condition over time, and from the fact that Φ is initially and finally 0, and the fact that Φ does not increase due to discrete events.In the next three lemmas, we provide simple bounds on the speed s a of qOA and the speed s o for OPT, which will be useful in this analysis.Lemma 3.7.Without loss of generality, we may assume that s o ≥ max Proof.OPT needs to complete at least w o (t 0 ,t) units of work by time t.As the function s α is convex, the energy optimal way to accomplish this is to run at a constant speed of w o (t 0 ,t)/(t − t 0 ) during [t 0 ,t].
Since OPT is optimum, it may run only faster (due to possible more jobs arriving in the future).
Proof.By the definition of qOA and d(t 0 ,t), we have that, Here the last inequality follows by Lemma 3.7 and the defintion of g 0 .
We are now ready to prove Theorems 3.1, 3.2 and 3.3.Let us first consider the easy case when w a (t 0 ,t 1 ) ≤ w o (t 0 ,t 1 ).
Case 1: Suppose that w a (t 0 ,t 1 ) ≤ w o (t 0 ,t 1 ).Now by definition, d(t 0 ,t 1 ) = 0 and g 0 = 0, and hence s a ≤ qs o by Lemma 3.9.Note that there is only one critical interval [t 0 ,t 1 ] and Thus to show (3.1) it suffices to show that q α ≤ c, which is easily verified for our choice of q and c in each of Theorems 3.1, 3.2 and 3.3.
Case 2: Henceforth we assume that w a (t 0 ,t 1 ) > w o (t 0 ,t 1 ).Note that g 0 > 0 in this case.As qOA follows EDF, it must work on some job with deadline at most t 1 , and hence w a (t 0 ,t 1 ) decreases at rate s a .For OPT, let s k o denote the speed with which OPT works on jobs with deadline in the critical interval (t k ,t k+1 ].We need to determine dΦ/dt 0 .To this end, we make the following observation.For k = 0, and by our assumption that w a (t 0 ,t 1 ) > w o (t 0 ,t 1 ), we have that where the last step follows from the second step using Observation 3.10.As the g i 's are non-increasing and ∑ k≥0 s k o = s o by definition, (3.2) implies that Thus to show the running condition (3.1), it is sufficient to show that Consider the left hand side of equation (3.3) as a function of s a while g 0 and s o are fixed.We note that it is a convex function of s a .Hence, to show (3.3), it is sufficient to show that it holds at the extreme possible values for s a , which by Lemma 3.8 and Lemma 3.9 are s a = qg 0 and s a = qg 0 + qs o .
For s a = qg 0 , the left hand side of (3.3) becomes Taking derivative with respect to s o , we see that this is maximized when . Substituting this value for s o and canceling g α 0 on both sides, it follows that it suffices to satisfy: Substituting s o = x • g 0 and canceling g α 0 on both sides, it suffices to satisfy We now fork our proofs of Theorems 3.1, 3.2 and 3.3.In each case we need to show that equations (3.5) and (3.7) hold.We first finish up the proof for Theorem 3.1.Recall that we set q = 2 − 1/α.We let 1) .With these choices of q, β and c, αq = 2α − 1. Substituting in equation (3.5), and dividing through by q α , we obtain that this equation is then equivalent to Taking the derivative with respect to x of the left side of equation (3.8), we can conclude that the maximum is attained at 1) .For this value of x, the left side of equation (3.8) evaluates to 0 and hence the result follows.Hence the running condition (3.3) is satisfied.So the competitive ratio is at most c = q α η α−1 with q = 2 − 1/α and η = 1 + α −1/(α−1) , which implies Theorem 3.1 holds.
To finish the proof of Theorem 3.2 we wish to determine the values of q and β so that the inequalities (3.5) and (3.7) hold with the minimum possible value of c. Plugging α = 3 into inequalities (3.5) and (3.7) we obtain: ≤ 0 , and We wrote a computer program to approximately determine the values of q and β that minimize c.The best values we obtained are q = 1.54, β = 7.78 and c = 6.73.It is easy to check that (3.5) is satisfied.The left hand side of (3.7) becomes −3.08x 3 + 10.96x 2 − 1.65x − 16.73, which can be shown to be negative by differentiation.A similar process yields Theorem 3.3.We set q = 1.46 and β = 2.7 and check that (3.5) and (3.7) are satisfied for these values to give c = 2.39.

General lower bound
The goal in this section is to prove the following theorem.We assume α is fixed and is known to the algorithm.We give an adversarial strategy for constructing a job instance based on the behavior of A. We demonstrate a schedule OPT whose energy usage is arbitrarily close to a factor of e α−1 /α less than the energy used by the schedule produced by A.
Adversarial strategy: Let ε > 0 be some small fixed constant.Work is arriving during [0, h], where 0 So the work that arrives during any time interval [u, v] is v u a(t)dt.All work has deadline 1.The value of h will be set by the adversary according to the action of A. Intuitively, if A spends too much energy initially, then h will be set to be small.If A does not spend enough energy early on, then h will be set to 1 − ε.In this case, A will have a lot of work left toward the end and will have to spend too much energy finishing this work off.To make this more formal, consider the function where b is some constant (which we will later set to 1/(α − 1) 1/α ).This is the total energy usage up to time t if A runs at speed Of course, A may run at speed other than s(t).We set h be the first time, satisfying 0 < h < 1 − ε, such that total energy usage of A up to time h is at least E(h).If no such time exists, then h = 1 − ε.
We break the lower bounding of the competitive ratio into three cases depending on whether h ∈ (0, 1 − 1/e], h ∈ (1 − 1/e, 1 − ε), or h = 1 − ε.The proofs of some inequalities are given in lemmas at the end of the section.The case that h ∈ (0, 1 − 1/e]: Since h < 1 − ε, the energy used by the algorithm E A is at least E(h) and hence Now consider the possible schedule OPT that runs at a constant speed there is always enough released work for OPT to process.Observe that the total processing done by OPT in [0, 1] is ln(1/(1 − h)), which equals the total work released, so OPT is feasible.
We can calculate a bound on the energy used by OPT as follows: where the inequality is proved in Lemma 4.2.
Combining our bounds on E A and E OPT , we can conclude that the competitive ratio in this case is at least which tends to e α−1 /α as ε tends to 0.
The case that h ∈ (1 − 1/e, 1 − ε): One possible schedule OPT is to run at speed s o (t) = a(t) for t ∈ [0, 1 − e(1 − h)] and run at a constant speed . Note that by simple algebra, we have 0 < 1 − e(1 − h) < h.We observe that s o (t) ≤ a(t) for all t ∈ [0, h], hence there is always enough released work for OPT to process during [0, h].To establish that OPT is feasible, we show that the total processing done by OPT equals the total work released, as follows.We now wish to bound the energy used by OPT.
By the fact that e x ≥ 1 + x for all x ≥ 0, we have that e α−1 ≥ α and α/e α−1 ≤ 1.Hence, we can loosen the above bound to: The following bound on E A from line (4.1)still holds in this case: Combining the bounds on E A and E OPT , we again conclude that the competitive ratio is at least which tends to e α−1 /α as ε tends to 0.
The case that h = 1 − ε: Note that the adversary ends the arrival of work at time 1 − ε and the total amount of work arrived is Also note that the total energy usage of A up to 1 − ε may be exactly E(1 − ε).
We first show that much of the work released is unfinished by A at time 1 − ε.To see this let s A (t) be the speed of the algorithm A at time t and consider the algorithm B that works at speed The energy consumed by B by time t ≤ 1 + ε is exactly which, by the definition of h, is at least the energy consumed by A by time t.We will prove in Lemma 4.3 that this implies the work processed by A by time 1 − ε is at most the work processed by B by time 1 − ε.
Hence the maximum amount of work completed by A by time 1 − ε is Hence, A has at least b units of work remaining at time 1 − ε.To complete this amount of work during where the last equality follows from setting b = 1/(α − 1) 1/α .Using the bound on E OPT from line (4.2), with h = 1 − ε, we find that there is a feasible schedule using energy at most Thus, the competitive ratio in this case is at least e α−1 /α.
To finish the proof of Theorem 4.1 we need the following two technical lemmas.
Lemma 4.2.For any h ∈ (0, Proof.Let us define Differentiating f (h), we have We can check easily by differentiation that for all h > 0, and the equality holds only at h = 1 − 1/e.Therefore, f (h) is non-positive, and f (h) ≤ f (0) = 0.The Lemma then follows.
Since F(y) ≤ 0, it follows that G(y) ≤ 0. As s B is monotonically increasing and positive, it follows that G(y)s B (y)/s B (y) α ≤ 0 for all y ∈ [0, x].Hence Applying integration by parts and noting that G (y) = αs B (y) α−1 (s A (y) − s B (y)), we obtain that The last equality follows from G(0) = 0. Since G(x) ≤ 0, we obtain the desired result.

Lower Bounds for qOA
Our goal in this section is to prove the following theorem.
Theorem 5.1.Let α > 2 be a fixed known constant.For any choice of q, the competitive ratio of qOA is at least .
Instance Definition: Let ε be some constant satisfying ε ∈ (0, 1).Consider the input job sequence where work is arriving during [0, 1 − ε] and the rate of arrival at time t is where β = 2/α.All work has deadline 1.Finally, a job is released at time 1 − ε with work ε 1−β and deadline 1.
We first give an upper bound on the energy used in the optimal energy schedule.Consider the schedule OPT that runs at speed a(t) during [0, 1 − ε] and then runs at speed 1/ε β during [1 − ε, 1].Clearly OPT completes all work by deadline 1.The energy usage of OPT, and hence an upper bound on the energy used by an optimal schedule, is then We want to calculate the energy usage of qOA during [1 − ε, 1], which is a lower bound to E qOA , as follows.Let s(t) be the speed of qOA at time t.We first determine s(t) for t ∈ [0, 1 − ε].The value of s(t) is the unique nonnegative solution to One can verify that satisfies the above equation by substituting it to the right hand side of the equation, as follows: We then bound the amount of work w that qOA has unfinished right after the last job arrives at time 1 − ε as follows: (5. 2) The speed s(t), for t ∈ [1 − ε, 1], is the unique solution to the equation By substitution one can verify that the solution is We can check that 1 1−ε s(t)dt = w, so qOA just finishes all the work at time 1.Hence, the energy usage of qOA during (5.3) As ε approaches 0, we can see from equation (5.2) that w approaches THEORY OF COMPUTING, Volume 8 (2012), pp.209-229 Hence, combining this with equation (5.3), the energy used by qOA at least approaches Combining this with our bound on the optimal energy from equation (5.1), and setting β = 2/α, we can conclude that the competitive ratio of qOA at least approaches We can see by differentiation that the above expression is increasing when q ≥ 2. Hence, to obtain a lower bound for the competitive ratio, we can assume q ≤ 2. We rewrite the lower bound for the competitive ratio as 1 2qα q(q + 2 α ) q + 2 α − 1 α (5.4) and focus on the term q(q + 2 α ) q + 2 α − 1 .
Setting this equal to 0, we obtain The unique positive solution for this equation is Plugging this value of q back into our lower bound for the competitive ratio in line (5.4), we finally conclude that the competitive ratio of qOA is at least .

Theorem 4 . 1 .
No deterministic online algorithm A can have a competitive ratio less than e α−1 /α.
runs at speed e • v(t) at time t, where Thus BKP can be viewed as computing a lower bound on the average speed in an online manner and running at e times that speed.
1 ,t 2 ) is the amount of work that has release time at least t 1 , deadline at most t 2 , and that has already arrived by time t.Clearly, if w(t 1 ,t 2 ) is the total work of jobs that are released after t 1 and have deadline before t 2 , then any algorithm must have an average speed of at least w(t 1 ,t 2 )/(t 2 − t 1 ) during [t 1 ,t 2 ].

1 .
Merging of two critical intervals:As qOA follows EDF it must work on jobs with deadline in [t 0 ,t 1 ], causing g 0 to decrease until it becomes equal to g 1 .At this point, the critical intervals [t 0 ,t 1 ] and [t 1 ,t 2 ] merge together.Now, Φ does not change by this merger as g 0 = g 1 at this point.
2. Splitting of a critical interval:As OPT works on some job with deadline t ∈ (t k ,t k+1 ], the quantityw a (t k ,t ) − w o (t k ,t ) t −t k may increase faster than w a (t k ,t k+1 ) − w o (t k ,t k+1 ) t k+1 − t k causing this interval to split into two critical intervals, [t k ,t ] and [t ,t k+1 ].This split does not change Φ as the density of the excess work for both of these newly formed intervals is g k .3. Formation of a new critical time: A job arrives with later deadline than any previous job, and a new critical time t h+1 is created.The potential Φ does not change because g h = 0.
Observation 3.10.For k > 0, g k increases at rate at most s k o /(t k+1 − t k ).For k = 0, g 0 changes at rate Proof.Let us first consider k > 0. The rate of change t k+1 − t k with respect to t 0 is 0.Moreover, as qOA does not work on jobs in w a (t k ,t k+1 ) and w o (t k ,t k+1 ) decreases at rate s k o , it follows that Lemma 4.3.Let s A (t) and s B (t) be non-negative speed functions for two algorithms A and B over a time interval [0, x].Assume s B (t) is positive, continuous, differentiable, and monotonically increasing.Then if A always has used no more energy than B, that is if