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Algorithmica

, Volume 81, Issue 9, pp 3725–3745 | Cite as

A Fully Polynomial-Time Approximation Scheme for Speed Scaling with a Sleep State

  • Antonios AntoniadisEmail author
  • Chien-Chung Huang
  • Sebastian Ott
Open Access
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Abstract

We study classical deadline-based preemptive scheduling of jobs in a computing environment equipped with both dynamic speed scaling and sleep state capabilities: Each job is specified by a release time, a deadline and a processing volume, and has to be scheduled on a single, speed-scalable processor that is supplied with a sleep state. In the sleep state, the processor consumes no energy, but a constant wake-up cost is required to transition back to the active state. In contrast to speed scaling alone, the addition of a sleep state makes it sometimes beneficial to accelerate the processing of jobs in order to transition the processor to the sleep state for longer amounts of time and incur further energy savings. The goal is to output a feasible schedule that minimizes the energy consumption. Since the introduction of the problem by Irani et al. (ACM Trans Algorithms 3(4), 2007), its exact computational complexity has been repeatedly posed as an open question (see e.g. Albers and Antoniadis in ACM Trans Algorithms 10(2):9, 2014; Baptiste et al. in ACM Trans Algorithms 8(3):26, 2012; Irani and Pruhs in SIGACT News 36(2):63–76, 2005). The currently best known upper and lower bounds are a 4 / 3-approximation algorithm and NP-hardness due to Albers and Antoniadis (2014) and Kumar and Shannigrahi (CoRR, 2013. arXiv:1304.7373), respectively. We close the aforementioned gap between the upper and lower bound on the computational complexity of speed scaling with sleep state by presenting a fully polynomial-time approximation scheme for the problem. The scheme is based on a transformation to a non-preemptive variant of the problem, and a discretization that exploits a carefully defined lexicographical ordering among schedules.

Keywords

Approximation algorithms Energy efficiency Polynomial-time approximation scheme 

1 Introduction

As energy efficiency in computing environments becomes more crucial, chip manufacturers are increasingly incorporating energy-saving functionalities to their processors. One of the most common functionalities is dynamic speed scaling, where the processor can adjust its speed dynamically. A higher speed yields a higher performance, but this performance comes at the cost of more energy consumption. On the other hand, a lower speed results in better energy-efficiency, but at the cost of performance degradation. In practice, it has been observed [7, 11] that the power consumption of the processor is approximately proportional to its speed cubed. However, even when the processor is idling, it consumes a non-negligible amount of energy just for the sake of “being active” (for example because of leakage current). Due to this fact, additional energy can be saved by incorporating a sleep state to the processor. A processor in a sleep state consumes zero (or negligible) energy; however, there is an extra energy cost when it is transitioned back to the active state.

This article studies the offline problem of minimizing energy consumption of a processor which is equipped with both speed scaling and sleep state capabilities. This problem is called speed scaling with sleep state, first introduced by Irani et al. [18].

Let us state the problem more formally. The given processor has two states: the active state, during which it can execute jobs and consume some energy, and the sleep state, during which no jobs can be executed but also no energy is consumed. We assume that a wake-up operation, that is a transition from the sleep state to the active state, incurs a constant energy cost \(C>0\), whereas transitioning from the active state to the sleep state is free of charge. Further, as in [2, 18], the power required by the processor in the active state is an arbitrary convex and non-decreasing function P of its speed s. In accordance to all previous work in the area, we make the necessary assumptions that the power function P is fixed and therefore not part of the input.1 We also assume that \(P(0)>0\), since (i) as already mentioned, real-world processors are known to have leakage current and (ii) otherwise the sleep state would be redundant. Further motivation for considering arbitrary convex power functions for speed scaling can be found, for example, in  [8].

The input is a set \(\mathcal {J}\) of n jobs. Each job j is associated with a release time \(r_j\), a deadline \(d_j\) and a processing volume \(v_j\). One can think of the processing volume as the number of CPU cycles that are required in order to completely process the job, so that if job j is processed at a speed of s, then \(v_j/s\) time-units are required to complete the job. We call the interval \([r_j,d_j)\) the allowed interval of job j, and say that job j is active at time point t if and only if \(t\in [r_j,d_j)\).2 Furthermore, we may assume without loss of generality that \(\min _{j\in \mathcal {J}}r_j=0\), and that \(v_{min}:= \min _{j\in \mathcal {J}}v_j\) is normalized to 1 (if \(v_{min}\ne 1\) is the case, we can scale the instance by dividing the \(r_j\)’s, \(d_j\)’s, and \(v_j\)’s by \(v_{min}\), and using the power function \(P(s) \cdot v_{min}\) along with the original wake-up cost C). Further, let \(d_{max}:=\max _{j\in \mathcal {J}} d_j\) be the latest deadline of any job.

A schedule is defined as a mapping of every time point t to the state of the processor, its speed, and the job being processed at t (or null if there is no job running at t). Note that the processing speed is zero whenever the processor sleeps, and that a job can only be processed when the speed is strictly positive. A schedule is called feasible when the whole processing volume of every job j is completely processed in j’s allowed interval \([r_j,d_j)\). Preemption of jobs is allowed.

The energy consumption incurred by schedule \(\mathcal {S}\) while the processor is in the active state, is its power integrated over time, i.e. \(\int P(s(t)) dt\), where s(t) is the processing speed at time t, and the integral is taken over all time points in \([0,d_{max})\) during which the processor is active under \(\mathcal {S}\). Assume that \(\mathcal {S}\) performs k transitions from the sleep state to the active state. (We will assume throughout the paper that initially, prior to the first release time, as well as finally, after the last deadline, the processor is in the active state. However, our results can be easily adapted for the setting where the processor is initially and/or eventually in the sleep state). Then the total energy consumption of \(\mathcal {S}\) is \(E(\mathcal {S}) := \int P(s(t))dt +kC\), where again the integral is taken over all time points at which \(\mathcal {S}\) keeps the processor in the active state. We are seeking a feasible schedule that minimizes the total energy consumption.

Observe that, by Jensen’s inequality, and by the convexity of the power function, it is never beneficial to process a job with a varying speed. Irani et al. [18] observed the existence of a critical speed\(s_{crit}\), which is the most efficient speed for processing jobs. This critical speed is the smallest speed that minimizes the function P(s) / s. Note that, by the convexity of P(s), the only case where the critical speed \(s_{crit}\) is not well defined, is when P(s) / s is always decreasing. However, this would render the setting unrealistic, and furthermore make the algorithmic problem trivial, since it would be optimal to process every job at an infinite speed. We may therefore assume that this case does not occur. Further, it can be shown (see [18]) that for any \(s\ge s_{crit}\), the function P(s) / s is non-decreasing.

1.1 Previous Work

The theoretical model for dynamic speed scaling was introduced in a seminal paper by Yao, Demers and Shenker [21]. They developed a polynomial time algorithm called YDS, that outputs a minimum-energy schedule for this setting. Irani, Shukla and Gupta [18] initiated the algorithmic study of speed scaling combined with a sleep state. Such a setting motivates the so-called race to idle technique: one saves overall energy by accelerating some jobs in order to transition the processor to the sleep state for longer periods of time (see [4, 13, 14, 20] and references therein for more information regarding the race to idle technique). Irani et al. developed a 2-approximation algorithm for speed scaling with sleep state, but the computational complexity of the scheduling problem has remained open. The first step towards attacking this open problem was made by Baptiste [9], who gave a polynomial time algorithm for the special case where the processor must execute all jobs at a fixed speed, and all jobs are of unit size. Baptiste’s algorithm is based on a clever dynamic programming formulation of the scheduling problem, and was later extended to (i) arbitrarily-sized jobs in [10], and (ii) a multiprocessor setting in [12].

More recently, Albers and Antoniadis [2] improved the upper bound on the approximation ratio of the general problem, by developing a 4 / 3-approximation algorithm. For the special case of agreeable deadlines and a power function of the form \(P(s) = s^\alpha + \beta \) (with constant \(\alpha >1\) and \(\beta >0\)), Bampis et al. [5] provided an exact polynomial time algorithm. With respect to the lower bound, [2] gave an NP-hardness reduction from the partition problem. The reduction uses a particular power function that is based on the partition instance, i.e., it is considered that the power function is part of the input. The reduction of [2] was later refined by Kumar and Shannigrahi [19], to show that the problem is NP-hard for any fixed, non-decreasing and strictly convex power function.

The online setting of the problem has also been studied. Irani et al. [18] gave a \((2^{2\alpha -2}\alpha ^\alpha + 2^{\alpha -1+2})\)-competitive online algorithm. Han et al. [15] improved upon this result by developing an \((\alpha ^\alpha +2)\)-competitive algorithm for the problem. Both of the above results assume a power function of the form \(P(s) = s^\alpha + \beta \), where \(\alpha >1\) and \(\beta >0\) are constants.

A more thorough discussion on the above scheduling problems can be found in the surveys [1, 17].

1.2 Our Contribution

We study the offline setting of speed scaling with sleep state. Since the introduction of the problem by Irani et al. [18], its exact computational complexity has been repeatedly posed as an open question (see e.g. [2, 10, 17]). The currently best known upper and lower bounds are a 4 / 3-approximation algorithm and NP-hardness due to [2] and [2, 19], respectively. In this paper, we settle the open question by presenting a fully polynomial-time approximation scheme.

At the core of our approach is a transformation of the original preemptive problem into a non-preemptive scheduling problem of the same type. At first sight, this may seem counterintuitive, especially as Bampis et al. [6] showed that (for the problem of speed scaling alone), for the same instance, the ratio of an optimal preemptive solution against an optimal non-preemptive solution can be very high. However, this does not apply in our case, as we consider the non-preemptive problem on a modified instance, where each job is replaced by a polynomial number of pieces. Furthermore, in our analysis, we make use of a particular lexicographic ordering that does exploit the advantages of preemption.

In order to compute an optimal schedule for the modified instance via dynamic programming, we require a number of properties that pieces must satisfy in a valid schedule. The definition of these properties is based on a discretization of the time horizon by a polynomial number of time points. Roughly speaking, we focus on those schedules that start and end the processing of each piece at such time points, and satisfy a certain constraint on the processing order of the pieces. Proving that a near-optimal schedule in this class of schedules exists is the most subtle part of our approach.

On one hand, the processing order constraint can be exploited by the DP; on the other hand, such a constraint is difficult to establish in an optimal schedule with the introduced indivisible volumes (since pieces of different jobs might have different volumes and cannot easily be interchanged). To get around this, we first ensure the right ordering in an optimal schedule for the preemptive setting, and then perform a series of transformations to a non-preemptive schedule with the above properties. Each of these transformations increases the energy consumption only by a small factor, and maintains the correct ordering among the pieces.

We remark that Baptiste [9] used a dynamic program of similar structure for the special case of unit-sized jobs and a fixed-speed processor equipped with a sleep state. His dynamic program is also based on a particular ordering of jobs, which, however, is not sufficient for our setting. Since we have pieces of different sizes, the swapping argument used in [9] fails.

In Sect. 2, we describe the YDS algorithm from [21] for the problem of speed scaling without a sleep state, and then show several properties that a schedule produced by YDS has for our problem of speed scaling with sleep state. We then, in Sect. 3, define a particular class of schedules that have a set of desirable properties, and show that there exists a schedule in this class, whose energy consumption is within a \((1+\epsilon )\)-factor from optimal. Finally, in Sect. 4, we develop an algorithm based on a dynamic program, that outputs, in polynomial time, a schedule of minimal energy consumption among all the schedules of the aforementioned class.

2 Preliminaries

We start by giving a short description of the YDS algorithm presented in [21]. For any interval I, let B(I) be the set of jobs whose allowed intervals are contained in I. We define the density of I as
$$\begin{aligned} dens(I) = \frac{\sum _{j\in B(I)} v_j}{|I|}. \end{aligned}$$
Note that the average speed that any feasible schedule uses during interval I is no less than dens(I). YDS works in rounds. In the first round, the interval \(I_1\) of maximal density is identified, and all jobs in \(B(I_1)\) are scheduled during \(I_1\) at a speed of \(dens(I_1)\), according to the earliest deadline first policy. Then the jobs in \(B(I_1)\) are removed from the instance and the time interval \(I_1\) is “blacked out”. In general, during round i, YDS identifies the interval \(I_i\) of maximal density and then processes all jobs in \(B(I_i)\) at a uniform speed of \(dens(I_i)\), removes the jobs in \(B(I_i)\) from the instance, and “shrinks” the interval \(I_i\) to length zero. YDS terminates when all jobs are scheduled, and its running time is polynomial in the input size.

We remark that the speed used for the processing of jobs can never increase between two consecutive rounds, i.e., YDS schedules the jobs by order of non-increasing speeds. Furthermore, all the jobs scheduled in each round i have their allowed intervals within \(I_i\).

Given any job instance \(\mathcal {J}\), let \(\text {FAST}(\mathcal {J})\) be the subset of \(\mathcal {J}\) that YDS processes at a speed greater than or equal to \(s_{crit}\), and let \(\text {SLOW}(\mathcal {J}):=\mathcal {J}\setminus \text {FAST}(\mathcal {J})\). The following lemma is an extension of a fact proven by Irani et al. [18].

Lemma 1

For any job instance \(\mathcal {J}\), there exists an optimal schedule (w.r.t. speed scaling with sleep state) in which
  1. 1.

    Every job in \(\text {FAST}(\mathcal {J})\) is processed according to YDS.

     
  2. 2.

    Every job \(k \in \text {SLOW}(\mathcal {J})\) is run at a uniform speed \(s_k< s_{crit}\), and the processor never (actively) runs at a speed less than \(s_k\) during \([r_k, d_k)\).

     
We call an optimal schedule with these properties a YDS-extension for \(\mathcal {J}\).

Proof

To break ties among different optimal schedules with equal energy consumption (which can occur when the power function is not strictly convex for any s), we introduce the pseudo cost function \(\int s(t)^2 dt\) (this idea was first used in [18]). Consider a minimal pseudo cost schedule Y, so that Y satisfies property 1 of the lemma, and minimizes the energy consumption among all schedules satisfying this property. It was shown in [18] that Y is optimal for instance \(\mathcal {J}\), and that under Y
It therefore remains to prove that the speeds \(s_k\) are no higher than \(s_{crit}\). For the sake of contradiction, assume that there exists a job \(j \in \text {SLOW}(\mathcal {J})\) which is processed at speed higher than \(s_{crit}\). Let \(\mathcal {I}\) be a maximal time interval, so that (i) \(\mathcal {I}\) includes at least part of the execution of j, and (ii) at any time point \(t \in \mathcal {I}\) the processor either runs strictly faster than \(s_{crit}\), or executes a job from \(\text {FAST}(\mathcal {J})\). Then there must exist a job \(k \in \text {SLOW}(\mathcal {J})\) (possibly \(k=j\)) which is executed to some extent during \(\mathcal {I}\), and whose allowed interval is not contained in \(\mathcal {I}\) (otherwise, when running YDS, the density of \(\mathcal {I}\) after the jobs in \(\text {FAST}(\mathcal {J})\) have been scheduled is larger than \(s_{crit}\), contradicting the fact that YDS processes all remaining jobs slower than \(s_{crit}\)). By the maximality of \(\mathcal {I}\), there exists some interval \(\mathcal {I'} \subseteq [r_k,d_k)\) right before \(\mathcal {I}\) or right after \(\mathcal {I}\), during which no job from \(\text {FAST}(\mathcal {J})\) is executed, and the processor either runs with speed at most \(s_{crit}\) or resides in the sleep state. The first case contradicts property (\(*\)), as k is processed during \(\mathcal {I}\) and thus at speed \(s_k > s_{crit}\). In the second case, we can use a portion of \(\mathcal {I'}\) to slightly slow down k to a new speed \(s'\), such that \(s_{crit}< s' < s_k\). The resulting schedule \(Y'\) has energy consumption no higher than Y, as P(s) / s is non-decreasing for \(s\ge s_{crit}\). Furthermore, if \(\mathcal {C}_p\) is the pseudo cost of Y, then \(Y'\) has pseudo cost \(\mathcal {C}_p-v_k s_k+v_k s' < \mathcal {C}_p\). This contradicts our assumptions on Y. \(\square \)

Clearly, in general a schedule produced by YDS only satisfies the first condition and is not a YDS-extension.

By the preceding lemma, we may use YDS to schedule the jobs in \(\text {FAST}(\mathcal {J})\), and need to find a good schedule only for the remaining jobs (which are exactly \(\text {SLOW}(\mathcal {J})\)). To this end, we transform the input instance \(\mathcal {J}\) to an instance \(\mathcal {J}'\), in which the jobs \(\text {FAST}(\mathcal {J})\) are replaced by dummy jobs. This introduction of dummy jobs bears resemblance to the approach of [2]. We then show in Lemma 2, that any schedule for \(\mathcal {J}'\) with a certain property, can be transformed to a schedule for \(\mathcal {J}\) without any degradation in the approximation factor.

Consider the schedule \(S_{\text {YDS}}\) that algorithm YDS produces on \(\mathcal {J}\). Let \(I_i=[y_i,z_i)\), \(i=1,\dots , \ell \) be the ith maximal interval in which \(S_{\text {YDS}}\) continuously runs at a speed greater than or equal to \(s_{crit}\), and let \(T_1,\dots , T_m\) be the remaining maximal intervals in \([0,d_{max})\) not covered by intervals \(I_1,I_2,\dots , I_\ell \). Furthermore, let \(\mathcal {T}:=\cup _{1\le k \le m}\ T_k\). Note that each maximal interval \(I_i\) or \(T_i\) may contain subintervals identified in different rounds of YDS, and thus more than one speed levels may be employed in each such maximal interval. Also note that the intervals \(I_p\) and \(T_q\) (\(1\le p\le \ell \) and \(1\le q\le m\)) partition the time horizon \([0,d_{max})\), and furthermore, by the way YDS is defined, every job \(j\in \text {FAST}(\mathcal {J})\) is active in exactly one interval \(I_i\), and is not active in any interval \(T_i\). On the other hand, a job \(j\in \text {SLOW}(\mathcal {J})\) may be active in several (consecutive) intervals \(I_i\) and \(T_{i'}\). We transform \(\mathcal {J}\) to a job instance \(\mathcal {J}'\) as follows:
  • For every job \(j\in \text {SLOW}(\mathcal {J})\), if there exists an i such that \(r_j \in I_i\) (resp. \(d_j\in I_i\)), then we set \(r_{j} := z_i\) (resp. \(d_{j} := y_i\)), else we keep the job as it is.

  • For each \(I_i\), we replace all jobs \(j\in \text {FAST}(\mathcal {J})\) that are active in \(I_i\) by a single job \(j^d_i\) (d stands for “dummy”) with release time at \(y_i\), deadline at \(z_i\), and processing volume \(v_i^d\) equal to the total volume that \(S_{\text {YDS}}\) schedules in \(I_i\), i.e. \(v_i^d = \sum _{j \in B(I_i)}v_j\).

Clearly, the above transformation can be done in polynomial time. Note that after the transformation, there is no release time or deadline in the interior of any interval \(I_i\). Furthermore, we have the following proposition:

Proposition 1

\(\text {FAST}(\mathcal {J}') = \{j_i^d: 1\le i\le \ell \}\) and \(\text {SLOW}(\mathcal {J}') = \text {SLOW}(\mathcal {J})\).

Proof

Since \(\mathcal {J}' = \{j_i^d: 1\le i\le \ell \}\cup \text {SLOW}(\mathcal {J})\), and furthermore \(\text {SLOW}(\mathcal {J}')\) and \(\text {FAST}(\mathcal {J}')\) are disjoint sets, it suffices to show that (i) \(\text {FAST}(\mathcal {J}') \supseteq \{j_i^d: 1\le i\le \ell \}\) and that (ii) \(\text {SLOW}(\mathcal {J}')\supseteq \text {SLOW}(\mathcal {J})\).

For (i), we observe that no job \(j_i^d\) can be feasibly scheduled at a uniform speed less than \(s_{crit}\). As YDS uses a uniform speed for each job, these jobs must belong to \(\text {FAST}(\mathcal {J}')\).

For (ii), consider the execution of YDS on \(\mathcal {J}'\). More specifically, consider the first round when a job from \(\text {SLOW}(\mathcal {J})\) is scheduled. Let \(\mathcal {I}\) be the maximal density interval of this round, and let \(\mathcal {J}_S\) and \(\mathcal {J}_d\) be the sets of jobs from \(\text {SLOW}(\mathcal {J})\) and \(\{j_i^d: 1\le i\le \ell \}\), respectively, that are scheduled in this round (note that \(\mathcal {I}\) contains the allowed intervals of these jobs). As the speed used by YDS is non-increasing from round to round, it suffices to show that \(dens(\mathcal {I})<s_{crit}\).

Consider a partition of \(\mathcal {I}\) into maximal intervals \(\varLambda _1, \dots , \varLambda _a\), s.t. each \(\varLambda _k\) is contained in some interval \(I_i\) or \(T_i\). Then
$$\begin{aligned} dens(\mathcal {I})= & {} \frac{\sum _{j \in \mathcal {J}_d}v_j}{|\mathcal {I}|} + \frac{\sum _{j \in \mathcal {J}_S} v_j}{|\mathcal {I}|} \\= & {} \sum \limits _{\varLambda _k \not \subseteq \mathcal {T}}\left( \frac{|\varLambda _k|}{|\mathcal {I}|} dens(\varLambda _k)\right) + \frac{\sum _{\varLambda _k \subseteq \mathcal {T}} |\varLambda _k|}{|\mathcal {I}|} \cdot \frac{\sum _{j \in \mathcal {J}_S} v_j}{\sum _{\varLambda _k\subseteq \mathcal {T}} |\varLambda _k|}\\\le & {} \left( \sum \limits _{\varLambda _k \not \subseteq \mathcal {T}} \frac{|\varLambda _k|}{|\mathcal {I}|}\right) dens(\mathcal {I}) + \left( 1-\sum \limits _{\varLambda _k \not \subseteq \mathcal {T}} \frac{|\varLambda _k|}{|\mathcal {I}|}\right) \cdot \frac{\sum _{j \in \mathcal {J}_S} v_j}{\sum _{\varLambda _k\subseteq \mathcal {T}} |\varLambda _k|}, \end{aligned}$$
since no \(\varLambda _k\) can have a density larger than \(dens(\mathcal {I})\) (because \(\mathcal {I}\) is the interval of maximal density). It follows that
$$\begin{aligned} dens(\mathcal {I}) \le \frac{\sum _{j \in \mathcal {J}_S} v_j}{\sum _{\varLambda _k\subseteq \mathcal {T}} |\varLambda _k|}\ . \end{aligned}$$
Furthermore, by the definition of \(\text {SLOW}(\mathcal {J})\), it is possible to schedule all jobs in \(\mathcal {J}_S\) during \(\mathcal {I} \cap \mathcal {T}\), at a speed slower than \(s_{crit}\) (since none of the steps in the transformation from \(\mathcal {J}\) to \(\mathcal {J}'\) reduce the time any job is active during \(\mathcal {T}\)). Together with the previous inequality, this implies \(dens(\mathcal {I}) < s_{crit}\).\(\square \)

The following lemma suggests that for obtaining an FPTAS for instance \(\mathcal {J}\), it suffices to give an FPTAS for instance \(\mathcal {J}'\), as long as we schedule the jobs \(j_i^d\) exactly in their allowed intervals \(I_i\).

Lemma 2

Let \(S'\) be a schedule for input instance \(\mathcal {J}'\), that (i) processes each job \(j_i^d\) exactly in its allowed interval \(I_i\) (i.e. from \(y_i\) to \(z_i\)), and (ii) is a c-approximation for \(\mathcal {J}'\). Then \(S'\) can be transformed in polynomial time into a schedule S that is a c-approximation for input instance \(\mathcal {J}\).

Proof

Given such a schedule \(S'\), we leave the processing in the intervals \(T_1,\dots , T_m\) unchanged, and replace for each interval \(I_i\) the processing of job \(j_i^d\) by the original YDS-schedule \(S_{\text {YDS}}\) during \(I_i\). It is easy to see that the resulting schedule S is a feasible schedule for \(\mathcal {J}\). We now argue about the approximation factor.

Let OPT be a YDS-extension for \(\mathcal {J}\), and let OPT\('\) be a YDS-extension for \(\mathcal {J}'\). Recall that \(E(\cdot )\) denotes the energy consumption of a schedule (including wake-up costs). Additionally, let \(E^I(S)\) denote the total energy consumption of S in all intervals \(I_1,\dots , I_\ell \) without wake-up costs (i.e. the energy consumption for processing or being active but idle during those intervals), and define similarly \(E^I(S')\), \(E^I(OPT)\), and \(E^I(OPT')\) for the schedules \(S'\), OPT, and \(\hbox {OPT}'\), respectively. Since \(S'\) is a c-approximation for \(\mathcal {J}'\), we have
$$\begin{aligned} E(S') \le c E(OPT'). \end{aligned}$$
Note that \(\hbox {OPT}'\) schedules exactly the job \(j_i^d\) in each \(I_i\) (using the entire interval for it) by Proposition 1, and thus each of the schedules S, \(S'\), OPT, and OPT\('\) keeps the processor active during every entire interval \(I_i\). Therefore
$$\begin{aligned} E(S) - E(S') = E^I(S) - E^I(S'), \end{aligned}$$
since S and \(S'\) have the same wake-up costs and do not differ in the intervals \(T_1,\)\( \dots , T_m\). Moreover,
$$\begin{aligned} E(OPT) - E(OPT') = E^I(OPT) - E^I(OPT'), \end{aligned}$$
as \(E(OPT) - E^I(OPT)\) and \(E(OPT') - E^I(OPT')\) are both equal to the optimal energy consumption during \(\mathcal {T}\) of any schedule that processes the jobs \(\text {SLOW}(\mathcal {J})\) in \(\mathcal {T}\) and resides in the active state during each interval \(I_i\) (including all wake-up costs of the schedule). Clearly, \(E^I(S) = E^I(OPT)\), and since both \(S'\) and OPT\('\) schedule exactly the job \(j_i^d\) in each \(I_i\) (using the entire interval for it), we have that \(E^I(S') \ge E^I(OPT')\). Therefore
$$\begin{aligned} E(S) - E(S') \le E(OPT) - E(OPT'). \end{aligned}$$
We next show that \(0 \le E^I(OPT) - E^I(OPT') = E(OPT) - E(OPT')\), which implies
$$\begin{aligned} E(S)&\le E(OPT) - E(OPT') + E(S')\\&\le c (E(OPT) - E(OPT')) + E(S') \\&\le c (E(OPT) - E(OPT')) + c E(OPT') \\&\le c E(OPT). \end{aligned}$$
Since YDS (when applied to \(\mathcal {J}\)) processes a volume of exactly \(v_i^d\) in each interval \(I_i\), the average speed of OPT in \(I_i\) is \(v_i^d / |I_i|\). On the other hand, OPT\('\) runs with a speed of exactly \(v_i^d / |I_i|\) during \(I_i\), and therefore \(E^I(OPT) \ge E^I(OPT')\). \(\square \)

3 Discretizing the Problem

After the transformation in the previous section, we have an instance \(\mathcal {J}'\). In this section, we show that there exists a “discretized” schedule for \(\mathcal {J}'\), whose energy consumption is at most \(1+\epsilon \) times that of an optimal schedule for \(\mathcal {J}'\). In the next section, we will show how such a discretized schedule can be found by dynamic programming.

Before presenting formal definitions and technical details, we here first sketch the ideas behind our approach.

A major challenge of the original problem is that we need to deal with an infinite number of possible schedules. We overcome this intractability by “discretizing” the problem as follows: (1) we break each job in \(\text {SLOW}(\mathcal {J}')\) into smaller pieces, and (2) we create a set of time points and introduce the additional constraint that each piece of a job has to start and end at these time points. The number of the introduced time points and job pieces are both polynomial in the input size and \(1/\epsilon \), which substantially reduces the amount of guesswork we have to do in the dynamic program. The challenge is how to find such a discretization and argue that it does not increase the optimal energy consumption by too much.

3.1 Further Definitions and Notation

We first define the set W of time points. Given an error parameter \(\epsilon > 0\), let \(\delta := \min \{\frac{1}{4}, \frac{\epsilon }{4} \frac{P(s_{crit})}{P(2s_{crit})-P(s_{crit})}\}\). Intuitively, \(\delta \) is defined in such a way that speeding up the processor by a factor \((1+ \delta )^3\) does not increase the power consumption by more than a factor \(1+\epsilon \) (see Lemma 5).

Let \(W' := \bigcup _{j \in \mathcal {J}'} \{r_j,d_j\}\), and consider the elements of \(W'\) in sorted order. Let \(t_i, 1\le i \le |W'|\) be the ith element of \(W'\) in this order. We call an interval \([t_i, t_{i+1})\) for \(1\le i \le |W'|-1\) a zone, and observe that every zone is either equal to some interval \(I_i\) or contained in some interval \(T_i\).

For each i in \(1,\dots , |W'|-1\), let x(i) be the largest integer j so that
$$\begin{aligned} (1+\delta )^j\frac{1}{4n^2s_{crit}(1+\delta ) \lceil 1/\delta \rceil }\le t_{i+1}-t_i. \end{aligned}$$
We are now ready to define the set W of time points as follows:
$$\begin{aligned} W&:= \ \ W' \underset{1\le r \le 16n^6 \lceil 1/\delta \rceil ^2(1+\lceil 1/\delta \rceil )}{\underset{0\le j \le x(i)}{\underset{i \text { s.t. }[t_i,t_{i+1})\subseteq \mathcal {T}}{\bigcup }}}\\&\quad \left\{ t_i + r\cdot \frac{(1+\delta )^j\frac{1}{4n^2s_{crit}(1+\delta ) \lceil 1/\delta \rceil }}{16n^6\lceil 1/\delta \rceil ^2 (1+\lceil 1/\delta \rceil )}, \right. \\&\quad \left. t_{i+1} - r\cdot \frac{(1+\delta )^j\frac{1}{4n^2s_{crit}(1+\delta ) \lceil 1/\delta \rceil }}{16n^6\lceil 1/\delta \rceil ^2 (1+\lceil 1/\delta \rceil )} \right\} . \end{aligned}$$
Let us explain how these time points in W come about. As we will show later (Lemma 3(2)), there exists a certain optimal schedule for \(\mathcal {J}'\) in which each zone \([t_i,t_{i+1})\subseteq \mathcal {T}\) contains at most one contiguous maximal processing interval, and this interval “touches” either \(t_i\) or \(t_{i+1}\) (or both). The geometric series
$$\begin{aligned} (1+\delta )^j\frac{1}{4n^2s_{crit}(1+\delta ) \lceil 1/\delta \rceil } \end{aligned}$$
of time points are used to approximate the ending/starting time of this maximal processing interval. For each guess of the ending/starting time, we split the guessed interval, during which the job pieces (to be defined formally immediately) are to be processed, into \(16n^6\lceil 1/\delta \rceil ^2 (1+\lceil 1/\delta \rceil )\) many intervals of equal length. An example of the set W for a given zone can be seen in Fig. 1.
Fig. 1

We assume that \(r=1\dots 8\) and that \(x(i)=2\). The red dashed points correspond to \(j=1\) and the blue dotted points to \(j=2\). For clarity, we drew the points defined from \(t_i\) and from \(t_{i+1}\) in two separate pictures. Note that for each j the number of points is the same and the points of the same color are at equal distance from each other (Color figure online)

Note that |W| is polynomial in the input size and \(1/\epsilon \).

Definition 1

We split each job \(j\in \text {SLOW}(\mathcal {J}')\) into \(4n^2\lceil 1/\delta \rceil \) equal sized pieces, and also consider each job \(j_i^d \in \text {FAST}(\mathcal {J}')\) as a single piece on its own. For every piece u of some job j, let \(job(u) := j\), \(r_u:=r_j\), \(d_u:=d_j\), and \(v_u:=v_j/(4n^2\lceil 1/\delta \rceil )\) if \(j\in \text {SLOW}(\mathcal {J}')\), and \(v_u:=v_j\) otherwise. Furthermore, let D denote the set of all pieces derived from all jobs in \(\mathcal {J}'\).

Note that \(|D|= \ell + |\text {SLOW}(\mathcal {J}')| \cdot 4n^2\lceil 1/\delta \rceil \) is polynomial in the input size and \(1/\epsilon \). We now define an ordering of the pieces in D.

Definition 2

Fix an arbitrary ordering of the jobs in \(\mathcal {J}'\) , s.t. for any two different jobs j and \(j'\), \(j\prec j'\) implies \(r_j \le r_{j'}\). Now extend this ordering to the set of pieces, s.t. for any two pieces u and \(u'\), there holds
$$\begin{aligned} u\prec u' \Rightarrow job(u) \preceq job(u'). \end{aligned}$$

We point out that any schedule for \(\mathcal {J}'\) can also be seen as a schedule for D, by implicitly assuming that the pieces of any fixed job are processed in the above order.

We are now ready to define the class of discretized schedules.

Definition 3

A discretized schedule is a schedule for \(\mathcal {J}'\) that satisfies the following two properties:
  1. (i)

    Every piece is completely processed in a single zone, and without preemption.

     
  2. (ii)

    The execution of every piece starts and ends at a time point from the set W.

     
A discretized schedule S is called well-ordered if and only if
  1. (iii)

    For any time point t, such that in S a piece u ends at t, S schedules all pieces \(u'\succ u\) with \(d_{u'}\ge t\) after t.

     

Finally, we define a particular ordering over possible schedules, which will be useful in our analysis.

Definition 4

Consider a given schedule. For every job \(j \in \mathcal {J}'\), and every \(x \le v_j\), let \(c_j(x)\) denote the earliest time point at which volume x of job j has been finished under this schedule. Furthermore, for any \(j \in \mathcal {J}'\), we define
$$\begin{aligned} q_j:=\int _{0}^{v_j} c_j(x)dx. \end{aligned}$$
Let \(j_1\prec j_2 \prec \dots \prec j_{|\mathcal {J}'|}\) be the jobs in \(\mathcal {J}'\). A schedule S is lexicographically smaller than a schedule \(S'\) if and only if it is lexicographically smaller with respect to the vector \((q_{j_1}, q_{j_2}, \dots , q_{j_{|\mathcal {J}'|}})\).

Observe that shifting the processing interval of any fraction of some job j to an earlier time point (without affecting the other processing times of j) decreases the value of \(q_j\).

3.2 Existence of a Near-Optimal Discretized Schedule

In this section, we first show that there exists a YDS-extension for \(\mathcal {J}'\) with certain nice properties (recall that a YDS-extension is an optimal schedule satisfying the properties of Lemma 1). We then explain how such a YDS-extension can be transformed into a well-ordered discretized schedule, and prove that the speed of the latter, at all times, is at most \((1+\delta )^3\) times that of the former. This fact essentially guarantees the existence of a well-ordered discretized schedule with energy consumption at most \(1+\epsilon \) that of an optimal schedule for \(\mathcal {J}'\). The transformation is depicted in Fig. 2.

Lemma 3

Let OPT be a lexicographically minimal YDS-extension for \(\mathcal {J}'\). Then the following hold:
  1. 1.

    Every job \(j_i^d\) is scheduled exactly in its allowed interval \(I_i\).

     
  2. 2.
    Every zone \([t_i,t_{i+1})\subseteq \mathcal {T}\) has the following two properties:
    1. (a)

      There is at most one contiguous maximal processing interval within \([t_i,t_{i+1})\), and this interval either starts at \(t_i\) and/or ends at \(t_{i+1}\). We call this interval the block of zone \([t_i,t_{i+1})\).

       
    2. (b)

      OPT uses a uniform speed of at most \(s_{crit}\) during this block.

       
     
  3. 3.

    There exist no two jobs \(j' \succ j\), such that a portion of j is processed after some portion of \(j'\), and before \(d_{j'}\).

     

Proof

  1. 1.

    Since \(\text {FAST}(\mathcal {J}') = \{j_i^d: 1\le i\le \ell \}\) (by Proposition 1), and OPT is a YDS-extension, it follows that each \(j_i^d\) is processed exactly in its allowed interval \(I_i\).

     
  2. 2.
    1. (a)

      Assume for the sake of contradiction that \([t_i,t_{i+1})\subseteq \mathcal {T}\) contains a number of maximal intervals \(N_1,N_2,\dots , N_\psi \) (ordered from left to right3) during which jobs are being processed, with \(\psi \ge 2\). Let \(M_1,M_2,\dots , M_{\psi '}\) (again ordered from left to right) be the remaining maximal intervals in \([t_i,t_{i+1})\), so that \(N_1,\dots , N_\psi \) and \(M_1,\dots , M_{\psi '}\) partition the zone \([t_i,t_{i+1})\). Furthermore, note that for each \(i=1,\dots , \psi '\), the processor is either active but idle or asleep during the whole interval \(M_i\), since otherwise setting the processor asleep during the whole interval \(M_i\) would incur a strictly smaller energy consumption.

      We modify the schedule by shifting the intervals \(N_i\), \(i = 2, \dots , \psi \) to the left, so that \(N_1, N_2, \dots , N_\psi \) now form a single contiguous processing interval. The intervals \(M_k\) lying to the right of \(N_1\) are moved further right and merge into a single (longer) interval \(M'\) during which no jobs are being processed. If the processor was active during each of these intervals \(M_k\), then we keep the processor active during the new interval \(M'\), else we transition it to the sleep state. We observe that the resulting schedule is still a YDS-extension (note that its energy consumption is at most that of the initial schedule), but is lexicographically smaller.

      For the second part of the statement, assume that there exists exactly one contiguous maximal processing interval \(N_1\) within \([t_i,t_{i+1})\), and that there exist two M-intervals, \(M_1\) and \(M_2\) before and after \(N_1\), respectively.

      We consider two cases:
      • The processor is active just before \(t_i\), or the processor is asleep both just before \(t_i\) and just after \(t_{i+1}\): In this case we can shift \(N_1\) left by \(|M_1|\) time units, so that it starts at \(t_i\). Again, we keep the processor active during \([t_i+|N_1|, t_{i+1})\) only if it was active during both \(M_1\) and \(M_2\). As before, the resulting schedule remains a YDS-extension, and is lexicographically smaller.

      • The processor is in the sleep state just before \(t_i\) but active just after \(t_{i+1}\): In this case we shift \(N_1\) by \(|M_2|\) time units to the right, so that its right endpoint becomes \(t_{i+1}\). During the new idle interval \([t_i,t_i+|M_1|+|M_2|)\) we set the processor asleep. Note that in this case the processor was asleep during \(M_1\). The schedule remains a YDS-extension, but its energy consumption becomes strictly smaller: (i) either the processor was asleep during \(M_2\), in which case the resulting schedule uses the same energy while the processor is active but has one wake-up operation less, or (ii) the processor was active and idle during \(M_2\), in which case the resulting schedule saves the idle energy that was spent during \(M_2\).

       
    2. (b)

      The statement follows directly from the second property of Lemma 1 and the fact that all jobs processed during \([t_i,t_{i+1})\) belong to \(\text {SLOW}(\mathcal {J}')\) and are active in the entire zone.

       
     
  3. 3.

    Assume for the sake of contradiction that there exist two jobs \(j' \succ j\), such that a portion of j is processed during an interval \(Z = [\zeta _1,\zeta _2)\), \(\zeta _2 \le d_{j'}\), and some portion of \(j'\) is processed during an interval \(Z' = [\zeta _1',\zeta _2')\), with \(\zeta _2'\le \zeta _1\). We first observe that both jobs belong to \(\text {SLOW}(\mathcal {J}')\). This follows from the fact that both jobs are active during the whole interval \([\zeta _1',\zeta _2)\), and processed during parts of this interval, whereas any job \(j_i^d\) (which are the only jobs in \(\text {FAST}(\mathcal {J}')\)) is processed exactly in its entire interval \([y_i,z_i)\) (by statement 1 of the lemma).

    By the second property of Lemma 1, both j and \(j'\) are processed at the same speed. We can now apply a swap argument. Let \(L:=\min \{|Z|,|Z'|\}\). Note that OPT schedules only \(j'\) during \([\zeta _2'-L,\zeta _2')\) and only j during \([\zeta _2-L,\zeta _2)\). Swap the part of the schedule OPT in \([\zeta _2'-L,\zeta _2')\) with the schedule in the interval \([\zeta _2-L, \zeta _2)\). Given the above observations, it can be easily verified that the resulting schedule (i) is feasible and remains a YDS-extension, and (ii) is lexicographically smaller than OPT.

     
\(\square \)

The next lemma shows how to transform the lexicographically minimal YDS-extension for \(\mathcal {J}'\) of the previous lemma into a well-ordered discretized schedule. This is the most crucial part of our approach. Roughly speaking, the transformation needs to guarantee that (1) in each zone, the volume of a job \(j \in \text {SLOW}(\mathcal {J}')\) processed is an integer multiple of \(v_j/(4n^2\lceil 1/\delta \rceil )\) (this is tantamount to making sure that each zone has integral job pieces to deal with), (2) the job pieces start and end at the time points in W, and (3) all the job pieces are processed in the “right order”. As we will show, the new schedule may run at a higher speed than the given lexicographically minimal YDS-extension, but not by too much.

Lemma 4

Let OPT be a lexicographically minimal YDS-extension for \(\mathcal {J}'\), and let \(s_{\mathcal {S}}(t)\) denote the speed of schedule \(\mathcal {S}\) at time t, for any \(\mathcal {S}\) and t. Then there exists a well-ordered discretized schedule F, such that at any time point \(t\in \mathcal {T}\), there holds
$$\begin{aligned} s_{F}(t)\le (1+\delta )^3s_{OPT}(t), \end{aligned}$$
and for every \(t\notin \mathcal {T}\), there holds
$$\begin{aligned} s_{F}(t)=s_{OPT}(t). \end{aligned}$$

Proof

Through a series of three transformations, we will transform OPT to a well-ordered discretized schedule F, while upper bounding the increase in speed caused by each of these transformations. More specifically, we will transform OPT to a schedule \(F_1\) satisfying (i) and (iii) of Definition 3, then \(F_1\) to \(F_2\) where we slightly adapt the block lengths, and finally \(F_2\) to F which satisfies all three properties of Definition 3. Each of these transformations can increase the speed by at most a factor \((1+\delta )\) for any \(t\in \mathcal {T}\) and does not affect the speed in any interval \(I_i\).

Fig. 2

An illustration of the three transformations. In Transformation 1, volume is shifted between the blocks so that each block contains an integer number of pieces for each job. Note how volume was shifted between the two star-patterned blocks. In Transformation 2, the block lengths are adapted, so that both block-endpoints align with W, see for instance brick-patterned block. Finally, in Transformation 3, some pieces within each block are acellerated while others decelerated in order to make sure that the execution of any piece starts and ends at a timepoint in W. See for example the leftmost pieces. We note that each of the three transformations is designed in a way so that no speed increases by more than an \((1+\delta )\)-factor

Transformation 1

(\(\mathrm {OPT} \rightarrow F_1\)): We will transform the schedule so that
  1. (i)

    For each job \(j\in \text {SLOW}(\mathcal {J}')\), an integer multiple of \(v_j/(4n^2\lceil 1/\delta \rceil )\) volume of job j is processed in each zone, and the processing order of jobs within each zone is determined by \(\prec \). Together with property 1 of Lemma 3, this implies that \(F_1\) (considered as a schedule for pieces) satisfies Definition 3(i).

     
  2. (ii)

    The well-ordered property of Definition 3 is satisfied.

     
  3. (iii)

    For all \(t\in \mathcal {T}\) it holds that \(s_{F_1}(t)\le (1+\delta )s_{OPT}(t)\), and for every \(t\notin \mathcal {T}\) it holds that \(s_{F_1}(t)=s_{OPT}(t)\).

     

Note that by Lemma 3, every zone is either empty, filled exactly by a job \(j_i^d\), or contains a single block. For any job \(j\in \text {SLOW}(\mathcal {J}')\), and every zone \([t_i,t_{i+1})\), let \(V^i_j\) be the processing volume of job j that OPT schedules in zone \([t_i,t_{i+1})\). Since there can be at most 2n different zones, for every job j there exists some index h(j), such that \(V^{h(j)}_j\ge v_j/(2n)\).

For every job \(j\in \text {SLOW}(\mathcal {J}')\), and every \(i\ne h(j)\), we reduce the load of job j processed in \([t_i,t_{i+1})\), by setting it to
$$\begin{aligned} \bar{\mathcal {V}}_j^i = \Big \lfloor V_j^i / \frac{v_j}{4n^2\lceil 1/\delta \rceil )}\Big \rfloor \cdot \frac{v_j}{4n^2\lceil 1/\delta \rceil }. \end{aligned}$$
Finally, we set the volume of j processed in \([t_{h(j)},t_{h(j)+1})\) to \(\bar{\mathcal {V}}_j^{h(j)} = v_j - \sum _{i\ne h(j)} \bar{\mathcal {V}}_j^i\). To keep the schedule feasible, we process the new volume of each non-empty zone \([t_i,t_{i+1})\subseteq \mathcal {T}\) in the zone’s original block \(B_i\), at a uniform speed of \(\sum _{j\in \text {SLOW}(\mathcal {J}')} (\bar{\mathcal {V}}_j^i)/|B_i|\). Here, the processing order of the jobs within the block is determined by \(\prec \).

Note that in the resulting schedule \(F_1\), a job may be processed at different speeds in different zones, but each zone uses only one constant speed level.

It is easy to see that \(F_1\) is a feasible schedule in which for each job \(j\in \text {SLOW}(\mathcal {J}')\), an integer multiple of \(v_j/(4n^2\lceil 1/\delta \rceil )\) volume of j is processed in each zone, and that \(\bar{\mathcal {V}}_j^i \le V_j^i\) for all \(i\ne h(j)\). Furthermore, if \(i = h(j)\), we have that \(\bar{\mathcal {V}}_j^i - V_j^i \le v_j/(2n\lceil 1/\delta \rceil )\), and \(V_j^i\ge v_j/(2n)\). It follows that \(\bar{\mathcal {V}}_j^i\le V_j^i+V_j^i/\lceil 1/\delta \rceil \le (1+\delta )V_j^i\) in this case, and therefore \(s_{F_1}(t) \le (1+\delta )s_{OPT}(t)\) for all \(t\in \mathcal {T}\). We note here, that for every job \(j_i^d\), and the corresponding interval \(I_i\), nothing changes during the transformation.

We finally show that \(F_1\) satisfies the well-ordered property of Definition 3. Assume for the sake of contradiction that there exists a piece u ending at some t, and there exists a piece \(u'\succ u\) with \(d_{u'}\ge t\) that is scheduled before t. Recall that we can implicitly assume that the pieces of any fixed job are processed in the corresponding order \(\prec \). Therefore \(job(u') \succ job(u)\), by definition of the ordering \(\prec \) among pieces. Furthermore, if \([t_k,t_{k+1})\) and \([t_{k'},t_{k'+1})\) are the zones in which u and \(u'\), respectively, are scheduled, then \(k'<k\), as \(k' = k\) would contradict \(F_1\)’s processing order of jobs inside a zone. Also note that \(d_{u'} \ge t_{k+1}\), since \(t\in (t_k,t_{k+1}]\), and \((t_k,t_{k+1})\) does not contain any deadline. This contradicts property 3 of Lemma 3, as the original schedule OPT must have processed some volume of \(job(u')\) in \([t_{k'},t_{k'+1})\), and some volume of job(u) in \([t_{k},t_{k+1})\).

Transformation 2

(\(F_1\rightarrow F_2\)): In this transformation, we slightly modify the block lengths, as a preparation for Transformation 3. For every non-empty zone \([t_i,t_{i+1})\subseteq \mathcal {T}\), we increase the uniform speed of its block until it has a length of \((1+\delta )^j\frac{1}{4n^2s_{crit}(1+\delta ) \lceil 1/\delta \rceil }\) for some integer \(j\ge 0\), keeping one of its endpoints fixed at \(t_i\) or \(t_{i+1}\). Note that in \(F_1\), the block had length at least \(\frac{1}{4n^2s_{crit}(1+\delta ) \lceil 1/\delta \rceil }\), since it contained a volume of at least \(1/(4n^2\lceil 1/\delta \rceil )\), and the speed in this zone was at most \((1+\delta )s_{crit}\). The speedup needed for this modification is clearly at most \((1+\delta )\).

As this transformation does not change the processing order of any pieces nor the zone in which any piece is scheduled, it preserves the well-ordered property of Definition 3.

Transformation 3

(\(F_2\rightarrow F\)): In this final transformation, we want to establish Definition 3(ii). To this end, we shift and compress certain pieces in \(F_2\), such that every execution interval starts and ends at a time point from W (this is already true for pieces corresponding to jobs \(j_i^d\)). The procedure resembles a transformation done in [16]. For any zone \([t_i,t_{i+1})\subseteq \mathcal {T}\), we do the following: Consider the pieces that \(F_2\) processes within the zone \([t_i, t_{i+1})\), and denote this set of pieces by \(D_i\). If \(D_i = \emptyset \), nothing needs to be done. Otherwise, let \(\gamma \) be the integer such that \((1+\delta )^\gamma \frac{1}{4n^2s_{crit}(1+\delta ) \lceil 1/\delta \rceil }\) is the length of the block in this zone, and let
$$\begin{aligned} \varDelta := \frac{1}{64n^8\lceil 1/\delta \rceil ^3 s_{crit}(1+\delta ) (1+\lceil 1/\delta \rceil )}. \end{aligned}$$
Note that in the definition of W, we introduced \(16n^6\lceil 1/\delta \rceil ^2(1+\lceil 1/\delta \rceil )\) many time points (for \(j=\gamma \) and \(r=1,\dots , 16n^6\lceil 1/\delta \rceil ^2(1+\lceil 1/\delta \rceil )\)) that subdivide this block into \(16n^6\lceil 1/\delta \rceil ^2(1+\lceil 1/\delta \rceil )\) intervals of length \(\varDelta \). Furthermore, since \(|D_i| \le 4n^3\lceil 1/\delta \rceil \), there must exist a piece \(u \in D_i\) with execution time \(\varGamma _u \ge 4n^3\lceil 1/\delta \rceil (1+\lceil 1/\delta \rceil ) \varDelta \). We now partition the pieces in \(D_i \setminus \{u\}\) into \(D^+\), the pieces processed after u, and \(D^-\), the pieces processed before u. First, we restrict our attention to \(D^+\). Let \(q_1,\ldots , q_{|D^+|}\) denote the pieces in \(D^+\) in the order they are processed by \(F_2\). Starting with the last piece \(q_{|D^+|}\), and going down to \(q_1\), we modify the schedule as follows. We keep the end of \(q_{|D^+|}\)’s execution interval fixed, and shift its start to the next earlier time point in W, reducing its uniform execution speed accordingly. At the same time, in order to avoid overlaps, we shift the execution intervals of all \(q_k,\ k<|D^+|\) by the same amount to the left (leaving their lengths unchanged). Eventually, we also move the execution end point of u by the same amount to the left (leaving its start point fixed). This shortens the execution interval of u and “absorbs” the shifting of the pieces in \(D^+\) (note that the processing speed of u increases as its interval gets shorter). We then proceed with \(q_{|D^+|-1}\), keeping its end (which now already resides at a time point in W) fixed, and moving its start to the next earlier time point in W. Again, the shift propagates to earlier pieces in \(D^+\), which are moved by the same amount, and shortens u’s execution interval once more. When all pieces in \(D^+\) have been modified in this way, we turn to \(D^-\) and apply the same procedure there. This time, we keep the start times fixed and instead shift the right end points of the execution intervals further to the right. As before, u “absorbs” the propagated shifts, as we increase its start time accordingly. After this modification, the execution intervals of all pieces in \(D_i\) start and end at time points in W.
To complete the proof, we need to argue that the speedup of piece u is bounded by a factor \((1+\delta )\). Since \(|D_i| \le 4n^3\lceil 1/\delta \rceil \), u’s execution interval can be shortened at most \(4n^3\lceil 1/\delta \rceil \) times, each time by a length of at most \(\varDelta \). Furthermore, recall that the execution time of u was \(\varGamma _u \ge 4n^3\lceil 1/\delta \rceil (1+\lceil 1/\delta \rceil ) \varDelta \). Therefore, its new execution time is at least \(\varGamma _u- 4n^3\lceil 1/\delta \rceil \varDelta \ge \varGamma _u - \frac{\varGamma _u}{1+\lceil 1/\delta \rceil }\), and the speedup factor thus at most
$$\begin{aligned} \frac{\varGamma _u}{\varGamma _u-\frac{\varGamma _u}{1+\lceil 1/\delta \rceil }} = \frac{1}{1-\frac{1}{1+\lceil 1/\delta \rceil }} \le 1 + \delta . \end{aligned}$$
Again, the transformation does not change the processing order of any pieces nor the zone in which any piece is scheduled, and thus preserves the well-ordered property of Definition 3.\(\square \)

We now show that the speedup used in our transformation does not increase the energy consumption by more than a factor of \(1+\epsilon \). To this end, observe that for any \(t\in \mathcal {T}\), the speed of the schedule OPT in Lemma 4 is at most \(s_{crit}\), by Lemma 3(2). Furthermore, note that the final schedule F has speed zero whenever OPT has speed zero. This allows F to use exactly the same sleep phases as OPT (resulting in the same wake-up costs). It therefore suffices to prove the following lemma, in order to bound the increase in energy consumption.

Lemma 5

For any \(s \in [0,s_{crit}]\), there holds
$$\begin{aligned} \frac{P\big ((1+\delta )^3 s\big )}{P(s)} \le 1+\epsilon . \end{aligned}$$

Proof

$$\begin{aligned} \frac{P\big ((1+\delta )^3 s\big )}{P(s)}&\overset{(1)}{\le } \frac{P\big ((1+4\delta ) s\big )}{P(s)} \\&= \frac{P(s) +4 \delta s \frac{P(s+4 \delta s)-P(s)}{4 \delta s} }{P(s)} \\&\overset{(2)}{\le } \frac{P(s) +4 \delta s \frac{P(s+s_{crit})-P(s)}{s_{crit}} }{P(s)} \\&\overset{(3)}{\le } \frac{P(s) +4 \delta s \frac{P(2s_{crit})-P(s_{crit})}{s_{crit}} }{P(s)} \\&\overset{(4)}{\le } 1+4 \delta \frac{s_{crit}}{P(s_{crit})} \cdot \frac{P(2s_{crit})-P(s_{crit})}{s_{crit}} \\&\overset{(5)}{\le } 1+\epsilon . \end{aligned}$$
In the above chain of inequalities, (1) holds since \(\delta \le \frac{1}{4}\) and P(s) is non-decreasing. (2) and (3) follow from the convexity of P(s), and the fact that \(4 \delta s \le s_{crit}\). Inequality (4) holds since \(s_{crit}\) minimizes P(s) / s (and thus maximizes s / P(s)), and (5) follows from the definition of \(\delta \). \(\square \)

We summarize the major result of this section in the following lemma.

Lemma 6

There exists a well-ordered discretized schedule with an energy consumption no more than \((1+\epsilon )\) times the optimal energy consumption for \(\mathcal {J}'\).

4 The Dynamic Program

In this section, we show how to use dynamic programming to find a well-ordered discretized schedule with minimum energy consumption. In the following, we discuss only how to find the minimum energy consumption of this target schedule, as the actual schedule can be easily retrieved by proper bookkeeping in the dynamic programming process.

Recall that D is the set of all pieces and W the set of time points. Let \(u_1, u_2, \ldots , u_{|D|}\) be the pieces in D, and w.l.o.g. assume that \(u_1\prec u_2\prec \ldots \prec u_{|D|}\).

Definition 5

For any \(k \in \{1,\ldots ,|D|\}\), and \(\tau _1 \le \tau _2\), \(\tau _1\), \(\tau _2 \in W\), we define \(E_k(\tau _1,\tau _2)\) as the minimum energy consumption during the interval \([\tau _1,\tau _2]\), of a well-ordered discretized schedule so that
  1. 1.

    all pieces \(\{u\succeq u_k: \tau _1<d_u\le \tau _2\}\) are processed in the interval \([\tau _1, \tau _2)\), and

     
  2. 2.

    the machine is active right before \(\tau _1\) and right after \(\tau _2\).

     

In case that there is no such feasible schedule, let \(E_k(\tau _1, \tau _2) = \infty \).

The DP proceeds by filling the entries \(E_k(\tau _1, \tau _2)\) by decreasing index of k. The base cases are
$$\begin{aligned} E_{|D|+1}(\tau _1, \tau _2) := \min \{ P(0)(\tau _2-\tau _1), C\}, \end{aligned}$$
for all \(\tau _1,\tau _2 \in W, \tau _1 \le \tau _2\). For the recursion step, suppose that we are about to fill in \(E_k(\tau _1,\tau _2)\). There are two possibilities.
  • Suppose that \(d_{u_k} \not \in (\tau _1, \tau _2]\). Then clearly \(E_k(\tau _1,\tau _2) = E_{k+1}(\tau _1,\tau _2)\).

  • Suppose that \(d_{u_k} \in (\tau _1, \tau _2]\). By definition, piece \(u_k\) needs to be processed in the interval \([\tau _1,\tau _2)\). We need to guess its actual execution period \([b,e) \subseteq [\tau _1,\tau _2)\), and process the remaining pieces \(\{u\succeq u_{k+1}: \tau _1<d_u\le \tau _2\}\) in the two intervals \([\tau _1, b)\) and \([e,\tau _2)\). We first rule out some guesses of [be) that are bound to be wrong.
    • By Definition 3(i), in a discretized schedule, a piece has to be processed completely inside a zone \([t_i,t_{i+1})\) (recall that \(t_i \in W'\) are the release times and deadlines of the jobs). Therefore, in the right guess, the interior of [be) does not contain any release times or deadlines; more precisely, there is no time point \(t_i \in W'\) so that \(b< t_i < e\).

    • By Definition 3(iii), in a well-ordered discretized schedule, if piece \(u_k\) ends at time point e, then all pieces \(u' \succ u_k\) with deadline \(d_{u'} \ge e\) are processed after\(u_k\). However, consider the guess [be), where \(e = d_{u'}\) for some \(u' \succ u_{k}\) (notice that the previous case does not rule out this possibility). Then \(u'\) cannot be processed anywhere in a well-ordered schedule. Thus, such a guess [be) cannot be right.

    By the preceding discussion, if the guess (be) is right, the two sets of pieces \(\{u\succeq u_{k+1}: \tau _1<d_u\le b\}\) and \(\{u\succeq u_{k+1}: e <d_u\le \tau _2 \}\), along with piece \(u_k\), comprise all pieces to be processed that are required by the definition of \(E_k(\tau _1, \tau _2)\). Clearly, the former set of pieces \(\{u\succeq u_{k+1}: \tau _1<d_u\le b\}\) has to be processed in the interval \([\tau _1,b)\); the latter set of pieces, in a well-ordered schedule, must be processed in the interval \([e,\tau _2)\) if [be) is the correct guess for the execution of the piece \(u_k\).

    We therefore have that if there exist \(b, e \in W\) with the properties stated under the min-operator, and \(E_k(\tau _1,\tau _2) = \infty \) otherwise.

Theorem 1

There exists a fully polynomial-time approximation scheme (FPTAS) for speed scaling with sleep state. Its time complexity is \(O(\frac{n^{32}}{\epsilon ^{17}} ( \log \frac{n^2 d_{\max }}{\epsilon })^4)\).

Proof

Given an arbitrary instance \(\mathcal {J}\) for speed scaling with sleep state, we can transform it in polynomial time to an instance \(\mathcal {J}'\), as seen in Sect. 2. We then apply the dynamic programming algorithm that was described in this section to obtain a well-ordered discretized schedule \(\mathcal {S'}\) of minimal energy consumption for instance \(\mathcal {J}'\). By Lemma 6, we have that \(\mathcal {S'}\) is a \((1+\epsilon )\)-approximation for instance \(\mathcal {J}'\). Furthermore, note that every discretized schedule (and therefore also \(\mathcal {S'}\)) executes each job \(j_i^d\) exactly in its allowed interval \(I_i = [y_i, z_i)\). This holds because there are no time points from the interior of \(I_i\) included in W, and any discretized schedule must therefore choose to run \(j_i^d\) precisely from \(y_i \in W\) to \(z_i \in W\). Therefore, by Lemma 2, we can transform \(\mathcal {S'}\) to a schedule \(\mathcal {S}\) in polynomial time and obtain a \((1+\epsilon )\)-approximation for \(\mathcal {J}\).

We next analyze the running time. The pre-processing of the instance and the YDS algorithm are easily dominated by the dynamic program. By construction in Sect. 3.1, \(|W|= O( \frac{n^7}{\epsilon ^4} \log \frac{n^2 d_{\max }}{\epsilon })\), \(|W'| = O(n)\) and \(|D| = O(\frac{n^3}{\epsilon })\). The total number of entries in the dynamic program is \(O(|D||W|^2)\). For each single entry of \(E_{k}(\tau _1, \tau _2)\), we need to check \(O({|W| \atopwithdelims ()2})\) possibilities. For each possibility, we need O(n) time. In sum, this gives the running time of \(O(n|D||W|^4)=O(\frac{n^{32}}{\epsilon ^{17}} ( \log \frac{n^2 d_{\max }}{\epsilon })^4)\). \(\square \)

Footnotes

  1. 1.

    It is assumed that we have access to an oracle which allows us to evaluate P(s) for any value of s and perform simple operations involving values returned by P. Similarily the critical speed (defined later) is a known parameter of P(s).

  2. 2.

    Unless stated differently, throughout the text an interval will always have the form \([\cdot ,\cdot )\).

  3. 3.

    For any two time points \(t_1<t_2\), we say that \(t_1\) is to the left of \(t_2\), and \(t_2\) is to the right of \(t_1\).

Notes

Acknowledgements

Open access funding provided by Max Planck Society.

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Authors and Affiliations

  1. 1.Saarland University and Max-Planck-Institut für InformatikSaarbrückenGermany
  2. 2.École Normale SupérieureParisFrance
  3. 3.Max-Planck-Institut für InformatikSaarbrückenGermany

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