Energy Efficient Scheduling for Real-Time Systems on Variable Voltage Processors æ

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1 Energy Efficient Scheduling for Real-Time Systems on Variable Voltage Processors æ Gang Quan Xiaobo (Sharon) Hu Department of Computer Science & Engineering University of Notre Dame Notre Dame, IN Abstract Energy consumption has become an increasingly important consideration in designing many realtime embedded systems. Variable voltage processors, if used properly, can dramatically reduce such system energy consumption. In this paper, we present a technique to determine voltage settings for a variable voltage processor that utilizes a fixed priority assignment to schedule jobs. Our approach also produces the minimum constant voltage needed to feasibly schedule the entire job set. Our algorithms lead to significant energy saving compared with previously presented approaches. 1 Introduction Energy consumption is one of the critical factors in designing battery-operated systems, such as portable personal computing and communication devices. To reduce system energy consumption, supply voltage reduction is the most powerful technique since power is a quadratic function of the voltage. However, if a single supply voltage is used, the amount of voltage reduction allowed may be severely limited due to the requirement of satisfying few tight timing constraints, which is often the case for real-time embedded systems. Recent advances in power supply circuits [2, 9] have enabled systems to operate under dynamically varying supply voltages. In such an environment, both power consumption and the speed of the system can be dynamically controlled. Judicious exploitation of this feature can dramatically improve the energy consumption of a real-time system. In this paper, we are interested in studying the following type of a real-time system implemented on a variable voltage processor. The real-time system consists of jobs with predefined release times, deadlines and required number of CPU cycles. Such jobs may either be aperiodic or be instances of periodic tasks. These jobs are scheduled by a preemptive scheduler based on some statically defined priorities, e.g., according to the rate-monotonic policy [7]. Such a fixed-priority assignment approach is used in most real-time scheduling algorithms [8]. If the jobs are executed by a variable voltage processor, the execution time of each job can vary depending on the speed determined by different voltage levels at different times. By setting the supply voltage to different values at different times, we can essentially build a voltage schedule. æ This research was supported in part by the National Science Foundation under Grant MIP

2 The challenge is to determine the voltage (or equivalently speed) schedule which lead to the minimal energy consumption. A number of papers have been published on similar topics. Off-line scheduling algorithm for nonpreemptive hard real-time tasks are discussed in [3, 6]. In [5, 6], more general variable voltage processor models are used assuming that processor voltage cannot change instantaneously or continuously. The more practical processor models make the problem much harder to solve. A heuristic approach is described in [5], and a linear programming formulation is introduced in [6]. Yao, Demers and Shenker [13] presented an OèN 2 è (or OèN log N è for a more sophisticated implementation), where N is the number of jobs to be scheduled, time off-line algorithm for finding the optimal voltage schedule. They assume that jobs are scheduled according to the earliest-deadline-first (EDF) scheduling policy [7]. Hong, Potkonjak and Srivastava described an on-line scheduling algorithm for real-time tasks on variable voltage processor, where it is assumed that the release times of jobs are not known apriori[4]. All of the above approaches employ the dynamic EDF priority assignment scheme for scheduling the jobs. Though the EDF policy is used in some real-time systems, fixed-priority assignments are adopted in most real-time scheduling algorithms of practical interest due to its low overhead and predictability [8]. Shin and Choi [11] presented a power conscious fixed priority scheduling scheme for hard real-time systems on a variable voltage processor. The approach makes use of a simple run-time checking mechanism. The processor can either be shut down (if there is no current active job) or adopt the speed such that the current active job finishes exactly at the release time of the next job. The advantage of the technique is its simplicity and hence can be readily incorporated into an operating system (OS) kernel. However, the on-line approach determines the processor speed based solely on the release time of the closest job to be executed. It cannot exploit the fact that the release times and deadlines of most real-time jobs, particularly the jobs corresponding to the instances of periodic tasks, are known off-line. Hence, it may not be able to fully utilize the benefit provided by a variable voltage processor. In [12], Shin, Choi, and Takayasu proposed an off-line algorithm to determine the lowest maximum processor speed to execute a periodic real-time task set on a variable speed processor. It is assumed that all the tasks start at the same time, and consequently the first job of each task would have the longest response time [7]. The algorithm finds the minimum processor speed that guarantees the schedulability of the first job for each task. Note that this approach can only be applied to the periodic tasks having the same starting time. Moreover, it is not difficult to see that the minimum processor speed can be further reduced after the completion of the first job. Though the algorithm in [12] can be combined with the approach in [11], it still fails to maximally explore the flexibility of a variable speed processor. In this paper, we present a technique to determine voltage schedules that result in more energy saving. Similar to that in [13], our technique is based on the assumption that the timing parameters of each job is known off-line. (For jobs whose timing parameters are not known, our technique can be readily combined with an on-line algorithm such as the one in [11].) Though we simply change the priority assignment from dynamic EDF in [13] to a fixed assignment, the problem becomes much harder to solve. We will provide some examples later to illustrate this. Two algorithms are given in the paper. The first one takes OèN 2 è time, where N is number of jobs to be scheduled, and finds the minimum constant speed needed to complete each job since constant voltage 2

3 tends to result in lower power consumption. The second algorithm, with OèN 3 è time complexity, builds on the first one and gives two results. First, the minimum constant voltage (or speed) needed to complete a set of jobs is obtained. This is an important parameter when designing systems with no sophisticated power management hardware but only simple on/off modes or where peak power consumption is a concern. Secondly, a voltage schedule is produced. We prove that this voltage schedule always results in lower energy consumption compared to using the minimum constant voltage and shutting down the system when it is idle. we show through experiments that the energy saving achieved by applying our algorithm is quite significant, especially compared to the algorithms presented in [11]. Note that our algorithm can be readily combined with on-line scheduling techniques such as the one in [11] to further improve energy consumption during run-time. The rest of the paper is organized as follows. Section 2 formulates the problem and gives some motivational examples. Two novel algorithms are presented in Section 3 and 4. Experimental results are discussed in Section 5 and Section 6 concludes the paper. 2 Preliminaries In this section, we first introduce the necessary notation and formulate the problem. Then, we review some known results and provide several motivational examples. 2.1 Problem formulation The system we are studying consists of N independent jobs, J = fj 1 ;J 2 ; æææ;j N g, arranged in the decreasing order of their statically assigned priorities. The following timing parameters are defined for each job J n : æ R n : the time at which job J n is ready to be executed, referred to as release time. æ D n : the time by which J n must be completed, referred to as deadline. æ C n : the maximum number of CPU cycles needed to complete job J n without any interruption, referred to as workload. Without loss of generality, we assume no special relationships among release times and deadlines of different jobs. It is not difficult to see that the above system model can be readily used to model task instances in periodic real-time systems, where R m and R n differ by some integer multiple of the task period if J m and J n belong to the same task. A single processor is used to execute the jobs in the system, and the processor can work at different voltage levels which can be continuously varied from ë0;v max ë. When the supply voltage v changes, the processor speed s changes proportionally. For simplicity, we will use processor speed and supply voltage interchangeably whenever applicable. The power consumption (P ) is a convex function of the processor speed, i.e., P = f èsè; where f 00 èsè ç 0; (1) 3

4 and s è v; 0 ç v ç V max : (2) Given a set of real-time jobs and a variable voltage processor introduced, different voltage values (or processor speeds) can be set at different times. We refer to a set of voltage values or speeds during the entire time interval when the job set J being executed as a voltage schedule. Our problem is then to determine a voltage schedule with which the lowest amount of energy is consumed and the jobs are all completed at or before their deadlines. Several observations are helpful in formulating our problem more formally. The authors of [13] presented a theorem regarding the best speed for a given set of jobs that must be completed within an interval. We restate the theorem in the following. Theorem 1 Assume that the power consumption of a variable voltage processor is a convex function of the processor speed. Given a set of jobs starting at t 0 and to be completed by t 1, the voltage schedule that employs a constant voltage in ët 0 ;t 1 ë is necessarily an optimal schedule in the sense that no other schedule consumes less energy to complete the jobs in time. Based on the above theorem, we can easily prove the following lemma which describes an important feature for any optimal voltage schedule. Lemma 1 An optimal voltage schedule for a job set J is defined on a set of time intervals in which the processor maintains a constant speed, and each of these intervals must start and end at either the release times or deadlines of the jobs. Proof: Suppose the curve Sètè in Figure 1 is the optimal voltage schedule for a given job set, and t 0 ;t 1 ; æææ;t m are the release times or deadlines of the jobs. Let s 1 ;s 2 ; æææ;s m be the constant speeds such that s i æ èt i, t i,1 è= Z ti t i,1 Sètèdt; i.e., the speeds that result in the same work load as Sètè within each interval. It follows that no job will violate its deadline while the processor running at the speeds s 1 ;s 2 ; æææs m : According to Theorem 1, the processor will consume less energy running at the speeds designated by s 1 ;s 2 ; æææ;s m than that running at the speed defined by Sètè. Thus an optimal schedule can only change its speed either at the release times or deadlines of some jobs. 2 According to Lemma 1, our voltage scheduling problem can be formally defined as follows: Definition 1 Given a job set J, find a set of intervals, ët k s;t k f ë, and a set of speeds, S = fsètk s;t k f è;k = 1; 2; æææ;kg, wheret k s and tk f are among the job release times and deadlines, and Sètk s ;tk f è is a constant speed, such that if the processor operates accordingly, all the jobs can be completed by their deadlines and no other voltage schedules can consume less energy. 2.2 Motivational examples Consider a simple real-time system with 3 jobs as shown in Figure2(a). The fixed priority assignment for the jobs are different from the EDF priority assignment where J 2 would have a higher priority than J 1. 4

5 Speed s 1 s 2 s 3 s 4 s m 1 s m S(t) t 3 t m 1 t 0 t 1 t 2 t 4 t m 2 t m Time Figure 1: Optimal energy saving scheduling for fixed-priority task set. Under the EDF policy, the best voltage schedule is Sè0; 8è = 13=8. However, it is not difficult to check that this voltage schedule would result J 2 missing its deadline if the jobs are scheduled according to the given priority. Thus, whenever the fixed priority assignment disagrees with the EDF assignment, the voltage schedule obtained based on the EDF priority assignment may not be applicable any more. For the same example, the off-line approach in [12] cannot be readily applied because the jobs do not start at the same time. If the on-line voltage scheduling algorithm proposed in [11] is used, we can obtain the voltage schedule as shown in Figure 2(b). This is a greedy approach as it only checks the release time of the next job to determine the current speed. A better voltage schedule is given in Figure 2(c). It is not difficult to verify that the voltage schedule in Figure 2(c) consumes less energy than that in Figure 2(b). If we assume P = s 2, the energy in Figure 2(c) is 22:5=27 = 83è of that in Figure 2b. From the above examples, it is clear that existing approaches are not able to determine the optimal voltage schedule for a given job set when a fixed-priority assignment is used. However, we find the rational behind the technique used in [13] is enlightening. The key to the algorithm in [13] is to find the minimum constant speed needed to finish certain subsets of all jobs. In the EDF priority assignment, this can be easily computed by P J SèR m ;D i 2J j C i n è= (3) D n, R m where D n é R m,andj j is the subset of jobs whose release times and deadline are both in ër m ;D n ë. In computing SèR m ;D n è, there is no need to include any jobs that are released in ër m ;D n ë, and have deadlines after D n, since there always have lower priorities than J n. In the fixed-priority assignment case, (3) is no longer valid due to the fact that a job J k released in ër m ;D n ë with deadline larger than D n may have a higher priority than J n. If J k does have a higher priority, it may preempt J n in ër m ;D n ë (depending on if J n is finished before or after J k 's release time). In Figure 2(a), J 2 and J 1 exhibit such a relationship. This uncertainty in the preemption relationship greatly increases the difficulty in finding the voltage schedules under the fixed-priority assignment scheme. In the 5

6 s J J J (a) C =2 1 C =6 2 C =5 3 t s R =2 1 1 D =6 R 2=0 D =4 R =3 2 2 D = / t (b) (c) Figure 2: An example real-time system with different voltage schedules. t following section, we present new observations and techniques to tackle such a problem. 3 Determining the Minimum Constant Speed for Each Job In this section, we present our approach to find the minimum constant speed needed to complete each job by each deadline. Finding such speeds is beneficial in two aspects. First, it can help us to determine the minimum overall constant speed for the entire job set J such that if the supply voltage is set for this speed, it will result in the minimum energy consumption compared to any other constant speed for J. Secondly, we can use it to derive a voltage schedule that leads to even lower energy consumption. Recall that the authors in [12] proposed a technique to compute the minimum constant speed to guarantee the schedulability for a periodic task system. Since all the tasks start at the same time, the problem reduces to finding the minimum constant speed for the first job of each task. This is a rather special case. We are dealing with a more general case where jobs can be released at any time. Let us denote the minimum constant speed needed to complete job J n by S n. We first use an example (see Figure3) to illustrate several critical properties that S n must possess. Suppose we want to find S 3 for J 3 in Figure3. If we use interval ë0; 10ë to compute S 3,thenS 3 = Së0; 10ë = 1. However, applying Së0; 10ë = 1 will cause an idle interval ë1; 2ë and J 3 to miss its deadline. On the other hand, if we set S 3 = Së3; 10ë = 5=7, J 3 can be completed by D 3. But in order for J 3 to start at t =3, the processor speed must be set to at least Së2; 3ë = 4. Otherwise,J 2 will prevent J 3 from finishing on time. Hence, Së3; 10ë is not the valid minimum constant speed for J 3. For this example, S 3 = Së2; 10ë = 9=8. If we let the processor speed during ë0; 2ë be 1=2, andp = s 2, it is easy to verify that the power consumption for this schedule is 10:6=20:1 = 52:7è of the one using S 3 = Së3; 10ë. To summarize, the minimum constant speed S n is computed based on some intervals which must have the following properties: 6

7 J J C 1=1 R 1 =0 D 1 =9 C 2=4 R 2 =2 D 2 =8 C 3=5 R 3=3 D 3 =10 J Figure 3: An example real-time system with three jobs. æ There is no idle time within the interval that S n corresponds to. æ Applying S n do not force other intervals to take higher speeds. æ The interval must begin and end at the release times or deadlines of some jobs. These properties play a key role in determining the interval to compute S n. When deriving the minimum speed of J n, we do not need to consider every job in J but only certain higher priority jobs whose execution may interfere with J n 's execution. The following definitions help us to limit the number of jobs to be considered. Definition 2 Time t is called a J n -scheduling point if t = R i ; 1 ç i ç n or t = D n. For the rest of the paper, when we refer to a time t, we always mean a scheduling point. According to Lemma 1, if the processor speed varies between two consecutive scheduling points, we can always find a constant speed in the corresponding interval which is more energy efficient. Definition 3 A J n -scheduling point t is called the earliest scheduling point of J n and denoted by T E ènè if it is the largest J n -scheduling point in ë0;r n ë that satisfies t ç D i if tér i ; 1 ç i ç n: ThelatesttimeofJ n by which J n must be completed is called the latest scheduling point of J n and is denoted by T L ènè. Based on the above definitions, T L ènè can be simply set to D n, while T E ènè can be obtained by checking each J n -scheduling point in the decreasing order starting from R n. It is not difficult to see that any higher priority jobs released prior to T E ènè or after T L ènè do not have any impact on the speed needed to complete J n provided that these jobs are finished by their deadlines. Thus, when computing S n, we only need to focus on the jobs released within ët E ènè;t L ènèë. Since speed is closely related with average workload, we introduce the definition of J n -intensity to capture the concept of average workload for job J n. 7

8 Definition 4 Let t a ;t b be two J n -scheduling points, J n -intensity in the interval ët a ;t b ë, denoted by I n èt a ;t b è, is defined to be P ni=1 æèj i è æ C i I n èt a ;t b è= ; (4) t b, t a where è 1 t æèj a ç R i ét b i è= (5) 0 otherwise J n -intensity has a very useful property which is summarized below. Lemma 2 Let t a ;t b, and t c be three J n -scheduling points, and t a ét b ét c.theni n èt a ;t b è, I n èt b ;t c è, and I n èt a ;t c è always satisfy one of the following two inequalities: I n èt a ;t b è ç I n èt a ;t c è ç I n èt b ;t c è; (6) or I n èt a ;t b è ç I n èt a ;t c è ç I n èt b ;t c è: (7) Proof: Let the workload in ët a ;t c ë, ët a ;t b ë,andët b ;t c ë be W, W 1,andW 2, respectively. Then W = W 1 + W 2 I n èt a ;t c è= W 1 + W 2 t a, t c I n èt a ;t b è= W 1 t b, t a I n èt b ;t c è= W 2 t c, t b : Without loss of generality, let I n èt a ;t b è ç I n èt a ;t c è, i.e., After some simple transformations, we have W 1 t b, t a ç W 1 + W 2 t c, t a : W 2 t c, t b ç W 1 + W 2 t c, t a : Thus (6) is true. Similarly, (7) can be proved to be correct. 2 Since having no idle time is one of the key properties required for S n, we give the following definition to precisely capture the idle time related concepts. Definition 5 Interval ët a ;t b ë is a J n -busy interval, if æ t a ;t b are J n scheduling points, and T E ènè ç t a ç R n ét b ç T L ènè. æ There is no idle time before the release time of any job within ët a ;t b ë if the constant speed I n èt a ;t b è is applied within the interval. For a J n -busy interval, we have the following lemma. 8

9 Lemma 3 An interval ët a ;t b ë is a J n -busy interval if and only if for every J n -scheduling point t 2 èt a ;t b ë. I n èt a ;tè ç I n èt a ;t b è (8) Proof: Let t 2 èt a ;t b ë.sincei n èt a ;tè ç I n èt a ;t b è,andi n èt a ;tè æ èt, t a è ç I n èt a ;t b è æ èt, t a è,the workload in ët a ;të is either not finished or finished at t if the processor speed is I n èt a ;t b è. Since this is true for any t 2 èt a ;t b ë, the processor is not idle within interval ët a ;t b ë. On the other hand, let the processor with constant speed I n èt a ;t b è be always busy during ët a ;të;t 2 èt a ;t b ë. Then, according to Definition 4, we have I n èt a ;tè æ èt, t a è ç I n èt a ;t b è æ èt, t a è, and thus I n èt a ;tè ç I n èt a ;t b è. 2 For job J n, it is not difficult to see that there may exist a number of J n -busy intervals. The largest one among them is particularly interesting and we give a definition for it below. Definition 6 A J n -busy interval ët s ;t f ë is called the J n -essential interval if for any J n -busy interval ët a ;t b ë, we have t s ç t a and t b ç t f : (9) The J n -intensity corresponding to the J n -essential interval possesses the properties of S n stated earlier in this section. This is summarized in the following important lemma. Lemma 4 The J n -essential interval, ët s ;t f ë, and the corresponding J n -intensity, I n èt s ;t f è, satisfy I n èt; t s è éi n èt s ;t f è; T E ènè ç tét s ; (10) I n èt s ;t f è éi n èt f ;tè; t f étç T L ènè; (11) and I n èt s ;t f è= min I n èt a ;t b è; (12) ët a;t b ë where ët a ;t b ë is any J n -busy interval satisfying (10) and (11). Furthermore, if I n èt s ;t f è is adopted as the processor speed during ët s ;t f ë, J n is completed by its deadline. Proof: We prove (10) by contradiction. Assume there exists t 2 ët E ènè;t s è, such that I n èt; t s è ç I n èt s ;t f è: Let So, I n èt m ;t s è = max I n èt i ;t s è; t m ;t i 2 ët E ènè;t s è: i I n èt m ;t s è ç I n èt s ;t f è: According to Lemma 2, for any t m étç t s,wehave, I n èt m ;tè ç I n èt m ;t s è ç I n èt s ;t f è: 9

10 and it is easy to show I n èt m ;tè ç I n èt m ;t f è: Similarly, for any t s étç t f, we can show that I n èt m ;tè ç I n èt m ;t f è: Thus for any t m étç t f,wehave I n èt m ;tè ç I n èt m ;t f è; andbylemma3,ët m ;t f ë is a J n -busy interval. Since t m é t s, this violates Definition 6. Inequality (11) can be proved similarly. Again, we prove (12) by contradiction. Suppose there exists another busy interval ët 0 a;t 0 bë such that I n èt 0 a;t 0 b è éi nèt s ;t f è; (13) where and I n èt; t 0 a è éi nèt 0 a;t 0 b è; t 2 ët Eènè;t 0 aè; (14) I n èt 0 a ;t0 b è éi nèt 0 b ;tè; t 2 èt0 b ;T Lènèë: (15) According to Definition 6, we have t s ç t 0 a ç R i ét 0 b ç t f: There are two cases needed to be considered. æ Case 1: t s ét 0 a ç R i ét 0 b ç t f: From Lemma 2 and (13), we have which contradicts (14). I n èt s ;t 0 a è ç I nèt s ;t f è ç I n èt 0 a;t 0 b è; æ Case 2: t s = t 0 a and t0 b ç t f. From Lemma 2, we have which contradicts (13). I n èt 0 a ;t0 b è=i nèt s ;t 0 b è ç I nèt s ;t f è; To show that J n is completed by its deadline, we only need to note that J n is finished by t f if I n èt s ;t f è is used as the processor speed within ët s ;t f ë,andt f ç D n. 2 According to Lemma 4, I n èt s ;t f è is the valid minimum constant speed S n. Thus, determining the minimum constant speed for each job now becomes determining the essential interval and corresponding intensity associated with each job. In the following, we present our algorithm, Algorithm 1, to search for J n -essential interval and compute S n. Our algorithm follows the basic principle laid down in Definition 6 but employs a little different search mechanism which on average takes less time than a straightforward implementation of Definition 6. 10

11 Algorithm 1 Construct the essential interval for a job 1: Input: Job set J = fj 1 ; :::; J N g,andt E ènè and T L ènè for job J n 2: Output: J n -essential interval ët s ;t f ë and S n 3: t 0 a = t b = R n ; 4: t a = T E ènè; 5: t 0 b = T Lènè; 6: while t a 6= t 0 a or t 0 b 6= t b do 7: t a = t 0 a; 8: t 0 b = t b; 9: for every J n -scheduling point in ët a ;T L ènèë do 10: Find t b such that Ièt a ;t b è is the minimum; 11: end for 12: for every J n -scheduling point in ët E ènè;t a ë do 13: Find t 0 a such that Ièt0 a ;t bè is the maximum; 14: end for 15: end while 16: t s = t a ;t f = t b ;S n = Ièt a ;t b è; For ease of understanding, the execution steps of Algorithm 1 are depicted in Figure 4. Algorithm 1 starts from R n and looks to the right of R n to find the scheduling point t 1 such that I n èr n ;t 1 è ç I n èr n ;tè for any scheduling point t in ër n ;T L ènèë (If a tie occurs, take the right most one). Then the algorithm searches the left hand side of R n to find t 2 such that I n èt 2 ;t 1 è ç I n èt; t 1 è for any scheduling point t in ët E ènè;r n ë (If a tie occurs, take the left most one). The algorithm continues in this fashion until the scheduling points on both sides of R n no longer change. To show that Algorithm 1 indeed produces the correct J n -essential t 4 T E(n) t t t s 2 R n t 1 t 3 t f T L (n) Figure 4: This illustration shows the sequence of steps taken to construct J n -essential interval according to Algorithm 1. interval, we only need to prove that t s and t f obtained from Algorithm 1 satisfies (9). The following lemmas and theorem are sufficient for this purpose. Lemma 5 At the end of each iteration of the while loop in Algorithm 1, ët a ;t b ë is a busy interval of J n. Proof: Refer to Figure 4. At the end of each iteration of the while loop in Algorithm 1, we get interval ët2;t1ë, ët 4 ;t 3 ë, and so on. Consider ët 2 ;t 1 ë. for t 2 èt 2 ;R n ë,sincei n èt 2 ;t 1 è ç I n èt; t 1 è, by Lemma 2, we have I n èt 2 ;tè ç I n èt 2 ;t 1 è. Similarly, for t 2 èr n ;t 1 ë,sincei n èt; t 1 è ç I n èr n ;t 1 è ç I n èt 2 ;t 1 è,so 11

12 I n èt 2 ;t 1 è ç I n èt 2 ;t 1 è. According to Lemma 3, ët 2 ;t 1 ë is a busy interval. Other intervals can be proved similarly. 2 Lemma 6 The final interval produced by Algorithm 1 satisfies I n èt; t s è éi n èt s ;t f è; T E ènè ç tét s ; (16) and I n èt f ;tè éi n èt s ;t f è; t f étç T L ènè: (17) Proof: Let T E ènè ç t é t s as shown in Figure 4. If I n èt; t s è ç I n èt s ;t f è, by Lemma 2, we have I n èt; t f è ç I n èt s ;t f è. This cannot be true since in Algorithm 1, I n èt s ;t f è is the maximum. Similarly, we can prove that I n èt f ;tè ç I n èt s ;t f è; t f étçt L ènè is not possible either. 2 Theorem 2 Algorithm 1 produces, in OèN 2 è time, the J n -essential interval. Proof: According to Lemma 4 and Lemma 5, we only need to prove that after the completion of Algorithm 1, there is no J n -busy interval ët a ;t b ë such that t a ét s or t b ét f : We use contradiction to prove the theorem. There are several cases needed to be considered. æ Case 1: t a ét s ç R n ét b ç t f.sinceët a ;t b ë and ët s ;t f ë are both busy intervals, I n èt a ;t s è ç I n èt s ;t b è ç I n èt s ;t f è; which is contradicted to (16). æ Case 2: t a ét s ç R n ét f ét b.sinceët a ;t b ë and ët s ;t f ë are both busy intervals, by Lemma 2,we have I n èt a ;t s è ç I n èt s ;t b è ç I n èt f ;t b è; and according to Lemma 6, we have I n èt f ;t b è éi n èt s ;t f è: So, I n èt a ;t s è éi n èt s ;t f è; which violates (16). æ Case 3: t s ç t a ç R n ét f ét b.sinceët a ;t b ë and ët s ;t f ë are both busy intervals, and according to Lemma 2, we have I n èt s ;t f è ç I n èt a ;t f è ç I n èt f ;t b è; which contradicts (17). 12

13 The worst case time complexity of Algorithm is OèN 2 è since the maximum number of scheduling points for J n is OèN è. 2 From the proof of Theorem 2, one may observe that Algorithm 1 only searches a subset of busy intervals whose starting points are the scheduling points in ët E ènè;r n ë. This is achieved by judiciously selecting only those scheduling points that may lead to the maximum speed. By repeatedly applying Algorithm 1, one can find the minimum speed for each job. 4 Determining the Global Voltage Schedule Based on the algorithm for searching the minimum constant speed of each job, we can find both the minimum constant speed needed to satisfy all job deadlines and a better voltage schedule to further improve the system energy consumption. In the following, we present the algorithm and prove some lemmas, which tackle the two problems simultaneously. We first introduce the concept of critical interval. Definition 7 The essential interval ët s ;t f ë with the largest J n -intensity is called the critical interval of job set J. The critical interval of J indicates the minimum constant speed, S = max n S n, needed for J to be feasible. If S n = I n èt s ;t f è is used in ët s ;t f ë, which guarantees to finish J n by its deadline, what happens to other jobs in this interval and the jobs elsewhere? Suppose that the critical interval of J corresponds to J n -essential interval ët s ;t f ë. We would like to investigate the impact of removing ët s ;t f ë from the overall execution time interval on the rest of the jobs. By removing the critical interval of J, we mean the following: 1. Remove from J the jobs associated with ët s ;t f ë,thatis,jobj n and all other jobs that have higher priorities than J n and are released within ët s ;t f ë. 2. Shrink the interval ët s ;t f ë into a single time point, i.e., reduce every time instant greater than t f by the amount of èt f, t s è.ifr i, T E èiè or T L èiè for any job J i is inside ët s ;t f ë before the reduction, it will be changed to the value of t s. For the remaining jobs, we can again find the critical interval. Repeatedly performing the above steps, we obtain a set of critical intervals as well as the corresponding speeds. We will show that these critical intervals form a valid, low-energy voltage schedule. We first summarize the above procedure in Algorithm 2. By applying Algorithm 2, we obtain a set of intervals T and their corresponding constant speeds S. In the following, we present two theorems describing the important characteristics of T and S. Theorem 3 GivenajobsetJ,letët k s ;tk f ë and Sètk s ;tk f è for 1 ç k ç K be the critical intervals and corresponding speeds output from Algorithm 2. Every job in J is guaranteed to be completed by its deadline if Sèt k s;t k f è is used in the corresponding interval ëtk s;t k f ë. The proof of Theorem3 is rather long and given in the Appendix. From Theorem 3, we conclude that the set of critical intervals and their associated speeds obtained by Algorithm 2 form a valid voltage schedule. Furthermore, the speed for each critical interval is the lowest constant speed possible for that interval. 13

14 Algorithm 2 Construct the set of critical intervals for J. 1: Input: Job set J = fj 1 ; :::; J n g 2: Output: A set of critical intervals T = fët 1 s ;t1 f ë; æææ; ëtk s ;tk f ë g, and a set of speeds S = fsèt k s;t k f èg for 1 ç k ç K 3: for every job J i 2J do 4: A; S 0 = ;; 5: Find T E èiè and T L èiè according to Definition 3; 6: Determine J i -essential interval ët s ;t f ë and minimum speed S i by Algorithm 1; 7: Add ët s ;t f ë to A and S i to S 0 ; 8: end for 9: k =1; 10: while A is not empty do 11: Select S j = S i ;i =1; æææ;nfrom S 0 and corresponding ët s ;t f ë from A; 12: Add ët s ;t f ë as ët k s ;tk f ë to T and S j as Sèt k s ;tk f è to S; 13: Remove ët s ;t f ë from A and S j from S 0 ; 14: for J m 2J do 15: 16: if m ç j and t k s ç R m ét k f then Remove J m from J ; 17: Remove S m from S and corresponding ët s ;t f ë from A; 18: else 19: if t k s ér m ç t k f ; then 20: R m = t k s ; 21: end if 22: if t k s ét E èmè ç t k f ; then 23: T E èmè =t k s; 24: end if 25: if t k s ét L èmè ç t k f ; then 26: T L èmè =t k s ; 27: end if 28: end if 29: end for 30: for every scheduling point tét k f do 31: t = t, èt k f, tk s è; 32: end for 33: for every J m 2J whose R m ;T E èmè or T L èmè has changed do 34: Re-construct and modify J m -essential interval ët s ;t f ë in A; 35: Re-compute and modify S m in S 0 ; 36: end for 37: k++; 38: end while 39: Output T and S; 14

15 However, we still do not know if any of the speeds from Algorithm 2 can be used as the overall minimum constant speed for the entire job set J. The following theorem helps answer this question. Theorem 4 Given a job set J, speeds Sèt k s;t k f è 2 S obtained by Algorithm 2 satisfy the following: Sèt 1 s;t 1 f è ç Sèt2 s;t 2 f è ç æææ çsètk s ;t K f è. The proof for this theorem is also given in the Appendix. Based on Theorem 3 and 4, we have the the following corollaries. Corollary 1 The first speed in the speed set produced by Algorithm 2 is the minimum constant speed that can be applied throughout the execution of all jobs such that no jobs violate their deadlines. Proof: According to Theorem 4, the first speed is the highest necessary processor speed among all the critical intervals. Note that each job is associated with one critical interval and is schedulable as long as the processor speed is no less than the minimum speed for the critical interval, so all the jobs are schedulable if the first speed in the speed set produced by Algorithm 2 is applied throughout the entire execution period. 2 Corollary 2 The voltage schedule obtained by Algorithm 2 always saves more energy than the one that applies the minimum constant speed when the processor is busy while shuts down the processor when it is idle. Proof: Let the minimum constant processor speed be S H, and the speed when it is shut down be S L.From Lemma 4, the processor speed by Algorithm 2 can be represented as Sètè =æètès L +è1, æètèès H ; 0 ç æètè ç 1: (18) Also, let the total execution time of the job set be T, and the the total time when processor running at the minimum constant speed be t 1. We want to prove that Z T 0 f èsètèèdt Z t1 Z T ç f ès H èdt + f ès L èdt (19) 0 t 1 = f ès H èt +èfès L è, f ès H èèèt, t 1 è: Since f èxè is a convex function as shown in (1), from (18), we have and thus, f èsètèè ç æètèf ès L è+è1, æètèèf ès H è; Moreover, since Z T Z T f èsètè ç 0 0 æètèf ès L èdt + Z T = f ès H èt +èf ès L è, f ès H èè Z T 0 Sètèdt = Z t1 0 S H dt + 0 Z T è1, æètèèf ès H èdt (20) Z T 0 æètèdt: t 1 S L dt; (21) 15

16 by (18) and (21), after several transformations, we have Z T 0 æètè =T, t 1 : (22) With (20) and (22), it is easy to see that (19) is correct. 2 Our approach to constructing a low-energy voltage schedule is a greedy approach since we strive to find the minimum constant speed during any critical interval. It guarantees to result the minimum peak power consumption. However, our algorithm may not always produce the minimum-energy voltage schedule. We will show in the following experimental section that the energy saving achieved by applying our algorithm is quite significant. 5 Experimental Results In this section, we apply our approach to several job sets and compare the performance of our research with the work proposed in [11] and [12]. For brevity, we use VSLP,LPFS, andlpps in the following to represent Algorithm 2, the algorithm in [11], and the one in [12], respectively. We assume that the power consumption of a processor is a quadratic function of the processor speed. The approaches in [11] and [12] are intended to apply to periodic tasks, therefore, we also construct our job sets from periodic tasks to ensure a fair comparison. Since different voltage scheduling approaches may benefit from the task timing parameters differently, a fair comparison needs to study a large spectrum of utilization factor values. In our experiments, we first randomly generate a set of periodic task sets, each of which contains five tasks. The period of each task is randomly selected from a uniform distribution between 10 to 50, the deadline of each task is assumed to equal its period, and the worst case execution time (WCET) is less than its period and is also randomly generated. According to Liu and Layland [7], the utilization bound for 5 tasks is approximately Hence, in our experiment, we partition the utilization factor values from 0.1 to 0.7 into intervals of length 0.1. To reduce statistical errors, the number of task sets with utilization values within each interval is no less than 20, and the average results are collected in Table 1. Next, we apply our approach to two real-world applications: CNC (Computerized Numerical Control) machine controller [10] and INS (Inertial Navigation System) [1], and the results are shown in Table 2. We have also investigated the scenario when the execution time of each job varies but is upper bounded by its WCET. We use the same algorithm as discussed in [11], i.e., extending the execution of the current job till its deadline or the arrival of the next job when there is no job in the ready queue. In our experiments, we assume that the best case execution time (BCET) for each task is half of its WCET. Similar to that in [11], the execution time for a task is assumed to be normally distributed within its best and worst case execution time. In Table 1 and Table 2, columns E LPFS, E LPPS,andE VSLP represent the power consumption by algorithms LPFS, LPPS, andvslp, respectively. To better present our results, in Table 1, we let E LPFS be 100 and normalize the other two correspondingly. Column Worst. Exec represents the cases when task execution times equal their WCETs, and column Norm. Exec represents the cases when the task execution times are normally distributed. 16

17 Table 1: Experimental results comparing the three voltage scaling approaches for tasks with fixed priorities. Worst. Exec Norm. Exec Utilization E LPFS E LPPS E VSLP E LPFS E LPPS E VSLP Worst. Exec Norm. Exec Systems Utilization E LPFS E LPPS E VSLP E LPFS E LPPS E VSLP CNC INS Table 2: Experimental results for two real-world real-time systems From the statistical data shown in Table 1, one can readily conclude that our voltage scheduling strategy always leads to more energy saving than the other two approaches, regardless if the job execution times equal to or less than their WCETs. The reason for this is that: when the ready queue is not empty, LPFS always uses the full speed to execute the jobs; LPPS is more efficient and uses the lowest maximum constant speed; in our approach, even lower speed is possible according to the voltage schedule obtained by Algorithm 2. Moreover, note that in Table 1, our algorithm can save more energy when the utilization of a task set is lower. The reason for this is that when the task utilization is low, our algorithm tends to find a constant speed which can be applied to relatively long intervals while still meet the deadline requirements for the jobs. When the task utilization is higher, the interval with constant speed becomes relatively shorter and saving becomes somewhat less. The results applying our approach to two real world systems, as shown in Table 2, also conform to our analysis above. Finally, we would like to point out that, even though LPPS can produce relative good results in these examples, it cannot be readily applied in non-periodic task systems, while VSLP and LPFS can easily work on any such systems. One way to apply LPPS in a non-periodic task system is to calculate the lowest maximum speed assuming all the jobs starting at the same time, which is rather pessimistic and thus will greatly reduce its efficiency in saving the power consumption. 6 Summary In this paper, we study the problem of determining the optimal voltage schedule for a real-time embedded system which is implemented on a variable voltage processor and employs a fixed priority scheme to schedule jobs. Two algorithms are presented in the paper. The first one takes OèN 2 è time, where N is the number of jobs to be scheduled, and finds the minimum constant speed needed to complete each job. The second algorithm, with OèN 3 è time complexity, builds on the first one and produces the following: æ the minimum constant voltage (or speed) needed to complete a set of jobs, and 17

18 æ a voltage schedule which always results in lower energy consumption compared to using the minimum constant voltage and shutting down the system when it is idle. The minimum constant voltage is an important parameter when designing systems with no sophisticated power management hardware but only simple on/off modes or where peak power consumption is a concern. The experimental results obtained from both a number of randomly generated and real-world real-time systems have shown the efficiency by applying our voltage schedule in saving energy. As we have point out earlier, the voltage schedule derived by our algorithm may not be optimal for some cases. Our future work is to improve the algorithms presented and strive to find an optimal voltage schedule. References [1] A.Burns, K. Tindell, and A. Wellings. Effective analysis for engineering real-time fixed priority schedulers. IEEE Transactions on Software Engineering, 21: , May [2] V. Gutnik and A. Chandrakasan. An efficient controller for variable supply-voltage low power processing. Symposium on VLSI Circuits, pages , [3] I. Hong, D. Kirovski, G. Qu, M. Potkonjak, and M. B. Srivastava. Power optimization of variable voltage core-based systems. Proceedings of Design Automation Conference, pages , [4] I. Hong, M. Potkonjak, and M. B. Srivastava. On-line scheduling of hard real-time tasks on variable voltage processor. International Conference on Computer-Aided Design, pages , [5] I. Hong, G. Qu, M. Potkonjak, and M. B. Srivastava. Synthesis techniques for low-power hard real-time systems on variable voltage processors. Proceedings of Real-Time Systems Symposium, pages , [6] T. Ishihara and H. Yasuura. Voltage scheduling problem for dynamically variable voltage processors. International Symposium on Low Power Electronics and Design, pages , Aug [7] C. L. Liu and J. W. Layland. Scheduling algorithms for multiprogramming in a hard real-time environment. Journal of the ACM, 17(2):46 61, [8] J. Liu. Real-Time Systems. Prentice Hall, NJ, [9] W. Namgoong, M. Yu, and T. Meng. A high-efficiency variable-voltage cmos dynamic dc-dc switching regulator. IEEE Internation Solid-State Circuits Conference, pages , [10] N.Kim, M. Ryu, S. Hong, M. Saksena, C. Choi, and H. Shin. Visual assessment of a real-time system design: a case study on a cnc controller. IEEE Real-Time Systems Sumposium, Dec [11] Y. Shin and K. Choi. Power conscious fixed priority scheduling for hard real-time systems. Design Automation Conference, pages , [12] Y. Shin, K. Choi, and T. Sakurai. Power optimization of real-time embedded systems on variable speed processors. International Conference on Computer-Aided Design, pages , [13] F. Yao, A. Demers, and S. Shenker. A scheduling model for reduced cpu energy. IEEE Annual Foundations of Computer Science, pages , Appendix We present the detailed proofs for Theorem 3 and 4 in this section. 7.1 Proof for Theorem 3 The following lemma is needed in proving Theorem 3. Lemma 7 Let T E ènè ç TE 0 ènè and T Lènè =TL 0 ènè in Algorithm 1, and let the two resultant J n-essential intervals be ët s ;t f ë and ët 0 s ;t0 f ë, respectively. Then I n èt s ;t f è ç I n èt 0 s;t 0 f è; 18

19 and t s ç t 0 s ét0 f ç t f. Proof: SinceT E ènè ç TE 0 ènè,andt Lènè =TL 0 ènè, from Algorithm 1, it is not difficult to see that t s ç t 0 s ét 0 f ç t f: If t s ç t 0 s ét 0 f = t f, by Lemma 2 and Lemma 3, we have I n èt s ;t 0 s è ç I nèt s ;t f è ç I n èt 0 s;t 0 f è: If t s ç t 0 s ét f ét 0 f, according to Lemma 6, we have I nèt f ;t 0 f è éi nèt 0 s;t 0 f è. By Lemma 2 and Lemma 3, it is easy to show that I n èt 0 s;t 0 f è éi nèt f ;t 0 f è ç I nèt s ;t f è: 2 To prove Theorem3, we concentrate on demonstrating that every job removed while removing the critical interval is fully executed within the interval and meets its deadline. Without loss of generality, consider a critical interval associated with job J n, i.e., ët k s;t k f ë = ët s;t f ë. From Theorem 2, we know that J n is schedulable. Consider the rest of the jobs that are removed when removing ët s ;t f ë. Note that only the jobs with priorities higher than that of J n and release time in ët s ;t f è are removed. For a removed job J i other than J n,ift s ç T E èiè ét L èiè ç t f,thej i -essential interval must be within ët s ;t f ë and have a lower speed requirement than I n èt s ;t f è,sojobj i must be schedulable. Otherwise, we only need to prove that if the original earliest and/or latest scheduling points T E èiè and T L èiè are moved to t s and/or t f, respectively, the constant speed for the resultant new J i -essential interval is still less than I n èt s ;t f è. Several cases need to be considered as shown in Figure 5. In each sub-figure of Figure 5, ët s ;t f ë is the critical interval, ët a ;t b ë is the original essential interval for J i,andët 0 a ;t0 b ë is the new essential interval for J i after the change of T E èiè and/or T L èiè. The dotted line represents the release time for job J i. æ t a ét s ét b ç t f : This case is shown in Figure 5(a). Since ët s ;t f ë is the critical interval, and earliest starting point of job J i have to be moved to t s. Note that moving T L èiè to t b will not change the original essential interval ët a ;t b ë. Therefore, according to Lemma 7, we have and thus J i is schedulable. I i èt 0 a ;t0 b è ç I ièt a ;t b è ç I n èt s ;t f è; æ t s ç t a é t b ç t f : This case is shown in Figure 5(b). Note that moving T E èiè to t s if T E èiè é t s and moving T L èiè to t f if T L èiè é t f do not change the initial J k -essential interval. Hence, J i is schedulable since I i èt a ;t b è=i i èt 0 a ;t0 b è ç I nèt s ;t f è. æ t a ç t s é t f ç t b : This case is shown in Figure 5(c). In this case, T E èiè is moved to t s and T L èiè is moved to t f. Suppose I i èt 0 a ;t0 b è é I nèt s ;t f è. According to Lemma 6, I i èt 0 b ;t f è é I i èt 0 a ;t0 b è. According to Definition 4, since ién,wehave I n èt 0 b ;t f è ç I i èt 0 b ;t f è éi i èt 0 a ;t0 b è éi nèt s ;t f è which contradicts I n èt s ;t f è ç I n èt 0 b ;t f è. Therefore, I i èt 0 a;t 0 b è ç I nèt s ;t f è,andj i is schedulable. 19

20 t a r t b ta ts tb t f (a) t s t a ta r (b) t b tb t f ta t a r t b tb t s (c) t f ta r t a t b tb t s t f (d) Figure 5: Original and new essential intervals for the jobs having arrive time within the critical interval 20

21 æ t s ç t a é t f é t b This case is shown in Figure 5(d). In this case, T L èiè is moved to t f and T E èiè is moved to t s if T E èiè é t s. With a similar proof as the previous case, we can show that I i èt 0 a;t 0 b è ç I nèt s ;t f è,andj i is schedulable. 7.2 Proof for Theorem 4 To prove Theorem 4, we only need to show that after removing a critical interval, the minimum speeds for jobs whose essential intervals have changed are never bigger than that of the removed critical interval. Let the critical interval ët s ;t f ë be a J n -essential interval, and J i be a job that R i, T E èiè or T L èiè need to be changed due to this removal. The corresponding J i -essential interval will be re-constructed. The priority of J i can be either higher or lower than that of J n. We deal with them separately as shown in Figure 6 and Figure 8. In both figures, ët s ;t f ë is the critical interval, ët 0 a;t 0 bë and ët00 a;t 00 b ë represent possible locations of the J i -essential before removing ët s ;t f ë, ët a ;t b ë is the J i -essential interval after removing ët s ;t f ë, and the dotted line represents the release time of J i. Note that after removing the critical interval, t s and t f become a single point. For the ease of reference, we still keep them apart. We will denote the intensity of an interval before (resp., after) removing the critical interval by I 0 (resp. I). For the critical interval, I and I 0 are always the same. 2 t a t a R i t b t b t" a t s (a) t f t" b t" a t a t a R i t b t b t" b t s (b) t f Figure 6: J i -essential intervals before and after the removal of critical interval ët s ;t f ë,wherej i has a higher priority than that of J n. Case 1: Job J i has a higher priority than J n (ién). This case is shown in Figure 6. In this case, R i must be outside the critical interval and either T E èiè or T L èiè is changed after removing ët s ;t f ë (see Algorithm 2). If R i é t s, T L èiè is moved to t s,asshownin Figure6(a). Similarly if R i ét f, T L èiè is moved to t f, as shown in Figure 6(b). In either case, according to Definition 6, it is easy to show that ët a ;t b ë is a busy interval contained in the original J i -essential interval 21

22 ët 0 a ;t0 bë (or (ët00 a ;t00 b ë). According to Lemma 4, I ièt a ;t b è = Ii 0 èt a;t b è ç Ii 0 èt0 a ;t0 b è. Since I nèt s ;t f è is the critical interval, we conclude I i èt a ;t b è ç I n èt s ;t f è. Case 2: Job J i has a lower priority than J n (ién). Let us first analyze the possible locations of the original J i -essential interval with respect to the critical interval. First, we show that ët 0 a;t 0 b ë can neither be totally contained in ët s;t f ë nor partially overlap with ët s ;t f ë.lett s ç t 0 a ét 0 b ç t f (as shown in Figure 7(a)) or t s ét 0 a ét f ét 0 b (as shown in Figure 7(c)). t s t f t a r t b (a) t a t s r t b (b) t f t s (c) t a t f r t b Figure 7: The essential interval ët 0 a ;t0 bë for lower priority jobs cannot be contained in or partially overlapped with the critical interval ët s ;t f ë. According to Definition 4 and Lemma 4, we have Ii 0 èt s;t 0 a è ç I nèt s ;t 0 a è ç I nèt s ;t f è ç Ii 0 èt0 a ;t0 b è,this contradicts Lemma 4. Let t 0 a ét s ét 0 b ét f (as shown in Figure 7(b)). Similarly, Ii 0 èt0 a ;t0 b è ç I i 0 èt s;t 0 b è ç I n èt s ;t 0 b è ç I nèt s ;t f è, this contradicts I n èt s ;t f è é Ii 0 èt0 a ;t0 bè. Therefore, to prove the theorem, we only need to consider the scenario where ët 0 a;t 0 bë contains the critical interval. Depending on the location of the J i -essential interval after the removal of the critical interval, the following cases exist. æ t a ét 0 a : According to Lemma 2 and Lemma 3, I ièt a ;t b è ç I i èt a ;t 0 a è=i i 0 èt a;t 0 aè. From Lemma 4, Ii 0 èt a;t 0 a è éi0 i èt0 a;t 0 b è ç I nèt s ;t f è. Thus, we have I i èt a ;t b è éi n èt s ;t f è. æ t b ét 0 b : Similar to the above proof, we have I ièt a ;t b è éi n èt s ;t f è. æ t 0 a ç t f ç t a and t f ç t 0 b ç t b: (Note that t s and t f become the same after removing the critical interval.) Refer to Figure 8(a). By Lemma 2 and Lemma 3, we have Ii 0 èt0 a;t 0 b è ç I i 0 èt a;t 0 b è=i ièt a ;t b è ç I i èt a ;t b è.sincei n èt s ;t f è ç I 0 i èt0 a;t 0 b è, therefore, we have I ièt a ;t b è ç I n èt s ;t f è. 22