Static-Priority Scheduling. CSCE 990: Real-Time Systems. Steve Goddard. Static-priority Scheduling

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1 CSCE 990: Real-Time Systems Static-Priority Scheduling Steve Goddard Static-priority Scheduling Real-Time Systems Static-Priority Scheduling - We now consider static-priority scheduling.» Under static-priority scheduling, different obs of a tas are assigned the same priority.» We will assume that tass are indexed in decreasing priority order, i.e., T i has higher priority than T if i <.» Notation: π i denotes the priority of T i. T i denotes the subset of tass with equal or higher priority than T i. Note: In some of the papers we will read, it is assumed no two tass have the same priority. (Is this OK?) Real-Time Systems Static-Priority Scheduling - 2 2

2 Rate-monotonic Scheduling (Liu and Layland) Priority Definition: Tass with smaller periods have higher priority. Example Schedule: Three tass, T = (3,0.5), = (4,), T 3 = (6,2). T Deadline-monotonic Scheduling (Leung and Whitehead) Priority Definition: Tass with smaller relative deadlines have higher priority. Same as rate-monotonic if each tas s relative deadline equals its period. Example Schedule: Let s change the RM example by giving a tighter deadline: T = (3,0.5), = (4,,2), T 3 = (6,2). T = T 3 T 2 = T T 3 = T 3 Real-Time Systems Static-Priority Scheduling - 3 Real-Time Systems Static-Priority Scheduling

3 Optimality of RM and DM (Section 6.4 of Liu) Proof: Theorem: Neither RM nor DM is optimal. Consider T = (2,) and = (5, 2.5). Total utilization is one, so the system is schedulable. However, under RM or DM, a deadline will be missed, regardless of how we choose to (statically) prioritize T and. Simply Periodic Systems Definition: A system of periodic tass is simply periodic if for every pair of tass T i and T in the system where p i < p, p is an integer multiple of p i. Theorem 6-3: A system T of simply periodic, independent, preemptable tass, whose relative deadlines are at least their periods, is schedulable on one processor according to the RM algorithm if and only if its total utilization is at most one. The details are left as an exercise. Real-Time Systems Static-Priority Scheduling - 5 Real-Time Systems Static-Priority Scheduling

4 Proof of Theorem 6-3 We wish to show: U T is schedulable. We prove the contrapositive, i.e., T is not schedulable U >. Assume T is not schedulable. Let J i, be the first ob to miss its deadline. T i t- ri, ri,+ this is the last idle instant for obs of T,, T i Real-Time Systems Static-Priority Scheduling - 7 Because J i, missed its deadline Proof (Continued) the demand placed on the processor in [t -, r i,+ ) by obs of tass T,, T i is greater than the available processor time in [t -, r i,+ ]. Thus, r t = available processor time in [t,r < demand placed on the processor in [t,r = i,+ i = i = (the number of ri,+ t e p obs of T released in [t,r )) e r i,+ t is an integer.] p Real-Time Systems Static-Priority Scheduling - 8 [Note : Because the systemis simply periodic, ] - i,+ - i,+ ) by obs of T,...,T - i,+ i 7 8

5 Proof (Continued) Thus, we have r i,+ Cancelling r < i.e., t i = e, p < U U. i < t t yields This completes the proof. i = r i,+ i,+ p e. Optimality Among Fixed-Priority Algs. Theorem 6-4: A system T of independent, preemptable periodic tass that are in phase and have relative deadlines at most their respective periods can be feasibly scheduled on one processor according to the DM algorithm whenever it can be feasibly scheduled according to any fixed-priority algorithm. Corollary: The RM algorithm is optimal among all fixed-priority algorithms whenever the relative deadlines of all tass are proportional to their periods. Real-Time Systems Static-Priority Scheduling - 9 Real-Time Systems Static-Priority Scheduling

6 Proof of Theorem 6-4 Suppose T,, T i are prioritized in accordance with DM. Suppose T i has a longer relative deadline than T i+, but T i has a higher priority than T i+. Then, we can interchange T i and T i+ and adust the schedule accordingly by swapping pieces of T i with pieces of T i+. Proof of Theorem 6-4 Suppose T,, T i are prioritized in accordance with DM. Suppose T i has a longer relative deadline than T i+, but T i has a higher priority than T i+. Then, we can interchange T i and T i+ and adust the schedule accordingly by swapping pieces of T i with pieces of T i+. T i T i+ T i+ T i T i+2 T i+2 By induction, we can correct all such situations. Real-Time Systems Static-Priority Scheduling - Real-Time Systems Static-Priority Scheduling - 2 2

7 Utilization-based RM Schedulability Test (Section 6.7 of Liu) Theorem 6-: [Liu and Layland] A system of n independent, preemptable periodic tass with relative deadlines equal to their respective periods can be feasibly scheduled on a processor according to the RM algorithm if its total utilization U is at most U RM (n) = n(2 /n ) Note that this is only a sufficient schedulability test. U RM (n) as a Function of n n U RM (n) ln truncated to three digits Real-Time Systems Static-Priority Scheduling - 3 Real-Time Systems Static-Priority Scheduling

8 Proof Setch for Theorem 6- We will assume that all priorities are distinct, i.e., p < p 2 < < p n. Note: The original proof for this theorem by Liu and Layland is incorrect. For a complete, correct proof, see Ed Overton s M.S. thesis on my web page. Overton s thesis also points out where the error is in Liu and Layland s proof. We will present our proof setch in two parts: First, we consider the special case where p n 2p. Then, we will remove this restriction. Special Case: p n 2p Definition: A system is difficult-to-schedule if it is schedulable according to the RM algorithm, but it fully utilizes the processor for some interval of time so that any increase in the execution time or decrease in the period of some tas will mae the system unschedulable. We see the most difficult-to-schedule system, i.e., the system whose utilization is smallest among all difficult-to-schedule systems. The proof for the special case p n 2p consists of four steps, described next. Real-Time Systems Static-Priority Scheduling - 5 Real-Time Systems Static-Priority Scheduling

9 Four Steps of the Proof Step : Identify the phases in the most difficult-toschedule system. Step 2: Define the periods and execution times for the most difficult-to-schedule system. Step 3: Show that any difficult-to-schedule system whose parameters are not lie in Step 2 has utilization that is at least that of the most difficultto-schedule system. Step 4: Compute an expression for U RM (n). Aside: Critical Instants Definition: A critical instant of a tas T i is a time instant such that: () the ob of T i released at this instant has the maximum response time of all obs in T i, if the response time of every ob of T i is at most D i, the relative deadline of T i, and (2) the response time of the ob released at this instant is greater than D i of the response time of some obs in T i exceeds D i. Informally, a critical instant of T i represents a worst-case scenario from T i s standpoint. Real-Time Systems Static-Priority Scheduling - 7 Real-Time Systems Static-Priority Scheduling

10 Critical Instants in Fixed-Priority Systems Theorem 6-5: [Liu and Layland] In a fixed-priority system where every ob completes before the next ob of the same tas is released, a critical instant of any tas T i occurs when one of its ob J i,c is released at the same time with a ob of every higher priority tas. We are not saying that T,, T i will all necessarily release obs at the same time, but if this does happen, we are claiming that the time of release will be a critical instant for T i. We give a different (probably more hand-waving) proof of Theorem 6-5 than that found in Liu. Proof of Theorem 6-5 Consider a system such that T,, T i all release obs together at some time instant t. Suppose t is not a critical instant for T i, i.e., T i has a ob released at another time t that has a longer response time than its ob released at t. Example: T T 3 Real-Time Systems Static-Priority Scheduling - 9 T 4 t Real-Time Systems Static-Priority Scheduling

11 Proof (Continued) Let t - be the latest idle instant for T,, T i at or before t. Let J be T i s ob released at t. Let t R denote the time instant when J completes. Example: Proof (Continued) If we (artificially) redefine J s release time to be t -, then t R remains unchanged (but J s response time may increase). Example: T T T 3 T 3 T 4 t - t t R Real-Time Systems Static-Priority Scheduling - 2 T 4 t - t t R Real-Time Systems Static-Priority Scheduling

12 Proof (Continued) If we (artificially) redefine J s release time to be t -, then t R remains unchanged (but J s response time may increase). Example: T Proof (Continued) Starting with T, let us left-shift any tas whose first ob is released after t - so that its first ob is released at t -. With each shift, T i s response time does not decrease. Why? Example: T Shift over T... T 3 T 3 T 4 t - t t R Real-Time Systems Static-Priority Scheduling - 23 T 4 t - t t R Real-Time Systems Static-Priority Scheduling

13 Proof (Continued) Starting with T, let us left-shift any tas whose first ob is released after t - so that its first ob is released at t -. With each shift, T i s response time does not decrease. Why? Proof (Continued) Starting with T, let us left-shift any tas whose first ob is released after t - so that its first ob is released at t -. With each shift, T i s response time does not decrease. Why? Example: Shift over... Example: T T T 3 T 3 T 4 t - t t R Real-Time Systems Static-Priority Scheduling - 25 T 4 t - t t R Real-Time Systems Static-Priority Scheduling

14 Proof (Continued) We have constructed a portion of the schedule that is identical to that which occurs at time t (when T,, T i all release obs together). Moreover, the response time of T i s ob released at t is at least that of T i s ob released at t. This contradicts our assumption that T i s ob released at t has a longer response time than T i s ob released at t. Thus, t is a critical instant. Step : Phases Bac to the proof of Theorem 6- Recall that Step is to identify the phases in the most difficult-to-schedule system. By Theorem 6-5, we can assume that each tas in the most difficult-to-schedule system releases its first ob at time 0. Real-Time Systems Static-Priority Scheduling - 27 Real-Time Systems Static-Priority Scheduling

15 Step 2: Periods and Execution Times By Theorem 6-5, we can limit attention to the first period of each tas. Step 2 (Continued) Let us define e = p + p for =, 2,, n- e n = p n 2 =,,n- e. We need to mae sure that each tas s first ob completes by the end of its first period. We will define the system s parameters so that the tass eep the processor busy from time 0 until at least p n, the end of the first period of the lowest-priority tas. T T 3 T n- T n Real-Time Systems Static-Priority Scheduling - 29 Real-Time Systems Static-Priority Scheduling

16 T T 3 T n- T n Step 2 (Continued) Notes: This tas system is difficult-to-schedule. (Why?) The processor is fully utilized up to p n. Step 3: Showing it s the Most D-T-S We still need to show that the system from Step 2 is the most difficult-to-schedule system. We must show that other difficult-to-schedule systems have equal or higher utilization. Other difficult-to-schedule systems can be obtained from the one in Step 2 by systematically increasing or decreasing the execution times of some of the tass. We show that any small increase or decrease results in a utilization that s at least as big as that of the original tas system. (Convince yourself that this argument generalizes.) Real-Time Systems Static-Priority Scheduling - 3 Real-Time Systems Static-Priority Scheduling

17 Step 3 (Continued) Let s increase the execution of some tas, say T, by ε, i.e., e = p 2 p + ε = e + ε. We can eep the processor busy until p n by decreasing some T s,, execution time by ε: e = e ε. Step 3 (Continued) e e e e The difference in utilization is: U U = + p p p p > 0 ε ε = p p since p < p T T T 3 T 3 T n- T n- T n T n Real-Time Systems Static-Priority Scheduling - 33 Real-Time Systems Static-Priority Scheduling

18 Step 3 (Continued) Let s decrease the execution of some tas, say T, by ε, i.e., e = p 2 p ε. We can eep the processor busy until p n by increasing some T s,, execution time by 2ε: e = e + 2ε. The difference in utilization is: Step 3 (Continued) 2ε ε U U = p p 0 since p 2p T T T 3 T 3 T n- T n- T n T n Real-Time Systems Static-Priority Scheduling - 35 Real-Time Systems Static-Priority Scheduling

19 Let U(n) = Define q U(n) = q adacent period ratio q following n -equations 2 2 q q q U(n) = n(2 + q Step 4: Calculate U RM (n) e p = p /p. Then, To find the minimum, we tae the partial derivative of U(n) with respect to each 2,, 2, 3,2 Solving these equations for q n adacent period ratios q /n n = 3,2 i ). denote the utilization of the system in Step q (+ ), q +, n,(n ) n,(n ) 2 + q q q and set the derivative to zero. for all (+ ), +, 2, 3,2 = n,(n ) n., 2,..., n., we find that U(n) is at its minimum when all the /n are equal to 2. Thus, This gives us the Removing the p n 2p Restriction Definition: The ratio q n, = p n /p is the period ratio of the system. We have proven Theorem 6- only for systems with period ratios of at most 2. To deal with systems with period ratios larger than 2, we show the following () Corresponding to every difficult-to-schedule n-tas system T whose period ratio is larger than 2 there is a difficult-toschedule n-tas system T whose period ratio is at most 2, and (2) T s utilization is at least T s. Real-Time Systems Static-Priority Scheduling - 37 Real-Time Systems Static-Priority Scheduling

20 Proof of () We show we can transform T step-by-step to get T. At each step, we find a tas T whose period is such that lp < p n (l +)p, where l is an integer that is at least 2. We modify (only) T and T n as follows. e T 0 p 2p (l-)p lp Proof of () Convince yourself that The resulting system is difficult-to-schedule. We eventually will get a system with a period ratio of at most 2. e T 0 p 2p (l-)p lp T n 0 e n pn T n 0 e n pn e = e T 0 p = lp e n = e n + (l-)e T n0 p n = p n e = e T 0 p = lp e n = e n + (l-)e T n0 p n = p n Real-Time Systems Static-Priority Scheduling - 39 Real-Time Systems Static-Priority Scheduling

21 Proof of (2) It suffices to loo at the difference between the utilization of the old and new system when one of the steps in the proof of () is applied. This difference is: e e ( l )e p lp p = l > 0 p ( )e p l n because lp < p. This concludes the proof of (2) and (finally!) the proof of Theorem 6-. n n Real-Time Systems Static-Priority Scheduling - 4 Other Utilization-based Tests The boo presents several other utilization-based schedulability tests. Some of these tests result in higher schedulable utilizations for certain inds of tas sets. For example, Theorem 6-3 considers systems in which we can partition the tass into subsets, where in each subset, tass are simply periodic. Other of these tests allow more detail into the model. For example, Theorem 6-7 considers multi-frame tass, which are tass where execution costs vary from ob to ob. (The motivation for this was MPEG video.) You will have a better understanding of why people are interested in utilization-based tests later, when we tal about intractability. Real-Time Systems Static-Priority Scheduling

22 Time-Demand Analysis (Section of Liu) Time-demand analysis was proposed by Lehoczy, Sha, and Ding. TDA can be applied to produce a schedulability test for any fixed-priority algorithm that ensures that each ob of every tas completes before the next ob of that tas is released. For some important tas models and scheduling algorithms, this schedulability test will be necessary and sufficient. Scheduling Condition Definition: The time-demand function of the tas T i, denoted w i (t), is defined as follows. w (t) i i = ei + = t p e for 0 < t p Theorem: A system T of periodic, independent, preemptable tass is schedulable on one processor by algorithm A if ( i:: ( t: 0 < t p i :: w i (t) t)) holds. This condition is also necessary for synchronous, real-world periodic tas systems and also real-world sporadic (= periodic here) tas systems. i Note: We are still assuming tass are indexed by priority. For any fixed-priority algorithm A that ensures that each ob of every tas completes by the time the next ob of that tas is released Real-Time Systems Static-Priority Scheduling - 43 Real-Time Systems Static-Priority Scheduling

23 Sufficiency Proof We wish to show: ( i:: ( t: 0 < t p i :: w i (t) t)) T is schedulable. We prove the contrapositive, i.e., T is not schedulable ( i:: ( t: 0 < t p i :: w i (t) > t)). Assume T is not schedulable. Let J i, be the first ob to miss its deadline. T i t- ri, ri,+ this is the last idle instant for obs of T,, T i Real-Time Systems Static-Priority Scheduling - 45 Because J i, missed its deadline Proof (Continued) at all instants t in (t -, r i,+ ], the demand placed on the processor in [t -, t) by obs of tass T,, T i is greater than the available processor time in [t -, t]. Thus, for any t in (t -, r i,+ ], t t = available processor time in [t, t] < demand placed on the processor in [t = i = i = (the number of t t p e obs of, t) by obs of T released in [t, t)) e T,...,T Real-Time Systems Static-Priority Scheduling - 46 i 45 46

24 Proof (Continued) To recapitulate, we have, for any t in (t -, r i,+ ], t t i < = t t p e Replacing t t - by t in (0, r i,+ t - ], we have t < i = Because p i r i,+ t -, we have ( i:: ( t : 0 < t p i :: w i (t ) > t )). Real-Time Systems Static-Priority Scheduling - 47 t e p Necessity and Efficiency The condition ( i:: ( t: 0 < t p i :: w i (t) t)) is necessary for» synchronous, real-world periodic tas systems, and» real-world sporadic (= periodic here) tas systems.» Why? For a given i, we don t really have to consider all t in the range 0 < t p i. Two ways to avoid this:» Iterate using t (+) := w i (t () ), starting with a suitable t (0), and stopping when, for some n, t (n) w(t (n) ) or t (n) > p i.» Only consider t = p, where =, 2,, i; =, 2,, min(p i, D i )/p. See Liu for an explanation of this. Real-Time Systems Static-Priority Scheduling

25 Fixed-Priority Tass with Arbitrary Response Times (Section 6.6 of Liu) The TDA scheduling condition is valid only if each ob of every tas completes before the next ob of that tas is released. We now consider a schedulability chec due to Lehoczy in which tass may have relative deadlines larger than their periods.» Note: In this model, a tas may have multiple ready obs. We assume they are scheduled on a FIFO basis. Busy Intervals Definition: A level-π i busy interval (t 0, t] begins at an instant t 0 when () all obs in T i released before this instant have completed, and (2) a ob in T i is released. The interval ends at the first instant t after t 0 when all obs in T i released since t 0 are complete. Example: T i t0 t Busy Interval Real-Time Systems Static-Priority Scheduling - 49 Real-Time Systems Static-Priority Scheduling

26 Busy Intervals (Continued) For any t that would qualify as the end of a level-π i busy interval, a corresponding t 0 exists.» Why? During a level-π i busy interval, the processor only executes tass in T i other tass can be ignored. Definition: We say that a level-π i busy interval is in phase if the first ob of all tass that execute in the interval are released at the same time. It Ain t So Easy For systems in which each tas s relative deadline is at most its period, we argued that an upper bound on a tas s response time could be computed by considering a critical instant scenario in which that tas releases a ob together with all higher-priority tass. In other words, we ust consider the first ob of each tas in an in-phase system. For many years, people ust assumed this approach would wor if a tas s relative deadline could exceed its period. Lehoczy showed that this fol wisdom that only each tas s first ob must be considered is false by means of a counterexample. Real-Time Systems Static-Priority Scheduling - 5 Real-Time Systems Static-Priority Scheduling

27 Lehoczy s Counterexample Consider: T = (70, 26), = (00, 62) [Note: the boo has a typo here.] Here s a schedule: T s seven obs have the following response times, respectively: 4, 02, 6, 04, 8, 06, 94. Note that the first ob s response time is not the longest. Bottom Line: We have to consider all obs in an in-phase busy interval. Real-Time Systems Static-Priority Scheduling - 53 General TDA Method Test one tas at a time starting with the highest-priority tas T in order of decreasing priority. For the purpose of determining whether a tas T i is schedulable, assume that all the tass are in phase and the first level-π i busy interval begins at time zero. While testing whether all the obs in T i can meet their deadlines (i.e., whether T i is schedulable), consider the subset T i of tass with priorities π i or higher. (i) If the first ob of every tas in T i completes by the end of the first period of that tas, chec whether the first ob J i, in T i meets its deadline. T i is schedulable if J i, completes in time. Otherwise, T i is not schedulable. (ii) If the first ob of some tas in T i does not complete by the end of the first period of the tas, do the following. (a) Compute the length of the in-phase level-π i busy interval by solving the equation t = =,,i t/p e iteratively, starting from t () = =,,i e until t (l+) = t (l) for some l. The solution t (l) is the length of the level-π i busy interval. (b) Compute the maximum response times of all t (l) /p i obs of T i in the in-phase level-π i busy interval in the manner described below and determine whether they complete in time. T i is schedulable if all of these obs complete in time; otherwise T i is not schedulable. Real-Time Systems Static-Priority Scheduling

28 Computing Response Times Computing the response time of T i s first ob is almost lie before. The time demand function is defined as follows. i t w i,(t) = ei + e for 0 < t w i,(t) p = The only difference from before. The maximum response time W i, of J i, is equal to the smallest t that satisfies the equation t = w i, (t). Can do this computation iteratively, as describe before. (Is termination a problem?) Response Times (Continued) Lemma 6-6: The maximum response time W i, of the -th ob of T i in an in-phase level-π i busy period is equal to the smallest value of t that satisfies the equation t = w i, (t + ( )p i ) ( )p i where i t w i, (t) = ei + e for ( )pi < t w p = The recurrence given in the lemma can be solved iteratively, as described before. (Once again, is termination a problem?) i, (t) Real-Time Systems Static-Priority Scheduling - 55 Real-Time Systems Static-Priority Scheduling

29 Example Let s apply Lemma 6-6 to our previous example: T = (70, 26), = (00, 62) T W 2, = minimum t s.t. t = w 2, (t) i t = e2 + e p = t = ??4 = = = 4 Yes! Real-Time Systems Static-Priority Scheduling - 57 Example (Continued) T = (70, 26), = (00, 62) T W 2,2 = minimum t s.t. t = w 2,2 (t + p ) p 2 i t + 00 = 2 e2 + e 00 p = t + 00 = ??02 = = = 02 Yes! Real-Time Systems Static-Priority Scheduling

30 Example (Continued) T = (70, 26), = (00, 62) Correctness of the General Schedulability Test T W 2,3 = minimum t s.t. t = w 2,3 (t + 2 p ) 2 p 2 i t = 3 e2 + e 200 p = t = ??6 = = = 6 Yes! Real-Time Systems Static-Priority Scheduling - 59 The general schedulability test hinges upon the assumption that the ob with the maximum response occurs within an in-phase busy interval. We must confirm that this is so. Aside: In steps (i) and (ii)-(b), the conclusion that T i is not schedulable is stated. In other words, Liu is presenting this as a necessary and sufficient schedulability test. Why is it OK to conclude that the test is necessary? Real-Time Systems Static-Priority Scheduling

31 Correctness (Continued) Correctness follows from several lemmas Lemma 6-7: Let t 0 be a time instant at which a ob of every tas in T i is released. All the obs in T i released prior to t 0 have been completed at t 0. Proof Setch: Let t - be the beginning of the latest busy interval. Demand created by obs of tass in T i released in [t -, t 0 ) must be fulfilled in [t -, t 0 ], or else U i exceeds. This is similar to the proof of Theorem 6-3. Note: To reach the stated conclusion about utilization, we must avoid introducing those nasty ceiling operators lie in the proof of the original TDA schedulability test. Why is this not a problem here? Real-Time Systems Static-Priority Scheduling - 6 Correctness (Continued) Lemma 6-8: When a system of independent, preemptive periodic tass is scheduled on a processor according to a fixed-priority algorithm, the time-demand function w i, (t) of a ob in T i released at the same time with a ob in every higher priority tas is given by i t w i,(t) = ei + e for 0 < t w i,(t). p = This should be pretty obvious to you by now Real-Time Systems Static-Priority Scheduling

32 Correctness (Continued) Correctness (Continued) Theorem 6-9: The response time W i, of the -th ob of T i executed in an in-phase level-π i busy interval is no less than the response time of the -th ob of T i executed in any level-π i busy interval. Lemma 6-0: The number of obs in T i that are executed in an in-phase level-π i busy interval is never less than the number of obs in this tas that are executed in a level-π i busy interval of arbitrary phase. This is ust the good old critical instant argument Intuitively, demand on the processor is maximal for a busy interval that is in phase. Thus, an in-phase busy interval should never be shorter than an arbitrary busy interval. Real-Time Systems Static-Priority Scheduling - 63 Real-Time Systems Static-Priority Scheduling

33 Wrapping Up Convince yourself that Lemmas 6-7 through 6-0 imply that we need only loo at the obs that execute within an in-phase busy interval. Thin carefully about necessity.» For which tas models is the schedulability test necessary?» For which is it not necessary? Real-Time Systems Static-Priority Scheduling