Rate-Monotonic Scheduling with variable. execution time and period. October 31, Abstract

Transcription

1 Rate-Monotonic Scheduling with variable execution time and period Oldeld Peter October 31, 1997 Abstract Abstract is something cannot be understood. 1 Rate Monotonic Model Let Ti be a task. Let Pi be the period of task Ti. Let ei be the execution time of Ti. 2 Extension Now, we extend the model to give more freedom on the denition of period and execution time of a task. Let us assume that a task Ti having dierent possible periods over the long term execution. That is instead of having a single period value Pi, we may have a list of possible periods like below: Pi1; Pi2; : : : ; Pin There is a list of probabilities that associated with each period, namely: i1; i2; : : : ; in That is, the probability of task Ti having a period of Pi1 is i1. And we can denote this relation by: 1

2 P rfp eriod = Pi1g = i1 P rfp eriod = Pi2g = i2 : : : P rfp eriod = Ping = in (We use P r to denote the symbol \probability of") Remark: Here, we make another assumption to x the number of dierent periods of each task to be n. That is every should have exactly n possible dierent period. It is, however, not a restriction, because we can set n to very big, and if a task do not have so many dierent periods, we can add arbitrary periods to it and give them a probability of happening to be 0. In order to illustrate the above notation, we give an example here: There are two tasks T 1 and T 2. Their execution time are: The periods of two tasks are: e 1 = 2 e 2 = 2 P 11 = 3; P 12 = 4 P 21 = 4; P 22 = 5 For notational convenience, we denote the periods as follow: P 1 = (P 11 ; P 12 ) = (3; 4) P 2 = (P 21 ; P 22 ) = (4; 5) And the probability of the periods are: 11 = 0:3; 12 = 0:7 21 = 0:5; 22 = 0:5 2

3 I1 = 3 - I2 = 4 - I3 = J1 = 4 Figure 1: Scheduling sequence of T 1, T 2 Base on the above settings, one of the possible outcome of task scheduling is depicted in Figure 1. From the gure, we can see that task T 1 can meet the rst three deadline, and T 2 can meet the rst two deadline. We can use I 1, I 2, I 3 to denote the rst three period of task T 1, and similarly, use J 1, J 2 to denote the rst two periods of task T 2. Now, we are going to estimate the probability that T 1 and T 2 will both meet the deadline in their next period (I 4, J 3 ). It is obvious that there are four possible arrangements of the next period: I 4 = 3; J 3 = 4 I 4 = 3; J 3 = 5 I 4 = 4; J 3 = 4 I 4 = 4; J 3 = 5 Figure 2 shows the rst possible arrangement, I 4 = 3; J 3 = 4, and we can see that task T 1 is unable to meet deadline (even if we rearrange T 1 to run rst, T 2 will not be able to meet deadline). And the probability of this case is: P rfi 4 = 3; J 3 = 4g = P rfi 4 = 3g P rfj 3 = 4g = 0:3 0:5 3

4 I3 = 3 - I4 = J3 = 4 Figure 2: First possible arrangment Figure 3 shows the second possible arrangement, I 4 = 3; J 3 = 5. Here, the task T 2 is unable to meet deadline. And the probability of this case is: P rfi 4 = 3; J 3 = 5g = P rfi 4 = 3g P rfj 3 = 5g = 0:3 0:5 I3 = 3 - I4 = J3 = 5 Figure 3: Second possible arrangment Figure 4 shows a schedulable arrangement that both T 1 and T 2 can meet deadline on the next coming period. The situation is that I 4 = 4; J 3 = 4, we can see that the probability of this case is: 4

5 P rfi 4 = 4; J 3 = 4g = P rfi 4 = 4g P rfj 3 = 4g = 0:7 0:5 I4 = 4 I3 = J3 = 4 Figure 4: Third possible arrangment Similarly, Figure 5 shows the last arrangment that both T 1 and T 2 can meet deadline, with I 4 = 4; J 3 = 5. The probability is given by: P rfi 4 = 4; J 3 = 4g = P rfi 4 = 4g P rfj 3 = 4g = 0:7 0:5 I3 = 3 - I4 = J3 = 5 Figure 5: Fourth possible arrangment 5

6 By combining the third and fourth cases, we can derive the probability that both T 1 and T 2 can meet deadline, the probability is: P rfi 4 = 4; J 3 = 4g P rfi 4 = 4; J 3 = 5g = 0:7 0:5 0:7 0:5 The result is quite a high probability. Intuitively we can do similar probability analysis on each period and obtain a long chain of probability that both tasks can meet deadline, in subsequence periods. However, the calculation is huge, and may be innite. Consider the example just mentioned, and recall the admission policy of Rate Monotonic (RM) algorithm: can T 1 and T 2 be accepted by the RM algorithm? Let us consider the T 1 rst, the period of T 1 is in general smaller than T 2 and therefore should be consider rst. To suit for the RM algorithm, we takes the period which is smaller, therefore: P 1 = 3. For T 2, we have P 2 = 4. The execution time of both tasks is still the same e 1 = e 2 = 2. To check for the necessary and sucient condition for T 1 to be accepted by RM algorithm, we check whether: e 1 P 1 It is true, since 2 3. Next, we have to check T 2, recall that we must nd some t that satisfy the following two conditions in order to accept T 2 : t P 2 t? d t ee 1 e 2 P 1 Substitute P 1 = 3; P 2 = 4; e 1 = e 2 = 2 into the equation, we got: t 4 t? d t 3 e2 2 However, it is not possible to nd any value for t satisfy these inequalities. It implies that T 1 and T 2 are not RM schedulable. 6

7 With this fact on hand, and the probability derived in the example, we can see that even a group of tasks that is not RM schedulable, it may have a high probability to meet their deadline, provide that their have a dynamic period, with an associated probability. Therefore, it is desirable to derive some method to calculate the probability of schedulable with the prescence of dynamic task period, or even execution time. In the near future, we will investigate the impact on dynamic execution time that can aect the probability of schedulable. And we will seek for the answer of calculation of schedulable probability in both dynamic period or execution time. At the very beginning, it is more favourable to separate them and analyse them one by one, to simplify the problem. A Threshold Probabilistic Schedulable The term schedulable often refer to a tight restriction that a group of tasks must meet their deadline over all the range of their execution time. However, with the probabilistic range of period or execution time, we cannot simply give a tight restriction on the schedulable principle. Instead, we propose a threshold based probabilistic schedulable scheme. Make it more concrete, given a group of tasks, with dynamic period and execution time, we say that they are probabilistic schedulable with a probability! (where 0! 1), if in long run, each task will have a probability! to meet its deadline. That is, for every task Ti, it will run for N periods (each period interval may not be the same), and over those N periods, there are at least N! periods that Ti can meet its deadline. Notice that when we say a task cannot meet a deadline at period k, it may have two meanings: 1) we simply do not run task Ti at period k, and the next period is adjacent to the deadline just missed, or 2) Ti run over the deadline at period k, and the start time of next period (k + 1) is just delayed to the point where Ti nished (time after the deadline of k). When we adopt to one of the above meaning of \cannot meet a deadline", it may turn out that the schedulable test to be dierent. B A rough probabilistic Schedulable test One of the possible, but not accurate, test for the probabilistic schedulable may be as follow: We got a group of tasks T 1 ; T 2 ; : : : Tn, and each of them associated with a range of periods. We may use the Rate Monotonic admission test to check whether the smallest period of each task can pass the test 7

8 or not. If not, we can use a larger periods, of a task, and check again. We continue to use a larger period of each task to check with the RM admission test. It stops until the point that we can nd a list of periods (one period associated with each task) that can satisfy the RM test. Then the schedulable probability may obtained by multipling the probability associated with those periods found above, and then we further calculate all larger periods combination to obtain a more slack probability of schedulable. We add them together and that is a rough estimation on the schedulable probability. References [1] Oldeld So and Peter Tam, Rate-Monotonic Scheduling Algorithm - a short report, Internal Technical Report RM1026-TR , 8