IN4343 Real Time Systems April 9th 2014, from 9:00 to 12:00

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1 TECHNISCHE UNIVERSITEIT DELFT Faculteit Elektrotechniek, Wiskunde en Informatica IN4343 Real Time Systems April 9th 2014, from 9:00 to 12:00 Koen Langendoen Marco Zuniga Question: Total Points: Score: This is a closed book exam You may use a simple calculator only (i.e. graphical calculators are not permitted) Write your answers with a black or blue pen, not with a pencil Always justify your answers, unless stated otherwise The exam covers the following material: (a) chapters 1-6, 8-9 of the book Hard Real-Time Computing Systems (3rd ed) by G. Buttazzo (b) the paper The Worst-Case Execution-Time Problem by Wilhelm et al. (except Section 6) (c) the paper Transforming Execution-Time Boundable Code into Temporally Predictable Code by P. Puschner (d) the paper Best-case response times and jitter analysis of real-time tasks by R.J. Bril, E.F.M. Steffens, and W.F.J. Verhaegh

2 IN4343 Real-Time Systems Cheat Sheet Liu and Layland (LL) bound Hyperbolic (HB) bound U RM lub = n(2 1/n 1) n (U i + 1) 2 i 1 WRi + AJ k Response Time Analysis WR i = + C k k=1 i 1 T k ( BRi AJ k BR i = + k=1 w + = max(w, 0) Processor Demand g(t 1, t 2 ) = d i t 2 T k r i t 1 g(0, L) = schedulability L D, g(0, L) L 1) + C k L + Ti D i D = {d k d k min(h, max(d max, L ))} H = lcm(t 1,..., T n ) ( D i )U i L = 1 U [ ( ) 2 1/n Polling Server schedulability U RM+P S lub = U s + n 1] U s + 1 n (U i + 1) 2 U s + 1 response time R a = C a + a + F a (T s C s ) ra a = T s r a F a = T s [ ( ) Deferrable Server schedulability U RM+DS Us + 2 1/n lub = U s + n 1] 2U s + 1 n (U i + 1) U s + 2 2U s + 1 response time R a = C a + a C 0 + F a (T s C s ) Ca C s 1 C 0 = min(c s (r a ), a ) ra Ca C 0 a = T s r a F a = 1 T s C s i Li NP scheduling level-i busy period L i = B i + C h N i = Li h=1 T h B i = max {C j} j>i i 1 response time s ik = B i + (k 1) + h=1 R ik = (s ik + ) (k 1) R i = max k [1,N i ] {R ik} ( sik T h ) + 1 C h Elastic Model utilization i U i = U i0 (U 0 U d ) E i E S where E S = E i

3 Question 1 [10 points] To avoid the intricacies of determining the WCET of a task, the code can be transformed into a single-path equivalent utilizing a predicated execution model. Unfortunately most real-world instruction set architectures only provide a few predicated instructions. As a consequence the transformation process becomes more complex. if (y!= 0) { norm = x / y; } else norm = 1; } (a) 5 points describe the complication(s) with the above code fragment if the only available predicated instruction is a conditional move. Solution: The non-predicated division will raise an exception when y = 0. (b) 5 points provide (pseudo) assembly code for the transformed code. Solution: See lecture slides. Note that resolving the exception by dividing by x when y==0, as in if (y==0) {y=x;} norm = x/y;, is not correct as x might be zero too. Question 2 [20 points] Given the following preemptable periodic tasks scheduled based on the Rate Monotonic Algorithm: τ τ τ (a) 5 points Is there a feasible schedule for these tasks? Solution: yes, see timeline of critical instant: (b) 10 points Derive the Best Case Response time for all tasks

4 Solution: From the optimal instant it follows BR i τ 1 2 τ 2 3 τ 3 13 (c) 5 points What is the maximum Activation Jitter we can insert in task τ 1 while still guaranteeing schedulability for the entire task set? Solution: Task 1 can have a max activation jitter of 1 unit; task 3 is the bottleneck with just one unit of slack in its period (from 23 to 24, see timeline above). This intuition can be verified using the equation for the response time analysis on the cheat sheet. Question 3 [20 points] Given the following preemptable periodic tasks running under the EDF algorithm: D i τ τ τ (a) 5 points What are the limitations of using the conditions D i to check schedulability? Describe the limitations of both conditions. and Solution: is too optimistic since it ignores the deadlines. This condition is thus not sufficient (and necessary). D i is too conservative as it assumes that the computation of the task takes until the deadline. This condition is thus sufficient, but not necessary. (b) 10 points Is there a feasible EDF schedule for this set of tasks? Solution: We need to check the processor demand until min(h, max(d max, L )) with H = 120, D max = 10, and L = 16.5.

5 demand g(0,4) 2 g(0,5) 4 g(0,9) 6 g(0,10) 9 g(0,13) 11 g(0,14) 13 (c) 5 points In general, what conditions in terms of the s, D i s and s would reduce the number of points that need to be checked for schedulability under the processor demand criteria? Describe at least three conditions. Solution: Three lines of attack: (i) reduce the length of the hyperperiod (making multiples of each other), (ii) reduce the maximum deadline D max, and (iii) reduce L by setting deadlines close to the periods and/or reducing the task utilizations (small ). Question 4 [30 points] When mixing periodic and aperiodic tasks one can make use of a priority server to schedule the aperiodic tasks. Consider the following periodic tasks and aperiodic jobs (under EDF): τ τ a i J J J (a) 3 points compute maximum the server utilization U S that can be used by a Total Bandwidth Server (TBS) without compromising the schedulability of the periodic tasks. Solution: U S = 1 U p = 2 9 (b) 8 points compute the response times for the three jobs when being served by (plain) TBS. Solution: Set server parameters to C s = 2, T s = 9. d i f i R i J J J (c) 8 points compute the response times for the three jobs when being served by optimized TBS.

6 Solution: Iterate with d i+1 = f i until convergence f 1 f 2 f 3 f 4 R i J J J (d) 3 points name the advantage(s) of a Constant Bandwidth Server over optimized TBS scheduling. Solution: When an aperiodic task introduces an overload only the CBS is compromised, not the periodic tasks. (e) 8 points compute the response times for the three jobs when being served by CBS. Solution: Be aware that the server handles aperiodic tasks one after the other, so job J 2 is effectively passed as (a i = 17, = 1) to the EDF scheduler. f i R i J J J Question 5 Considering the set of tasks below: [10 points] T min i T max i E i τ τ (a) 10 points Use the Elastic Period Method to derive the periods s that would guarantee the schedulability of these tasks under the Rate Monotonic Algorithm (LL bound). Assume that the initial period is Ti min for both tasks. Solution: From the LL bound we know that the utilization (U d ) has to be (two tasks). Using the elastic bound method we get: U 1 = 5 8 (( ) 0.828) 1 4 = U 2 = 6 10 (( ) 0.828) 3 4 = Then T 1 = and T 2 =