1. A Delay Composition theorem for Real-Time Pipelines P. Jayachandran T. Abdelzaher Presenter: Sina Meraji

2. Outline Problem and our goal Delay Composition theorem Pipeline Reduction

3. Problem We have: A set of tasks and a multistage pipeline Goal: Decide the schedulability of tasks in the pipeline

4. Goal bound the end-to-end delay of a job in a multistage pipeline as a function of the execution times of higher-priority jobs in the pipeline

5. Delay Composition A job-additive component that is proportional to the sum of invocation execution times on a single stage A stage-additive component that is proportional to the number of stages

6. Delay Composition No assumptions on the scheduling policy No assumption on periodicity of the task set No assumption on whether different invocations of the same task have the same priority.

7. Delay Composition Multi stage pipeline system Equivalent single-stage system

8. Assumptions Tasks arrive to the system and require execution on a set of resources Consider individual task invocations in isolation (job) All the jobs require processing on all the stages and in the same order

9. Assumptions N: number of stages A i,j : the arrival time of job J i at stage j A i : arrival time of the job to the entire system D i : end to end deadline of J i

10. Assumptions C i,j : Stage Execution Time S i,j : Stage Start Time F i,j : Stage Finish Time J 1 : job whose delay is to be estimated

11. Assumptions S: denote the set of all higher-priority jobs that have execution intervals in the pipeline between J 1 ’s arrival and finish time (s include j 1 ) C i,max1 : largest stage execution time C i,max2 : second largest stage execution times

12. Pipeline Delay Composition Theorem the end-to-end delay of a job J 1 in an N- stage pipeline

13. Example

14. Six stage pipeline using EDF Computation time of each task on each stage is 1 T1=9, T2=6

15. Solution partition the end-to-end deadline of each task into per-stage deadlines Per-stage deadline T1=1.5 T2=1 T2 has 0 slack it runs first T1 needs 2 time units per stage

16. Apply Theorem T1 can be preempted by at most 2 invocations of T2 S contain 2 invocations of T2 with the invocation of T1

17. Apply Theorem A2>A1: C eq2 =2 A2<A1: C eq2 =1 A1: C eq1 =1 4

18. Apply Theorem Second summation is equal to 5 because of 5 stages Total delay: 4+5=9 Lower than 12 schedulable

19. Pipeline Reduction Reduce the pipeline scheduling problem to an equivalent single stage S wc : worst-case set of higher priority jobs that delay or preempt job J 1 S bef : jobs with A i <=A 1 S after : jobs with A i >A 1

20. Pipeline Reduction

21. replacing each pipeline job J i in S bef by an equivalent single stage job (with the same priority and deadline) of execution time equal to C i,max1 replacing each pipeline job J i in S after by an equivalent single stage job of execution time equal to C i,max1 + C i,max2 adding a lowest-priority job, J ∗ e of execution time equal to

22. Example Rate Monotonic Scheduling There can be at most one invocation of each higher-priority task Ti in set Sbef

23. Example

24. Task (of lowest priority), with a computation time equal to: Task, each has the same period and deadline as one Ti in original set and execution time

25. Example According to Liu & layland bound:

26. Thanks