A Delay Composition theorem for Real-Time Pipelines

P. Jayachandran T. Abdelzaher

Presenter:Sina Meraji

Outline

Problem and our goal

Delay Composition theorem

Pipeline Reduction

Problem

We have: A set of tasks and a multistage

pipeline

Goal: Decide the schedulability of tasks in

the pipeline

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

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

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.

Delay Composition

Multi stage pipeline system

Equivalent single-stage system

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

Assumptions N: number of stages

Ai,j: the arrival time of job Ji at stage j

Ai: arrival time of the job to the entire system

Di: end to end deadline of Ji

Assumptions

Ci,j: Stage Execution Time

Si,j: Stage Start Time

Fi,j: Stage Finish Time

J1: job whose delay is to be estimated

Assumptions

S: denote the set of all higher-priority jobs that have execution intervals in the pipeline between J1’s arrival and finish time (s include j1)

Ci,max1: largest stage execution time

Ci,max2: second largest stage execution times

Pipeline Delay Composition Theorem

the end-to-end delay of a job J1 in an N-stage pipeline

Example

Example

Six stage pipeline using EDF

Computation time of each task on each stage is 1

T1=9, T2=6

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

Apply Theorem

T1 can be preempted by at most 2 invocations of T2

S contain 2 invocations of T2 with the invocation of T1

Apply Theorem

A2>A1: Ceq2=2

A2<A1: Ceq2=1

A1: Ceq1=1

4

Apply Theorem

Second summation is equal to 5 because of 5 stages

Total delay: 4+5=9

Lower than 12 schedulable

Pipeline Reduction

Reduce the pipeline scheduling problem to an equivalent single stage

Swc: worst-case set of higher priority jobs that delay or preempt job J1

Sbef: jobs with Ai<=A1

Safter: jobs with Ai>A1

Pipeline Reduction

Pipeline Reduction replacing each pipeline job Ji in Sbef by an equivalent

single stage job (with the same priority and deadline) of execution time equal to Ci,max1

replacing each pipeline job Ji in Safter by an equivalent single stage job of execution time equal to Ci,max1 + Ci,max2

adding a lowest-priority job, J∗ e of execution time equal to

1

1, )(

N

jjii CMax

Example

Rate Monotonic Scheduling

There can be at most one invocation

of each higher-priority task Ti in set Sbef

Example

Example

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

i

N

Jjiiie CCC

1

1,1max,

* )(max

*eT

*iT

2max,1max,*

iii CCC

Example According to Liu & layland bound:

Thanks