real-time workload models with e cient analysis · the digraph real-time (drt) task model adaptive...

Real-Time Workload Models with Efficient AnalysisAdvanced Course, 3 Lectures, September 2014

Martin Stigge

Uppsala University, Sweden

Fahrplan

1 DRT Tasks in the Model HierarchyLiu and Layland and Sporadic TasksFrames and BranchingThe Digraph Real-Time (DRT) Task ModelAdaptive Variable-Rate (AVR) Tasks

2 Feasibility Analysis of DRTFeasibility TheoremDemand PairsTest TerminationEvaluation

3 Static Priority Schedulability Analysis of DRTResponse-Time AnalysisRequest FunctionsRefinement AlgorithmEvaluation

Martin Stigge Workload Models + Analysis 2

Fahrplan

Problem Overview

Workload Model

Task A

Task B

Task C

High priority

Medium priority

Low priorityC

Scheduler Model

EDF/Static Prio/...

Task At

Task Bt

Feasible? Schedulable? Response times?

Our Setting:• DRT tasks

• Static Priorities

• Precise Test

The Liu and Layland (L&L) Task Model(Liu and Layland, 1973)

• Tasks are periodic• Job WCET e• Period p (implicit deadline)

(e, p)e e

• Advantages: Well-known model; efficient schedulability tests

• Disadvantage: Very limited expressiveness

A Hierarchy of Models

Liu & Layland(e, d = p)

efficient

difficult

The Sporadic Task Model(Mok, 1983)

• Deadlines 6= periods; releases sporadic

• Each tasks defined by:• Job WCET e• Relative deadline d• Minimum inter-release delay p

(e, d , p)e e

t6 p 6 p

d d...

The General Multiframe (GMF) Task Model

• Behavior is not always periodic

Frame 0 Frame 1 Frame 2 Frame 3 Frame 0

• Task is split into frames, each with own• Execution time e(j)

• Inter-release separation p(j)

• Deadline d (j) for the job

The General Multiframe (GMF) Task Model

• Behavior is not always periodic

Frame 0 Frame 1 Frame 2 Frame 3 Frame 0

• Task is split into frames, each with own• Execution time e(j)

• Inter-release separation p(j)

• Deadline d (j) for the job

The General Multiframe (GMF) Task Model (cont.)(Baruah et al., 1999)

• Tasks cycle through job types• Vector for WCET (e(1), . . . , e(n))• Vector for deadlines (d (1), . . . , d (n))• Vector for minimum inter-release delays (p(1), . . . , p(n))

p(1)p(2)

p(4)p(5)

;e(1) e(2)

6 p(1) 6 p(2)

d (1) d (2)

sporadic(e, d , p)

multiframe(ei , d = p)

generalized multiframe (GMF)(ei , di , pi )

efficient

difficult

The Non-Cyclic GMF Task Model(Moyo et al., 2010)

• Frame order unknown a priori

• Syntax similar to GMF:• Vector for WCET (e(1), . . . , e(n))• Vector for deadlines (d (1), . . . , d (n))• Vector for minimum inter-release delays (p(1), . . . , p(n))

;e(1) e(2)

6 p(1) 6 p(2)

d (1) d (2)

The Recurring Branching (RB) Task Model(Baruah, 1998)

• Introduces branching structures

• Tree for tasks• Vertices J: jobs to be released (with WCET and deadline)• Edges (Ji , Jj): minimum inter-release delays p(Ji , Jj)• General period parameter P

Period P = 57

The Recurring Branching (RB) Task Model(Baruah, 1998)

• Introduces branching structures

• Tree for tasks• Vertices J: jobs to be released (with WCET and deadline)• Edges (Ji , Jj): minimum inter-release delays p(Ji , Jj)• General period parameter P

Period P = 57

The Recurring Real-Time (RRT) Task Model(Baruah, 1998)

• Compact Branching representation

• Directed acyclic graph (DAG) for tasks• Vertices J: jobs to be released (with WCET and deadline)• Edges (Ji , Jj): minimum inter-release delays p(Ji , Jj)• General period parameter P

Period P = 57

sporadic(e, d , p)

recurring branching (RB)(tree, p)

recurring RT (RRT)(DAG, p)

non-cyclic GMF(order arbitrary)

non-cyclic RRT(DAG, pi )

efficient

difficult

Restrictions of RRT

• Tasks are still recurrent

• Always revisit source J1

• No cycles allowed!

• Consequences:

• No local loops• Not compositional

(for modes etc.)

Restrictions of RRT

• Consequences:

(for modes etc.)

Restrictions of RRT

• Consequences:

(for modes etc.)

The Digraph Real-Time (DRT) Task Model(S. et al., 2011)

• Generalizes periodic, sporadic, GMF, RRT, . . .

• Directed graph for each task• Vertices v : jobs to be released (with WCET and deadline)• Edges (u, v): minimum inter-release delays p(u, v)

v1〈2, 5〉

〈1, 8〉v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

DRT: Semantics

v1〈2, 5〉

〈1, 8〉v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Path π = (v4)Path π = (v4, v2)Path π = (v4, v2, v3)

0 10 20 28

DRT: Semantics

v1〈2, 5〉

〈1, 8〉v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Path π = (v4)

Path π = (v4, v2)Path π = (v4, v2, v3)

DRT: Semantics

v1〈2, 5〉

〈1, 8〉v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Path π = (v4)

Path π = (v4, v2)

Path π = (v4, v2, v3)

0 10 20 28

DRT: Semantics

v1〈2, 5〉

〈1, 8〉v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Path π = (v4)Path π = (v4, v2)

Path π = (v4, v2, v3)

0 10 20 28

sporadic(e, d , p)

Digraph (DRT)(arbitrary graph)

efficient

difficult

sporadic(e, d , p)

efficient

difficult

Strongly (co)NP-hard

Pseudo-Polynomial

Extended DRT (EDRT)

• Extends DRT with global delay constraints

• Directed graph for each task• Vertices v : jobs to be released (with WCET and deadline)• Edges (u, v): minimum inter-release delays p(u, v)• k global constraints (u, v , γ)

Theorem (S. et al., 2011)

For k-EDRT task systems with bounded utilization, feasibility is

1 decidable in pseudo-polynomial time if k is constant, and

2 strongly coNP-hard in general.

Extended DRT (EDRT)

• Extends DRT with global delay constraints

• Directed graph for each task• Vertices v : jobs to be released (with WCET and deadline)• Edges (u, v): minimum inter-release delays p(u, v)• k global constraints (u, v , γ)

For k-EDRT task systems with bounded utilization, feasibility is

1 decidable in pseudo-polynomial time if k is constant, and

2 strongly coNP-hard in general.

sporadic(e, d , p)

k-EDRT(k constraints)

Extended DRT (EDRT)(constraints)

efficient

difficult

Strongly (co)NP-hard

Pseudo-Polynomial

DRT Examples

〈e, d〉

Sporadic Task

〈e, p〉

Sporadic Task(implicit deadline)

〈e1, d1〉

〈e2, p2〉

Sporadic Taskwith 2 modes

GMF Task

Branching Task

DRT Examples

〈e, d〉

Sporadic Task

〈e, p〉

〈e1, d1〉

〈e2, p2〉

GMF Task

Branching Task

DRT Examples

〈e, d〉

Sporadic Task

〈e, p〉

〈e1, d1〉

〈e2, p2〉

GMF Task

Branching Task

DRT Examples

〈e, d〉

Sporadic Task

〈e, p〉

〈e1, d1〉

〈e2, p2〉

GMF Task

Branching Task

DRT Examples

〈e, d〉

Sporadic Task

〈e, p〉

〈e1, d1〉

〈e2, p2〉

GMF Task

Branching Task

Adaptive Variable-Rate (AVR) Tasks

• Rate-Adaptive Tasks (Buttazzo et al., DATE 2014)

• Variable Rate-dependent Behaviour (VRB) (Davis et al., RTAS 2014)

• Adaptive Variable-Rate (AVR) Tasks (Biondi et al., ECRTS 2014)

• . . .

(Biondi, ECRTS 2014)

AVR Tasks: Execution Modes

(Biondi, ECRTS 2014)

AVR Business

(This morning in Pisa)

Fahrplan

Feasibility Theorem

Theorem

For a task set τ , the following three properties are equivalent.

1 Task set τ is feasible.

2 Task set τ is EDF schedulable.

3 The following condition holds: ∀t > 0 :∑

T∈τ dbfT (t) 6 t

What is dbfT (t)? How to compute it?

Feasibility Theorem

Theorem

T∈τ dbfT (t) 6 t

What is dbfT (t)? How to compute it?

Demand-Bound Functions

dbfT (t): Maximal demand in any window of size t

0 10 20 28 35 43

5 1 3...

Demand: 5 + 1 + 3 = 9

Feasibility Test

Theorem (Feasibility Theorem)

T∈τ dbfT (t) 6 t∑dbfT (t)

1 How to calculate dbfT (t)?

2 How to check existence of violating t?

Feasibility Test

dbf for Sporadic Tasks

Sporadic task T = (e, d , p)∑dbfT (t)

dbfT (t) = max

⌊t − d

⌋· e}

dbf for Sporadic Tasks

Sporadic task T = (e, d , p)∑dbfT (t)

dbfT (t) = max

⌊t − d

⌋· e}

dbf for DRT: From G (T ) to dbfT

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Path Abstraction: Demand Pair

0 10 20 30 40 50 600369

dbfT (t)

〈9, 43〉

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

0 10 20 30 40 50 600369

dbfT (t)

〈9, 43〉

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

0 10 20 30 40 50 600369

dbfT (t)

〈9, 43〉

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

0 10 20 30 40 50 600369

dbfT (t)

〈9, 43〉

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

0 10 20 30 40 50 600369

dbfT (t)

〈9, 43〉

Demand Pairs

Formally:

• Given path π = (π0, . . . , πl)

• Execution demand: e(π) :=∑l

i=0 e(πi )

• Deadline: d(π) :=∑l−1

i=0 p(πi , πi+1) + d(πl)

• 〈e(π), d(π)〉 is a demand pair for π

dbfT (t) = max {e | 〈e, d〉 demand pair with d 6 t}

How to compute all demand pairs?

• Enumerate paths: Too expensive! (Exponential..)

• Better: Iteration using abstraction

Demand Pairs

Formally:

• Given path π = (π0, . . . , πl)

• Execution demand: e(π) :=∑l

i=0 e(πi )

• Deadline: d(π) :=∑l−1

i=0 p(πi , πi+1) + d(πl)

• 〈e(π), d(π)〉 is a demand pair for π

dbfT (t) = max {e | 〈e, d〉 demand pair with d 6 t}

How to compute all demand pairs?

• Enumerate paths: Too expensive! (Exponential..)

• Better: Iteration using abstraction

Demand Triples• Idea: Start with 0-paths (one vertex), extend stepwise• We need: Abstraction which

1 allows to extend paths,2 contains demand pair information,3 without visiting/storing all paths

• Idea: Demand triples• Execution demand e(π)• Deadline d(π)• Last vertex πl

• Demand triple 〈e(π), d(π), πl〉 is another abstraction!

v1〈2, 5〉

〈1, 8〉v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Path (v4)

; 〈5, 10, v4〉

Path (v4, v2)

; 〈6, 28, v2〉

Path (v4, v2, v3)

; 〈9, 43, v3〉

v1〈2, 5〉

〈1, 8〉v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Path (v4)

; 〈5, 10, v4〉

Path (v4, v2)

; 〈6, 28, v2〉

Path (v4, v2, v3)

; 〈9, 43, v3〉

v1〈2, 5〉

〈1, 8〉v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Path (v4)

; 〈5, 10, v4〉

Path (v4, v2)

; 〈6, 28, v2〉

Path (v4, v2, v3)

; 〈9, 43, v3〉

v1〈2, 5〉

〈1, 8〉v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Path (v4)

; 〈5, 10, v4〉

Path (v4, v2)

; 〈6, 28, v2〉

Path (v4, v2, v3)

; 〈9, 43, v3〉

Iterative Procedure

• Create all demand triples up to some D:

1 Start with all 0-paths, i.e., 〈e(v), d(v), v〉 for all vertices v

2 Pick some stored demand triple 〈e, d , u〉3 Create new demand triple:

• Choose successor vertex v of u• e′ = e + e(v)• d ′ = d − d(u) + p(u, v) + d(v)• 〈e′, d ′, v〉 is new demand triple!

4 Store 〈e′, d ′, v〉 if• not stored yet, and• d ′ 6 D

5 Repeat from 2 until no change

• Efficient procedure!• Note: Actual paths never stored• Optimizations: Discard non-critical triples along the way

• Exercise: What’s dbfτi (26) for graph on previous slide?

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

Optimization: Discard dominated triples

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

〈6, 30, v4〉〈6, 20, v4〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

〈6, 30, v4〉〈6, 20, v4〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Example

0 10 20 30 40 50 60 700

dbfT (t)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

(e, d , v) < (e ′, d ′, v) ⇐⇒ e > e ′ ∧ d 6 d ′

Feasibility Test Revisited

Calculating the Bound

∑dbfT (t)

dbf(t)

Linear bou

nd for dbf(t)

• Linear bound for dbf(t)• Slope: Less than 1

• Intersection with t gives bound D

• Check only up to D

“Most dense” cycle

dbf(t) 6 t · U(τ) + esum

∑dbfT (t)

dbf(t)

Linear bou

nd for dbf(t)

∑dbfT (t)

dbf(t)

Linear bou

nd for dbf(t)

Complexity Result

For DRT task systems τ with a utilization bounded by any c < 1,feasibility can be decided in pseudo-polynomial time.

Pseudo-polynomial time = Tractable/efficient

Evaluation: Runtime vs. Utilization

0% 20% 40% 60% 80% 100%Task Set Utilization

Setting:

• Randomly generated task sets

• 1-30 tasks, 5-10 vertices per task, branching degree 1-3, . . .

Fahrplan

Complexity of Schedulability Tests

• Pseudo-polynomial schedulability tests possible?

EDF SP

L&L YesGMF YesDRT Yes

EDRT No

• ∗ = Takada & Sakamura, 1997

• Flawed!

• Replaced by:

For GMF task systems, the schedulability problem for static priorityschedulers is strongly coNP-hard.

EDF SP

L&L Yes YesGMF Yes Yes∗

DRT Yes ?EDRT No ?

• Flawed!

• Replaced by:

EDF SP

L&L Yes YesGMF Yes Yes∗No!DRT Yes No

EDRT No No

• Flawed!

• Replaced by:

Hardness Result: Proof Sketch

3-PARTITION instance I :

m bins 3m items

Possible to (exactly) fit all items? (strongly NP-hard)

Reduction to GMF schedulability:

One GMF task

for each item

Packing possible ⇐⇒ Busy interval

Thus: τ(I ) unsched. ⇐⇒ I ∈ 3-PARTITION

Hardness Result: Proof Sketch

3-PARTITION instance I :

m bins 3m items

Possible to (exactly) fit all items? (strongly NP-hard)

Reduction to GMF schedulability:

One GMF task

for each item

Packing possible ⇐⇒ Busy interval

Thus: τ(I ) unsched. ⇐⇒ I ∈ 3-PARTITION

Model Hierarchy Revisited

coNP-hard

sporadicMF

k-EDRT

EDF SP

coNP-hard

sporadicMF

k-EDRT

Response-Time Analysis (RTA)

Use RTA for SP Schedulability Analysis.

Response time

Standard RTA for static priorities + periodic/sporadic tasks:

Rj = Cj +∑

i∈hp(j)

Problem: Path Combinations

Response time

Combinatorial Explosion!

Problem: Path Combinations

Response time

Combinatorial Explosion!

Intuition: No Task-Local Worst Cases

Task Thigh:v1〈2, 5〉 v2 〈6, 10〉

Which path is worse: (v1, v2) or (v2, v1)?

It depends! T1: u〈2, 7〉 versus T2: w〈4, 10〉

Thigh:

0 2 4 6 8 10 12

Task Thigh:v1〈2, 5〉 v2 〈6, 10〉

Thigh:

0 2 4 6 8 10 12

Task Thigh:v1〈2, 5〉 v2 〈6, 10〉

Thigh:

0 2 4 6 8 10 12

Task Thigh:v1〈2, 5〉 v2 〈6, 10〉

Thigh:

0 2 4 6 8 10 12

Request Functions

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

rf (t)

t0 5 10 15 20 25 30 35 40

10 rf (v4,v2,v3)

rf π(t) := max{e(π′) | π′ is prefix of π and p(π′) < t

}Martin Stigge Workload Models + Analysis 45

Request Functions (cont.)

Useful for deriving response time:

RSP(v , r̄f ) = min

{t > 0 | e(v) +

∑T ′>T

rf (T ′)(t) 6 t

}RSP(v) = max

r̄f ∈RF (τ)RSP(v , r̄f )

Combinatorial Explosion?!

Request Functions (cont.)

Useful for deriving response time:

RSP(v , r̄f ) = min

{t > 0 | e(v) +

∑T ′>T

rf (T ′)(t) 6 t

}RSP(v) = max

r̄f ∈RF (τ)RSP(v , r̄f )

Combinatorial Explosion?!

Abstract Request Functions

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

rf (t)

t0 5 10 15 20 25 30 35 40

10 rf (v4,v2,v3)

rf (v5,v4,v2)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

rf (t)

t0 5 10 15 20 25 30 35 40

10 rf (v4,v2,v3)

rf (v5,v4,v2)

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

rf (t)

t0 5 10 15 20 25 30 35 40

10 rf (v4,v2,v3)

rf (v5,v4,v2)

Abstraction Tree

rf 1 rf 2

rf 3 rf 4

Define an abstraction tree per task:

• Leaves are concrete rf

• Each node: maximum function of child nodes

• Root is maximum of all rf

Allows stepwise refinement!

Abstraction Tree

rf 1 rf 2

rf 3 rf 4

Abstraction Tree

rf 1 rf 2

rf 3 rf 4

Abstraction Tree

rf 1 rf 2

rf 3 rf 4

Abstraction Tree

rf 1 rf 2

rf 3 rf 4

Abstraction Tree

rf 1 rf 2

rf 3 rf 4

Refinement Algorithm

r̄f = (rf (T1), rf (T2), rf (T3))Tuple:

RSP(v , r̄f ) = 23Response time:

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Initialization:• Most abstract functions

Each iteration:

• Replace functions along abstraction trees

Termination:

• All functions are concreteOR:

• Estimate is already safeIn T1:

rf ′ rf ′′

rfIn T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

Using: RSP(v , r̄f ) = min{t > 0 | e(v) +

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

rf ′ rf ′′

rfIn T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

rf ′ rf ′′

rfIn T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

• Estimate is already safe

In T1:

rf ′ rf ′′

In T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

In T1:

rf ′ rf ′′

In T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

In T1:

rf ′ rf ′′

In T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

rf ′ rf ′′

rfIn T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

rf ′ rf ′′

rfIn T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

rf ′ rf ′′

In T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

rf ′ rf ′′

In T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

rf ′ rf ′′

In T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

rf ′ rf ′′

rfIn T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

In T1:

rf ′ rf ′′

rfIn T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

r̄f 1 = (rf (T1), rf (T2), rf (T3))

r̄f 2 = (rf ′(T1), rf (T2), rf (T3))

r̄f 3 = (rf ′′(T1), rf (T2), rf (T3))

→ 18

→ 21

r̄f 2 = (rf (T1), rf (T2), rf (T3))

r̄f 4 = (rf (T1), rf ′(T2), rf (T3))

r̄f 5 = (rf (T1), rf ′′(T2), rf (T3))

→ 20

→ 17

Each iteration:

Termination:

In T1:

rf ′ rf ′′

rfIn T2:

rf ′ rf ′′

(23, r̄f 1)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(18, r̄f 3)

(21, r̄f 2)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

(20, r̄f 4)

(18, r̄f 3)

(17, r̄f 5)

∑T ′>T rf (T ′)(t) 6 t

Evaluation: Run-time Scaling

0% 20% 40% 60% 80% 100%Task Set Utilization

1-30 tasks with 5-10 vertices each, branching degree 1-3

Evaluation: Precision Improvement

0% 20% 40% 60% 80% 100%Fraction of improved RT estimates

t Type AType B

Type A: lower parameter varianceType B: higher parameter variance

Generality

• Exact solution for NP-hard problem• Efficient method• Iterative refinement

• General!• Apply to any combinatorial problem• ... with monotonic abstraction lattice

rf 1 rf 2 rf 5

rf 3 rf 4

Fahrplan

Summary

coNP-hard

sporadicMF

EDF SP

coNP-hard

sporadicMF

k-EDRT

Summary

coNP-hard

sporadicMF

k-EDRT

EDF SP

coNP-hard

sporadicMF

k-EDRT

Summary

coNP-hard

sporadicMF

k-EDRT

coNP-hard

sporadicMF

k-EDRT

Summary

coNP-hard

sporadicMF

k-EDRT

EDF SP

coNP-hard

sporadicMF

k-EDRT

Summary

coNP-hard

sporadicMF

k-EDRT

EDF SP

coNP-hard

sporadicMF

k-EDRT

Thanks!

Backup Slides Coming Up . . .

Path Abstractions: SP + EDF

Path Abstractions: Static Priorities

v1〈2, 5〉

〈1, 8〉 v3 〈3, 8〉

v4 〈5, 10〉

〈1, 5〉

rf (t)

t0 5 10 15 20 25 30 35 40

10 rf (v4,v2,v3)

rf π(t) := max{e(π′) | π′ is prefix of π and p(π′) < t

}Martin Stigge Workload Models + Analysis 58

Path Abstractions: EDF

wf π(t, t ′) := max{e(π′) | π′ is prefix of π,

p(π′) < t and d(π′) 6 t ′}.

Path Abstractions: EDF

wf π(t, t ′) := max{e(π′) | π′ is prefix of π,

p(π′) < t and d(π′) 6 t ′}.

real-time workload models with e cient analysis · the digraph real-time (drt) task model adaptive...

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