real-time scheduling - penn engineeringlee/10cis541/papers/... · 2010. 1. 15. · real-time...

1

Real-Time Scheduling

Introduction to Real-Time

Review

� Main vocabulary� Definitions of tasks, task invocations, release/arrival time,

absolute deadline, relative deadline, period, start time, finishtime, …

� Preemptive versus non-preemptive scheduling

� Priority-based scheduling

� Static versus dynamic priorities

� Utilization (U) and Schedulability� Main problem: Find Bound for scheduling policy such that

U < Bound � All deadlines met!

� Optimality of EDF scheduling� BoundEDF = 100%

2

Schedulability Analysis of Periodic Tasks

� Main problem:� Given a set of periodic tasks, can they meet their deadlines?

� Depends on scheduling policy

� Solution approaches� Utilization bounds (Simplest)

� Exact analysis (NP-Hard)

� Heuristics

� Two most important scheduling policies� Earliest deadline first (Dynamic)

� Rate monotonic (Static)

Schedulability Analysis of Periodic Tasks

� Main problem:� Given a set of periodic tasks, can they meet their deadlines?

� Depends on scheduling policy

� Solution approaches� Utilization bounds (Simplest)

� Exact analysis (NP-Hard)

� Heuristics

� Two most important scheduling policies� Earliest deadline first (Dynamic)

� Rate monotonic (Static)

3

Utilization Bounds

� Intuitively:

� The lower the processor utilization, U, the easier it is to meet deadlines.

� The higher the processor utilization, U, the more difficult it is to meet deadlines.

� Question: is there a threshold Ubound such that

� When U < Ubound deadlines are met

� When U > Ubound deadlines are missed

Example (Rate-Monotonic Scheduling)

Task 1P1=2

C1=1

Task 2P2=3C2=1.01

%3.833

01.1

2

1

2

2

1

1 ≈+=+=P

C

P

CU

0 1 2 3 4 5 6 time

?

100%

0

U

� Question: is there a threshold Ubound such that� When U < Ubound deadlines are met


4

Example(Rate-Monotonic Scheduling)

Task 1P1=2

C1=1

Task 2P2=3C2=1.01

%3.833

01.1

2

1

2

2

1

1 ≈+=+=P

C

P

CU

0 1 2 3 4 5 6 time

?

100%

0

U



Example(Rate-Monotonic Scheduling)

Task 1P1=2

C1=1

Task 2P2=3C2=1.01

%3.833

01.1

2

1

2

2

1

1 ≈+=+=P

C

P

CU

0 1 2 3 4 5 6 time

Missed deadline!

Unschedulable

?

100%

0

U

83.3%



5

Another Example(Rate-Monotonic Scheduling)

Task 1P1=2

C1=1

Task 2P2=6C2=2.4

%906

4.2

2

1

2

2

1

1 =+=+=P

C

P

CU

0 1 2 3 4 5 6 time

?

100%

0

U

83.3%




Task 1P1=2

C1=1

Task 2P2=6C2=2.4

%906

4.2

2

1

2

2

1

1 =+=+=P

C

P

CU

0 1 2 3 4 5 6 time

?

100%

0

U

83.3%



6


Task 1P1=2

C1=1

Task 2P2=6C2=2.4

%906

4.2

2

1

2

2

1

1 =+=+=P

C

P

CU

0 1 2 3 4 5 6 time

Schedulable!

?

100%

0

U

83.3%90%




Task 1P1=2

C1=1

Task 2P2=6C2=2.4

%906

4.2

2

1

2

2

1

1 =+=+=P

C

P

CU

0 1 2 3 4 5 6 time

Schedulable!

?

100%

0

U

83.3%90%



Schedulability depends on task set!No clean utilization threshold between schedulable and unschedulable

task sets!

7

A Conceptual View of Schedulability

Utilization

Task Set

SchedulableUnschedulable



∑=i i

i

P

C


Utilization

Task Set


� Modified Question: is there a threshold Ubound such that� When U < Ubound deadlines are met

� When U > Ubound deadlines may or may not be missed

?

All green area (sche

dulable)

∑=i i

i

P

C

8


Utilization

Task Set




?


dulable)

U < Ubound is a sufficient but not necessaryschedulabilitycondition

∑=i i

i

P

C


Utilization

Task Set




?


dulable)

Equivalent question:What’s the lowest utilization of an unschedulable task set?

∑=i i

i

P

C

9


Utilization

Task Set




?


dulable)

Equivalent question:What’s the lowest utilization of an unschedulable task set?

critically(Called the Utilization Bound, U

bound)

∑=i i

i

P

C

Solution Approach: Look at Critically-Schedulable Task Sets

Utilization

Task Set� Modified Question: is there a threshold Ubound such that� When U < Ubound deadlines are met


?


dulable)

Case (a) Case (b)Utilization increases with x

Utilization decreases with x

Find some task set parameter xsuch thatCase (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x

Thus U(x) is minimum when x=xo

Find U(xo)

10

Deriving the Utilization Bound for Rate Monotonic Scheduling

� Consider a simple case: 2 tasks

Find some task set parameter xsuch that

Case (a): x<xo � U(x) decreases with xCase (b): x>xo � U(x) increases with x


Find U(xo)



P1

P2

Task 1

Task 2




Find U(xo)

11



C1

P1

P2

Task 1

Task 2




Find U(xo)



C1

P1

P2

Task 1

Task 2

Critically schedulable

total = C2




Find U(xo)

12



C1

P1

P2

Task 1

Task 2

?




Find U(xo)



C1

P1

P2

Task 1

Task 2

1

1

22 P

P

PP

−




Find U(xo)

13


� Consider these two sub-cases:

Case (a):

C1

P1

P2

Task 1

Task 2

Case (b):

P1

P2

C1

1

1

22 P

P

PP

− 1

1

22 P

P

PP

−




Find U(xo)



Case (a):

C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−≤

Case (b):

P1

P2

C1

1

1

221 P

P

PPC

−>




Find U(xo)

14



Case (a):

C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−≤

Case (b):

P1

P2

C1

1

1

221 P

P

PPC

−>

C1Find some task set parameter x

such that



Find U(xo)



Case (a):

C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−≤

Case (b):

P1

P2

C1

1

1

221 P

P

PPC

−>


such that



Find U(xo)

15



Case (a):

C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−≤

Case (b):

P1

P2

C1

1

1

221 P

P

PPC

−>


such that



Find U(xo)



Case (a):

C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−≤

Case (b):

P1

P2

C1

1

1

221 P

P

PPC

−>

+

−= 1

1

2122

P

PCPC

−=

1

2112 )(

P

PCPC

−

−+=+= 11

1

2

1

2

2

1

2

2

1

1

P

P

P

P

P

C

P

C

P

CU

−+

=

1

2

1

2

2

1

1

2

2

1

P

P

P

P

P

C

P

P

P

PU

16



Case (a):

C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−≤

Case (b):

P1

P2

C1

1

1

221 P

P

PPC

−>

+

−= 1

1

2122

P

PCPC

−=

1

2112 )(

P

PCPC

−+

=

1

2

1

2

2

1

1

2

2

1

P

P

P

P

P

C

P

P

P

PU

C1

�� UC

1�� U

−

−+=+= 11

1

2

1

2

2

1

2

2

1

1

P

P

P

P

P

C

P

C

P

CU


� The minimum utilization case:

C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−=

−

−+= 11

1

2

1

2

2

1

P

P

P

P

P

CU

17



C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−=

−

−+= 11

1

2

1

2

2

1

P

P

P

P

P

CU

0

1

2

=

PP

d

dU

83.0≈⇒ U

21

2 =⇒P

P

−

−

−+= 11

1

2

1

2

1

2

1

2

2

1

P

P

P

P

P

P

P

P

P

PU

11

2 =

⇒

P

P

1

2

1

2

1

2 21

1

PP

PP

PP

U

−

−

+=⇒

Note that C1 = P2 – P1

−=

1

2

1

211

P

P

P

PPC

Generalizing to N Tasks

C1 = P2 – P1

C2 = P3 – P2

C3 = P4 – P3

…

...3

3

2

2

1

1 +++=P

C

P

C

P

CU

18


C1 = P2 – P1

C2 = P3 – P2

C3 = P3 – P2

…

0

1

2

=

PP

d

dU0

2

3

=

PP

d

dU0

3

4

=

PP

d

dU …

...3

3

2

2

1

1 +++=P

C

P

C

P

CU


C1 = P2 – P1

C2 = P3 – P2

C3 = P3 – P2

…

0

1

2

=

PP

d

dU0

2

3

=

PP

d

dU0

3

4

=

PP

d

dU …

...3

3

2

2

1

1 +++=P

C

P

C

P

CU

n

i

i

P

P 11 2=⇒ +

−=⇒ 12

1nnU

19


C1 = P2 – P1

C2 = P3 – P2

C3 = P3 – P2

…

0

1

2

=

PP

d

dU0

2

3

=

PP

d

dU0

3

4

=

PP

d

dU …

...3

3

2

2

1

1 +++=P

C

P

C

P

CU

n

i

i

P

P 11 2=⇒ +

−=⇒ 12

1nnU

2ln→∞→ Un

Periodic Tasks

Periodic Task Scheduling

Rate Monotonic EDF

Bound Optimality Bound Optimality

20

Periodic Tasks


Rate Monotonic EDF


Coming Up


Rate Monotonic EDF


21

Rate Monotonic Continued

� Rate monotonic scheduling is the optimal fixed-priority scheduling policy for periodic tasks.

� Optimality (Trial #1):



� Optimality (Trial #1): If any other fixed-priority scheduling policy can meet deadlines, so can RM.

22



� Optimality (Trial #1): If any other fixed-priority scheduling policy can meet deadlines, so can RM




23







24



� Optimality (Trial #2): If any other fixed-priority scheduling policy can meet deadlines in the worst case scenario, so can RM.

� How to prove it?



� Optimality (Trial #2): If any other fixed-priority scheduling policy can meet deadlines in the worst case scenario, so can RM.

� How to prove it?

� Consider the worst case scenario

� If someone else can schedule then RM can

25

The Worst-Case Scenario

� Q: When does a periodic task, T, experience the maximum delay?

� A: When it arrives together with all the higher-priority tasks (critical instance)

� Idea of Proof� If some higher-priority task does not arrive together with T, aligning the arrival times can only increase the completion time of T

Proof (Case 1)

Task 1

Task 2

Case 1: higher priority task 1 is running when task 2 arrives

26

Proof

Task 1

Task 2


� shifting task 1 right will increase completion time of 2

Proof

Task 1

Task 2


� shifting task 1 right will increase completion time of 2

Task 1

Task 2

27

Proof (Case 2)

Task 1

Task 2

Case 2: processor is idle when task 2 arrives

Proof (Case 2)

Task 1

Task 2


� shifting task 1 left cannot decrease completion time of 2

28

Proof (Case 2)

Task 1

Task 2

Task 1

Task 2


� shifting task 1 left cannot decrease completion time of 2

Optimality of Rate Monotonic

� If any other policy can meet deadlines so can RM

Policy X meets deadlines?

29

Optimality of Rate Monotonic

� If any other policy can meet deadlines so can RM

Policy X meets deadlines?

� RM meets deadlines

YES

Coming Up


Rate Monotonic EDF


30

Coming Up


Rate Monotonic EDF


Utilization Bound of EDF

� Why is it 100%?

� Consider a task set where:

� Imagine a policy that reserves for each task i a fraction fi of each clock tick, where fi = Ci /Pi

1=∑i i

i

P

C

Clock tick

31


� Imagine a policy that reserves for each task i a fraction fi of each time unit, where fi = Ci /Pi

� This policy meets all deadlines, because within each period Pi it reserves for task i a total time

� Time = fi Pi = (Ci / Pi) Pi = Ci (i.e., enough to finish)

Clock tick


� Pick any two execution chunks that are not in EDF order and swap them

32



� Still meets deadlines!



� Still meets deadlines!

� Repeat swap until all in EDF order� EDF meets deadlines

33

Periodic Tasks


Rate Monotonic EDF


Done


Rate Monotonic EDF


34

Exercise:Know Your Worst Case Scenario

� Consider a periodic system of two tasks

� Let Ui = Ci /Pi (for i = 1,2)

� What is the maximum value of:

Πi(1+Ui)

for a schedulable system?



C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−=

−

−+= 11

1

2

1

2

2

1

P

P

P

P

P

CU

−

−

−+= 11

1

2

1

2

1

2

1

2

2

1

P

P

P

P

P

P

P

P

P

PU

11

2 =

⇒

P

P

−=

1

2

1

211

P

P

P

PPC

( )

−=

1

2112

P

PCPC

35



C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−=

−

−+= 11

1

2

1

2

2

1

P

P

P

P

P

CU

−

−

−+= 11

1

2

1

2

1

2

1

2

2

1

P

P

P

P

P

P

P

P

P

PU

11

2 =

⇒

P

P

−=

1

2

1

211

P

P

P

PPC

C1

P1

P2

Task 1

Task 2

C2



C1

P1

P2

Task 1

Task 2

1

1

221 P

P

PPC

−=

−

−+= 11

1

2

1

2

2

1

P

P

P

P

P

CU

−

−

−+= 11

1

2

1

2

1

2

1

2

2

1

P

P

P

P

P

P

P

P

P

PU

11

2 =

⇒

P

P

−=

1

2

1

211

P

P

P

PPC

C1

P1

P2

Task 1

Task 2

C2

112

121

CPP

PCC

=−

=+

36

Solutions

( )∏ =+

=+

=+=+

=+

=+=+

−=−=

−=

i iU

P

P

P

PC

P

CU

P

P

P

PC

P

CU

PPCPC

PPC

21

211

11

2

2

1

2

22

2

22

1

2

1

11

1

11

21112

121

Critically

Schedulable

Schedulable ( )∏ ≤+i iU 21

Solutions

( )∏ =+

=+

=+=+

=+

=+=+

−=−=

−=

i iU

P

P

P

PC

P

CU

P

P

P

PC

P

CU

PPCPC

PPC

21

211

11

2

2

1

2

22

2

22

1

2

1

11

1

11

21112

121

Critically

Schedulable


37

Solutions

( )∏ =+

=+

=+=+

=+

=+=+

−=−=

−=

i iU

P

P

P

PC

P

CU

P

P

P

PC

P

CU

PPCPC

PPC

21

211

11

2

2

1

2

22

2

22

1

2

1

11

1

11

21112

121

Critically

Schedulable


Solutions

( )∏ =+

=+

=+=+

=+

=+=+

−=−=

−=

i iU

P

P

P

PC

P

CU

P

P

P

PC

P

CU

PPCPC

PPC

21

211

11

2

2

1

2

22

2

22

1

2

1

11

1

11

21112

121

Critically

Schedulable


38

The General Case

( )∏ ==+

=+

=+=+

=+

=+=+

−=

−=

−

+

+

inn

n

i

nn

nn

n

n

n

i

i

i

ii

i

i

i

nn

iii

P

P

P

P

P

P

P

PU

P

P

P

PC

P

CU

P

P

P

PC

P

CU

PPC

PPC

22

...1

211

11

2

1

12

3

1

2

1

1

1

1

Critically

Schedulable


The General Case

( )∏ ==+

=+

=+=+

=+

=+=+

−=

−=

−

+

+

inn

n

i

nn

nn

n

n

n

i

i

i

ii

i

i

i

nn

iii

P

P

P

P

P

P

P

PU

P

P

P

PC

P

CU

P

P

P

PC

P

CU

PPC

PPC

22

...1

211

11

2

1

12

3

1

2

1

1

1

1

Critically

Schedulable


39

The General Case

( )∏ ==+

=+

=+=+

=+

=+=+

−=

−=

−

+

+

inn

n

i

nn

nn

n

n

n

i

i

i

ii

i

i

i

nn

iii

P

P

P

P

P

P

P

PU

P

P

P

PC

P

CU

P

P

P

PC

P

CU

PPC

PPC

22

...1

211

11

2

1

12

3

1

2

1

1

1

1

Critically

Schedulable


The General Case

( )∏ ==+

=+

=+=+

=+

=+=+

−=

−=

−

+

+

inn

n

i

nn

nn

n

n

n

i

i

i

ii

i

i

i

nn

iii

P

P

P

P

P

P

P

PU

P

P

P

PC

P

CU

P

P

P

PC

P

CU

PPC

PPC

22

...1

211

11

2

1

12

3

1

2

1

1

1

1

Critically

Schedulable


40

The Hyperbolic Bound for Rate Monotonic Scheduling

� A set of periodic tasks is schedulable if:

( )∏ ≤+i iU 21



� It’s a better bound than

� Example:

� A system of two tasks with U1=0.8, U2=0.1

( )∏ ≤+i iU 21

( )∑ −≤i

n

inU 12 /1

41



� It’s a better bound!

� Example:


� Liu and Layland bound: U1+U2 = 0.9 > 0.83

( )∏ ≤+i iU 21



� It’s a better bound!

� Example:


� Liu and Layland bound: U1+U2 = 0.9 > 0.83

� Hyperbolic bound (U1+1)(U2+1) =1.8 x 1.1=1.98 < 2

( )∏ ≤+i iU 21

42

Scheduling Taxonomy


Rate Monotonic EDF


Scheduling Taxonomy


Rate Monotonic EDF


Hyperbolic Bound

real-time scheduling - penn engineeringlee/10cis541/papers/... · 2010. 1. 15. · real-time...

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