1Periodic TaskScheduling

2

Problem formulation

For each periodic task, guarantee that:

each job tik is activated at rik = (k-1)Ti each job tik completes within dik = rik + Di

ti (Ci, Ti) job tik

rik dik

815

Rate Monotonic (RM) Each task is assigned a fixed priority

proportional to its rate [Liu & Layland 73].

0

500 10025 75

tA

tB

0

tC

40 80

100

16

How can we verify feasibility?

Each task uses the processor for a fractionof time:

i

ii T

CU =

Hence the total processor utilization is:

=

=n

i i

ip T

CU

1

Up is a misure of the processor load

917

A necessary condition

If Up > 1 the processor is overloadedhence the task set cannot be schedulable.

However, there are cases in which Up < 1but the task is not schedulable by RM.

18

An unfeasible RM schedule

0 9 18

6 120 183

3 6 12

9

15

15

deadline miss

t1

t2

944.094

63

=+=pU

10

19

Utilization upper bound

833.093

63

=+=pU

0 9 18

6 120 183

3 6 12

9

15

15

t1

t2

NOTE: If C1 or C2 is increased,t2 will miss its deadline!

20

A different upper bound

184

42

=+=pU

The upper bound Uub depends on thespecific task set.

0

4 120 8 16

t1

t24 128 16

11

21

The least upper bound

1

G

Uub

Ulub

. . .

22

A sufficient condition

If Up Ulub the task set is certainlyschedulable with the RM algorithm.

If Ulub < Up 1 we cannot say anythingabout the feasibility of that task set.

NOTE

12

23

Ulub for RM

In 1973, Liu and Layland proved that for aset of n periodic tasks:

( )12 /1lub -= nRM nU

for n Ulub ln 2

24

RM Schedulability

010203040

50607080

90100

1 2 3 4 5 6 7 8 9 10

69%

n

CPU%

13

25

RM Guarantee Test

We compute the processor utilization as:

=

=n

i i

ip T

CU

1

( )12 /1 - np nU Guarantee Test (only sufficient):

26

Basic Assumptions

A1. Ci is constant for every instance of tti

A2. T i is constant for every instance of tti

A3. For each task, Di = Ti

A4. Tasks are independent: no precedence relations no resource constraints

14

27

RM Optimality

RM is optimal among all fixed priorityalgorithms:

If there exists a fixed priority assignmentwhich leads to a feasible schedule for G, thenthe RM assignment is feasible for G.

If G is not schedulable by RM, then it cannotbe scheduled by any fixed priority assignment.

28

Critical InstantFor any task ti, the longest response time occurswhen it arrives together with all higher priority tasks.

t1

t2R2

t1

t2R2

17

33

Earliest Deadline First (EDF)

Each job receives an absolute deadline:

di,k = ri,k + Di

At any time, the processor is assigned to thejob with the earliest absolute deadline.

Under EDF, any task set can utilize theprocessor up to 100%.

34

EDF Example

0 9 18

6 120 183

3 6 12

9

15

15

t1

t2

94.094

63

=+=pUDi = Ti

18

35

The RM unfesible schedule

0 9 18

6 120 183

3 6 12

9

15

15

deadline miss

t1

t2

944.094

63

=+=pU

36

EDF Optimality

EDF is optimal among all algorithms:

If there exists a feasible schedule for G, thenEDF will generate a feasible schedule.

If G is not schedulable by EDF, then it cannotbe scheduled by any algorithm.

19

37

EDF Optimality [Dertouzos 74]

s

tEt fE dE dk

tk

tE

Transforming ss in ss

s(t) = s(tE)

s(tE) = s(t)fk = fE dE dk

Feasibility is preserved

38

EDF schedulability

In 1973, Liu and Layland proved that for aset of n periodic tasks:

1lub =EDFU

This means that a task set G is schedulableby EDF if and only if

Up 1

22

43

RM vs. EDF

Its more efficient

It reduces context switches

EDF

It is simpler to implement oncommercial operating systems

More predictable during overloads

RM

44

Extension to tasks with D < T

ri,k di,k

Ci

tti

Di

Ti

ri,k+1

Deadline Monotonic:pi 1/Di (static)

Earliest Deadline First:pi 1/di (dynamic)

Scheduling algorithms

1Handling sharedresources

Problems caused bymutual exclusion

2

Critical sectionst2t1

globlalmemory buffer

write readx = 3;y = 5;

a = x+1;b = y+2;c = x+y;

int x;int y;

wait(s)

signal(s)

wait(s)

signal(s)

23

Blocking on a semaphore

CS

t1 t2

CS

p1 > p2

t1

t2

D

It seems that the maximum blockingtime for t1 is equal to the length ofthe critical section of t2, but

4

Schedule with no conflicts

BCT

priority

SCT

MT

35

Conflict on a critical section

BCT

priority

SCT

MT

B

6

Conflict on a critical section

BCT

priority

SCT

MT

B

47

Priority Inversion

A high priority task is blocked by a lower-priority task a for an unbounded interval of time.

SolutionIntroduce a concurrency control protocol foraccessing critical sections.

8

Resource Access Protocols

Non Preemptive Protocol (NPP)

Highest Locker Priority (HLP)

Priority Inheritance Protocol (PIP)

Priority Ceiling Protocol (PCP)

Stack Resource Policy (SRP)

815

Problem with HLP

CS

test

t1

CS

t2

t1

t2

p1p2

t1 blocks just in case ...

16

Priority Inheritance Protocol[Sha, Rajkumar, Lehoczky, 90]

A task in a CS increases its priority only ifit blocks other tasks.

A task in a CS inherits the highest priorityamong those tasks it blocks.

PCS = max {Pk | tk blocked on CS}

917

Schedule with PIPpriority

t1

t2

t3p1

p3

direct blocking

push-through blocking

18

Types of blocking Direct blocking

A task blocks on a locked semaphore

Push-through blockingA task blocks because a lower priority taskinherited a higher priority.

BLOCKING:a delay caused by a lower priority task

10

19

Identifying blocking resources

A task ti can be blocked by thosesemaphores used by lower priority tasksand

directly shared with ti (direct blocking) or

shared with tasks having priority higher thanti (push-through blocking).

Theorem:ti can be blocked at most onceby each of such semaphores

20

Examplepriority

B Ct1t2t3

A

C

DB

A

D

t1 can be blockedonce by t2 (on A2 or C2) andonce by t3 (on B3 or D3)

t2 can be blockedonce by t3 (on B3 or D3)

t3 cannot be blocked