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Chapter-7 L11: "Embedded Systems - Architecture,

Programming and Design" , Raj Kamal, Publs.: McGraw-Hill,

Inc.1

INTER-PROCESS COMMUNICATION

AND SYNCHRONISATION:

LessonLesson--11:11: Priority InversionPriority InversionProblem and Deadlock SituationsProblem and Deadlock Situations

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Inc.2

1.1. Priority Inversion

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Inc.3

Assume

Priorities of tasks be in an order such that

taskIhighest priority, taskJa lower, andtaskKthe lowest priority.

Only tasksIand Kshare the data andJdoes

not share data with K.

Also let tasksIand Kalone share a

semaphore sikand notJ.

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Inc.4

Few tasks share a semaphoreFew tasks share a semaphore

Why do only a few tasks share a

semaphore? Can't all share a semaphore? AnswerWorst-case latency becomes too

high and may exceed the deadline if all

tasks are blocked when one task takes asemaphore.

The worst-case latency will be small only if

the time taken by the tasks that share theresources is relevant

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Inc.5

At an instant t0, suppose task K takes sik , it

does not block taskJ and blocks only the

taskI. This happens because only tasks I and K

share the data andJdoes not and I is blocked

at instance t0 due to wait for some message

and sik.

Consider the problem that now arises on

selective sharing between KandI.

Priority Inversion SituationPriority Inversion Situation

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Inc.6

At next instant t1, let task Kbecome ready

first on an interrupt.

Now, assume that at next instant t2, taskIbecomes ready on an interrupt or getting the

waiting message.

At this instant, Kis in the critical section.

Therefore, taskI cannot start at this instant

due to K being in the critical region.


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Programming and Design" , Raj Kamal, Publs.: McGraw-Hill,Inc.

7

Now, if at next instant t3, some action

(event) causes the unblocked higher than K

priority taskJto run. After instant t3, running task J does not

allow the highest priority task I to run

because K is not running, and therefore K

can't release sikthat it shares withI.


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8

Further, the design of taskJmay be such that

even when sikis released by taskK, it may not

letIrun. [Jruns the codes as if it is in criticalsection all the time after executing DI.] TheJ

action is now as ifJhas higher priority than

I. This is because K, after entering the

critical section and taking the semaphore the

OS let the J run, did not share the priorityinformation about Ithat taskI is of higher

priority thanJ.


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9

The priority information of another higher-

priority task I should have also been

inherited by K temporarily, if Kwaits forIbutJdoes not andJruns when Khas still not

finished the critical section codes.

This did not happen because the given OS

was such that it didnt provide for temporary

priority inheritance in such situations. Above situation is also called a priority

inversion problem


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10

Some OSes provide for priority inheritance

in these situations and thus priority

inheritance problem does not occur whenusing them.

A mutex should be a mutually exclusive

Boolean function, by using which the critical

section is protected from interruption in such

a way that the problem of priority inversiondoes not arise.

Mutex is automatically provided in certain

RTOS so that it the priority inversion

OS Provision forOS Provision for temporary priority

inheritance in such situations

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11

Mutex is automatically provided with

priority inheritance by task on taking it incertain OSes so that it the priority inversion

problem does not arise and certain OSes

provides for selecting priority inheritance aswell as priority sealing options.

OS Provision forOS Provision for temporary priority

inheritance in such situations

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12

2.2. Deadlock Situation

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13

Priorities of tasks be such that taskHis of

highest priority, taskIa lower priority and

taskJthe lowest.

Two semaphores, SemTok1 and SemTok2.

TasksIandHhave a shared resourcethrough SemTok1 only.

TasksIandJhave two shared resources

through two semaphores, SemTok1 andSemTok2.

Assume

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14

At a next instant t1, being now of a

higher priority, the taskHinterrupts thetaskIandJafter it takes the semaphore

SemTok1, and thus blocks bothIandJ.

Deadlock Situation

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15

In between the time interval t0 and

t1, the SemTok1 was released but

SemTok2 was not released during the

run of taskJ. But the latter did not

matter as the tasksIandJdont shareSemTok2.

Deadlock Situation

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16

At an instant t2, ifHnow releases the

SemTok1, lets the taskItake it.

Even then it cannot run because it is alsowaiting for taskJto release the SemTok2.

The taskJis waiting at a next instant t3, for

eitherHorIto release the SemTok1 because

it needs this to again enter a critical section

in it. Neither taskIcan run after instant t3 nor

taskJ.

Deadlock Situation

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17

There is a circular dependency established

betweenIandJ.

Deadlock Situation Solution

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On the interrupt byH, the taskJ,before

exiting from the running state, should have

been put in queue-front so that later on, it

should first take SemTok1, and the taskIput

in queue next for the same token, then the

deadlock would not have occurred

Deadlock Situation solution

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19

SummarySummary

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We learnt Priority becomes inverted and deadlock

(circular dependency) develops in certain

situations when using semaphores.

Certain OSes provide the solution to this

problem of semaphore use by ensuring thatthese situations do not arise during the

concurrent processing of multitasking

operations .

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21

End ofEnd ofLesson11 of Chapter 7 onLesson11 of Chapter 7 on

Priority Inversion Problem and DeadlockPriority Inversion Problem and Deadlock

SituationsSituations

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