Concurrency - Aalborg Universitethomes.et.aau.dk/akh/2010/operatingsystem-10_files/LECT-4-CLASS.pdf · CHAPTER 5 Fall 2010 - DE5 5 Operating System Concerns What are the issues o/s

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ConcurrencyConcurrency

Dr. D. M. Akbar Hussain

Department of Electronic SystemsDE5 1

ConcurrencyConcurrency

EExecution of multiple processes .xecution of multiple processes .

Multi-programming:Management of multiple processes within a multiple processes within a uniuni--

processor systemprocessor system, every system has this support, whether big, small or complex.

Multi-processing:Management of multiple processes within a multi-processor system, servers and works

stations



stations. ((Shared Memory)Shared Memory)

Distributed Processing:Management of multiple processes executing on

number of distributed computer systems, for example clusters.

((Do not Share Memory)Do not Share Memory)

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Interleaving:

Concurrency PrincipleConcurrency Principle

Interleaving:

1. In a single processor case multiple processes areinterleaved in time to provide the illusionillusion ofofsimultaneoussimultaneous executionexecution of these processes.

2. Although, it is not really parallel processing but there arebenefitsbenefits usingusing suchsuch techniquetechnique apart from having



benefitsbenefits usingusing suchsuch techniquetechnique, apart from havingoverheads involved in switching of these processes.

Concurrency PrinciplesConcurrency Principles

Issues:

1. Basically, it is not possible to predict the speed predict the speed of execution of processes.

2. Optimal allocation Optimal allocation of resources is not possible by the o/s.3. Sharing of global variablesglobal variables.



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Pseudo Code for Demonstration:S h t li ti di i t f th

Concurrency PrinciplesConcurrency Principles

Suppose we have two or more applications reading input from the keyboard and putting the result on the screen. It would make sense to have the same procedure for all these applications and obviously it will be loaded into the global address space.

Pseudo Code:function_test ( string charact){



{readin (input_variable, keyboard);charact = input_variable;readout (charact, display);exit;}

Where is the problem

P1

Process P1 has called function_testand have just read (X) (X) the input and j ( )( ) p

being interrupted,

so so charactcharact = X;= X;

P2

Process P2 has called function_test and allowed to run till end with input “Y”“Y”,

e.g., e.g., charactcharact = Y.= Y.function_test ( string charact){readin (input_variable, keyboard);h t i t i bl



P1When P1 resume it will display

what ?

charact = input_variable;readout (charact, display);exit;}

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How to Achieve Mutual ExclusionHow to Achieve Mutual Exclusion



Basics of Mutual ExclusionBasics of Mutual Exclusion

1. Only oneone processprocess is allowed into its critical section.

2 A process which may stop execution in non critical section mustmust notnot2. A process which may stop execution in non critical section mustmust notnotaffect/interfereaffect/interfere withwith othersothers..

3. A process should notnot bebe delayeddelayed indefinitelyindefinitely requiring access to criticalsection.

4. Any process mustmust bebe allowedallowed toto enterenter intointo criticalcritical sectionsection withoutwithout delaydelay, ifno process is in critical section.



5. Noo assumptionsassumptions aboutabout speedspeed oror numbersnumbers of CPUs to be considered.

6. A process should remain in the critical section forfor aa finitefinite timetime onlyonly..

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Operating System ConcernsOperating System Concerns

What are the issues o/s has to consider for concurrency:

1 O/S t k t k k t k f ll th ( ll it i d th h PCB’ )1. O/S must keep track keep track of all the processes (normally it is done through PCB’s).

2. O/S must allocate and deallocate and de--allocate allocate the resources for the processes. Which includes processor time, memory, files and i/o devicesprocessor time, memory, files and i/o devices.

3.3. Protection of resources Protection of resources against unintended interference from other processes.

4 Result of a process must be independent of the speed independent of the speed at which the



4. Result of a process must be independent of the speed independent of the speed at which the execution is taking place relative to the speed of other concurrent processes.

Operating System ConcernsOperating System Concerns

Degree of Awareness Relationship Influence on others ProblemsUnaware of others Competition 1. Result is independent of others

2. Timing may be affected1. Mutual Exclusion2. Deadlock3. Starvation

Indirectly aware of others Cooperation by sharing 1. Result is not independent of others2. Timing may be affected

1. Mutual Exclusion2. Deadlock3. Starvation4. Data Coherence

Directly aware of others Cooperation by communication 1. Result is not independent of others2. Timing may be affected

1. Deadlock2. Starvation



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Mutual Exclusion has cost ?Mutual Exclusion has cost ?

Mutual ExclusionMutual Exclusion

DeadlockDeadlock StarvationStarvation



DeadlockDeadlock

Suppose two processes P1 P1 and P2 P2 are to accomplish a task by using two resources R1 R1 and R2R2As shown here, What happened next ?

P1P1 P2P2



R1R1 R2R2Waiting (for ever)Waiting (for ever)

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P1P1 P3P3P2P2

P2 is StarvedP2 is Starved

StarvationStarvation

RR



P1P1 P3P3 P1P1 P3P3 P1P1 P3 ……………………………….P3 ……………………………….

In addition to deadlock and starvation another requirement another requirement may also be necessary, for example, suppose an application needs that a must condition should always be in placed say: a = b. a = b. In other words any process updating aa must update b b or vice versa,

Data CoherencyData Coherency

say for example a = b = 20;say for example a = b = 20;P1: P1: a = a a = a -- 10;10;

b = b b = b -- 10;10;P2:P2: b = b * 2;b = b * 2;

a = a * 2;a = a * 2;

Which is fine if the consistency is achieved, but consider the following situation and both processes impose mutual exclusion;

11 1010



P1: P1: a = a a = a -- 10;10;P2:P2: b = b * 2;b = b * 2;P1:P1: b = b b = b -- 10;10;P2:P2: a = a * 2;a = a * 2;

Which clearly demonstrate that a = b does not hold any more.

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DekkersDekkers Algorithm for MEAlgorithm for ME

Process 0 Process 1

Critical Section



Critical Section

Turn 0/1

DekkersDekkers algorithmalgorithm

IgloIglo ((eskimoiskeskimoisk snehyttesnehytte))



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There is a strict alternation ruleThere is a strict alternation rule

DekkersDekkers algorithmalgorithm

There is a strict alternation rule.There is a strict alternation rule.

ME is guaranteedME is guaranteed

ConsequencesConsequences::

1.1. Execution is dictated by the slowest of the two.Execution is dictated by the slowest of the two.



yy

2.2. If one is lost the other is blocked for ever.If one is lost the other is blocked for ever.

Modification 1Modification 1



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Dictated by the slowestslowest ofof thethe twotwo..

NoNo guaranteeguarantee ofof MutualMutual ExclusionExclusion:: Because both can setthere flags at the same time to truetrue after one processreached at the end and setting others flag to false, so at thatpoint both flags becomes false so now there is a chance to setboth flags to true (meaning both can enter in their critical



g ( gsection).


Own Flag is set outside the while loop & then checking others FlagOwn Flag is set outside the while loop & then checking others Flag



Dictated by the slowest of the two.Dictated by the slowest of the two.Guarantee Mutual Exclusion.Guarantee Mutual Exclusion.Deadlock,Deadlock, if both has set the Flag to True at the same instantif both has set the Flag to True at the same instant

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Modification 3Modification 3Own Flag is set outside the while loop, then checking others Flag & Own Flag is set outside the while loop, then checking others Flag &

resetting resetting its own flag & its own flag & introducing Delayintroducing Delay



Dictated by the slowest of the twoDictated by the slowest of the twoGuarantee Mutual ExclusionGuarantee Mutual ExclusionDeadlock ?Deadlock ?Almost thereAlmost there

Correct SolutionCorrect Solution

AsAs wewe havehave observedobserved thatthat knowingknowing thethe statestate ofof otherotherprocessesprocesses isis notnot enoughenough toto makemake ourour problemproblemworkable,workable, basicallybasically oneone needsneeds anan orderorder thethe waywayprocessesprocesses proceedsproceeds..



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Dictated by the slowest of the two.Dictated by the slowest of the two.Guarantee Mutual Exclusion.Guarantee Mutual Exclusion.No Deadlock.No Deadlock.

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Peterson’s solution is similar but more elegant as the text says:

Peterson’s SolutionPeterson’s Solution



Mutual exclusion can also be implemented in Hardware by using eithereitherinterruptinterrupt handlinghandling oror throughthrough specialspecial machinemachine instructionsinstructions::

Hardware Mutual ExclusionHardware Mutual Exclusion

interruptinterrupt handlinghandling oror throughthrough specialspecial machinemachine instructionsinstructions::

InterruptInterrupt DisablingDisabling::Just before entering the criticalcritical region,region, forcing interrupt disabling willensure ME:

Potentially system performance is degradeddegraded..NoNo supportsupport forfor multimulti--processorprocessor system architecture.



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Machine InstructionMachine Instruction

1. Basically twotwo actionsactions areare performedperformed inin oneone cyclecycle forthese type of instruction atomically, for example, readingreadingandand writingwriting.

2. Therefore, one can use them as they cannotcannot bebe interruptedinterruptedduring these actions.




Test and Set InstructionTest and Set Instruction

Boolean testsettestset (int a){

if (a == 0) a = 1; return true;else return false;

}



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Exchange InstructionExchange Instruction


Exchanges Exchanges the contents of a register with a memory location, obviously during this, access to the memory location is blocked.access to the memory location is blocked.

void exchangeexchange (int register, memory){ int temp;

temp = memory; memory = register;



register = temp;}

n n number of processesbolt bolt Shared variable (int)Function_MEFunction_ME (process)(process){while true

TestsetTestset for MEfor ME

while true {while { (!testset (boltbolt))

be happy and do nothingbe happy and do nothing}

Enter Critical SectionEnter Critical Sectionset bolt to 0;set bolt to 0;do remaining part}



}}

InitiallyInitially boltbolt isis setset toto 00,, anyany processprocess findingfinding itit zerozero willwill enterenter intointo thethe criticalcriticalsection,section, obviouslyobviously allall othersothers willwill gogo intointo ““bebe happyhappy looploop””.. OnceOnce thatthat processprocessleavesleaves itsits criticalcritical sectionsection anyany otherother processprocess findingfinding boltbolt zerozero willwillimmediatelyimmediately gogo intointo itsits criticalcritical sectionsection andand soso onon..

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n n number of processesboltbolt Shared variable (int)Function_MEFunction_ME (process)(process){keykey local variable (int)

Exchange for MEExchange for ME

while true {key = 1;while { (keykey != 0)

exchangeexchange (key, boltkey, bolt); ”be happy until you get bolt zerobe happy until you get bolt zero”}

Enter Critical SectionEnter Critical Sectionexchange (key, bolt);exchange (key, bolt);do remaining part



g p}

}

InitiallyInitially boltbolt isis setset toto 00,, anyany processprocess findingfinding itit zerozero willwill enterenter intointo thethe criticalcritical section,section,butbut thisthis timetime itit willwill bebe throughthrough thethe locallocal variablevariable “key”“key” andand thethe boltbolt exchangeexchangemechanismmechanism ofof thethe exchangeexchange instructioninstruction.. AfterAfter completingcompleting itsits job,job, boltbolt isis againagain setset toto 00throughthrough thethe exchangeexchange instructioninstruction..

Plus/Minus of Machine Plus/Minus of Machine InstructionInstruction

1.1. Easy to implementEasy to implement2.2. Support for any number of processorsSupport for any number of processors3.3. Multiple Critical section support (By using different variables)Multiple Critical section support (By using different variables)

1.1. Busy waiting (performance degradation)Busy waiting (performance degradation)2.2. Starvation is possibleStarvation is possible33 Deadlock is possibleDeadlock is possible



3.3. Deadlock is possibleDeadlock is possible

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Design of an o/s as sequential processes with a reliable mechanism of support

SemaphoresSemaphores

g q p ppfor cooperation.Dijkstra proposed a solutionsolution basedbased onon thethe fundamentalfundamental principleprinciple ofofcooperationcooperation basedbased onon signalssignals, so that a process can be forcedforced toto stopstop at arequired place and later starts when instructed through a signal.

These signalssignals areare calledcalled semaphoressemaphores, to transmit a signal the processexecutes signal (s) and to receive a signal it executes wait (s). Therefore, ifthe corresponding signals are not transmitted the process is suspended till



the corresponding signals are not transmitted the process is suspended tillthe transmission takes place.

It is initialized to a non negative valuenon negative value

Properties of SemaphoresProperties of Semaphores

It is initialized to a non negative value.non negative value.

Wait operation decrements semaphore valueWait operation decrements semaphore value, if it goes negative the corresponding process is blocked.blocked.

Signal operation Signal operation increments semaphore value, if the value is not positiveif the value is not positive, process can be unblocked unblocked from the suspended list.



Signal and Wait are atomicatomic..

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Semaphore could be declared as a simple structure having a variable declaration, may be an integer for count and another variable/structure for a process queue.

it ( ) i l ( )

Semaphore Semaphore PrimitivesPrimitives

Critical Section (if count is zero)

wait (s)count = count - 1;

if count < 0put the process in queueblock the process

signal (s)count = count + 1;if count ≤ 0

get a process from the queueunblock the process and put it in ready queue



The order of removal from the queue is not defined, FIFO etc., the only requirement is that no process should wait indefinitely in the queue.

Binary SemaphoresBinary Semaphores

waitBwaitB (s)(s)if count == 1if count == 1

count = 0;count = 0;elseelse

put the process in queueput the process in queueblock the processblock the process

signalBsignalB (s)(s)if queue is emptyif queue is empty

count = 1;count = 1;elseelse

get a process from the queueget a process from the queueput it in ready queueput it in ready queue



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Process A

s = 1 C D B

Ready ListSuspended ListProcess C

s = 0 D B A

Ready ListSuspended List

Semaphore Mechanism (Strong s)Semaphore Mechanism (Strong s)

Semaphore Semaphore

Process B

s = 0 A C D


Semaphore

Process D

s = -3 B A C


Semaphore

Process D

s = -1 A C B


S h

Process D

s = -2 C B A


S h



Semaphore

Process D

s = 0 B A C


Semaphore

Semaphore

The following pseudo ops can be used for the mutual exclusion problem:

There could be n processesn processes

Each process executes wait executes wait before entering to its critical section

ME Using SemaphoresME Using Semaphores

Positive value means Positive value means it can enter into critical section

If the value becomes negative, process is suspendedbecomes negative, process is suspended

Semaphore s = 1; n = “number of processes”;

Process ()

loop

it ( )it ( )



wait (s)wait (s)

critical sectioncritical section

signal (s)signal (s)

remaining executionremaining execution

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1

•

1

•

00

LockLock AA BB CC

Wait (lock) Normal Execution

ME Using SemaphoresME Using Semaphores

••

0

•

0

•

0

•

0

•

0

•

0

•

B

B

C

C

Wait (lock)

Wait (lock)

signal (lock) Blocked



0

•

0

•

1

•

1

•

signal (lock)

signal (lock)

Critical Section

ProduceProduce waitB(s)waitB(s) InsetInset n = n + 1n = n + 1 If n == 1If n == 1 signalB (delay)signalB (delay) signalB (s)signalB (s)

Producer Consumer (Binary)Producer Consumer (Binary)

ProduceProduce waitB(s)waitB(s) InsetInset n n + 1n n + 1 If n 1If n 1 signalB (delay)signalB (delay) signalB (s)signalB (s)

Critical SectionCritical Section

First Time only



waitB(s)waitB(s) TakeTake n = n - 1n = n - 1 consumeconsume waitB (delay)waitB (delay)signalB (s)signalB (s)waitB(delay)waitB(delay) If n == 0If n == 0

Correct Solution ?

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Producer Consumer s n Delay

1 1 0 0

2 waitB (s) 0 0 0

3 n=n+1 0 1 0

4 If n==1 then signalB (delay) 0 1 1

5 signalB (s) 1 1 1

Possible ScenarioPossible Scenario

5 signalB (s) 1 1 1

6 waitB (delay) 1 1 0

7 waitB (s) 0 1 0

8 n=n-1 0 0 0

9 signalB (s) 1 0 0

10 waitB (s) 0 0 0

11 n=n+1 0 1 0

12 If n==1 then signalB (delay) 0 1 1

13 signalB (s) 1 1 1

14 If n==0 then waitB (delay) 1 1 1



15 waitB (s) 0 1 1

16 n=n-1 0 0 1

17 signalB (s) 1 0 1

18 If n==0 then waitB (delay) 1 0 0

19 waitB (s) 0 0 0

20 n=n-1 0 -1 0

21 signalB (s) 1 -1 0

ProduceProduce waitB(s)waitB(s) InsetInset n = n + 1n = n + 1 If n = 1If n = 1 SignalB (delay)SignalB (delay) signalB (s)signalB (s)

Producer Consumer (Binary)Producer Consumer (Binary)

ProduceProduce waitB(s)waitB(s) InsetInset n n + 1n n + 1 If n 1If n 1 SignalB (delay)SignalB (delay) signalB (s)signalB (s)

Critical SectionCritical Section

First Time only



waitB(s)waitB(s) TakeTake

m = nm = n

consumeconsume WaitB (delay)WaitB (delay)signalB (s)signalB (s)waitB(delay)waitB(delay) If m = 0If m = 0

n = n - 1n = n - 1

Correct Solution

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Previously for this problem an infinite buffer was consider, now for realistic situations one has bounded buffer (BB) some thing like a circular buffer, therefore, it can be implemented in the following way:

Producer: Consumer:

Producer Consumer (Bounded Buffer)Producer Consumer (Bounded Buffer)

Loop Loop

produce if (in == out) {do nothing, remain here}

if ((in + 1)%n) == out {do nothing, remain here} take[out]

put[in] out = (out + 1)%n

in = (in + 1)%n consume

out in



Filled slots

[1] [2] [3] [4] [5] [6] [n]

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Concurrency - Aalborg Universitethomes.et.aau.dk/akh/2010/operatingsystem-10_files/LECT-4-CLASS.pdf · CHAPTER 5 Fall 2010 - DE5 5 Operating System Concerns What are the issues o/s