07. Deadlock

8/12/2019 07. Deadlock

1/51

SIT222Operating SystemsSession 07. Deadlock

8/12/2019 07. Deadlock

2/51

Outline

Introducing deadlock

Deadlock detection and recovery Deadlock avoidance Deadlock prevention

2SIT222 Session 7

8/12/2019 07. Deadlock

3/51


What is deadlock?

Two or more processes are blocked waiting foradditional resources Those resources are held by other processes

that are also waiting, i.e., The processes are waiting for each other

No processes can continue/progress until theresource/s held by the other/s are released

3SIT222 Session 7

8/12/2019 07. Deadlock

4/51


We already identified a possible deadlock using semaphoreslast week:

P 0 P 1 wait ( A ); wait ( B);wait ( B); wait ( A );

signal ( A ); signal ( B);signal ( B) signal ( A );

Deadlock is possible in this example where: P 0 holds resource A and is waiting for resource B P 1 holds resource B and is waiting for resource A Both wait indefinitely

4SIT222 Session 7

8/12/2019 07. Deadlock

5/51


This occurs because the resources in the previousexample are non-preemptable It is not possible to forcibly remove the resources

from the process without causing an error in theprocess

E.g., semaphores protect critical regions,taking semaphores away would allow twoprocesses in the critical region resulting in raceconditions

Non-preemptable resources also include resourcessuch as the console, files, printer, keyboard, etc.

5SIT222 Session 7

8/12/2019 07. Deadlock

6/51


Note however that there are also preemptableresources

Can be safely taken away from a process andgiven to another We have already seen two preemptable resources:

CPU is switched between processes regularlywithout causing any problems

CPU state is saved and restored each switch

Memory using virtual memory and pagereplacement, physical memory can be sharedbetween processes even when demand exceedsphysical capacity

6SIT222 Session 7

8/12/2019 07. Deadlock

7/51


There are four conditions that are required for deadlock1. Mutual exclusion condition Each resource is either

currently assigned to exactly one process or isavailable.2. Hold and wait condition Processes currently holding

resource that were granted earlier can request newresources.

3. No preemption condition Resources previouslygranted cannot be forcibly taken away from a process.They must be explicitly released by the process holdingthem.

4. Circular wait condition There must be a circular chainof two or more processes, each of which is waiting fora resource held by the next member of the chain.

7SIT222 Session 7Ref: Tanenbaum, A.S., Modern Operating Systems 3/e, Pearson Education, 2008: Section 6.2.1, pp438.

8/12/2019 07. Deadlock

8/51


Example problem: Dining Philosophers A table is laid out for n philosophers to dine

There are n plates of spaghetti There are n forks to eat the spaghetti Two forks are required to eat spaghetti Philosopher alternates between thinking and eating

Philosopher considers philosophical problems(thinking)

Philosopher picks up the left fork, then the right fork Philosopher eats some spaghetti (eating) Philosopher drops both forks Repeat

8SIT222 Session 7

8/12/2019 07. Deadlock

9/51


Here we see places forfive philosophers

Deadlock conditions: Mutual exclusion

a fork can only beheld by onephilosopherotherwise it mustbe available

9SIT222 Session 7

8/12/2019 07. Deadlock

10/51



Deadlock conditions: Hold and wait a

philosopher maypick up one forkthen try to pick upanother

10SIT222 Session 7

8/12/2019 07. Deadlock

11/51



Deadlock conditions: No preemption

there is no processdefined to remove afork from aphilosopher

11SIT222 Session 7

8/12/2019 07. Deadlock

12/51



Deadlock conditions: Circular wait if

each philosopherpicks up the leftfork, no philosophercan get a secondfork to start eating

12SIT222 Session 7

8/12/2019 07. Deadlock

13/51


Solution 1: Allow only four

(n-1) philosophersto sit at the table

At least onephilosopher willbe able to pickup second fork,eat, and returnto thinking

13SIT222 Session 7

8/12/2019 07. Deadlock

14/51


Solution 2: Allow a philosopher

to pick up only twoforks at a time

Requires asemaphore orsimilarmechanism toprotect criticalregion

14SIT222 Session 7

8/12/2019 07. Deadlock

15/51


Solution 3: Use an asymmetric

solution, e.g., Philosophers are

numbered Odds pick up left

fork first Evens pick upright fork first

15SIT222 Session 7

8/12/2019 07. Deadlock

16/51


There are four ways to deal with deadlock:1. Do nothing allows deadlock to occur and

manually recover2. Detection and recovery deadlocks occur with

automatic actions to recover3. Avoidance manage allocation of resources to

avoid deadlock4. Prevention eliminate one of the fourconditions for deadlock

16SIT222 Session 7

8/12/2019 07. Deadlock

17/51


The first solution initially seems quite strange: Do nothing allow deadlocks to occur andmanually recover

This can be quite valid however, e.g., consider thefollowing scenario

Deadlock occurs only once a year

System crashes on a daily basis What is the cost of introducing other deadlocksolutions instead of handling deadlock asanother system crash?

17SIT222 Session 7

8/12/2019 07. Deadlock

18/51

Deadlock detection and recovery

Deadlock detection is possible by producing aresource graph:

Boxes = resources Circles = processes

18SIT222 Session 7

8/12/2019 07. Deadlock

19/51


Deadlock detection is possible by producing aresource graph:

(a) Holding a resource, (b) Requesting a resource,and (c) deadlock (circular wait)

19SIT222 Session 7

8/12/2019 07. Deadlock

20/51


Deadlocks are apparent wherever there is a cycle inthe graph, e.g.,

20SIT222 Session 7

8/12/2019 07. Deadlock

21/51


Three approaches to recovery: Preemption (for preemptable resources)

Take resource/s from one process and give it toanother, which will complete and free resources

RollbackTake checkpoints/snapshots of processesregularly and rewind a process in the deadlock,

freeing unused resources at the last snapshot Process terminationTerminate one or more processes to free theirresources (preferably one that can be re-run!)

21SIT222 Session 7

8/12/2019 07. Deadlock

22/51


Deadlock detection and recovery has problemshowever

What is the cost of the chosen recovery option/s? e.g., lost computation time when terminating aprocess

In undertaking recovery, is it possible deadlockmay occur again

e.g., terminating one process allows

computation to continue for a short whilebefore deadlock occurs a second/third/etc. time If the recovery requires intervention by a person,

who will do this? Will someone need to behired/paid to do this?

22SIT222 Session 7

8/12/2019 07. Deadlock

23/51

Deadlock avoidance

It is possible to avoid deadlock by choosingwhether to allocate resources to a process or toblock their request, suspending the process

Requires the following information Number of resources available Number of resources allocated

Number of resources required (maximumnumber required)

23SIT222 Session 7

8/12/2019 07. Deadlock

24/51

Deadlock avoidance

Definitions: Safe state: Where there exists a sequence of

execution of processes sharing resources suchthat deadlock will not occur

Unsafe state: Where there does not exist such asequence and deadlock will occur if processes

Do not free the resources they have Require their maximum resources tocomplete

24SIT222 Session 7

8/12/2019 07. Deadlock

25/51

Deadlock avoidance

The Bankers Algorithm is used to determine whichrequests for resources can be fulfilled Begin with:

Resources already allocated to processes(Allocated matrix)

Maximum resources required by processes(Maximum matrix)

The number of resources currently available(Available vector)

Example

25SIT222 Session 7

8/12/2019 07. Deadlock

26/51

Deadlock avoidance

Allocation Max Available A B C A B C A B C

P 0 0 1 0 7 5 3 3 3 2P 1 2 0 0 3 2 2P 2 3 0 2 9 0 2P 3 2 1 1 2 2 2

P 4 0 0 2 4 3 3 Calculate the resources required (matrix need) for

each process to complete (= Max Allocation)

26SIT222 Session 7

8/12/2019 07. Deadlock

27/51

Deadlock avoidance

Allocation Max Need Available A B C A B C A B C A B C

P 0 0 1 0 7 5 3 7 4 3 3 3 2P 1 2 0 0 3 2 2P 2 3 0 2 9 0 2P 3 2 1 1 2 2 2

P 4 0 0 2 4 3 3 Example:

P 0 : Max (7 5 3) Allocation (0 1 0) = Need (7 4 3)

27SIT222 Session 7

8/12/2019 07. Deadlock

28/51

Deadlock avoidance


P 0 0 1 0 7 5 3 7 4 3 3 3 2P 1 2 0 0 3 2 2 1 2 2P 2 3 0 2 9 0 2P 3 2 1 1 2 2 2

P 4 0 0 2 4 3 3 Example:


28SIT222 Session 7

8/12/2019 07. Deadlock

29/51

Deadlock avoidance


P 0 0 1 0 7 5 3 7 4 3 3 3 2P 1 2 0 0 3 2 2 1 2 2P 2 3 0 2 9 0 2 6 0 0P 3 2 1 1 2 2 2

P 4 0 0 2 4 3 3 Example:


29SIT222 Session 7

8/12/2019 07. Deadlock

30/51

Deadlock avoidance


P 0 0 1 0 7 5 3 7 4 3 3 3 2P 1 2 0 0 3 2 2 1 2 2P 2 3 0 2 9 0 2 6 0 0P 3 2 1 1 2 2 2 0 1 1

P 4 0 0 2 4 3 3 Example:


30SIT222 Session 7

8/12/2019 07. Deadlock

31/51

Deadlock avoidance


P 0 0 1 0 7 5 3 7 4 3 3 3 2P 1 2 0 0 3 2 2 1 2 2P 2 3 0 2 9 0 2 6 0 0P 3 2 1 1 2 2 2 0 1 1

P 4 0 0 2 4 3 3 4 3 1 Example:


31SIT222 Session 7

8/12/2019 07. Deadlock

32/51

Deadlock avoidance

Allocation Need Available A B C A B C A B C

P 0 0 1 0 7 4 3 3 3 2P 1 2 0 0 1 2 2P 2 3 0 2 6 0 0P 3 2 1 1 0 1 1

P 4 0 0 2 4 3 1 Now identify processes which can be allocated their

remaining resources and be allowed to complete

32SIT222 Session 7

8/12/2019 07. Deadlock

33/51

Deadlock avoidance


P 0 0 1 0 7 4 3 3 3 2P 1 2 0 0 1 2 2P 2 3 0 2 6 0 0P 3 2 1 1 0 1 1

P 4 0 0 2 4 3 1 Either of these two could (need < available),

returning their (allocated) resources to the pool

33SIT222 Session 7

Need (1 2 2) is lessthan available (3 3 2)

Need (0 1 1) is lessthan available (3 3 2)

8/12/2019 07. Deadlock

34/51

8/12/2019 07. Deadlock

35/51

Deadlock avoidance


P 0 0 1 0 7 4 3 5 3 2P 1 0 0 0 0 0 0P 2 3 0 2 6 0 0P 3 2 1 1 0 1 1

P 4 0 0 2 4 3 1 Repeat until all processes can complete

Sequence: P 1

35SIT222 Session 7

8/12/2019 07. Deadlock

36/51

Deadlock avoidance


P 0 0 1 0 7 4 3 7 4 3P 1 0 0 0 0 0 0P 2 3 0 2 6 0 0P 3 0 0 0 0 0 0


Sequence: P 1 P 3

36SIT222 Session 7

8/12/2019 07. Deadlock

37/51

Deadlock avoidance


P 0 0 0 0 0 0 0 7 5 3P 1 0 0 0 0 0 0P 2 3 0 2 6 0 0P 3 0 0 0 0 0 0


Sequence: P 1 P 3 P 0

37SIT222 Session 7

8/12/2019 07. Deadlock

38/51

Deadlock avoidance


P 0 0 0 0 0 0 0 10 5 5P 1 0 0 0 0 0 0P 2 0 0 0 0 0 0P 3 0 0 0 0 0 0


Sequence: P 1 P 3 P 0 P 2

38SIT222 Session 7

8/12/2019 07. Deadlock

39/51

8/12/2019 07. Deadlock

40/51

Deadlock avoidance

By using the Bankers Algorithm, we determinedthat the system is in a safe state because there is asequence in which processes can complete withoutdeadlock: P 1 P 3 P 0 P 2 P 4 P 1 can complete, freeing its resources P 2 can then complete, freeing its resources There are now enough resources for P 4 to

complete, i.e., the system is in a safe state!

40SIT222 Session 7

8/12/2019 07. Deadlock

41/51

Deadlock avoidance

The fact that other sequences are possible does notmatter, there is at least one

The system does not need to execute according toany determined sequence Instead, when a request for a resource comes in

Assume the request is granted Use the Bankers Algorithm to determine if the

system is in a safe state If yes, grant the resource If no, block the process until more resources

are available

41SIT222 Session 7

8/12/2019 07. Deadlock

42/51

Deadlock avoidance


P 0 0 1 0 7 4 3 3 3 2P 1 2 0 0 1 2 2P 2 3 0 2 6 0 0P 3 2 1 1 0 1 1

P 4 0 0 2 4 3 1 Consider what happens if a request arrives from P 0

for three type B resources

42SIT222 Session 7

Subtract three type Bresources from Needand add to Allocation

8/12/2019 07. Deadlock

43/51

Deadlock avoidance


P 0 0 4 0 7 1 3 3 0 2P 1 2 0 0 1 2 2P 2 3 0 2 6 0 0P 3 2 1 1 0 1 1

P 4 0 0 2 4 3 1 Assume the request is granted

43SIT222 Session 7

8/12/2019 07. Deadlock

44/51

Deadlock avoidance


P 0 0 4 0 7 1 3 3 0 2P 1 2 0 0 1 2 2P 2 3 0 2 6 0 0P 3 2 1 1 0 1 1

P 4 0 0 2 4 3 1 Clearly, the system is in an unsafe state, no

process can now complete its execution

44SIT222 Session 7

8/12/2019 07. Deadlock

45/51

Deadlock avoidance

Unfortunately there is a major problem toovercome to successfully apply avoidance: How do you know how many resources a

process will need to complete its work?i.e., how to determine the Maximum matrix?

45SIT222 Session 7

8/12/2019 07. Deadlock

46/51

Deadlock prevention

Deadlock prevention is somewhat straightforward Recall the conditions for deadlock:

1. Mutual exclusion condition Each resource is eithercurrently assigned to exactly one process or is available.

2. Hold and wait condition Processes currently holdingresource that were granted earlier can request newresources.

3. No preemption condition Resources previously grantedcannot be forcibly taken away from a process. They mustbe explicitly released by the process holding them.

4. Circular wait condition There must be a circular chain oftwo or more processes, each of which is waiting for aresource held by the next member of the chain.

All you have to do is eliminate one!

46SIT222 Session 7

8/12/2019 07. Deadlock

47/51

Deadlock prevention

1. Mutual exclusion condition Each resource iseither currently assigned to exactly one process

or is available.

For preemptable resources, this is already the case For non-preemptable resources, there can often be

other solutions, e.g., virtualize the resource: Two processes printing will cause corrupt output

Spool jobs to a daemon which then prints jobsin sequence to the printer

Not always possible!!!

47SIT222 Session 7

8/12/2019 07. Deadlock

48/51

8/12/2019 07. Deadlock

49/51

Deadlock prevention

3. No preemption condition Resourcespreviously granted cannot be forcibly takenaway from a process. They must be explicitlyreleased by the process holding them.

Not a problem for preemptable resources theyare just preempted!

For non-preemptable resources, again virtualizationmay provide a solution Remember, not always possible!

49SIT222 Session 7

8/12/2019 07. Deadlock

50/51

Deadlock prevention

4. Circular wait condition There must be a circular chainof two or more processes, each of which is waiting fora resource held by the next member of the chain.

Only allow a process to hold one resource at any one time Impose an ordering on resources, i.e., number every

resource, two alternatives: A process can only request resources in order of the

sequence More generally, a process holding one resource can only

request resources appearing later in the sequence

50SIT222 Session 7

8/12/2019 07. Deadlock

51/51

Summary

Introducing deadlock Deadlock detection and recovery Deadlock avoidance Deadlock prevention

Documents

07. Deadlock