Mrs. K. Nithyanandakumari

Assistant Professor of Computer Science

Bodinayakanur.

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OPERATING SYSTEMS

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DEADLOCKS

DEADLOCKS

 EXAMPLES:

You can't get a job without experience; you can't get experience without a job.

BACKGROUND:

The cause of deadlocks: Each process needing what another process has. This results from sharing resources such as memory, devices, links.

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Under normal operation, a resource allocations proceed like this::

• Request a resource (suspend until available if necessary ).
• Use the resource.
• Release the resource.

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• Traffic only in one direction.
• Each section of a bridge can be viewed as a resource.
• If a deadlock occurs, it can be resolved if one car backs up (preempt resources and rollback).
• Several cars may have to be backed up if a deadlock occurs.
• Starvation is possible.

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DEADLOCKS

Bridge Crossing Example

NECESSARY CONDITIONS

ALL of these four must happen simultaneously for a deadlock to occur:

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DEADLOCK CHARACTERISATION

Mutual exclusion

One or more than one resource must be held by a process in a non-sharable (exclusive) mode.

Hold and Wait

A process holds a resource while waiting for another resource.

No Preemption

There is only voluntary release of a resource - nobody else can make a process give up a resource.

Circular Wait

Process A waits for Process B waits for Process C .... waits for Process A.

A visual ( mathematical ) way to determine if a deadlock has, or may occur.

G = ( V, E ) The graph contains nodes and edges.

V Nodes consist of processes = { P1, P2, P3, ...} and resource types

{ R1, R2, ...}

E Edges are ( Pi, Rj ) or ( Ri, Pj )

An arrow from the process to resource indicates the process is requesting the resource. An arrow from resource to process shows an instance of the resource has been allocated to the process.

Process is a circle, resource type is square; dots represent number of instances of resource in type. Request points to square, assignment comes from dot.

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RESOURCE ALLOCATION GRAPH

P_i

R_j

P_i

R_j

P_i

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RESOURCE ALLOCATION GRAPH

Resource allocation graph

with a deadlock.

Resource allocation graph

with a cycle but no deadlock.

HOW TO HANDLE DEADLOCKS – GENERAL STRATEGIES

There are three methods:

Ignore Deadlocks:

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Ensure deadlock never occurs using either

Prevention Prevent any one of the 4 conditions from happening.

Avoidance Allow all deadlock conditions, but calculate cycles about to happen and stop dangerous operations..

Allow deadlock to happen. This requires using both:

Detection Know a deadlock has occurred.

Recovery Regain the resources.

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DEADLOCK STRATEGIES

Most Operating systems do this!!

 Do not allow one of the four conditions to occur.

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Mutual exclusion:

a) Automatically holds for printers and other non-sharables.

b) Shared entities (read only files) don't need mutual exclusion (and aren’t susceptible to deadlock.)

c) Prevention not possible, since some devices are intrinsically non-sharable.

Hold and wait:

a) Collect all resources before execution.

b) A particular resource can only be requested when no others are being held. A sequence of resources is always collected beginning with the same one.

c) Utilization is low, starvation possible.

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DEADLOCK PREVENTION

 Do not allow one of the four conditions to occur.

No preemption:

a) Release any resource already being held if the process can't get an additional resource.

b) Allow preemption - if a needed resource is held by another process, which is also waiting on some resource, steal it. Otherwise wait.

Circular wait:

a) Number resources and only request in ascending order.

b) EACH of these prevention techniques may cause a decrease in utilization and/or resources. For this reason, prevention isn't necessarily the best technique.

c) Prevention is generally the easiest to implement.

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DEADLOCK PREVENTION

If we have prior knowledge of how resources will be requested, it's possible to determine if we are entering an "unsafe" state.

Possible states are:

Deadlock No forward progress can be made.

Unsafe state A state that may allow deadlock.

Safe state A state is safe if a sequence of processes exist such that there are enough resources for the first to finish, and as each finishes and releases its resources there are enough for the next to finish.

The rule is simple: If a request allocation would cause an unsafe state, do not honor that request.

NOTE: All deadlocks are unsafe, but all unsafes are NOT deadlocks.

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DEADLOCK AVOIDANCE

NOTE: All deadlocks are unsafe, but all unsafes are NOT deadlocks.

7: Deadlocks

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SAFE

DEADLOCK

UNSAFE

Only with luck will processes avoid deadlock.

O.S. can avoid

deadlock.

DEADLOCK AVOIDANCE

Let's assume a very simple model: each process declares its maximum needs. In this case, algorithms exist that will ensure that no unsafe state is reached.

EXAMPLE:

There exists a total of 12 tape drives. The current state looks like this:

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In this example, < p1, p0, p2 > is a workable sequence.

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Suppose p2 requests and is given one more tape drive. What happens then?

There are multiple instances of the resource in these examples.

DEADLOCK AVOIDANCE

A method used to determine if a particular state is safe. It's safe if there exists a sequence of processes such that for all the processes, there’s a way to avoid deadlock:

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The algorithm uses these variables:

Need[I] – the remaining resource needs of each process.

Work - Temporary variable – how many of the resource are currently available.

Finish[I] – flag for each process showing we’ve analyzed that process or not.

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need <= available + allocated[0] + .. + allocated[I-1] <- Sign of success

Let work and finish be vectors of length m and n respectively.

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Safety Algorithm

DEADLOCK AVOIDANCE

1. Initialize work = available

Initialize finish[i] = false, for i = 1,2,3,..n

2. Find an i such that:

finish[i] == false and need[i] <= work

If no such i exists, go to step 4.

3. work = work + allocation[i]

finish[i] = true

goto step 2

4. if finish[i] == true for all i, then the system is in a safe state.

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Safety Algorithm

DEADLOCK AVOIDANCE

Do these examples:

Consider a system with: five processes, P0 🡪 P4, three resource types, A, B, C.

Type A has 10 instances, B has 5 instances, C has 7 instances.

At time T0 the following snapshot of the system is taken.

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Is the system

in a safe state?

Safety Algorithm

1

3

4

2

0

0

P4

1

1

0

1

1

2

P3

0

0

6

2

0

3

P2

0

2

0

0

0

2

P1

2

3

3

3

4

7

0

1

0

P0

C

B

A

C

B

A

C

B

A

🡪

Avail

🡨

🡪

Req

🡨

🡪

Alloc

🡨

Max Needs = allocated + can-be-requested

DEADLOCK AVOIDANCE

Do these examples:

Now try it again with only a slight change in the request by P1.

P1 requests one additional resource of type A, and two more of type C.

Request1 = (1,0,2).

Is Request1 < available?

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Produce the state chart as if the request is Granted and see if it’s safe.

(We’ve drawn the chart as if it’s granted.

DEADLOCK AVOIDANCE

Safety Algorithm

Can the request be granted?

Need an algorithm that determines if deadlock occurred.

Also need a means of recovering from that deadlock.

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SINGLE INSTANCE OF A RESOURCE TYPE

• Wait-for graph == remove the resources from the usual graph and collapse edges.
• An edge from p(j) to p(i) implies that p(j) is waiting for p(i) to release.

DEADLOCK DETECTION

SEVERAL INSTANCES OF A RESOURCE TYPE

Complexity is of order m * n * n.

We need to keep track of:

available - records how many resources of each type are available.

allocation - number of resources of type m allocated to process n.

request - number of resources of type m requested by process n.

Let work and finish be vectors of length m and n respectively.

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DEADLOCK DETECTION

1. Initialize work[] = available[]

For i = 1,2,...n, if allocation[i] != 0 then

finish[i] = false; otherwise, finish[i] = true;

2. Find an i such that:

finish[i] == false and request[i] <= work

If no such i exists, go to step 4.

3. work = work + allocation[i]

finish[i] = true

goto step 2

4. if finish[i] == false for some i, then the system is in deadlock state.

IF finish[i] == false, then process p[i] is deadlocked.

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DEADLOCK DETECTION

EXAMPLE

We have three resources, A, B, and C. A has 7 instances, B has 2 instances, and C has 6 instances. At this time, the allocation, etc. looks like this: 

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Is there a sequence that will allow deadlock to be avoided?

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Is there more than one sequence that will work?

DEADLOCK DETECTION

EXAMPLE

 Suppose the Request matrix is changed like this. In other words, the maximum amounts to be allocated are initially declared so that this request matrix results.

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USAGE OF THIS DETECTION ALGORITHM

Frequency of check depends on how often a deadlock occurs and how many processes will be affected.

Is there now a sequence that will allow deadlock to be avoided?

DEADLOCK DETECTION

So, the deadlock has occurred. Now, how do we get the resources back and gain forward progress?

PROCESS TERMINATION:

• Could delete all the processes in the deadlock -- this is expensive.
• Delete one at a time until deadlock is broken ( time consuming ).
• Select who to terminate based on priority, time executed, time to completion, needs for completion, or depth of rollback
• In general, it's easier to preempt the resource, than to terminate the process.

RESOURCE PREEMPTION:

• Select a victim - which process and which resource to preempt.
• Rollback to previously defined "safe" state.
• Prevent one process from always being the one preempted ( starvation ).

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DEADLOCK RECOVERY

COMBINED APPROACH TO DEADLOCK HANDLING:

• Type of resource may dictate best deadlock handling. Look at ease of implementation, and effect on performance.

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• In other words, there is no one best technique.

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• Cases include:

Preemption for memory,

Preallocation for swap space,

Avoidance for devices ( can extract Needs from process. )

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DEADLOCK RECOVERY

THANK YOU

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