Deadlock MCQ Quiz in मराठी - Objective Question with Answer for Deadlock - मोफत PDF डाउनलोड करा

Necessary Conditions for Deadlock are as follows

• Mutual exclusion: At least one resource must be held in a non-sharable mode; that is, only one process at a time can use the resource. If another process requests that resource, the requesting process must be delayed until the resource has been released.
• Hold and wait: A process must be holding at least one resource and waiting to acquire additional resources that are currently being held by other processes.
• No preemption: Resources cannot be preempted; that is, a resource can be released only voluntarily by the process holding it, after that process has completed its task.
• Circular wait: A set {P0, P1, ..., Pn} of waiting processes must exist such that P0 is waiting for a resource held by P1, P1 is waiting for a resource held by P2, ..., Pn−1 is waiting for a resource held by Pn, and Pn is waiting for a resource held by P0.

Therefore pre-emptive is not required condition for a deadlock

Here mutex = 1

first, take Process 1, that is, P₁

P(mutex);

CS

V(mutex);

This code will run. mutex = 0 and P₁ will move into the critical section.

After this, try to enter the P₈ in the critical section, it will run the code :

V(mutex);

CS

V(mutex);

In this case, it will changes mutex into 1, and P₈ will enter into the critical section. After this, exit the P₈ from the critical section, it will change mutex to 2. Then again enter P₈. In this way, it increases the mutex value always.

In this way, if Mutex becomes 8 then all the 8 processes can simultaneously enter the critical section.

Available = {0, 3, 0, 2}

The given system is not in a safe state.

Let's consider the peak demand situation of resources (A, B, C)=(3,4,6).

To find the possibility of deadlock (not deadlock-free), we have to find at least one resource allocation which results in a deadlock.

There will be a deadlock in which each process wants a resource for completion of execution. 

The resource allocation which results in a deadlock is assigning each process the number of resources equal to one less than peak demand.

Therefore, process A gets 2 resources, B gets 3 resources and C gets 5 resources.

Total number of resources = 2+3+5=10

By having 1 more resource of the same type, the state will be deadlock free.

Therefore, the minimum value of 'm' that ensures that deadlock will never occur is 11.

The correct answer is option 1, option 2, and option 4.

Concept:

When a deadlock is found, deadlock recovery takes place. When a deadlock is found, our system stops operating, and when the stalemate is resolved, our system resumes operation.

As a result, after detecting a deadlock, a method/way must be required to recover that impasse in order to restart the system. The procedure is known as deadlock recovery.

• Deadlock recovery through preemption.
• Deadlock recovery through rollback.
• Deadlock recovery through killing processes.

Explanation:

Banker’s algorithm is a deadlock avoidance mechanism, not a recovery mechanism.

• Abort all deadlock processes,
• Abort one process at a time,
• Roll back return to some safe state, and restart process for that state are 3 different recovery mechanisms.

Hence the correct answer is option 1, option 2, and option 4.

The allocation and request matrix for the above resource allocation graph is:

Cycle P1 → R1 → P3 → R2 → P1 
Here, available resources are = {0, 0}

So, with this, we can satisfy the need for P2 or P4.

Suppose the need for P2 is satisfied, then P2 will release its resources.

Now avail become = 1 0

The need for P4 or P1 can be satisfied. Suppose P4 release the resources.

New avail = 1, 1

It can satisfy the need of both P1 and P3.

So, P1 need is satisfied, its resources are released.

New avail = 1 2

P3 needs will also be satisfied with this.

So, there is no deadlock in this even with a cycle in the resource-allocation graph.  

The correct answer is option 3

Data:

Total number of instances of resource Type 1= 5

Total number of instances of resource Type 2= 4

type 1 allocated = 1+1+2=4

type 2 allocated = 1+1+1=3

type 1 available =total - type 1 allocated =5-4=1

type21 available =total - type 2 allocated =4-3=1

Need =Max-Used

Available(Type1, Type 2) = (1, 1)

Explanation:

Need in the processes are

• (1,2)
• (2,1)
• (2,3)

and Available is (1,1)

So, now it can't fulfill any of the one need in the processes

Because of this, it is in an unsafe state.

Concept:

Deadlock can occur If any process gets < needed (requested) resource

Max resource per process to be in deadlock = max demand – 1 

For 6 process, max resource to be in deadlock = (7 – 1) + (5 – 1) + (3 – 1) + (8 - 1) + (4 - 1) + (6 - 1) =27 

For 6 process, min resource not to be in a deadlock: 27 + 1 =28

Tips and Tricks:

$\sum _i^n{P}_i-n+1=(7+5+3+8+4+6)-6+1=28$

where Pi is the max resource needed by the ith process and n is the total number of process

Data:

Remaining Need = Maximum need – Current need

Total resources = 12

Total resources allocated = 5 + 2 + 2 = 9

Available resources = 12 – 9 = 3

Calculation:

Available resource ≥ remaining need of p₁

Using 3 resources, p₁ gets executed first. It frees 2 resources.

Available resources = 3 + 2 = 5

Available resource ≥ remaining need of p₀

Using 5 resources, p₀ gets executed first. It frees 5 resources.

Available resources = 5 + 5 = 10

Available resource ≥ remaining need of p₂

Now, p₂ can be executed easily. Hence, (p₁, p₀, p₂) is a safe sequence.

Concepts:

Deadlock prevention can be done by eliminating any of the below condition:

1. Mutual Exclusion

2. Hold and wait

3. No preemption

4. Circular Wait.

In deadlock prevention, the request for resources is not always granted if the resulting state is safe.

Deadlock avoidance can be done with Banker’s Algorithm.

Banker’s Algorithm

Banker’s Algorithm is resource allocation and deadlock avoidance algorithm which test all the request made by processes for resources, it checks for the safe state, if after granting request system remains in the safe state it allows the request and if there is no safe state it doesn’t allow the request made by the process. Deadlock avoidance requires knowledge of resource requirements a priori

Comparison

Deadlock prevention is more restrictive than deadlock avoidance.

Therefore statement 1 is false