Test: Deadlock- 1 - Computer Science Engineering (CSE) MCQ

Dead lock occurs when each of the 3 user processes hold one resource and make simultaneous demand for another. If there are 4 resources one of the 3 user processes will get the fourth instance of the resource and release one or both of the resource it is currently holding after using.

Deadlock will never occur for a single resource.

2 process can never lead to deadlock as the peak time demand of 6 (3 + 3) tape drives can be satisfied. But 3 processes can lead to a deadlock if each hold 2 drives and then demand one more.

Data:

Maximum resource requirement of Process A = 3

Maximum resource requirement of Process B = 4

Maximum resource requirement of Process C = 6

Concepts:

Deadlock can occur If any process gets available resource < demanded resource

Max resource for process A to be in deadlock = needed - 1 = 3 - 1 = 2

Max resource for process B to be in deadlock = needed - 1 = 4 - 1 = 3

Max resource for process C to be in deadlock = needed - 1 = 6 - 1 = 5

If one resource is added, any of the n processes can take it a finish its execution leaving behind the allocated resource and hence the system will be deadlock free.

Calculation:

The minimum value of m that ensures that deadlock will never occur = (2 + 3 + 5) + 1 = 11

∴ m = 11

At least one process will be holding 2 resources in case of a simultaneous demand from, all the processes. That process will release the 2 resources, thereby avoiding any possible deadlock.

Assume m = 5 and n = 10
That means, 5 processes are sharing 10 resources and worst case will be if everyone is demanding equal number of cases.
So, for deadlock to be there every process must be holding 2 resources and seeking 1 more resource. This will make total demand to 15, which is nothing but 10 + 5 (m + n).
However, as given maximum demand is always less than (m + n), so we can say there will never be a deadlock.

The system is not in deadlock state, if the following sequence of fulfillment of request of the processes is made.
P₂→ P_o→ P₁→ P₃

Dead lock can not occure since first P₃ will complete then P₂ then P₁ will complete.

Inconsistency: message received by P_j from P_i that is not shown as sent by P_i. Deadlock resolution: always require the abortion of one or more processes. Deadlock avoidance: requires the recognition of unsafe state.
Transit: message sent by P_i to P_j that does not belong to the set of messages received by P_j.

If three additional resources are allocated P3 then its need becomes zero. And the system will get seven resources back from P3. These resources can be allocated to either P1 or P2 to become the system in safe state.

The time complexity of the banker’s algorithm as a function of number n of processes and m of resources is O(n²xm).

<P₁, P₃, P₄, P₂, P₀> is the safe sequence.

To avoid deadlock, we must request the devices in ascending order of enumeration. This ensures that circular wait does not hold. The correct sequence is 1, 4, 12, i.e. R₂, R₃, R₁.

In (a) the processes P₁, P₂ and P₃ are in deadlock state. P₂ waiting on P₃. Which is held by P₃. P₁ is waiting on ft, which is held by P₂. While P₃ in waiting on R₂ which is held by P₁ and P₂. 
In (b) we can observe that P₄ can release instance o f R₂, which can be allocated to either P₁ or P₃, breaking the cycle. Deadlock does not occur in (b).