###### Question 14212 – Syntax-tree-and-context-flow-graph

November 18, 2023###### Question 13879 – UGC NET JRF November 2020 Paper-2

November 18, 2023# Question 8559 – Operating-Systems

An operating system uses the Banker’s algorithm for deadlock avoidance when managing the allocation of three resource types X, Y, and Z to three processes P0, P1, and P2. The table given below presents the current system state. Here, the Allocation matrix shows the current number of resources of each type allocated to each process and the Max matrix shows the maximum number of resources of each type required by each process during its execution.

There are 3 units of type X, 2 units of type Y and 2 units of type Z still available. The system is currently in a safe state. Consider the following independent requests for additional resources in the current state:

REQ1: P0 requests 0 units of X, 0 units of Y and 2 units of Z REQ2: P1 requests 2 units of X, 0 units of Y and 0 units of Z

Which one of the following is TRUE?

Correct Answer: B

Question 48 Explanation:

Lets take 1st case,

After allowing Req 1,

Available: X=3, Y=2, Z=0

With this we can satisfy P1’s requirement. So available becomes: X=6, Y=4, Z=0.

Since Z is not available, neither P0’s nor P2’s requirement can be satisfied. So, it is an unsafe state.

Lets take 2nd case,

After allowing Req 2,

Available: X=1, Y=2, Z=2

With this we can satisfy any one of P1’s or P2’s requirement. Lets first satisfy P1’s requirement. So Available now becomes:

X=6, Y=4, Z=2

Now with the availability we can satisfy P2’s requirement. So Available now becomes,

X=8, Y=5, Z=3

With this availability P0 can also be satisfied. So, hence it is in safe state.

So from above two cases Req 1 cannot be permitted but Req 2 can be permitted.

After allowing Req 1,

Available: X=3, Y=2, Z=0

With this we can satisfy P1’s requirement. So available becomes: X=6, Y=4, Z=0.

Since Z is not available, neither P0’s nor P2’s requirement can be satisfied. So, it is an unsafe state.

Lets take 2nd case,

After allowing Req 2,

Available: X=1, Y=2, Z=2

With this we can satisfy any one of P1’s or P2’s requirement. Lets first satisfy P1’s requirement. So Available now becomes:

X=6, Y=4, Z=2

Now with the availability we can satisfy P2’s requirement. So Available now becomes,

X=8, Y=5, Z=3

With this availability P0 can also be satisfied. So, hence it is in safe state.

So from above two cases Req 1 cannot be permitted but Req 2 can be permitted.

Only REQ1 can be permitted.

Only REQ2 can be permitted.

Both REQ1 and REQ2 can be permitted.

Neither REQ1 nor REQ2 can be permitted.

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