A system has four processes and five allocatable resources. The current allocation and maximum needs are as follows:

            Allocated    Maximum           Available
Process A    1 0 2 1 1    1 1 2 1 3    0 0 x 1 1
Process B    2 0 1 1 0    2 2 2 1 0    
Process C    1 1 0 1 0    2 1 3 1 0    
Process D    1 1 1 1 0    1 1 2 2 1    
The smallest value of x for which the above system in safe state is __________.

(A) 1
(B) 3
(C) 2
(D) Not safe for any value of x.


Answer: (D)

Explanation:

            Allocated    Maximum           Available      Need 
Process A    1 0 2 1 1    1 1 2 1 3    0 0 x 1 1    0 1 0 0 2 
Process B    2 0 1 1 0    2 2 2 1 0                 0 2 1 0 0
Process C    1 1 0 1 0    2 1 3 1 0                 1 0 3 0 0
Process D    1 1 1 1 0    1 1 2 2 1                 0 0 1 1 1 
The smallest value of x for which the above system in safe state is __________.

For x = 1 process D will execute and free 1 1 2 2 1 instances. Now none of the other process will execute.
let x = 2 then process D will execute and free 1 1 3 2 1 instances. Now process C will execute and free 2 2 3 3 1 instances. With these free instances process B will execute, but process A will not execute because 5 resources needs 2 instances which will never be satisfied. That’s why system is not in safe state.
So, option (D) is correct.

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