Last Updated :
26 Oct, 2018
Consider the following system:
Process Allocation Maximum Available
A B C A B C A B C
P0 0 1 2 2 2 3 x 4 2
P1 1 3 1 1 4 2
P2 2 0 1 2 4 2
P3 4 1 1 5 4 3
P4 1 1 1 1 3 3
Find the minimum value of x for deadlock free system?
(A) 0
(B) 1
(C) 3
(D) The system is always in unsafe state.
Answer: (A)
Explanation: According to Banker\’s algorithm,
Need matrix = maximum matrix – Allocation matrix
We need minimum A = 0, B = 1, and C = 1 to execute a process (i.e., P1). Process P1 P2 and P3 can easily executed With available resources. Process P1 execute first
Then available resources will be A = 2, B = 8, and C = 2. Now process P0 will execute and available resources A = 4, B = 10, and C = 5. Then P2, P3 and P4 will run in sequence.
Hence, x = 0 is correct, so option (A) is true.
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