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There will be more than one safe sequence (4 to be precise), try with banker's algorithm..

After P3 more than one option possible as P2,P1,P4 all can be satisfied so many combinations possible hence the answer. Hope it helps!

D) more than 2 safe sequence.

First P3 shall complete and then other 3 processes in any order.

safe sequences:

(P3 P1 P2 P4), (P3 P1 P4 P2)

(P3 P2 P4 P1), (P3 P2 P1 P4)

(P3 P4 P2 P1), (P3 P4 P1, P2)
by
(D)More than 2 safe sequences exist.

Explanation: After creating the Request and allocated table from the request allocated graph given above we can see R1, R2 and R3 are left with only 0,1 and 0 resources respectively. With this, we only can satisfy the need for P3. After P3 executes it releases 1 instance of R2 and one instance of R3 making the available resources of R1, R2, and R3 as 0,2 and 1 respectively. With this, we can satisfy the requirement of all the processes so we can schedule any one of them so there will be more than two safe sequences possible.
Yess more than 1 safe sequence possible.