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asked in Operating System by Active (4.4k points) | 158 views

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Looking at available resources, we only have 2 possible options i.e To satisfy the request of either P1 or P3. Lets consider both cases one by one ..

Case 1 : Initially request of P1 is satisfied....

When P1 release all resources, available resources will become (5,3,2)
After P1 we have 2 option i.e we can satisfy the request of P3 or P4..
Case 1.a : If we satisfy the request of P3 first, After P3 completes its execution, available resources will becomes (7,4,3).. its enough to satisfy request of remaining processes i.e P4, P2, P0 in any order.
So number of possible safe sequence will be
P1 -> P3 -> ( P4, P2, P0 ) i.e 3 !  =  6 possible safe sequences

Case 1.b : If after P1 if we decide to satisfy the request of P4, available resources after P4 completes its execution will be (5,3,4)..after P4 only possibility is P3, But as soon as P3 completes its execution we have enough resources we can satisfy the need of either P0 or P2, in any order.
So number of possible safe sequence will be
P1 -> P4 -> P3 -> ( P0 , P2) ie. 2! = 2 possible safe sequences

Case 2 : Initially request of P3 is satisfied....

Using similar logic you will get sequences as
P3 -> P1 -> (P4, P0, P2) i.e 3! = 6 possible safe sequences
P3 -> P4 -> P1 -> (P2, P0) i.e 2! = 2 possible safe sequences

So total number of safe sequences possible is 6+2+6+2 = 16

answered by Loyal (7.5k points)
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@stblue plz see this this how option C that is 3 just after 2 is possible can u explain????

@rajoramanoj  u can check the answer now, I have answered it based on what i know.

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