ME-OS test

Question

How can it be (d)??

Comments

how can u start giving resources with P3 if need is 011 and available is 230?
b and c cant be true

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available = ( 2 3 0) with this check whose need can be satisfied.

p1's need (0 2 0) can be satisfied and p3's need (0 1 1) can't be satisfied.

then p1 executes and releases it's resources (3 0 2 ) ===>current available = (5 3 2)

p3 or p4 need can be satisfied.==> p1--> p3 -->p0-->p4-->p2

option A.

Answer 1

Here we have to remember 2 things :

a)  A process P_i can be allocated resources maintaining the safety property iff  need(P_i)  <=  available for each type of resources (in this case we have to check for each of A , B and C )

b) After this process which satisfies the above property and is completed execution we set its flag == FALSE indicating that it will not come again unless other processes are services.So after execution of it , release of its resources allocated to it is nescessary.So available no of resources will obviously increase.So we do ,

Available = Available + Allocation(P_i) for each of A,B and C resource type.

Now let us check for option B) firstkeeping in mind the above mentioned points

Initially available =  [ 2   3    0]

Now need(P₃)    =   [ 0   1    1]

So we can see need(P₃)  for type C is 1 which is less than available resource of type C which is 0.

So here itself the criteria for selection of P₃ fails.

So P₃ cannot come first in safe sequence.

Hence both B and C are false.

Therefore the correct option should be A)

Comments

Thanks for confirming