Zeal Testseries

Question

I’m getting 2 as the answer, but the correct one given by them is 5.

Somebody please confirm.

Answer 1

Given the available Resources(A,B,C,D) =  0 0 1 1

1. We can service the need of P3 which is 0 0 1 0, so once P3 is executed, it releases all the resources it was holding making total available as 1 1 3 2.
2. Next we can service the need of P1 which is 1 0 0 0, so once P1 is executed, it releases all the resources it was holding making total available as 3 2 4 3.
3. Next the need of P2 is 0 2 X 2, Now to make the system unsafe we need to have value of X > 4 as 4 is the current availability of Resource C. Hence the smallest value of X = 5.

Comments

@Sunnidhya Roy Why should I add  0 0 1 1 to 1 1 2 1?

After allocating to p3, I’m left with 0 0 0 1. So I should be adding  0 0 0 1 to 1 1 2 1. Making the total 1 1 2 2. Right?

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@DAWID15 I will try to explain using an analogy. Suppose we have 10 pens. You currently hold 5 of them. So Pens left to be used = 5. Now you realize you need another 3 pens so now you have 8 pens and pens left to be used = 2. After your work is done you give away all the pens so total comes back to 8+2=10.

Now try to map this to the above problem why after allocating 0 0 1 0 to P3 which already has 1 1 2 1 allocated to it, we are left with 0 0 0 1 available resources but P3 now has 1 1 3 1 resources allocated to itself and when P3 releases them new available resources = current available resources( 0 0 0 1) + what P3 already holds(1 1 3 1) = 1 1 3 2.

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@Sunnidhya Roy You are right. I messed up. I overlooked “ALLOCATION” Column as “MAX ALLOCATION” and that’s why I was making the mistake.

Anyways thanks for clarifying me :-)

Answer 2

Answer: 5

The need table suggests the future need. 

Now the available resources in terms of $(A,B,C,D) = 0,0,1,1$

We can first satisfy $P3$ (b/c it needs $0,0,1,0$). Once $P3$ is satisfied it gives all the resources that it is currently allocated with, i.e. $1,1,2,1$. Thus, new available = $0,0,1,1 + 1,1,2,1 = 1,1,3,2$

With new available = $1,1,3,2$ we can satisfy $P1$ (b/c it needs $1,0,0,0$). Once $P1$ is satisfied it gives all the resources that it is currently allocated with, i.e. $2,1,1,1$. Thus, new available = $1,1,3,2 + 2,1,1,1 = 3,2,4,3.$

Now we to satisfy P2, P2 needs $0,2,X,2$. We can see that our available resources in $C = 4.$ So, with $4$ we can safely satisfy $P2$’s needs, and thus the system will be safe.

But the question asks 

the smallest value of X for which the system will not be in safe state

This will happen only if X is such that it can’t be satisfied with our current available $C$ i.e $4$. Hence $X$ must be $5$.