s) A system has four processes and five types of allocatable resources. The current allocation and maximum needs are as follows: Allocated Maximum Available Process A 2 1 0 2 2 4 2 2 3 3 3 2 x 2 3 Process B 3 1 1 0 2 3 3 6 1 2 Process C 2 1 0 2 1 3 2 3 3 1 Process D 1 1 0 1 0 1 2 3 2 1 What is the smallest value of x for which this is a safe state? Show all steps.

The smallest value of x is 5 which leads to a safe state.

Explanation:

Solution

Given that:

Process Available Maximum Request = Max-Available

A [2,1,0,2, 2] [4, 2,2, 3, 3] [2,1,2,1,1]

B [3,1, 1, 0,2] [3,3,6,1,2] [0,2,5,1,0]

C [2,1,0,2,1 ] [3,2,3,3,1] [1,1,3,1,0]

D [1, 1, 0, 1, 0 ] [1, 2, 3, 2,1 ] [0,1,3,1,1]

Available = 3,2, x, 2,3 ⇒ x has to be determined.

Now

consider x=1 then Available = 3,2,1,2,3

It can't satisfy A, B, C, D since the minimum value of x among those is 2

Consider x=2 then Available = 3,2,2,2,3

It can't satisfy B, C, D since the minimum value of x among those is 3

Thus

consider x=3 then Available = 3,2,3,2,3

It can't satisfy D since the minimum value of x among those is 5

Then

consider x=5 then Available = 3,2,5,2,3

It can satisfy A, B, C, D

Therefore, the minimum value of x is 5. So, that it leads to a safe state.