Question

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.

Answer 1

Answer:

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.