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.
