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:

ProcessAvailableMaximumRequest = 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.

i mean, there is a lot of homeless people. but it is a very busy place and always in need for people who can be busy so it's not impossible to find one.

it's d. out with the planting and harvest of the crops.