Answer:
Answer: (b) As request(P1) < Available, we can grant request
(c) As request(P4) < Available, we can grant request
Explanation:
Need Matrix
Need [i, j] = Max [i, j] – Allocation [i, j]
A B C D
P0 2 2 1 1
P1 2 1 3 1
P2 0 2 1 3
P3 0 1 1 2
P4 2 2 3 3
Available = (3 3 2 1)
1. Need(P0) < Available so, P0 can take all resources
Available = (3 3 2 1) + (2 0 0 1) (Allocation of P0) = (5 3 2 2)
2. Need(P3)<Available so, P3 will go next
Available = (5 3 2 2) + (1 3 1 2) = (6 6 3 4)
Like wise next P4, P1, P2 will get resources.
So safe sequence is P0 , P3, P4, P1, P2
b.
Request from P1 is (1 1 0 0) and Available is (3 3 2 1)
As request(P1) < Available, we can grant request
c.
Request from P4 is (0 0 2 0) and Available is (3 3 2 1)
As request (P4) < Available, we can grant request
Question:
The question is incomplete. The snapshot of a system was not added. See below the remaining part of the question and the answers.
Allocation Max Available Available
ABCD ABCD ABCD
PO 2001 4212 3321
P1 31 21 5252
P2 2103 2316
P3 1312 1424
P4 1432 3665
Answer:
(a) safe sequence is P0 , P3, P4, P1, P2
(b) The request can be granted
(c) The request can be granted
Explanation:
a.
Need Matrix
Need [i, j] = Max [i, j] – Allocation [i, j]
A B C D
P0 2 2 1 1
P1 2 1 3 1
P2 0 2 1 3
P3 0 1 1 2
P4 2 2 3 3
Available = (3 3 2 1)
1. Need(P0) < Available so, P0 can take all resources
Available = (3 3 2 1) + (2 0 0 1) (Allocation of P0) = (5 3 2 2)
2. Need(P3)<Available so, P3 will go next
Available = (5 3 2 2) + (1 3 1 2) = (6 6 3 4)
Like wise next P4, P1, P2 will get resources.
So safe sequence is P0 , P3, P4, P1, P2
b.
Request from P1 is (1 1 0 0) and Available is (3 3 2 1)
As request(P1) < Available, we can grant request
c.
Request from P4 is (0 0 2 0) and Available is (3 3 2 1)
As request(P4) < Available, we can grant request
