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IRISSAK [1]
3 years ago
8

A. illustrate that the system is in a safe state by demonstrating an order in which the processes may complete.

Business
2 answers:
Rashid [163]3 years ago
8 0

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

Nuetrik [128]3 years ago
6 0

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

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