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bearhunter [10]
3 years ago
7

Illustrate that the system is in a safe state by demonstrating an order in which the threads may complete.If a request from thre

ad T4 arrives for (2, 2, 2, 4), can the request be granted immediately?If a request from thread T2 arrives for (0, 1, 1, 0), can the request be granted immediately?If a request from thread T3 arrives for (2, 2, 1, 2), can the request be granted immediately?
Computers and Technology
1 answer:
Kamila [148]3 years ago
5 0

Answer:

a. safe sequence is T2 , T3, T0, T1, T4.

b. As request(T4) = Available, so the request can be granted immediately

c. As request(T2) < Available, so the request can be granted immediately

d. As request(T3) < Available, so the request can be granted immediately.

Explanation:

It will require matrix

[i, j] = Max [i, j] – Allocation [i, j]

A B C D

T0 3 3 3 2

T1 2 1 3 0

T2 0 1 2 0

T3 2 2 2 2

T4 3 4 5 4

Available = (2 2 2 4)

1. Need(T2) < Available so, T2 can take all resources

Available = (2 2 2 4) + (2 4 1 3) (Allocation of T2) = (4 6 3 7)

2. Need(T3)<Available so, T3 will go next

Available = (4 6 3 7) + (4 1 1 0) = (8 7 4 7)

Like wise next T0, T1, T4 will get resources.

So safe sequence is T2 , T3, T0, T1, T4.

(Note, there may be more than one safe sequence).

Solution b.

Request from T4 is (2 2 2 4) and Available is (2 2 2 4)

As request(T4) = Available, so the request can be granted immediately.

Solution c.

Request from T2 is (0 1 1 0) and Available is (2 2 2 4)

As request(T2) < Available, so the request can be granted immediately.

Solution d.

Request from T3 is (2 2 1 2) and Available is (2 2 2 4)

As request(T3) < Available, so the request can be granted immediately.

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