| Process | Max Need | current Usage |
| P3 | 8 | 3 |
| P4 | 5 | 1 |
| P5 | 8 | 2 |
| Available=4 |
|
|
1st we determine the Need for each process.
Max need = Need +Allocated
Need = Max Need – Allocated
| Process | Max Need | Current usage | Need = Max Need - current usage | |
| p3 | 8 | 3 | 5 |
|
| p4 | 5 | 1 | 4 |
|
| p5 | 8 | 2 | 6 |
|
Now our table will become such as
| Process | Current usage | Need | Available |
| P3 | 3 | 5 |
|
| P4 | 1 | 4 |
|
| P5 | 2 | 6 |
|
In 1ST ITERATION
The need of process P4 is
(Need P4) 4 ≤ 4 (Available)
| Process | Current usage | Need | Available |
| P3 | 3 | 5 | 5 |
| P4 | 1 | 4 |
|
| P5 | 2 | 6 |
|
Safe sequence <P4>
In 2nd iteration
(Need of P3) 5 ≤ 5(Available)
| Process | Current usage | Need | Available |
| P3 | 3 | 5 | 5 |
| P4 | 1 | 4 | 8 |
| P5 | 2 | 6 |
|
Safe sequence <P4,P3>
In 3rd Iteration
(Need of p5) 6 ≤ 8(Available)
| Process | Current usage | Need | Available |
| P3 | 3 | 5 | 5 |
| P4 | 1 | 4 | 8 |
| P5 | 2 | 6 | 10 |
Safe sequence <P4, P3, P5> Required Ans.
Q k " is poori kinaat main jitni jaldi Rab Raazi hota
hai " itni jaldi koi Raazi nahi hota .
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