December2015 Cs Q29
| Allocated | Maximum | Available | |
| Process A | 1 0 2 1 1 | 1 1 2 1 3 | 0 0 x 1 1 |
| Process B | 2 0 1 1 0 | 2 2 2 1 0 | |
| Process C | 1 1 0 1 0 | 2 1 3 1 0 | |
| Process D | 1 1 1 1 0 | 1 1 2 2 1 |
The smallest value of x for which the above system in safe state is .............
- Option : D
- Explanation : If Process A’s Maximum need is 1 1 2 1 2 instead of 1 1 2 1 3, then answer will be x=1 The needs matrix is as follows: 0 1 0 0 1 0 2 1 0 0 1 0 3 0 0 0 0 1 1 1 If x is 0, available vector will be 0 0 0 1 1, we have a deadlock immediately. If x is 1, available vector will be 0 0 1 1 1, now, process D can run to completion. When it is finished, the available vector is 1 1 2 2 1. Now A can run to complete, the available vector then becomes 2 1 4 3 2. Then C can run and finish, return the available vector as 3 2 4 4 2. Then B can run to complete. Safe sequence D A C B.
