GATE - 1998 | OS | A computer has six tape drives, with n processes competing for
Posted on 28-02-2022
GATE - 1998 [Operating System]
Question:
A computer has six tape drives, with n processes competing for them. Each process may need two drives. What is the maximum value of n for the system to be deadlock free?
|
A |
6 |
|
B |
5 |
|
C |
4 |
|
D |
3 |
Solution:
Option (B) is Correct.
Each process needs 2 drives. So for deadlock just give each process one drive.
So total 6 process can be given 1 drive each and can cause deadlock.
So to break deadlock just reduce 1 process.
So maximum no. of process for the system to be deadlock free is 5.
Thank You