GATE - 1998 | OS | A  computer has six tape drives,  with n  processes competing for

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

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