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Aptitude | GATE CS 1998 | Question 32

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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


Answer: (B)

Explanation: Given tape drive = 6 and each process may need 2 drive.

When we give 1 drive to 1 process then total process will be 6 but in this case there will definitely deadlock occur because every process contain 1 drive and waiting for another drive which is hold by other process therefore when we reduce 1 process then system to be deadlock free.

Hence maximum value of n = 6 – 1 = 5.

Option (B) is correct.

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Last Updated : 28 Mar, 2020
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