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UGC-NET | NTA UGC NET 2019 June – II | Question 3

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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) 5 (B) 4 (C) 3 (D) 6

Answer: (A)

Explanation: Each process needs 2 drives. So for deadlock just give each process one drive. So total 6 processes can be given 1 drive each and can cause deadlock. So to break the deadlock just reduce 1 process. So maximum no. of processes for the system to be deadlock-free is 5.

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Last Updated : 02 Dec, 2021
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