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# NET 4

CBSE NET DECEMBER 2015 PAPER III
OPERATING SYSTEM QUESTION
Q. Consider a system with twelve magnetic tape drives and three processes p1,p2 and p3. Process p1 requires maximum ten tape drives, process p2 may need as many as four tape drives and p3 may need upto nine tape drives. Suppose that at time t1, process p1 is holding five tape drives, process p2 is holding two tape drives and process p3 is holding three tape drives. At time t1, system is in :
(1) Safe state
(2) Unsafe state
(4) Starvation state
Ans: (3)
Exaplanation:
 Processes Maximum needs Current needs P1 10 05 P2 04 02 P3 09 03
Total number of magnetic tapes = 12
Already allocated magnetic tapes = 10
So, number of free magnetic tapes = 12 – 10 =2
From the above table,
Process P2 requires 2 more tapes for completion and since two tapes are free, they are allocated to process P2.Then, number of free tape drives = 4.
Since process P1 is allocated 5 tape drives, and has a maximum of 10, P1 may then request 5 more tape drives. Since only 4 tape drives are available, so, process P1 must wait.
Similarly, process P3 may request an additional 6 tape drives and have to wait.
Waiting of P1 and P3 will resilt in a deadlock. So the system is in deadlocked state.