UGC NET Computer Science December 2019 | Question 78

Question 78

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Suppose a system has 12 magnetic tape drives and at time t₀. three processes are allotted tape drives out of their need as given below:

Process	Maximum Needs	Current Needs
P0	10	5
P1	4	2
P2	9	2

At time t₀, the system is in safe state. Which of the following is safe sequence so that deadlock is avoided?

1. 1. ( P₀, P₁, P₂ )
2. 2. ( P₁, P₀, P₂ )
3. 3. ( P₂, P₁, P₀ )
4. 4. ( P₀, P₂, P₁ )

Answer : 2. (P₁, P₀, P₂ )

Explanation

Out of Total 12 magnetic tape drives, 9 magnetic tape drives are allocated. So There are 3 remaining magnetic tape drives.

Process	Maximum Needs	Allocated resources	Remaining Needs
P0	10	5	5
P1	4	2	2
P2	9	2	7

With available 3 magnetic tape drives, process P₁ requirements of 2 magnetic tape drives can be fulfilled. So, after completion of process P₁, it releases total 4 magnetic tape drives.

Now, with available tap drives are 4 + 1 = 5,
Process P₀ requirements of 5 magnetic tape drives can only be fulfilled. So, after completion of process P₀, it releases total 10 magnetic tape drives.

Now, available tap drives are 10,
Process P₂ requirements of 9 magnetic tape drives can be fulfilled.
Process P₂ will completed wihtout deadlock.

The only safe sequence that avoide deadlock is (P1, P0, P2). So, that system is in safe state for given process needs.

So, option 2 is correct answer

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