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Consider 3 processes P0 ,P1, P2 to be scheduled as per the SRTF algorithm. the process P0 is known to be scheduled first ad and when P0 is running 5 units of time, the process P2 has arrived. When the process P2 has run 2 units of time , the process P1 has arrived and completed running in 4 units of time. Then minimum burst time of P0 is............(in units).

Given answer is 12 ..

I think it should be 10 pls verify what should be correct answer..
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brother, i said, when it is Arrival time also equal,

( but i didn't mention in my comment, sorry for that )
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Yes, the answer will be 13.

p0 p2 p1
13 7 4

If we take p0 = 12 BT then After 5 Unit of time p0 = 7 And p2 = 7 same BT. Then Here According to the process id 

p0 will execute again Which is against the question. So P0 will be 13.

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ok got it. thanks.
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