Operating system

Question

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..

Comments

brother, i said, when it is Arrival time also equal,

( but i didn't mention in my comment, sorry for that )

---

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.

---

ok got it. thanks.