ISRO CSE 2020 | Question: 31

Question

Three CPU-bound tasks, with execution times of $15,12$ and $5$ time units respectively arrive at times $0,t$ and $8$, respectively. If the operating system implements a shortest remaining time first scheduling algorithm, what should be the value of $t$ to have $4$ context switches? Ignore the context switches at time $0$ and at the end.

• A. $0<t<3$
• B. $t=0$
• C. $t<=3$
• D. $3<t<8$

Comments

no at t=3 there are 3 context switching

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@Rutuja Desai Because at time t = 3, P1 and P2 have equal remaining times (12), so P1 continues (no switch).
We need a situation where P2 preempts P1

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Not a good question is does not mention how to break tie

Answer 1

P1 comes at 0 and executes till t now look at the burst times for the process p1 and p2  so for a context switch to take place burst time for p2 has to be less than p1 12<15-t , t<3 so t value can be anything between 0 to 3 option A is the correct answer

Comments

At    t=0 , there there are only 3 context switches

So only at  t=2. Is correct answer.

Which is option (a) 0 < t<3

Answer 2

Also fails at t = 0. at t=0,context switches=3 Option B,C eliminated.

 

But at t = 2, we get 4 context switches. So, Option A

Comments

But we are getting 1 context switch for t=1.