ME test series os

Question

Consider 3 processes P₀, P₁ and P₂ to be scheduled as per the SRTF algorithm. The process ‘P₀’ is known to be scheduled first and when ‘P₀ has been running ‘6’ units of time, then the process ‘P₂’ has arrived. The process ‘P₂’ has run for ‘4’ unit of time,then the process ‘P₁’ has arrived and completed running in ‘5’ units of time. Then the minimum burst time of ‘P₀’ is _______ (in units).

Answer 1

Let the initial burst time of $P_0=x,P_2=y.$

After completing $6$ unit of time $P_0$ burst time $=x-6$

Process $P_2$ arrived and prempt process $P_0$ which illustrates that at this point of time remaining burst time of $P_0(x-6) >y. -(i) $

After completing $4$ unit of time $P_1$ arrived and preempts i.e same logic  $y-4>5-(ii)$

Using $(ii):$
$\Rightarrow$  $y-4>5$

$\Rightarrow$  $y-4\geq6$

Minimum  is asked so ,
$\Rightarrow$  $y=10$

Using eqn $(i):$

$\Rightarrow$  $(x-6)>10$

$\Rightarrow$  $(x-6)\geq11$

$\Rightarrow$  $\color{darkblue}{x=17}$

Comments

If I take x = 16 and y=9 after P2 completes 4 time unit, then P2 has 5 left and P1 comes with 5 time unit. SRTF will chose P1 because of lower id, isn't it. That was my whole premise which lead me to the answer of 16 rather than 17. Otherwise 17 is the answer and how you explained it.

---

Yes! this is absolutely correct when question says "if burst time is same for two processes then schedule the one with lower process-id" but since nothing is mentioned here , we should go for best possible case.

---

But the question also does not mentions otherwise too. Should this be done for whenever they do not mention anything in the question even for the simpler questions ?

---

I always follow what question mentions,if there is some information that is missing I prefer the real computation that is going on in computer science.
Here there is no such information regarding priority based on identifier or arrival therefore preferred according to algorithm.