P4 burst time is less than all the process then why you have taken po process first

As you said we have to check shortest burst time then p4 has burst time 2 then you have taken p0

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Nice video sir, understood very well

Sir meru shortest burst time will be executed first Ani chepparu kadha Mari problem lo p0 having 3 and P1 having 6 kadha sir mari Gantt chart lo p0 tarvatha P1 ni arrive chesaru enti sir p0 having 3 but shortest job first is p4 that having burst time 2only p4ni p1tarvatha arrive chesaru Mari sjf antey processes which is having shortest burst time ani chepparu kadha pls answer my question sir....

To whoever who is confused as of to why, after P0, the p1 process is taken in consideration, the reason is as follows : "Firstly, at 0th second, only one process is available in the ready queue, which is p0. So, p0 will be executed first. Secondly, the time taken i.e Burst time for the p0 i.e the total time taken to complete the execution of p0 process is 3 seconds. So, in these 3 seconds of execution, look at the arrival times column. Which process has loaded into the ready queue during the 3 seconds? Only p1 process is there in the ready queue in the 3 seconds. That is why, p1 process was considered and not p4. Because, look at p4's arrival time. It is going to arrive into the ready queue at 8th second, but only 3 seconds have been passed. To make it simpler for you to understand what is going on, think about two scenarios happening simultaneously "In your left, there is ready queue. In this ready queue, the process are being loaded. In your right hand, the CPU is being assigned to the Process from the left hand, and being executed. During the time when the CPU executes the process, the processes continue to be loaded into the left hand i.e ready queue" I hope I have put it in more sensible words and in more simpler words for you to understand and I hope that I have cleared your doubt. Gimme a like if I did :> And aswell, thank you sir.

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very Good explanation sir

Thanks a lot sir no more words to express my gratitude sir 🥺✨

But p4 has least burst time than p1.then why did you consider p1 first?

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P1 burst time is high compare to others then y taken sir

Informative

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Tq so much sir your teaching way good keep going sir

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Meeru cheppali anukuntunna content and explanation good sir. But meeru total english kaakundaa telugu lo kuda chepthey baaguntundhi sir. It my openion sir...

I just love your teaching sir!!!

Very good Explain sir !!

Sir?? Can a sjf preemptive work if we r given a data without arraival time???

Thank u sir....

Thankyou sir

Thank you man, but i've one problem that i can't solve, maybe you can help me: We have to use Shortest Process Next with aging technique, estimate the execution time of T3 knowing the previous T0, T1, T2: (some examples) 1. T0=10ms T1=30us T2=0,015ms 2. T0=200us T1=300ns T2=0,015ms 3. T0=0,05ms T1=30ns T2=40us

Thank You

thank you so much.. your better than my prof hahaha

My question is in SJF non preemptive A.T is all 0 then Gantt chart is ?

What if arrival time is non 0 then which will consider 1st

Good teacher sir...but p1 have highest B.T compare to p4 then y taken p1 first

How can we know that the processes are in ready queue or not ? sir Plz reply i have exam tomorrow

Thank u a lot sir 😊😊😊

why have you taken p1 .....u said to write based on burst time right ....its burst time is 6 how is it possible

Sir y did you take p1 that have a highest burst time