At A.T 0 only p0 in ready queue. So, cpu completed p0 execution.now completetion time is 3 . At 3 only P1 in ready queue, i.e P2, p3, p4 are not in ready queue. Hope ur doubt clarified.Plz subscribe to the channel and if possible share with your friends. Thanks in advance.
Thanks for every video sir after completion of my exams I will share with all my contacts to subscribe... Now I completed 75% of syllabus because of you .. It is very easy to understand😉
You're Teaching is awsome sir. Belive me, You are helping a lot of students For better scorings in their examinations. Already subscribed your channel from 2 accounts. Thankyou For all your lectures.
Meeru cheppali anukuntunna content and explanation good sir. But meeru total english kaakundaa telugu lo kuda chepthey baaguntundhi sir. It my openion sir...
Already I upload a video on this in my Telugu channel. refer that , type computer Pantulu operating systems. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.
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.
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....
At A.T 0 only p0 in ready queue. So, cpu completed p0 execution.now completetion time is 3 . At 3 only P1 in ready queue, i.e P2, p3, p4 are not in ready queue. Hope ur doubt clarified Navya. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.
Hai, plz go through my videos once. Definitely you will like more videos. Thanks for your encouragement and support towards our channel. Share my sessions with your friends and subscribe to our channel.Keep on following my sessions. Thank you so much.
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
But At O Arrival time , P4 doesn't exist in the ready queue(main memory). Plz subscribe to the channel and if possible share with your friends. Thanks in advance...
Ready queue contains list of processes ready for execution. initially all programs will be stored in hard disk i.. job queue and at the time of execution os loads the programs from hard disk to main memory ie ready queue.All the best. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.
P4 burst time is less than all the process then why you have taken po process first
At A.T 0 only p0 in ready queue. So, cpu completed p0 execution.now completetion time is 3 . At 3 only P1 in ready queue, i.e P2, p3, p4 are not in ready queue. Hope ur doubt clarified.Plz subscribe to the channel and if possible share with your friends. Thanks in advance.
Nice video sir, understood very well
Glad to hear that ledu aggi petta. Plz subscribe to the channel and if possible share with your friends. Thanks
Thanks for every video sir after completion of my exams I will share with all my contacts to subscribe... Now I completed 75% of syllabus because of you .. It is very easy to understand😉
Many many thanks please share it
As you said we have to check shortest burst time then p4 has burst time 2 then you have taken p0
You're Teaching is awsome sir.
Belive me, You are helping a lot of students For better scorings in their examinations.
Already subscribed your channel from 2 accounts. Thankyou For all your lectures.
Welcome
Thanks a lot sir no more words to express my gratitude sir 🥺✨
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thank you teacher, greetings from Chile
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I have been looking someone like you .
you did it
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Tq so much sir your teaching way good keep going sir
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Sir tqsm
Ur teaching was best n we r getting better way than our cls lectures 😇tqsm
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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...
Already I upload a video on this in my Telugu channel. refer that , type computer Pantulu operating systems. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.
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.
Nice. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.
I just love your teaching sir!!!
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very Good explanation sir
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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....
At A.T 0 only p0 in ready queue. So, cpu completed p0 execution.now completetion time is 3 . At 3 only P1 in ready queue, i.e P2, p3, p4 are not in ready queue. Hope ur doubt clarified Navya.
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Very good Explain sir !!
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Informative
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Thank u sir....
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Sure sir.....thank u
thank you so much.. your better than my prof hahaha
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Thankyou sir
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P1 burst time is high compare to others then y taken sir
At first we have to take in at then ...2
I think you got logic...?
But p4 has least burst time than p1.then why did you consider p1 first?
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Thank You
Hai, plz go through my videos once. Definitely you will like more videos. Thanks
for your encouragement and support towards our channel. Share my sessions with your friends and subscribe to our channel.Keep on following my sessions. Thank you so much.
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
Sir?? Can a sjf preemptive work if we r given a data without arraival time???
yes, we can preempt its work Gayathri. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.
Good teacher sir...but p1 have highest B.T compare to p4 then y taken p1 first
But At O Arrival time , P4 doesn't exist in the ready queue(main memory). Plz subscribe to the channel and if possible share with your friends. Thanks in advance...
How can we know that the processes are in ready queue or not ? sir Plz reply i have exam tomorrow
Ready queue contains list of processes ready for execution. initially all programs will be stored in hard disk i.. job queue and at the time of execution os loads the programs from hard disk to main memory ie ready queue.All the best. Plz subscribe to the channel and if possible share with your friends. Thanks in advance.
Thank u a lot sir 😊😊😊
My question is in SJF non preemptive A.T is all 0 then Gantt chart is ?
Refer this video for srtf
ruclips.net/video/Tcuyr91Y_Ro/видео.html
Then both sjf and srtf become same
What if arrival time is non 0 then which will consider 1st
Refer to this video
ruclips.net/video/Tcuyr91Y_Ro/видео.html
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why have you taken p1 .....u said to write based on burst time right ....its burst time is 6 how is it possible
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Sir y did you take p1 that have a highest burst time
It's a text book problem only. Plz subscribe to the channel and if possible share with your friends. Thanks in advance..