Clear and concise! Thanks for posting this video. For those asking for the Average Waiting Time (given in the video description): P1: 4 P2: 1 + 5 + 6 = 12 P3: 2 + 7 + 4 = 13 P4: 4 P5: 10 P6: 8
plz note down the arrival time of each process p1--- 6-2=4 p2---1+(9-4)+(17-11)=12 p3---2+(13-6)+(19-15)=13 p4---4 p5---5+(18-13)=10 p6---15-7=8 TA=(4+12+13+4+10+8)/6=8.5
@Mifta. Thankyou so much for sharing info with us. It is indeed a great job u r doing, coz trust me I searched for nearly an hour or more to find good tutorial for RR scheduling. No one could make me understand properly. But u wer jus fab.Thanks once again God bless you nd dnt forget to upload more videos.. 😜
In the Request Queue the 2nd P1 is added before P4 because the BT of P1 is still remaining... and then we add newly arrived processes. But applying the same logic and considering the case of 2nd P2 then P5 and the P3.. where P5 arrives at 6 , then why the request queue can't be ..., P2, P3 , P5, ...? Nice video by the way :D
Pease one more tutorial for calculating the Turn around time and waiting time. I know you write it in the description but i can't understand that. . . Thanks in advance.
I am not geting this part To check your answer, waiting time for P1 --- 4 P2 --- 1+5+6=12 P3 --- 2+7+4=13 P4 --- 4 P5 --- 10 P6 --- 8 ould you please do a video explaining the calculations for AWT. Happy new year ....
Stanley Nyadzayo plz note down the arrival time of each processp1--- 6-2=4 p2---1+(9-4)+(17-11)=12 p3---2+(13-6)+(19-15)=13 p4---4 p5---5+(18-13)=10 p6---15-7=8 TA=(4+12+13+4+10+8)/6=8.5
Mifta Sintaha can i execute the first remaining p1 after executing p6 or do i have to execute the remaining p1 after executing p3 as u showed in the tutorial?
hey mifta my ques is at t=2 when first time quantum is over, the process p0 will again join the queue and process p3 is also ariving at t=2. so which one will be inserted first and why?
what if two processes come at the same arrival time? should we check burst time?or we should server the process who came into first and put it in the queue first?
Clear and concise! Thanks for posting this video.
For those asking for the Average Waiting Time (given in the video description):
P1: 4
P2: 1 + 5 + 6 = 12
P3: 2 + 7 + 4 = 13
P4: 4
P5: 10
P6: 8
The Request Queue really helps to keep things in order.
Thank you! I didn't imagine that in just less than 5 minutes, I've learn Round Robin.
thank you so much...have exams tomorrow and this explanation was exactly what i needed..
I Love the way you explain it. Truly you should teach at University. Many Thanks for this amazing Tutorials. God bless.
plz note down the arrival time of each process
p1--- 6-2=4
p2---1+(9-4)+(17-11)=12
p3---2+(13-6)+(19-15)=13
p4---4
p5---5+(18-13)=10
p6---15-7=8
TA=(4+12+13+4+10+8)/6=8.5
p5 = 5? or 6
how do you get this numbers? im so confused right now
Its not arrival time, it's the waiting time of each process.
Your videos really helped me do my assignment :)
Thank you!
Thank you Mifta for these videos. Your channel is so far the best i could find on youtube.. subb!!
thanks alot i watch ur all scheduling video and u help me alot ... really thanks ..now i m prepared for my exam ..thank u.😃
@Mifta. Thankyou so much for sharing info with us. It is indeed a great job u r doing, coz trust me I searched for nearly an hour or more to find good tutorial for RR scheduling. No one could make me understand properly. But u wer jus fab.Thanks once again God bless you nd dnt forget to upload more videos.. 😜
Very nice Presentation, I have midterm exam on 6th and I have an Edge. THanks
Your voice is really pleasant to hear. it is really trippy . where are u from ??
thanks a lot for the these... helped me a lot in my OS exam.... keep up the good work!
In the Request Queue the 2nd P1 is added before P4 because the BT of P1 is still remaining... and then we add newly arrived processes.
But applying the same logic and considering the case of 2nd P2 then P5 and the P3..
where P5 arrives at 6 , then why the request queue can't be ..., P2, P3 , P5, ...?
Nice video by the way :D
Best RR teaching!
You explained this really well. Thanks
thanks, very helpfull. i"m from indo but i dont really understand when indonesian explain about this. thank you so much
Fast and clear, thanks a lot!!
Thankyou mam! Very helpful for Operating System exam! keep up the good work
thanks alot seriously. thank u keep on doing these kinda videos :D
Thank you Mifta, well explained.
your tutorials helped me alot :) thanks mam
TAT = CompletionTime - (SubmissionTime - ArrivalTime)*
* ST - AT is the Initial waiting time
Your channel was really helpful. Thank you so much!
it would be nice to see in the video how to calculate avg turn and waiting time. Thanks!)
Its in one of the other scheduling videos, I think its the SJF scheduling video.
You saved my life !! Thank you soooooooooo much !!
Great tutorial! So clearly explained.... Thank you!
Thank you very much ma'am Awesome and outstanding toturial.👍
u explain very well..!thanks alot for posting these videos..!can u please upload some videos on semaphores algorithms...!
can you do preemptive priority scheduling ?
your videos are nice easy to understand thank you
Great video lecture series! Thanks a lot....
Feeling great to know that you are a great fan of Naruto.. :)
what if my time quantum is 3, and the next to arrive is 4? how's that?
Amazing mifta easily understand ...... thanku so much
Very helpful presentation. Thank you so much. :)
Can I know the software you're using?? Thanks for the video.
Pease one more tutorial for calculating the Turn around time and waiting time.
I know you write it in the description but i can't understand that. . .
Thanks in advance.
lady, i think higher prirority is chosen first as due to process of ageing
you are an angel to me
mam u saved my lie in exam
becouse
exept few .... paper is so tough...........
I am not geting this part To check your answer, waiting time for
P1 --- 4
P2 --- 1+5+6=12
P3 --- 2+7+4=13
P4 --- 4
P5 --- 10
P6 --- 8
ould you please do a video explaining the calculations for AWT. Happy new year ....
WAITING TIME: SERVICE TIME - ARRIVAL TIME
P1 --- (0-0 [given] )+(6-2 [using gantt chart] )=4
P2 --- (2-1)+(9-4)+(17-11)=12
P3 --- (4-2)+(13-6)+(19-15)=13
P4 --- (8-4)=4
P5 --- (11-6)+(18-13)=10
P6 --- (15-7)=8
Stanley Nyadzayo plz note down the arrival time of each processp1--- 6-2=4
p2---1+(9-4)+(17-11)=12
p3---2+(13-6)+(19-15)=13
p4---4
p5---5+(18-13)=10
p6---15-7=8
TA=(4+12+13+4+10+8)/6=8.5
archisha tomar can u show me the turn around time?
Very helpful. thank you ma'am. please keep it up
can I ask something hope u answer asap :) what if p2 a.t is 2 and p3 a.t is 1 where should I do 1st ? thanks in advance
thanks !!!!!!!, one question, if time quantom isnt specified, should we assume it defaulted to 1 ? thanks
*****
i know im asking if the question doesnt specify it
chill it will specify
if its not specified then its not Round Robin
Thank you!! Helped me a lot.
please provide priority preemptive and priority non-preemptive algorithm video mam....
Mifta Sintaha can i execute the first remaining p1 after executing p6 or do i have to execute the remaining p1 after executing p3 as u showed in the tutorial?
i did try to understand but i don t get it why p1......p3 then we go back to p1 plz zxplain more
Thank you sister 🇮🇳 🇮🇳 🇮🇳 💯💯
Ukinnam Nag sanger ka agisuro! Haan nga kasta ti kuna ni sir jayson! Nagangdod ka nga bakit
hey mifta my ques is at t=2 when first time quantum is over, the process p0 will again join the queue and process p3 is also ariving at t=2. so which one will be inserted first and why?
it's round robin thus p3 comes first and then p0
10x dear i can't understand this thing at college but now i know how to fix it so 10x a lot
Thank you mam !! Best explanation !!
Thanks a lot Mifta...
What if along with Burst Time and Arrival Time; Priority is also mentioned.
Will that affect my way of solving the Sum.
explained well. thank you.
what if two processes come at the same arrival time?
should we check burst time?or we should server the process who came into first and put it in the queue first?
nicely explained thanks
if priority is also given whether in that condition quantum will fully executed??
thnx i have understood within no time thnx a lot
JUST PERFECT. THANKS.
Here the AWT or waiting time & TAT is not done but overall it was very nice
how would priority affect solving the problem?
Please upload the multilevel scheduling problem :) !!
you are so good.
thank you so much.
mam,thank you so much...God bless you!!!
nice explanation...............................
kindly please provide the students with the ending part that is avg. waiting time.
thank you....tht was really helpful
fab work i love it
It was very helpful ☺✌
what if you have 2 processes have the same arrival time which one do you queue first?
Thanks so much.
waiting time for process P5=12
what if there are 2 CPU's?
Thanks a lot! :)
would a dispatcher time for switching processors effect these results ? please someone help :(
Thanks!
heiii goood job.....please make the sound less pitchy.....thenks
i think waiting time for P2 is like 2+(9-4)+(17-11)=2+5+6=13 am i right?
No . ( 2-1)+(9-4)+(17-11)=1+5+6=12
thank you
plz also describe turn around time :) plz
P4 used only one unit (8 to 9).. TQ given was 2. So didn't we have one unit left there? Why didn't you use that?
Damn! This is so confusing !
+Adhiraj Mathur You do not need to consider the TQ if your Burst Time is less than TQ. :)
grt, u just crack it.
thank u very very much
mam do u have notes to upload for round robin
Thank you mam
CAN U GIVE A PROPER CODE FOR THIS
how p4= 4 ??
now i gt d drill!!!!! thanks alot
thank u
briliant
tysm maam !!!
need to calculate" waiting time and turn around time"
+Sushmita Roy Check the description box please :)
thanks :)
How to calculate the turnaround time..??
Suvodip Ganguly Waiting time + burst time
Soham Mondal thank u..
:P Mention Not
Soham Mondal it is spelt dont mention...:-P
Suvodip Ganguly *don't....kemon holo ? :D
thank youuuuuu
great!!!!!
Awesome
your given p5 waiting time was wrong
check again..its right