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waiting time is not correct : we should follow some steps to avoid wrong calculation : > A.T= Arrival Time > B.T= Burst Time > C.T= Completion Time >> T.T = Turn around Time = C.T - A.T >> W.T = Waiting Time = T.T - B.T if we apply these steps on the example at the video we will get waiting time results as follows : P1 = ( 134 - 0 ) - 53 = 81 P2 = ( 37 - 0 ) - 17 = 20 P3 = ( 162 - 0 ) - 68 = 94 P4 = ( 121 - 0 ) - 24 = 97 avr. waiting time = (81 + 20 + 94 + 97) / 4 = 73 let's calculate it manually to make sure that these steps are correct . for example , P1 timeline : start stop wait_A resume stop wait_B resume finish 0----------20------waiting------77----------97-------waiting------121---------134 let' sum the periods of waiting time : wait_A = 77-20 = 57 wait_B = 121 - 97 = 24 >> total wait time for P1 = wait_A + wait_B = 57 + 24 = 81 >> as we calculated before
This is how I would prefer to calculate it, but the lectures' way of calculating is also correct and give the same answer. The lecturs' way of calculating requires just a bit more to keep track of in your head, which can result in errors when calculating complex scheduling schemes.
Suppose process P1 has a period of 45 units and a burst time of 30 units. If 97% of the total CPU resources are being utilized to run P1 and another process P2, calculate the period of process P2.(Assume that the burst for P2 is a third of P1).what this value? sir
Not easy to understand by junior...never use large numbers like 53,17,68 in BT...using small no.like 0,9,4,3 is enough for junior.Using of large number make junior student burdens ...from my opinion... and if it available like CT,TT,RT it will be complicated..
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When i have any query i use to firstly search your channel.....Thank you so much sir for your teaching
P3 waiting time is 134. I didn't get exactly. There is another waiting time value which takes P3 at 154.
No, waiting time, in essence, means the total time a process waits. At 154, P3 is not waiting. It's already running, since 134.
yes bro
Thank youso mich sir
I have solved my confusion by seeing your video
Thank You Sir, The video was very comprehensive
thanks so much sir for the great explanation
What about calculating context switches, do you have a tutorial for that? Please assist
very straight forward and clear thank you
Sir, can please explain what should the sequence in Gnatt chart if B.T is less than time quantum
helpful.
we are following sir
Why did they leave out the 162 for P3
waiting time is not correct :
we should follow some steps to avoid wrong calculation :
> A.T= Arrival Time
> B.T= Burst Time
> C.T= Completion Time
>> T.T = Turn around Time = C.T - A.T
>> W.T = Waiting Time = T.T - B.T
if we apply these steps on the example at the video we will get waiting time results as follows :
P1 = ( 134 - 0 ) - 53 = 81
P2 = ( 37 - 0 ) - 17 = 20
P3 = ( 162 - 0 ) - 68 = 94
P4 = ( 121 - 0 ) - 24 = 97
avr. waiting time = (81 + 20 + 94 + 97) / 4 = 73
let's calculate it manually to make sure that these steps are correct .
for example , P1 timeline :
start stop wait_A resume stop wait_B resume finish
0----------20------waiting------77----------97-------waiting------121---------134
let' sum the periods of waiting time :
wait_A = 77-20 = 57
wait_B = 121 - 97 = 24
>> total wait time for P1 = wait_A + wait_B = 57 + 24 = 81 >> as we calculated before
You just calculated the waiting time in a slightly different way. Both calculations are correct.
This is how I would prefer to calculate it, but the lectures' way of calculating is also correct and give the same answer. The lecturs' way of calculating requires just a bit more to keep track of in your head, which can result in errors when calculating complex scheduling schemes.
Suppose process P1 has a period of 45 units and a burst time of 30 units. If 97% of the total CPU resources are being utilized to run P1 and another process P2, calculate the period of process P2.(Assume that the burst for P2 is a third of P1).what this value? sir
best explanation
thank you sir for your help, i realy appreciate you.
Not easy to understand by junior...never use large numbers like 53,17,68 in BT...using small no.like 0,9,4,3 is enough for junior.Using of large number make junior student burdens ...from my opinion...
and if it available like CT,TT,RT it will be complicated..
U r right sir !
thank u king
In this problem how to calculate turn around time .
When time quantam = infinity, round robin becomes FCFS..
What will happen when time quantam = 0?
Sir claritiga kanipinchatladhu mi screen but your teaching is very well sir thank you
👍👍👍👍
at 154 P3 did not need to wait and not 134...but concept grasped like!
SIR PLZ CAN YOU EXPLAIN THE THEORY OF ALL THESE SCHEDULING
AV of p3 =154-60
thanks
didn't get that how to get waiting time here
ohh...................I got it at second watching the video thank you so much
The ans is 73 is it correct?
Yup
thanks so much sir for the great explanation