Before starting it I solved it myself and my Gann chart is same as yours and even the answers but let suppose for P1 I took total time as 49 as it got completed after that time and subtracted 11 the total burst time as it is also the no. Of millisecond process executed and subtracted zero as it is arrival time. Igot the same answer by following this.
If they had arrived at the same time then P2 would have executed first. P1 arrived at 0 while P4 arrived at 2, meaning that P1 was the only process needing to run for those 2 seconds before P4 arrived and interrupted it. P2 arrived at 5 so after those 5 milliseconds (2 for P1, then 3 for P4) P2 would arrive and be given the CPU due to its priority of 0.
Thank you so much!! Idk if we can call this coincidence or what but you uploaded this video today and tomorrow I've got my O/S mid semester examination xD Thank you!! #gg explanation :3
@@vennila1423 P4 ARRİVAL TİME İS 2ms AND HAS HİGHER PRİORİTY THAN P1 HENCE P1 İS REMOVED AND P4 İS FED İNTO THE CPU FOR EXECUTİON TİLL THE ARRİVAL OF PROCESS P1 i.e, 5ms... DİD YOU GET İT??
to calculate turn around time of each process just find out the competion time and arrival time of each process and subtract arrival time from the competition time to get the result
Consider the set of 5 process whose arrival time and burst time are given below in milliseconds. Process ID Arrival time Burst time P1 0 4 P2 1 2 P3 2 1 P4 3 3 P5 4 2 if the CPU scheduling policy is Round Robin with time quantum = 2 unit. Calculate average waiting time and turn around time. Q.3 Consider a reference st Consider the set of 5 process whose arrival time and burst time are given below in milliseconds. Process ID Arrival time Burst time P1 0 4 P2 1 2 P3 2 1 P4 3 3 P5 4 2 if the CPU scheduling policy is Round Robin with time quantum = 2 unit. Calculate average waiting time and turn around time. plzz solve this
During the class it was going above my head🤣 but now I understood clearly 🥳 thank you so much
😂
same
Now I am able to solve these kind of questions.Thank you so much for your video.
Thank you is small word for your explanation sir...
Before starting it I solved it myself and my Gann chart is same as yours and even the answers but let suppose for P1 I took total time as 49 as it got completed after that time and subtracted 11 the total burst time as it is also the no. Of millisecond process executed and subtracted zero as it is arrival time. Igot the same answer by following this.
Please tell me is it also correct way??
As for me it is easy.
Please Reply. :(
Yes
To find the average waiting time,can you first find the turnaround time to make it easier?
Sir i just wanted to let you know that in question they have asked premptive but u have used the non-premptive approach
11:31 I think that the waiting time of P4 as 33-5-2=26 => 28.6 => 29
I know it's been a while, but here is my answer:
P4 had only 3ms executed. The first 2ms was P1.
Thank you so much sir I was literally stucked at this question and couldn't solve it.
What happens if 2 processes have equal priority, Dear Sir ?
fcfs algo
Then follow fcfs algorithm
bro dropping a banger after every tutorial😁
what happens when 2 processes has the same priority?, would the one that arrived in the CPU start first?
thnx a lot
5. What is the total no. of queues required to perform Priority Scheduling?
Thank you so much sir
after execution for specific time does the scheduler decrease the priority of the process
On my question um given size(msec) is it the same as burst time
how did p4 executed for 3 times? isnt it 2?
Thank you sir
'time till first put into cpu execution' > 'total waiting time'
Thank you so much!
Why the way calculating waiting time every people is always different
Thankyou sir
but in the qtn it says 0 has the highest priority, thought P2 would execute first
If they had arrived at the same time then P2 would have executed first. P1 arrived at 0 while P4 arrived at 2, meaning that P1 was the only process needing to run for those 2 seconds before P4 arrived and interrupted it. P2 arrived at 5 so after those 5 milliseconds (2 for P1, then 3 for P4) P2 would arrive and be given the CPU due to its priority of 0.
@@jaredmccune7443 thanks
Thank you sir ❤
Thank you so much!! Idk if we can call this coincidence or what but you uploaded this video today and tomorrow I've got my O/S mid semester examination xD Thank you!! #gg explanation :3
Very helpful
thank you so much 👏👏👏👏👏
What priority does 0 have here?
explain how you got the 2 and 5 seconds
in Gant chart
@@vennila1423 P4 ARRİVAL TİME İS 2ms AND HAS HİGHER PRİORİTY THAN P1 HENCE P1 İS REMOVED AND P4 İS FED İNTO THE CPU FOR EXECUTİON TİLL THE ARRİVAL OF PROCESS P1 i.e, 5ms...
DİD YOU GET İT??
sir please solve it with higher the number higher the priority
Waiting time formula is turn around time - burst time he is doing something different
how do I calculate its Turn Around Time ?, can u help me sir?
to calculate turn around time of each process just find out the competion time and arrival time of each process and subtract arrival time from the competition time to get the result
aapka solution ka ans glt nikl hai
kyunki p4 sirf 2 baar execute hu hai
Consider the set of 5 process whose arrival time and burst time are given below in milliseconds.
Process ID Arrival time Burst time
P1 0 4
P2 1 2
P3 2 1
P4 3 3
P5 4 2
if the CPU scheduling policy is Round Robin with time quantum = 2 unit. Calculate average
waiting time and turn around time.
Q.3 Consider a reference st Consider the set of 5 process whose arrival time and burst time are given below in milliseconds.
Process ID Arrival time Burst time
P1 0 4
P2 1 2
P3 2 1
P4 3 3
P5 4 2
if the CPU scheduling policy is Round Robin with time quantum = 2 unit. Calculate average
waiting time and turn around time.
plzz solve this
🤣🤣🤣
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