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Effective access time cannot be lesser than the memory access time when there is no page fault. It should come out to be 101 ms instead of 91 ms. This mistake has occurred because you have not added the usual (when there is no page fault) access time to the overhead in case of page fault before multiplying it by 0.1. Also, comprehensive explanation otherwise. This was a great help.
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Tanks for video. I hade a question about "page fault overlaod" in 10:35 : shoulden´t it be 110? because "Momory access time" includes in page fault overload.
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Page fault overload process already includes the memory access time. So, if you put 1 instead of (1-p), you count the memory access time in page fault overload process twice
Great explanation. It made perfect sense. I’m slightly curious about the test question though. Wouldn’t it actually be .9*100 + .1*110 which is 101? Because 10% of the time we have page fault overhead which adds to the already existing access time right? You could also think about it as 100% of the time, memory access is 100ms, but then 10% of the time there will be an extra 10ms which means it’s 1*100 + .1*10 = 101
Thank you very much, well explanation. I have a question. Is there a way to predict or count the number of page faults? either statically or at run-time?
Your EAT calculation is wrong. The EAT is coming to be lower than memory access time after 10% page fault. The correct calculation would bb 0.9 * 100 + 0.1 *(100 + 10) = 101ms
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Why EAT is shorter than memory access time??? Why it seems the more page fault happens, the shorter EAT is?? I understand the formula, but are the numbers realistic?
10:54 I believe your numerical problem example is not the best example. If memory access time is 100ms, how can the page fault overhead be 10ms? Page fault overhead = memory access to page table + memory access to secondary storage + updating page table in memory + restarting the instruction so if memory access time is itself 100ms, then page fault overhead needs to be at least 4x the memory access. Your formula is right, but the example doesn't make any sense.
Thanks for the tutorial, however, i have an issue with something. Imagine if all pages were in hdd and no pages are in memory. Then the access time is: 0*100 + 1* 10 = 10 ms But if everything was in memory already: 1*100+0*10 = 100 How is accessing pages in memory slower than retrieving from hdd?
@@yxlxfxf You need to use the probability since not all pages will be in the secondary memory. it might be there not there. the answer should be greater or equal to 100 and less than 110
Bhai ye page fault waale question ka answer galat hai...... Answer 101 hoga.... ((0.9*100)+(0.1*(100+10)))=101... In this question 100 second will be memory access time + 10 seconds will be penalty for page fault....
Good question, well you have applied the formula incorrectly buddy. the second part is (probability of failure * the overhead time) and not the total time with overhead. the overhead time generally itself this the time taken to perform all the page fault activities itself :-)
@@SimpleSnippets dear, We will only get to know that demanded page is missing in the primary memory when we will access it..... So we should add access time to page fault time...... This is my last argument..... Please resolve it.......
By the videos logic, it would be slower to access pages if all pages were already in the frame: (1-0)*100+(0*10) = 100 > 91 I agree with Shubham. We still do memory access for the small portion of cases with page fault, but we also waste time retrieving them from hdd. So it should be ((1-p)* (t memory)) + p * (t memory + overhead)
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Effective access time cannot be lesser than the memory access time when there is no page fault. It should come out to be 101 ms instead of 91 ms. This mistake has occurred because you have not added the usual (when there is no page fault) access time to the overhead in case of page fault before multiplying it by 0.1.
Also, comprehensive explanation otherwise. This was a great help.
It was really clean how you explained the thrashing from what's needed to understand how the thrashing happens !
Your explanation was superrb.....
I liked it a lot....
Just resolve my query
Thanks for the response buddy. I guess you used the formula wrong :-)
hey! there.Thank you so much. I could understand every concept of OS a day before exam cuz of your videos.
Most welcome buddy. I would request you to please share the videos and channel with your friends too. Thats the biggest help and support for the channel 😇
Tanks for video.
I hade a question about "page fault overlaod" in 10:35 : shoulden´t it be 110?
because "Momory access time" includes in page fault overload.
Fantastic explanation. Thank you, sir!
Thanks so much you explain computing so well!
Thank you so much Thomas, really glad to hear this from you. Please do subscribe to this channel & share the content with your friends too. That will be the biggest help & support 😇
Rather than (1-P)* + P*, EAT should be = + P*.
This is because irrespective of page fault or not, memory access time is going to be there always.
Page fault overload process already includes the memory access time. So, if you put 1 instead of (1-p), you count the memory access time in page fault overload process twice
@@kemalkubilayylmaz4142The page fault overload in the example should be 110ms instead of 10ms which is used in the example calculation.
Thank you! This video was very helpful!
Glad it was helpful!
Hey man , thanks for such an awesome explanation !!!
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Clear cut and simple boss...
Thank you Deepam 🤟 I try to keep it short simple and sweet 😇
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excellent explanation
Amazing video/explanation man.
Glad you liked it!
Well explained, thank you
Thank you so much Vikas. Please share the video with your friends as well. That will be really helpful and the best support ✌
very nice explanation brother.
thank you mr indian sirmen.
This helped me a lot, thank you!
Most welcome ✌️
Great explanation. It made perfect sense.
I’m slightly curious about the test question though. Wouldn’t it actually be .9*100 + .1*110 which is 101? Because 10% of the time we have page fault overhead which adds to the already existing access time right? You could also think about it as 100% of the time, memory access is 100ms, but then 10% of the time there will be an extra 10ms which means it’s 1*100 + .1*10 = 101
Thank you very much, well explanation.
I have a question. Is there a way to predict or count the number of page faults? either statically or at run-time?
Thank you!!!
Your EAT calculation is wrong. The EAT is coming to be lower than memory access time after 10% page fault.
The correct calculation would bb 0.9 * 100 + 0.1 *(100 + 10) = 101ms
Yes, you're right
Nice explanation
The explanation was very good. Can you give the notes of page fault also in the way you have given for paging
Thank you so much, Sir, for this video! You are amazing! SUBSCRIBED :D
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Best ever
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Thanks for sharing info. Just one question.. how paging (VM) created in secondary memory?
in step 4 is the MMU that takes p3 3 to the main memory or OS ?
Everything is good,but please make a subject wise playlist
Like paging videos in Operating systems playlist
Is there a way I can download the diagram, I haven't found any other diagrams that come close to the level of detail you present.
Take a screenshot for now if that helps
Why EAT is shorter than memory access time??? Why it seems the more page fault happens, the shorter EAT is??
I understand the formula, but are the numbers realistic?
10:54 I believe your numerical problem example is not the best example. If memory access time is 100ms, how can the page fault overhead be 10ms?
Page fault overhead = memory access to page table + memory access to secondary storage + updating page table in memory + restarting the instruction
so if memory access time is itself 100ms, then page fault overhead needs to be at least 4x the memory access. Your formula is right, but the example doesn't make any sense.
Currently watching this in the bathroom during the exam😭
Why are you missing TLB memory ?
Thanks for the tutorial, however, i have an issue with something.
Imagine if all pages were in hdd and no pages are in memory.
Then the access time is: 0*100 + 1* 10 = 10 ms
But if everything was in memory already:
1*100+0*10 = 100
How is accessing pages in memory slower than retrieving from hdd?
I thought the same thing, I think the page fault access time should be memory access time + page fault overhead = 110 ms
@@yxlxfxf Exactly. That hasn't been addressed properly. You have to add the memory access time with the overhead.
@@yxlxfxf You need to use the probability since not all pages will be in the secondary memory. it might be there not there. the answer should be greater or equal to 100 and less than 110
@@julesbamwe6061 The average memory access time would be between 100 and 110, I was referring to the time for one single page fault
@@yxlxfxf exact... Then u are right in that case. Obvious 110.
No page found, finds secondary storage, bring into free frame, update page table, restart instruction
Love from Pakistan bro❤️
Can you cover all the concept in operating system concepts by abraham silberschatz
Bhai ye page fault waale question ka answer galat hai......
Answer 101 hoga....
((0.9*100)+(0.1*(100+10)))=101...
In this question 100 second will be memory access time + 10 seconds will be penalty for page fault....
Good question, well you have applied the formula incorrectly buddy. the second part is (probability of failure * the overhead time) and not the total time with overhead. the overhead time generally itself this the time taken to perform all the page fault activities itself :-)
@@SimpleSnippets dear,
We will only get to know that demanded page is missing in the primary memory when we will access it.....
So we should add access time to page fault time......
This is my last argument.....
Please resolve it.......
By the videos logic, it would be slower to access pages if all pages were already in the frame:
(1-0)*100+(0*10) = 100 > 91
I agree with Shubham.
We still do memory access for the small portion of cases with page fault, but we also waste time retrieving them from hdd. So it should be ((1-p)* (t memory)) + p * (t memory + overhead)
Semaphore daal do yaar