Second Chance Page Replacement Algorithm in OS Solved Problem
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- čas přidán 18. 04. 2019
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--------------------------------------------------------------------------------------------- Second Chance Page replacement algorithm is a modified form of the FIFO page replacement algorithm, fares relatively better than FIFO at little cost for the improvement.
It works by looking at the front of the queue as FIFO does, but instead of immediately paging out that page, it checks to see if its referenced bit is set.
If it is not set, the page is swapped out. Otherwise, the referenced bit is cleared.
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Bro can you provide NOT RECENTLY USED(NRU) Page replacement algorithm with example
@@kamasanisaikumar2607 I surely can will check up on this one too. Check the full playlist though ✌️😊
At second last reference bit of 3 is (1) and at last reference bit of 5 is (1).
@Akash Okash
@@SimpleSnippets Sorry for this comment but you make mistake.
In this algorithm we've *victim* and we've a pointer at it.
Since we use page , the reference bit becomes 1.
For exemple : sequence : 2321... for three frame f1,f2,f3.
Even we use 2 in frame 1 , he's reference bit becomes 1. In f2 , we use page 3 and his reference bit becomes 1,....
*Here the true algorithm : second chance.*
When we want to replace a page , we must loot at on his reference bit .
If he's 1 or 0.
If he's reference bit are 0, we replace it. And we go in another frame, this frame becomes the victim.
But if he's reference bit are 1, we change it at 0 and give it a second chance and we go at the next victim.
Have a good day please.✍️🙏
At second last reference bit of 3 is (1) and at last reference bit of 5 is (1).
ty
@Salman Ali because we are applying FIFO here..3 just came in and 5 and 4 came before 3 so we cannot replace 3.
yeah
yessss its true
In last at the time of replacing 3 it will be set value (1)
will ubplease tell that why for the last two requests 3 and 5 we do not set reference bit to 1 they are also repeating na so
Thank you so much for creating this good vid. You saved my life!
Most welcome my friend. I'm happy it helped you 😊 I would request you to please share the video with your friends too ✌️
Would be great to show special cases like what happens if all 3 pages have read-bit set to 1 and all get second chance.
Superb clarity
Just finished the entire playlist. It was awesome, but just a question shouldn't the last references of 3 and 5 have (1) as reference?
yep
It doesn't matter
what happenes if i try to insert , for example, a 3 when 3 already has the reference of (1). Will it get the reference of (2)?
no, dumbass
All youtubers ignoring galvin book in galvin book it's said that when refrence bit comes to 0 of a process then arrival time reset to current time means that process will be not considered as first come it will mean now this will be put at last in the queue so it will not be replaced.
Great lecture sir
Thank you!
Clean & nice explanation. Thank you.
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pointer is not visible clearly ,making little bit difficult to focus where u r pointing
one of the best i have seen.
Thanks bro 😊✌️ please do share the video with your friends 😉 #teamsimplesnippets
thank you, this was very helpful
That's great to know buddy, please do Subscribe to our channel as we have a lot tech educational content which you'll surely like. Also please do share our channel and videos with your friends too, that's the biggest help and support you can provide buddy ✌
What will happen when a page has reference bit equals to 1 and a request comes for that same page. The reference bit can't increase to 2 can it? Or is this scenario impossible? What happens here?
Reference bit is always set to 0 or 1 and nothing other than that
hi thank you very much for this video ! With the actual logic couldnt we implement a "third-chance page replacement algorithm ? and if so wouldnt it be better?
nth chance - where n is the unused counts it was requested.
What will happen when each page has reference bit equals to 1 and a request comes for a page, which is not in the memory?
If i not wrong...we will follow fifo but, every page will get a second chance as reference bit of every page is 1....then which page we will remove?
Good question. If every page has reference bit 1 then we use the fifo rule again to remove the one that came in first and not give it a second chance ✌️
@@SimpleSnippets Thanks😀
Most welcome buddy ✌️
thank u love from nepal
Most welcome bro ✌️ from India 🇮🇳
here at last second step reference of 3 will set because it is requested again
Correct !
I think the teacher somehow got confused.
Second chance page replacement works by using a circular buffer, not a timestamped queue, the explanation and solution are incorrect. the page faults would be 6, and the end frames would be {2,5,3} and second_chance_bits as {1,0,1}
3:26 10:59 sooo, this is like a hybrid of FIFO & LFU (least frequently used)
10:29*
bro got 217K AND STILL TEACHES WRONG
totally wrong does not work this way sorry
stop copying other videos. come up with at least different numbers
Thank you!
thank you, this was very helpful
Glad it was helpful!