Paging (OS)
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- Опубліковано 5 сер 2024
- In this video, we will discuss the basics of paging, how we might implement paging, followed by some examples of how to calculate the number of memory frames a computer has based on the specifications, how many bits are needed for displacement into a page in the page table etc.
This video was developed for the CSCE 311 Operating Systems class at the University of South Carolina in response to the COVID-19 pandemic.
watching this 1 hour before Operating System Final Exam
Results plz
@@insist1754 got A
@@firdausfadzil4857 well done
same lol
give tips
Every other explanation I could find was highly technical. This was the first I could find that actually made sense. Thank you!
Great video, really informative and well explained. Thank you
Very underrated video. Helped me a lot. Thank you!
the only video which cleared the concept for me
thank you casey
You helped me out with my operating systems class, thank you!!
THANK YOU A LOT. This is the video that finally made me understand the topic
Vous expliquez superbement bien, j'ai pu compléter ma compréhension.
incredible video, thank you so much for your efforts!
thanks a lot for these lectures!
Thank you very much for the exercise, it was very useful!
loved the video! thank you so much!! :)
excellent video, thank you!
Thanks very much for this explanation 👏👏👏
Thank you, the example is very easy to understand
crystal clear explanation, thanks
Really helpful thanks !
watching this 2 hour before Operating System Final Exam. incredible video😉🥰
Did you pass ?
@@b213videoz yes
@@PoetrybyEraj6 Congratulations!
@@b213videoz Thanks
In question 4 (at 23:48 to be precise), each entry in the page table is going to hold a frame number at max, right? We know that we have 2^{32} frames (done at 17:30) in our system. Now, we have 2^{36} entries in the page table with each entry holding a possible 32 bit value referencing the frame (we don't really need 64 bits for this), giving us the total size of page table as 2^{36} * ( 32 bits), right? You have it as 2^{36} * ( 64 bits). Where am I wrong?
appreciate it a lot =)
Great Explanation
thank you soo much
I been struggling to understand so bad. This video was super helpful
any idea how 4KB was displayed on 12 bits? i d\think she did a miscalculation
Hi,
Great vid! Can anybody explain why we even need a page offset to begin with, I get as she explains it is to 6:54 "get to the spot in memory we are looking for". Like what, surely where a frame starts thats where the process' page start(like i se thats not the case, but why).
Current understanding is: A page is a division of a process (a process is splitt evenly into pages). But where a process starts in RAM why is it stored with an offset from the a frames start.
Thank you ma'am you have saved me
We have an exam 2mrw ... Pray for us
But you said number of entries in page table means number of pages! which was 2^36. Later, on question 4, you calculated the number of pages 2^27! which one is wrong? is number of entries in the page number not equal to number of pages in the table? or there was something wrong with calculation?
nice one !
Super clear explanation of concepts. Great video
she deserves more subs
Thanks Dera Darling🤗
MCs and cs students before main exam 😂
Thank you
Question about Q4: don’t we need 36*2 bits per entry to store the mapping from page number to frame number (36 bits each)?
Thanks for the helpful video!
something also looks off to me about her calculations
thank you
I think the mathematical explanation is all good except that paging actually doesn't guarantee internal fragmentation will be avoided. of course external fragmentation would no longer be an issue but internal fragmentation will always be there.For example, if page size
is 2,048 bytes, a process of 72,766 bytes will need 35 pages plus 1,086 bytes. It
will be allocated 36 frames, resulting in internal fragmentation of 2,048 − 1,086
= 962 bytes. In the worst case, a process would need n pages plus 1 byte. It
would be allocated n + 1 frames, resulting in internal fragmentation of almost
an entire frame. [Source: Operating system concepts by Abraham Silberschatz ...]
Thanks
0:45 so I'm not alone confusing "contiguous" with "continuous" 😊
27:45
How is 1GB × 512B = 512GB
Why is the result unit is GB ?
What’s an Offset , Address , Indexing ??
Watching at 2x speed, 15 mins before the exam.
What university year do you study this ?
can offset be zero?
Yes.
Wish me luck Bois, I have OS exam tomorrow
For people who may be confused, what she is referring to as 'virtual memory' is actually called logical memory. Virtual memory is a whole different thing
For Question 3, around 22:50 timestamp, the virtual memory is answered as 2^48, however, why does it matter if it takes 36 bits to index into page table? if the machine is a 64 bit machine, then the virtual memory has to be 2^64 isn't it? and that gets divided into 4kb pages.....or that question now ignores that machine is 64 bit and makes it into a 48 bit machine?
On x64 48 bits are used for paging if I’m not mistaking , not whole 64
can u give us the slide
and thnxs
Casey's cheeks
Brace yourself, cash refund incoming