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thank yu

you are the best .... i hope your account keeps growing.

Please upload a video on How to thank you very much😂

she speaks so slow that at 1.5x speed still sound totally normal lol

I have exam time but I can't concentrate i am watching shorts evry time.i try to study.

this is one of the greatest videos that explain Contiguous Memory Allocation & Frafmentation, Thanks

Thanks it really helped

hands down the best explanation of TLB in TH-cam

amazing thank you. you explained things very clearly

Спасибо! Хорошо что есть Русский перевод

Super nerdy and I love it.

27:45 How is 1GB × 512B = 512GB Why is the result unit is GB ?

Brace yourself, cash refund incoming

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?

wow thank you!!!

i love you from gilgit baltistan

Thank you for this video. Very well explained.

16:04 I do not get why these calculations are made.... If there is a fixed size for each of these partitions, we need only have an array of partition sizes, order the pointers to those partitions according to their sizes and compare the process size until we find the first one that's equal in size or larger than the process. If it's occupied, go on to the next one. Why waste computing cycles checking for every single partition?

What’s an Offset , Address , Indexing ??

I think no of bits required for page no is equal to the log of size of page table

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

thank you !

This exemple is wrong. In the classic second chance algorithm after inserting new data into a frame the reference bit must be set to 1. When starting at frame 0 after inserting 4 into the frame we have to set the reference bit to 1 in this frame. There is also a pointer showing which frame is the candidate frame (victim frame) to insert data at the next page fault. So a the the beginning all the reference bits a set to 0 and the pointer is set to frame 0. Inserting 4. Setting reference bit to 1. Moving pointer to the next frame (in this case frame 1). When there is no page fault the frame where the value is locate will have it bit set to 1 if not already. The pointer will still be set to the last frame in date.

0:45 so I'm not alone confusing "contiguous" with "continuous" 😊

THANK YOU A LOT. This is the video that finally made me understand the topic

❤

thank you so much, i cannot understand when i study at school but you help me a lót

thank you soo much , im glad that you included all the conditions in one question itself

That was good! Thanks Mam!

crystal clear explanation, thanks

the only video which cleared the concept for me thank you casey

Thank you for such an easily understandable video. I find the Dinosaur book tough to understand for some concepts

Thank you ma'am you have saved me

Thanks for the explanation

Great explanation <3

thank you so much 🙏

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.

We have an exam 2mrw ... Pray for us

You are the best'!!! Keep up the good work. I passed my Operating Systems quiz thanks to you. I can't thank you enough!!!! Greatings from Turkey !!!

Your videos are really useful, greetings from lebanon

Thanks for this detailed explanation

Thank you

Explained in a very crisp and concise manner. Thank you.

MCs and cs students before main exam 😂

If we have 2^47 entries, and each entry being 8 bytes long, isn't the size of the virtual memory be 2^47 x 2^3 = 2^50 bytes instead of 2^59?

Thank you, the example is very easy to understand

Everything about this seems wrong and confusing as hell.

saving me the day before my final

Excellent explanation! The best one in TH-cam. The exercises are very clear. Thank you!

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