My professor has worked at top companies like intel and NVIDIA and lectures this class for 15+ years but cannot make it simple those concepts just as much as you sir! Thank you for sharing these videos for free.
So to clarify, especially for the physical address bits: Virtual mem offset= 4KB = 2^12, so 12 bit offset for both virtual and physical. 32bit machine, 32-12 = 20 bits for virtual address RAM is 256MB, 256 = 2^28, 28 * 12 = 16 bits for the physical address
Dan, the video is a little confusing. On a given computer, physical addresses always have the same number of bits, no matter how much RAM is installed. For example, on a 32-bit system (having 32 address lines), physical addresses always have 32 bits. With 256 MB RAM installed, the highest 4 bits of all *valid* physical addresses would be zero. With 512 MB RAM installed, the highest 3 bits of valid addresses would always be zero. Etc. In other words, installing more/less of RAM does not change the size of page table entries.
@Rahul R The details of the virtual addressing system depend on the CPU architecture. The architecture may allow the OS to choose the page size and/or page table size. TMI for this comment. The Wikipedia article "Page (computer memory)" has a table titled "Page sizes among architectures" that has good info on this.
Great video and a very great explanation. To Clarify, The presentation is using MebiByte (MIB) and Kibibyte (KiB) and it's not using Megabyte (MB) and KilloByte (KB). so it's 256 MiB = 268, 435, 456 Byte and you can check that since 2^28 also equals 268, 435, 456. 1 KB = 1000 Byte 1 KiB = 1024 byte = 1.024 KB 1MiB = 1024 KiB 1MiB = 1048.58 KB So if the page size was 4 KB it would had contained 4,000 entry but since page size is 4 KiB, it contains 4,096 entry (as you can see in the slides)
The video says the first 20 bits of the virtual address give the "virtual page number". That's not exactly how it works. For example, on Intel x86, there are *two levels* of tables. The high-order 10 bits of a virtual address are used as an index into the first table (called the page directory). That gives the physical address of a page table. Then, the next 10 bits of the virtual address are used as an index into that page table. That gives the physical address of the memory page. Finally, the lowest 12 bits of the virtual address are used as the byte offset into the page.
Thank You! Really Good Explanation. While studying these concepts when I was in college days, I had so many doubts and I thought I understood, but I wasn't
Actually he is right since size of ram is 256MB if we assume 1byte==1word then there are 256M words 2^8*2^20=2^28 words in ram . To access these many words randomly we need 28 address lines(buses). There for we can say 256MB RAM==28-bit physical address space.
Nice and clear. However, there is no reason to store the virtual page number, since it is exactly equal to the index that finds that entry in the page table.
is Physical Address Extension referred to after this one ? It would be great to see how it fits in. See that with PAE 32bit virtual memory can address even more than 4gb of physical memory. From what I understand it was a commercial decision by hardware driver makers to kill the idea 32 bit O/S's by not supporting that larger address space.
I'm confused: If we are assuming word-addressable memory, then why do we need 12 bits for the offset? That many bits for the offset lets us identify individual bytes in the 4KB page, but shouldn't we have just enough offset bits to identify words within the page? So 10 offset bits?
@Dylan F What you say would be true for a word machine but, per Wikipedia, those are "largely only of historical interest." This video assumes byte addressing.
Wasn't it that each VA had 1024 words (each word 4 bytes), so address 4 should map to physical address 4bytes+4bytes+4bytes (first 3 addresses) = 12bytes, so the offset should be 12 bytes, and address 4096 + 12 = 4108, not 4100 in the last video
this is one of the best explanations on memory management. Thank you for doing this. I have a doubt at 1:16. Why does the last 12 bits not get translated? in the previous video @David had mentioned that a page get translated to a page of equal size on ram. In this video, With the introduction of offsets, it became a bit confusing for me. please help. thanks again
@Harpreet Saini Good question. My understanding is that, on Linux, disabling swapping does *not* disable virtual memory (addressing), because VM is still needed for process memory isolation. Even without swapping, Linux can run multiple processes and it still runs each process in its own virtual address space, to prevent any process spying on or corrupting other processes. So I think that VM works the same with swapping disabled except that, when physical memory is exhausted (when all page frames are in use), the OS can't page out dirty pages. Note that it can still discard read-only, file-backed pages (such as program code), and it can still run the out-of-memory algorithm ("OOM killer") to free up memory by killing some process. How does VM isolate process address spaces? Each process has its own set of page tables and the OS makes sure that no physical page frame is used by more than one process. (Except in the case of shared memory, where processes intentionally want to access the same memory).
QUICK question, this part wasn't too clear to me, but once the remaining 20 gets translated through the map to the PA, you say the PA becomes 16.. IS this because we're assuming as in the last video, that we're mapping from VA 4, to 20 - 4 = 16? Is that why the PA here was 16???
okay i had the same exact question and i figured it out! how he figures out the size of each memory and the offset and the page address bits: Question is: What happens to a 32-bit Machine with a 256MB RAM abd 4KB pages. from what we are given: it says 32 bit machine that means that the virtual machines total bits are 32. VM: ----------------------------------- ----------------------------------- then they tell us that the RAM is 256MB: 256= 2^8 and M indicates 2^20 -> there is a table for this. where K=2^10 , M=2^20, G=2^30 . . . and so on. so 256MB is same as (2^8)(2^20)B adding the powers we get 2^28 B for the RAM note that 2^28 B is same as 256MB but since memory is measured in terms of bytes the power of 2 of that Byte indicates the bits kinda like a rule. now we know that RAM: ----------------------------------- ----------------------------------- now we need to know the page offset? how do we figure that out? simple if we know the size of the page which the question gives us as 4KB now 4KB is same as (2^2)(2^10)B-> (2^12)B B is a standard now so the power of 2 will be the bits needed that means the page offset is 12 and its same for both VM and RAM VM: ----------------------------------- ---- --------------------------------------- < (32-12)bits > | VM address |pageOffset RAM: ----------------------------------- ---- --------------------------------------- < (28-12)bits >| RAMaddress |pageOffset
@@asdf256 I edited my reply. I made a typo there. Instead of 2^3 it's 2^8 which makes sense as 2^8 is actually 256. And that would answer your other question too. So 2^8 * 2^20 = 2^28.
hi do you offer some more private online lectures which is going more in detail with paging? I need this for my exercise and i never understood it and my university doesnt explains it clearly
Thankyou very much. Really great explaination you taught every part of the topic. The people who disliked this video maybe were not smart enough to figure out obvious info themselves.
@7.28 you are actually wrong. For a 32 bit system, maximum 4GB of RAM is accessible, beyond that is inaccessible. So, Page table will not be helpful if we are going beyond 32 bit address lines.
+purple elf I have a question, so, for example as you said, well let's take an easier example like 32 BITS assumes 4 gb of memory, because 2^32 = 4294967296, 4294967296 BYTES are approximately 4 gigs. So, if we're converting BYTES to gigs to render that answer, why do we say a 32 BIT system? It seems to me we take the amount of BITS and bring 2 to the power of that and suddenly THAT answer becomes the BYTE amount, which we convert to gigs to figure how much memory MIPS promises us via our Virtual address.. but that makes no mathematical sense, because 8 bits is a byte
Purple elf thanks so much for the response. In regards to the commodore example, I understand completely how you came up with 64 kb, but I'm confused as to the 32 bit 4 gig example.. So for the commodore example, 2^16 * 8 = 524288 bits of addressing space converted to kb = 64 kb address space, GREAT, that works out for me, I understand. BUT, the 32 bit system example.. not so much: 2 to the power of 32 = 4294967296 bytes 4294967296 / 32 = 134217728 BITS 134217728 bits converted to kilobytes = 16777.216 kb of addressing space, but we know we have 4 kb of addressing space, not over 16 thousand.. o.O help? lol thanks
At least I know my math isn't that terrible.. haha thanks bud, IOP is input output per second right? Geez with 2^4096 wouldn't the addressing space per page's OFFset be enormous ? All those unused bytes
you could have explained what's the functions of the physical and virtual memories.. It's kind of hard to learn until you know what it is your learning lol
While my professor just shows us this crap on a slide and just speeds through it, you actually show it and explain it, thank you.
bruh i could have just watched this video all semester
These videos helped me get a full understanding of this material in about 2 hours, it took my professor two months to attempt and fail to teach it....
My professor has worked at top companies like intel and NVIDIA and lectures this class for 15+ years but cannot make it simple those concepts just as much as you sir! Thank you for sharing these videos for free.
So to clarify, especially for the physical address bits:
Virtual mem offset= 4KB = 2^12, so 12 bit offset for both virtual and physical.
32bit machine, 32-12 = 20 bits for virtual address
RAM is 256MB, 256 = 2^28, 28 * 12 = 16 bits for the physical address
28-12**
Dan, the video is a little confusing.
On a given computer, physical addresses always have the same number of bits, no matter how much RAM is installed. For example, on a 32-bit system (having 32 address lines), physical addresses always have 32 bits. With 256 MB RAM installed, the highest 4 bits of all *valid* physical addresses would be zero. With 512 MB RAM installed, the highest 3 bits of valid addresses would always be zero. Etc.
In other words, installing more/less of RAM does not change the size of page table entries.
4KB != 2^12 so it does not cause 12 bit offset
@@chrispearson6356 how the page table size are determined? whether it depends on 32 bit os or 64 bit os like that?
@Rahul R The details of the virtual addressing system depend on the CPU architecture. The architecture may allow the OS to choose the page size and/or page table size. TMI for this comment. The Wikipedia article "Page (computer memory)" has a table titled "Page sizes among architectures" that has good info on this.
one of the most underrated channel on TH-cam :( I feel bad for students who do not know this series due to the lack of popularity
You probably heard this a thousand times, but your explanations are so beautifully precise and helpful! Thanks so much!
I got confused by VA for almost 2 month.. and now this whole stuff becomes clear because of this video. Thanks so much
This video is slightly off topic in accordance to my assignment and yet it still taught me more about my assignment than upper division lectures.
By far the best lecture on TH-cam. Thank You!!!
please, give this guy a medal
Great video and a very great explanation. To Clarify, The presentation is using MebiByte (MIB) and Kibibyte (KiB) and it's not using Megabyte (MB) and KilloByte (KB).
so it's 256 MiB = 268, 435, 456 Byte and you can check that since 2^28 also equals 268, 435, 456.
1 KB = 1000 Byte
1 KiB = 1024 byte = 1.024 KB
1MiB = 1024 KiB
1MiB = 1048.58 KB
So if the page size was 4 KB it would had contained 4,000 entry
but since page size is 4 KiB, it contains 4,096 entry (as you can see in the slides)
Sir I had wasted three days ...bcz I didn't see it video.. hatsoff for ur level of teaching 🙏
Thank you... I have a chance to pass my computer systems final.
Michael Siderius You're welcome to look at the full class online at test.scalable-learning.com, enrollment key YRLRX-25436
+David Black-Schaffer Hey, Nyckeln verkar inte fungera, skulle du kunna skicka mig en ny? THX!
+Oskar Friberg make sure you are on test.scalable-learning.com for the trial course and not www.scalable-learning.com
Thx! I got in, you're awesome!
I tried to watch many videos to understand this topic and i learned nothing .. but your playlist is great , thank you for your great explanation
each page size is 4096, hence 2^12 is 4096. that's how we get the number 12. So 12 bits is sufficient to access an arbitrary byte within a page
I am very appreciate this list of videos. I get confused by reading book and this video really explains everything so clear!
Thank you.
This explanation is better than university professor
How did you come up with a 28 bit address space from 256MB of RAM?
correction : "physical memory pages" are called frames.
Page frame
Very great video and this is 9 years old video.
The video says the first 20 bits of the virtual address give the "virtual page number". That's not exactly how it works. For example, on Intel x86, there are *two levels* of tables. The high-order 10 bits of a virtual address are used as an index into the first table (called the page directory). That gives the physical address of a page table. Then, the next 10 bits of the virtual address are used as an index into that page table. That gives the physical address of the memory page. Finally, the lowest 12 bits of the virtual address are used as the byte offset into the page.
The explanation is great, and the video is even better, is so clear!! Ty for your work
Thanks David very much, you answered the question which bothered me for years
Thank You! Really Good Explanation. While studying these concepts when I was in college days, I had so many doubts and I thought I understood, but I wasn't
this is gold. Made perfect sense.
Although the Video talks about KB and MB, it means KiB and MiB.
A really youtube hero!!!!
thank you very much! this really helps!
wish i have found your channel earlier.
how exactly does 256MB of ram give 28 bits?
yeah wtf it doesn't
I guess there's some mistake. Maybe 256MegaBITS he meant idk
Actually he is right since size of ram is 256MB if we assume 1byte==1word then there are 256M words 2^8*2^20=2^28 words in ram . To access these many words randomly we need 28 address lines(buses).
There for we can say 256MB RAM==28-bit physical address space.
This series is very instructive! Thanks!
Hi - thanks for the vid. Could someone give me a bit more insight as to why we always need an offset?
Nice and clear. However, there is no reason to store the virtual page
number, since it is exactly equal to the index that finds that entry in
the page table.
is Physical Address Extension referred to after this one ? It would be great to see how it fits in. See that with PAE 32bit virtual memory can address even more than 4gb of physical memory. From what I understand it was a commercial decision by hardware driver makers to kill the idea 32 bit O/S's by not supporting that larger address space.
I'm confused: If we are assuming word-addressable memory, then why do we need 12 bits for the offset? That many bits for the offset lets us identify individual bytes in the 4KB page, but shouldn't we have just enough offset bits to identify words within the page? So 10 offset bits?
@Dylan F What you say would be true for a word machine but, per Wikipedia, those are "largely only of historical interest." This video assumes byte addressing.
your teaching skills are great Thank you
Wasn't it that each VA had 1024 words (each word 4 bytes), so address 4 should map to physical address 4bytes+4bytes+4bytes (first 3 addresses) = 12bytes, so the offset should be 12 bytes, and address 4096 + 12 = 4108, not 4100 in the last video
unless each address is 1 byte, although you said each one was 4 bytes = 1 word
Does every process have one common virtual address space?
this is one of the best explanations on memory management. Thank you for doing this. I have a doubt at 1:16. Why does the last 12 bits not get translated? in the previous video @David had mentioned that a page get translated to a page of equal size on ram. In this video, With the introduction of offsets, it became a bit confusing for me. please help. thanks again
when swap (virtual memory) in linux is not in use means used memory of swap space is 0 then any TLB is in working mode or not ? please clarify.
@Harpreet Saini Good question. My understanding is that, on Linux, disabling swapping does *not* disable virtual memory (addressing), because VM is still needed for process memory isolation. Even without swapping, Linux can run multiple processes and it still runs each process in its own virtual address space, to prevent any process spying on or corrupting other processes.
So I think that VM works the same with swapping disabled except that, when physical memory is exhausted (when all page frames are in use), the OS can't page out dirty pages. Note that it can still discard read-only, file-backed pages (such as program code), and it can still run the out-of-memory algorithm ("OOM killer") to free up memory by killing some process.
How does VM isolate process address spaces? Each process has its own set of page tables and the OS makes sure that no physical page frame is used by more than one process. (Except in the case of shared memory, where processes intentionally want to access the same memory).
0:35 calculation is true only if the machine in question has byte-level addressing.
QUICK question, this part wasn't too clear to me, but once the remaining 20 gets translated through the map to the PA, you say the PA becomes 16.. IS this because we're assuming as in the last video, that we're mapping from VA 4, to 20 - 4 = 16? Is that why the PA here was 16???
what if I have 64 bit address with page size of 4 kilo bytes and the system having a 4 GB RAM?
Totally amazing and helpful tutorials. Thank you
shouldnt 256 MB translate to 2^28 bytes instead of 2^28 bits (since its B not b).
okay i had the same exact question and i figured it out!
how he figures out the size of each memory and the offset and the page address bits:
Question is:
What happens to a 32-bit Machine with a 256MB RAM abd 4KB pages.
from what we are given:
it says 32 bit machine that means that the virtual machines total bits are 32.
VM:
-----------------------------------
-----------------------------------
then they tell us that the RAM is 256MB:
256= 2^8
and M indicates 2^20 -> there is a table for this.
where K=2^10 , M=2^20, G=2^30 . . . and so on.
so 256MB is same as (2^8)(2^20)B
adding the powers we get
2^28 B for the RAM
note that 2^28 B is same as 256MB
but since memory is measured in terms of bytes
the power of 2 of that Byte indicates the bits
kinda like a rule.
now we know that
RAM:
-----------------------------------
-----------------------------------
now we need to know the page offset?
how do we figure that out?
simple if we know the size of the page which the question gives us as 4KB
now 4KB is same as (2^2)(2^10)B-> (2^12)B
B is a standard now so the power of 2 will be the bits needed
that means the page offset is 12 and its same for both VM and RAM
VM:
-----------------------------------
----
---------------------------------------
< (32-12)bits > |
VM address |pageOffset
RAM:
-----------------------------------
----
---------------------------------------
< (28-12)bits >|
RAMaddress |pageOffset
@@nimragauhar4838 How does 2^3 = 256?
@@nimragauhar4838 Further, how does (2^3) * (2^20) = 2^28 ???
@@asdf256 I edited my reply. I made a typo there. Instead of 2^3 it's 2^8 which makes sense as 2^8 is actually 256. And that would answer your other question too. So 2^8 * 2^20 = 2^28.
@@nimragauhar4838 Thank you so much! I was so confused how he got those values!
How does he get 28 bit physical address? Anyone care to show the calculation?
hi do you offer some more private online lectures which is going more in detail with paging? I need this for my exercise and i never understood it and my university doesnt explains it clearly
Thankyou very much.
Really great explaination you taught every part of the topic.
The people who disliked this video maybe were not smart enough to figure out obvious info themselves.
Beautifully explained!
@7.28 you are actually wrong. For a 32 bit system, maximum 4GB of RAM is accessible, beyond that is inaccessible. So, Page table will not be helpful if we are going beyond 32 bit address lines.
So is it 1 byte per word?
Very helpful ! Great presentation :)
What about logical Adress??
how did you convert the 256mb of ram to 28 bits physical address?
256 MB = 256 * 1 MB = (2^8) * (2^20) = 2^28 bytes (locations) in memory. So we need 28 bits to address all of them.
i lke ur teaching and the presentation! :) thank you
nice series,thanks for the videos
Why do 20bits tell you it goes to virtual number 0, while 16bits tell you to go to physical page number 1?
this is great, thank you so much!
A big thanks to your sir.. These are best Memory Related Videos I have seen.. Please keep it going!!!! :0
If had sixteen plus gigs of virtual r.a.m; would it made a substantially difference??
***** What?
+purple elf I have a question, so, for example as you said, well let's take an easier example like 32 BITS assumes 4 gb of memory, because 2^32 = 4294967296, 4294967296 BYTES are approximately 4 gigs. So, if we're converting BYTES to gigs to render that answer, why do we say a 32 BIT system? It seems to me we take the amount of BITS and bring 2 to the power of that and suddenly THAT answer becomes the BYTE amount, which we convert to gigs to figure how much memory MIPS promises us via our Virtual address.. but that makes no mathematical sense, because 8 bits is a byte
Purple elf thanks so much for the response. In regards to the commodore example, I understand completely how you came up with 64 kb, but I'm confused as to the 32 bit 4 gig example.. So for the commodore example,
2^16 * 8 = 524288 bits of addressing space converted to kb = 64 kb address space, GREAT, that works out for me, I understand.
BUT, the 32 bit system example.. not so much:
2 to the power of 32 = 4294967296 bytes
4294967296 / 32 = 134217728 BITS
134217728 bits converted to kilobytes = 16777.216 kb of addressing space, but we know we have 4 kb of addressing space, not over 16 thousand.. o.O help? lol thanks
At least I know my math isn't that terrible.. haha thanks bud, IOP is input output per second right? Geez with 2^4096 wouldn't the addressing space per page's OFFset be enormous ? All those unused bytes
Lol so You're telling me with that much address space it would be able to store data telling you if you've opened the picture file before or not
Great video, thank you.
are 'page table' the same as 'translation table'
you ask too many questions
Wait so we don't need to translate the page offset?🤔🤦
be quiet
amazing explanation!
I really like the questions part
amazing job !
i want to give you a hug
that was very helpful thank you
can david sir or anyone solve a problem based on offset please?
Best. Super clear. Thanks so much.
very good tutorial
thank you very much
One thing I’m still not clear on is, is the page offset translated?
Just kidding. Great explanation!
Thank you!
Thank you sir!
Thank you
bro u r goat
This is great
am I the only one who watches these videos because wants to know how this all works "underneath" from programmer perspective?
Good Pretsentation
you could have explained what's the functions of the physical and virtual memories.. It's kind of hard to learn until you know what it is your learning lol
Thanks a lot man!
thank you ...
The upper 20 bits of virtual memory gets translated. Nothing more in this video.
Oookkk... We got that... "OFFSET IS NOT TRANSLATED".....
yes, it's the only thing I understood
vow vow vow !!!
Finally a normal English not Indian guy saying "hay doday a gonna dalk aboud sbsbabshsh"
but 32bit computer doesn't support 8GB of ram, ryt?
Thank you very much