In a 3-level paging system, when every page size is 512 bytes and an entry size is 4 bytes, we can know that every single page has 128 entries. So we can design 7 bits (log128) of the VPN(Virtual Page Number) as an index in the third level page table, which makes the page table fit perfectly in a page. What happens if we just design 6 bits as an index? I think on third level page table, some pages could not be correctly "transfer" to corresponding physical frame number because their page number can not be represented by a six-digit index. For example, when the OS "cut" the whole page table into 3 parts, each of which has 128 entries, and one entry happens to be the 100th entry of second part. How can we use 6 digits to index into 100 to get the right frame number in physical address? Or the "cut" process is smart enough to put just 64 entries and waste another 64 entries in every page? Or we actually can use 8 digits to index into 128 entries table, using just 7 digits and ignore one digit? This problem really confuses me a lot because even if the third level page table perfectly fits into one page, the first and the second level page table sometimes do not fit perfectly at all.
I got the answer from Stack Exchange and here is the same question https://cs.stackexchange.com/questions/103454/what-does-it-mean-the-outer-level-page-table-need-not-be-page-aligned.
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Consider a paging system with the page table being stored in memory. The logical address space used is 32 bit and the page size is 8KB. This will result in a very large page table(s) and therefore the system uses hierarchical paging with two levels. The number of entries in the outer page table is 256.
Specify the number of bits in each of the three fields composing the logical address namely, the outer page, the inner page, and the offset.
I found some information on finding the page offset, Page offset = log2(page size in bytes), so for this case, it would be 13, but I haven't found much information on how to find the number of bits for the outer page and inner page. Can anyone shine some light on this problem for me?
Thank you.
I might not be entirely correct, but since VPN to PPN translation is one of my favorite parts from OS, I decided to share my understanding. Maybe this picture can help to understand how is virtual address translated in the physical address.
In this example page directory contains 1024 entries, so you will need 10 bits, to be able to define which entry you need. This entry contains address of the inner table. Then, as the inner page table also contains 1024 entries, once you know the address of it, similarly you still need to find the index of its entry which holds the physical page address. So next 10 bits are used for calculating that index. Finally, when the page table entry gives you the physical address of the page, offset gives the exact physical address. If this is not very clear I can go into more details.
In your case, as you have 8KB pages, as you said last 13 bits will be used to calculate the offset. If the outermost page table contains 256 entries then you will need 8 bits (log2(256)) to be able to determine the index of its entry. Then it depends on the number of the entries of the inner table. Or if the size of the entry is defined, the number of entries can be calculated from it. If we assume that left 11 bits are entirely used for the inner table, than it would have to contain 2048 entries, as based on my understanding one instance of page table fits and fills one physical page.
The lowest bits of a logical address will be used for "offset in 8192-byte page". You'd need 13 bits for that (because 1 << 13 = 8192 or because log2(8192) = 13).
The highest bits of a logical address will be used for "index into 256-entry outer page table". You'd need 8 bits for that (because 1 << 8 = 256 or because log2(2562) = 8).
If a logical address is 32 bits and the lowest 13 bits and highest 8 bits are used for other things; how many bits are left over for the index into the inner page table?
I solved some question, where the page table entry size needed only 26 bits - 22 for the physical address, and 4 for dirty bits and such. However it was rounded up to 32 - because 26 is not a power of 2. Must be something simple I'm missing but why do we have to do that? Thanks!
I think here that you need to realize that the page table entry needs to accessed like any other piece of data. Typically, this means that it needs to fit into a byte or a word.
Now bytes only hold 8 bits, so that is not enough room. For many machines (and I suspect, your machine too), words are 32 bits.
Thus the page table entry is allocated 32 bits of space.
Number of entries in a page table entry and the size are two things. Obviously the size is equal to the number of entries times size of a single entry. Page table entry is there to tell you which virtual page maps to which physical page. Which means, the number of entries you need in a page table entry is, number of virtual page you have. which can be calculated by dividing the total addressable space by the size of a page. (For example, 32 bit address and a 4k page size gives us 2 to the power 20 entries), virtual part of an entry will be 20 bits. Size of a physical part entry is determined by the available physical memory. Usually the page size remain same. By this way you can calculate the bits needed for a single entry. Then you can multiply this by the number of entries and you have the total size.
A processor uses 36 bit physical addresses and 32 bit virtual addresses with a page frame size of 4 KBytes. Each page table entry size is of 4 bytes. A three level page table is used for virtual to physical address translation, where the virtual address is used as follows :
bits 30 - 31 are used to index into the first level page table
bits 21 - 29 are used to index into the second level page table
bits 12 - 20 are used to index into the third level page table
bits 0 - 11 are used as offset within the page
The number of bits required for addressing the next level page table ( or page frame ) in the page table entry of the first , second and third level page tables are respectively ?
This was a question asked in GATE 2008.
My Analysis: The maximum number of page frames = (physical address size) / ( page size ) = 2^36 / 2^12 = 2^24.
Thus, 24 bits will be enough to index the page numbers in the 3rd level page table. Now we have to find out how many page tables will be there in the 3rd level. Its given that 9 bits are used to index into the 3rd level page tables. So there are 2^9 page tables in level 3. That means 2^32 virtual space is contained in 2^9 page tables so entries per page table= 2^32/2^9 = 2^23. So 23 bits are required in an l2 page table entry to index the entries of a particular page table in 3rd level page table. 2^9 page tables are there in L2 from L1 page table we need to get to any one of these 2^9 page tables. So 9 bits are required in L1.
This analysis somehow doesn't seem right. I am very much confused. Can someone please explain the concepts ?
At first we have physical address of 2^36, and page size is 2^12. Hence 2^24 should be the number of pages. and you figured that right.
Since given 9 bits for 3rd level page table, ie 2^9 and then 4 bytes per entry, so 2^11. Now 2^36/2^11 would give 2^25. Hence 25 bits (25 bits at second level table).
Now again given 9 bits for second level page table. so the same logic applies again
2^36/2^11, which again is 2^25.(25 bits at first level table).
hence the answer must be
25,25,24
One thing that you have to understand is, that though virtual address space is 2^32, we need to map the entire 2^36 entries and not 2^32.Also, each page can take care of 2^9 * 2^2, and you didn't consider this 2^2 as well. The same thing holds at the higher level as well
You should have a look at this question
and this wikipedia entry, especially the figure.
In this question, there are 36 bits of physical memory, out of which 12 bits are designated for the offset. The offset is not going to change in any level (since it is the size of the page). Therefore in any level there will be 24 frame bits required for the page table. Since there are 36 bits of physical memory, you cannot increase the physical memory size by increasing the number of bits in the page table.
I had this problem in an exam today:
Suppose you have a computer system with a 38-bit logical address, page size of 16K, and 4 bytes per page table entry.
How many pages are there in the logical address space? Suppose we use two level paging and each page table can fit completely in a frame.
For the above mentioned system, give the breakup of logical address bits clearly indicating number of offset bits, page table index bits and page directory index bits.
Suppose we have a 32MB program such that the entire program and all necessary page tables (using two level paging) are in memory. How much memory (in number of frames) is used by the program, including its page tables?
How do I go about solving such a problem? Until now I would have thought page size = frame size, but that won't happen in this case.
Here is what I think:
Since the page size is 16K, my offset will be 17 bits (2^17 = 16K). Now how do I divide the rest of the bits, and what will be the frame size? Do I divide the rest of the bits in half?
238 / 16384 = 16777216 pages.
On one hand, the remaining 38-log216384=24 bits of address may be reasonable to divide equally between the page directory and page table portions of the logical address since such a symmetry will simplify the design. On the other hand, each page table should have the same size as a page so they can be offloaded to the disk in exactly the same way as normal/leaf pages containing program code and data. Fortunately, in this case using 12 bits for page directory indices and page table indices gets us both since 212 * 4 bytes of page table entry size = 16384. Also, since a page address always has 14 least significant bits set to zero due to the natural page alignment, only 38-14=24 bits of the page address need to be stored in a page table entry and that gives you 32-24=8 bits for the rest of control data (present, supervisor/user, writable/non-writable, dirty, accessed, etc bits). This is what we get assuming that the physical address is also not longer than 38 bits. The system may have slightly more than 38 bits of the physical address at the expense of having fewer control bits. Anyway, everything fits. So, there, 38=12(page directory index)+12(page table index)+14(offset).
32MB/16KB = 2048 pages for the program itself. Each page table covers 212=4096 pages, so you will need about 2048/4096=0 page tables for this program. We round this up to 1 page table. Then there's also the page directory. 2048+1+1=2050 is how many pages are necessary to contain the entire program with its related pages tables in the memory.
Example question from a past operating system final, how do I calculate this kind of question?
A computer has a 64-bit virtual address space and 2048-byte pages. A page table entry takes 4 bytes. A multi-level page table is used because each table must be contained within a page. How many levels are required?
How would I calculate this?
Since page table must fit in a page, page table size is 2048 bytes and each entry is 4 bytes thus a table holds 2048/4=512 entries. To address 512 entries it requires log2(512)=9 bits. The total number of bits available to encode the entry for each page level is 64-log2(2048)=53 bits (the number of bits of address space minus the page offset bits). Thus the total number of levels required is 53/9=6 (rounded up).
The x86-64 default page table size is 4096 bytes, each page table must fit in a page and a page table entry is 8 bytes. Current CPUs only implement 48 bits of virtual address space. How many page table levels are required?
Logical Address bit=64,
Number of page will be= 2^64/2048 = 2^64/2^11 = 2^53
Pages we have entry sine of page table= 4 Byte ,
Number of Entry in 1 Page will be= 2048/4=>512,
bit To represent one Entry=Log(512)=9bit,
and bit for Page is= 53bit
Therefore Number of Level =53/9=>6 Level Page Table