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.
I have a homework question during studying for a test:
You have a new device with logic addresses space of 32 bit and physical addresses space of 34 bit. Size of page is 8KB (2^13 B). Calculate size of page table (of single level).
There are 2^32 / 2^13 = 2^19 entries. And what is the size of entry, I have to assume that it's 4B or I have a way to calculate it?
As you have already mentioned no. of enteries will be 2^32/2^13, but the page size can be anything (predefined), also you can have operating systems supporting multiple page sizes, hence to best of my understanding you can't calculate the page size through this as it is predefined and virtual memory can be any big.
There are 2^19 entries now each of these entry will be 32 Bit wide so space occupied = (2^19) * (32) Bits
An homeword question:
Given system with 32 bit virtual addresses and 16 bit physical addresses.
Assume that a page size is 1K.
Calculate the size of the page table.
truly I have no idea how to start so I ask here for hint and directions.
Thanks in advance.
I found this example.
Consider a system with a 32-bit logical address space. If the page
size in such a system is 4 KB (2^12), then a page table may consist of
up to 1 million entries (2^32/2^12). Assuming that
each entry consists of 4 bytes, each process may need up to 4 MB of physical address space for the page table alone.
What is the meaning of each entry consists of 4 bytes and why each process may need up to 4 MB of physical address space for the page table?
A page table is a table of conversions from virtual to physical addresses that the OS uses to artificially increase the total amount of main memory available in a system.
Physical memory is the actual bits located at addresses in memory (DRAM), while virtual memory is where the OS "lies" to processes by telling them where it's at, in order to do things like allow for 2^64 bits of address space, despite the fact that 2^32 bits is the most RAM normally used. (2^32 bits is 4 gigabytes, so 2^64 is 16 gb.)
Most default page table sizes are 4096 kb for each process, but the number of page table entries can increase if the process needs more process space. Page table sizes can also initially be allocated smaller or larger amounts or memory, it's just that 4 kb is usually the best size for most processes.
Note that a page table is a table of page entries. Both can have different sizes, but page table sizes are most commonly 4096 kb or 4 mb and page table size is increased by adding more entries.
As for why a PTE(page table entry) is 4 bytes:
Several answers say it's because the address space is 32 bits and the PTE needs 32 bits to hold the address.
But a PTE doesn't contain the complete address of a byte, only the physical page number. The rest of the bits contain flags or are left unused. It need not be 4 bytes exactly.
1) Because 4 bytes (32 bits) is exactly the right amount of space to hold any address in a 32-bit address space.
2) Because 1 million entries of 4 bytes each makes 4MB.
Your first doubt is in the line, "Each entry in the Page Table Entry, also called PTE, consists of 4 bytes". To understand this, first let's discuss what does page table contain?", Answer will be PTEs. So,this 4 bytes is the size of each PTE which consist of virtual address, offset,( And maybe 1-2 other fields if are required/desired)
So, now you know what page table contains, you can easily calculate the memory space it will take, that is: Total no. of PTEs times the size of a PTE.
Which will be: 1m * 4 bytes= 4MB
Hope this clears your doubt. :)
The page table entry is the number number of bits required to get any frame number . for example if you have a physical memory with 2^32 frames , then you would need 32 bits to represent it. These 32 bits are stored in the page table in 4 bytes(32/8) .
Now, since the number of pages are 1 million i.e. so the total size of the page table =
page table entry*number of pages
=4b*1million
=4mb.
hence, 4mb would be required to store store the table in the main memory(physical memory).
So, the entry refers to page table entry (PTE). The data stored in each entry is the physical memory address (PFN). The underlying assumption here is the physical memory also uses a 32-bit address space. Therefore, PTE will be at least 4 bytes (4 * 8 = 32 bits).
In a 32-bit system with memory page size of 4KB (2^2 * 2^10 B), the maximum number of pages a process could have will be 2^(32-12) = 1M. Each process thinks it has access to all physical memory. In order to translate all 1M virtual memory addresses to physical memory addresses, a process may need to store 1 M PTEs, that is 4MB.
Honestly a bit new to this myself, but to keep things short it looks like 4MB comes from the fact that there are 1 million entries (each PTE stores a physical page number, assuming it exists); therefore, 1 million PTE's, which is 2^20 = 1MB. 1MB * 4 Bytes = 4MB, so each process will require that for their page tables.
size of a page table entry depends upon the number of frames in the physical memory, since this text is from "OPERATING SYSTEM CONCEPTS by GALVIN" it is assumed here that number of pages and frames are same, so assuming the same, we find the number of pages/frames which comes out to be 2^20, since page table only stores the frame number of the respective page, so each page table entry has to be of atleast 20 bits to map 2^20 frame numbers with pages, here 4 byte is taken i.e 32 bits, because they are using the upper limit, since page table not only stores the frame numbers, but it also stores additional bits for protection and security, for eg. valid and invalid bit is also stored in the page table, so to map pages with frames we need only 20 bits, the rest are extra bits to store protection and security information.
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.