Determine page table size for virtual memory

Consider a virtual memory system开发者_运维知识库 with a 38-bit virtual byte address, 1KB pages and 512 MB of physical memory. What is the total size of the page table for each process on this machine, assuming that the valid, protection, dirty and use bits take a total of 4 bits, and that all the virtual pages are in use? (assume that disk addresses are not stored in the page table.)

Well, if the question is simply "what is the size of the page table?" irrespective of whether it will fit into physical memory, the answer can be calculated thus:

First physical memory. There are 512K pages of physical memory (512M / 1K). This requires 19 bits to represent each page. Add that to the 4 bits of accounting information and you get 23 bits.

Now virtual memory. With a 38-bit address space and a 10-bit (1K) page size, you need 2²⁸ entries in your page table.

Therefore 2²⁸ page table entries at 23 bits each is 6,174,015,488 bits or 736M.

That's the maximum size needed for a single-level VM subsystem for each process.

Now obviously that's not going to work if you only have 512M of physical RAM so you have a couple of options.

1. You can reduce the number of physical pages. For example, only allow half of the memory to be subject to paging, keeping the other half resident at all time. This will save one bit per entry, not really enough to make a difference.

2. Increase the page size, if possible. A 1K page on a 38-bit address space is the reason for the very chunky page tables. For example, I think the '386, with its 32-bit address space, uses 4K pages. That would result in a million page table entries, far less than the 260 million required here.

3. Go multi-level. A bit more advanced but it basically means that the page tables themselves are subject to paging. You have to keep the first level of page tables resident in physical memory but the second level can go in and out as needed. This will greatly reduce the physical requirements but at the cost of speed, since two levels of page faults may occur to get at an actual process page (one for the secondary paging tables then one for the process page).

Let's look a little closer at option 3.

If we allow 32M for the primary paging table and give each entry 4 bytes (32 bits: only 23 are needed but we can round up for efficiency here), this will allow 8,388,608 pages for the secondary page table.

Since each of those secondary page table pages is 1K long (allowing us to store 256 secondary page table entries at 4 bytes each), we can address a total of 2,147,483,648 virtual pages.

This would allow 8,192 fully-loaded (i.e., using their entire 28-bit address space) processes to run side by side, assuming you have a fair chunk of disk space to store the non-resident pages.

Now obviously the primary paging table (and the VM subsystem, and probably a fair chunk of the rest of the OS) has to stay resident at all times. You cannot be allowed to page out one of the primary pages since you may well need that page in order to bring it back in :-)

But that's a resident cost of only 32M of the 512M for the primary paging table, much better than the (at a minimum, for one fully-loaded process) of 736M.

size of the page table= total no of page table entries*size of the page table entry

STEP 1:FINDING THE NO OF ENTRIES IN PAGE TABLE:

no of page table entries=virtual address space/page size

=2^38/2^10=2^28

so there are 2^28 entries in the page table

STEP2:NO OF FRAMES IN PHYSICAL MEMORY:

no of frames in the physical memory=(512*1024*1024)/(1*1024)=524288=2^19

so we need 19 bits and additional 4 bits for valid, protection, dirty and use bits totally 23 bits=2.875 bytes

size of the page table=(2^28)*2.875=771751936B=736MB

1KB pages = 2^10, 512MB = 2^29 => Offset = 29 - 10 = 19 bit.

virtual includes two part: page frame + offset => page frame + dirty bit = 38 - 19 = 29 bit. 29 bit includes 4 bit dirty (above) => 25 bit for real page frame, each page frame has 10 bit long.

So, page table size: 2^25 * 10 = 320M.

Hope this correct.