2
votes

Given : 32 bit virtual address, 4KiB per page, 4B - size of page table entrie, 1 - GiB size of address space

For a process is using 1GiB address space, find out size of 1-level page table and 2-levels page table, when page table directory has got 1024 entries. For 2-level page table - What is max and min size of page table?

So... I know that page offset is 12 bits and page number is 20 bits. So we got 2^20 entries, so total page table size should be 4.2 MB for 1-level page table.

But how can I calculate size of 2-levels page table? any tips?

1
Actually, there is no answer to this question under the parameters given.user3344003

1 Answers

2
votes

The two level case may be examined thus:

  1. There will be one page at level one.
  2. There will be enough pages for the required address space at level two.

Now each level 2 page can map 2^10 * 2^12 bytes since it has 2^10 entries and each of those points to a page of 2^12 bytes. This yields 2^22 bytes.

Now your target address space is 2^30 (1GB) so you need 2^30/2^22 or 2^8 level 2 pages. In addition, you need one level one page.

The total used is 1 + 2^8 or 257 pages or 1028 KB