Multi-level page tables - hierarchical paging

Question

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?

Solution 1

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?

Solution 2

• 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

Comments

[Bobby S]

4096/8 = 512 entries; To address 512 entries log2(512) = 9 bits. The total number of bits to encode each page level is 48 - log2(4096) = 36 bits. Thus the total number of levels required is 36/9 = 4.

[Bobby S]

Since the 2048 is in bytes, shouldn't it be converted to bits, which would make it 50/9 = 6 rounded up?

[bsantos]

No, because you what to know how many bits are required to address 2048 bytes and not bits.

[Hans]

I might be answering in 2014, but big THANKS!!

[Jeff Hammond]

The 2048 is for 2K pages? What OS uses those? Don't you want to use 4K pages here?