gateforum CO&A cache memory Level 2 Q17 closed

Question

On a system with 32 bit addresses and 4KB pages, how many levels are required in multilevel page table (assume that each entry in the page table takes 4 bytes of storage)?

a. 2     b. 3      c.1        d. None of these

Comments

IF WE ASSUME OUTER LEVEL PAGE IS FILLED INTO EXACTLY ONE PAGE THEN ANS IS 2

Answer 1

Initially we have 2^32 byte LAS and page size = 2^12 byte

So no of page table entry needed = 2^32/s^12 = 2^20.

so size of 1st level page table=2^20 * 2^2 = 2^22 byte

 now again do paging on this page table

So no of page table entry needed = 2^22/s^12 = 2^10.

size of 2nd level page table is=(2^10 /) * 2^2 = 2^12

this last page table can be fit into single page as page size is also 2^12.

therefore 2 level paging is required.

Answer 2

3 levels

Comments

Can you explain how you got that?

Answer 3

only 2 level paging is required