#GATE #PAGING OS

Question

Consider a paged virtual memory system with 32-bit virtual address and 1KB pages. Each PTE requires 32 bits. It is desired to limit the page table size to one page.  how many levels of page tables are required?   Tables at two of the levels have 2^8 entries and at one level have 2^6 entries(8+8+6=22). If the top level has 2^6 entries then how many pages are possible?  If the middle level page table has  2^6 entries how many pages are possible?  If the bottom level page table has  2^6 entries how many pages are possible?

Answer 1

If the top level has 2^6 entries , the total pages in page table will be = 1 + 2^6 + 2^14

if the middle level page table has  2^6 entries, number of page table = 1+ 2^8 + 2^14

If the bottom level page table has  2^6 entries , number of page table = 1+ 2^8 + 2^16

Answer 2

LA = 32 bits
Page Size = 1KB
PTE = 4 Bytes

No. of pages in LAS = $2^{32}$/$2^{10}$ = $2^{22}$ = 4 million

Size of first level table = $2^{22}$ * 4B = $2^{24}$B = 16MB

Size of second level page table = (16MB/1KB)*4B = 64KB

Size of third level page table = (64KB/1KB)*4B =  256B < 1KB

It will be 3 levels paging system.

LA= 32 bits, 32 = |6|8|8|10|   

I didn't understand your question further, what do you mean by "How many pages are possibe"?

Comments

1 KB pages means pages of size 1KB

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can you help me with further parts of the same questn?