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Consider a machine with byte addressable memory 32 bits virtual addresses, 32 bits physical addresses and 4 KB page size. If a two-level page table system is used where each page table occupies one page and page table entries of 4 B each.

How to calculate the size of 1st level page table & 2nd level page table(s) ? Also how to calculate memory overhead?

3 Answers

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Page size=4KB

PTE=4B

Virtual Memory=2^32

Physical Memory=2^32

Offset=4kb => 2^12 so offset=12 bits

As given that each page table occupy one page

so for first level =4KB/4B =2^10 thus 10 bits required

similarly for second page also 10 bits required such that 10+10+12=32(virtual memory bits)

Memory overhead= memory occupied by first level page table + memory occupied by second level page table (one page) out of 1024 pages

= 2^10*4+2^10*4

=8KB

Please verify the solution,

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