madeeasy OS 2019- Memory overhead multilevel paging

Question

for memory overhead in Multi level paging, for innermost table only 1 page size shall be counted na? and NOT the complete page table size?
please explain the concept, thanks!

Comments

No..overhead of memory means what is the extra amount of memory that is needed apart from the given memory(i.e. memory required by all page tables is the overhead)

But here they have explicitly specified what overhead is --> Outer page table size +one page of inner page table.

 

$\frac{2^{32}}{2^{12}}$==>$2^{20}$ entries in inner page table with each entry size = 4B ==> $2^{20}*4 B$

==>$\frac{2^{22}}{2^{12}}$ =>$2^{10}$ pages in outer page table 

=> overhead = $2^{10}*4 + 2^{10}*4$

(but not getting this statement "page table is divided into 1K pages each has 1k size)

 

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yeah but as they mentioned one page of inner page table so why to take complete page table size for inner then?

thanks for the ans though!

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@Hemanth_13 how one page of inner page table you calculate??

 

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@Ramij 

page table size = no. Of pages * page table entry.

no. Of pages = process size / page size.

So, no. Of pages = 2^32 / 2^12 = 2^20 pages

note: no. Of pages in innermost page table = no. Of pages in process.

page table size = 2^20 * 2^2 B = 2^22 B >= page size.

Note: page table divides till page size>=page table size.

So again division will happen and new table will be form and so on.

 

Answer 1

here virtual address size is 32 bits and 4KB page size is used so we get an offset of 12 bits.

page table is divided into 1K pages each with size of 1K and page table entry size is 4B each.

so remaining 20 bits of VA is used for these 1K pages each of 1K size .so overhead (memory needed) for level 1 page table is 1K*4 B = 4KB.

once level 1 PT(page table)  is accessed we just need to access 1 PT of level 2 (not all PT of level 2), which need an overhead of 1K*4 B =4KB.

so total overhead is 4KB + 4KB = 8KB.

....(correct me if i am wrong)