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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!

asked in Operating System by Junior (651 points) | 174 views
+1

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)

 

0
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.

 

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