MadeEasy Test Series: Operating System - Memory Management

Question

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, then the memory overhead for this machine is ________ KB

please explain.

Answer 1

As each page table occupies one page and size of page table entry is $4B$.

Number of entries in page table = $4KB/4B$ = $1024$ . Each entry of first level page table will point to a second level page table.

This means $1024$  $second -level- page -tables$ and $1$ $ first -level- page- table$ Hence total memory overhead is = $1025X4KB$ = $4100KB$

Comments

but instead, if we use single level paging, then memory overhead will be equal to (2^32/2^12)*4B=2^22B=4096KB which is less than 4100B , then what is the need of using 2 level paging

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Why this solution is given in test series ? Please explain it.

1024 * 4KB + 1024 * 4KB = 8KB

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The solution provided by made easy is wrong. I also answered 4100KB.

Answer 2

I don't know the meaning of memory overhead. My guess extra space required by the page tables. Please help out if i'm wrong on that. The calculation on my solution is right though.

Comments

i think u are correct... it should be 5000KB

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$4 \text{MB}$ + $4KB = 4100KB$

Answer 3

It's clearly mentioned in the question that each page table occupies one page.

The usual way of solving is giving us an inner page table of size 4MB > PS. Thus, this approach is not asked in the question.

Hint: 1024 outer page table entries should point to 1024 inner page tables. Thus, outer page table size = 1024 entries * 4B entry size

And inner page table size = (1024 entries * 4B entry size) * 1024 such tables

Thus the overhead is : 1024 * 4B + 1024 * 4B * 1024 = 4 KB + 4096 KB = 4100 KB

 

Comments

From $1024$ inner(2^nd level) page tables only $1$ table will be stored in memory right, then why are you storing all the $1024$ tables?...and it’s not possible to store all 2^ndlevel PTs in memory as $1024*1024*4B = 4MB >$ Size of frame $(=4KB)$. 

So, the answer should be $8KB$ memory overhead I guess.

please correct me if I’m wrong.