Testbook

Question

Couldnt attempt the question. Note that the solution given has powers of 2 instead of 214 and 215 and so on.

Comments

@sushant, I think page size and frame size both being different is not possible.

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I didnt get the quesion. First of all, how to find size of virtual memory? 

And what about the statement:

The page table is divided into 8k pages each of size 16KB

What can you interpret about the statement? Are they referring to innermost page table or outermost?

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Each level having 8K pages and page size=16KB.

No of pages in outer level=No of pages in inner level=8K.

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That would answer in Megabytes.

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outer page table size=No of pages*PTE

                              =8K*4B

                              =32KB.

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it should be    

#pages * page size

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Page table size=No of pages*PTE. //PTE= no of entries in a page of page table..

not page size.

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But they have given page size, right? Also you have assumed PTE = 4B.

How, can the #page table entries = 4B ?  :)

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@sushant

page table size=no of entries×PTE

And here what they want to say is that a single level page is divided into 8K pages and 2 level paging is implemented.

So we 8k entries in outer level and each of 32 bit which gives 32KB

But when I tried to calculate inner page table I couldn't match any options I tried frame size as 4KB as well as 16KB.

And moreover I don't think page and frame size could be different. Even if possible options are not matching for inner table for me.