NIELIT 2017 July Scientist B (CS) - Section B: 33

Question

A computer uses $46$-bit virtual address, $32$-bit physical address, and a three–level paged page table organization. The page table base register stores the base address of the first-level table $(T1)$, which occupies exactly one page. Each entry of $T1$ stores the base address of a page of the second-level table $(T2)$. Each entry of $T2$ stores the base address of a page of the third-level table $(T3).$ Each entry of $T3$ stores a Page Table Entry ($PTE$). The $PTE$ is $32$ bits in size. The processor used in the computer has a $1$ MB $16$ way set associative virtually indexed physically tagged cache. The cache block size is $64$ bytes.

What is the size of a page in $KB$ in this computer?

• A. $2$
• B. $4$
• C. $8$
• D. $16$

Comments

Previous year

https://gateoverflow.in/379/gate2013-52

Answer 1

Ans is 8 KB . 

You have to assume the page size as 2^x and hence in VA it will take x bits which will leave the rest as (46-x )

Now that indicates that Page size is 2^(46-x) * 4B ……...{ as PTE is 4B } 

But since this is not a single level page mechanism therefore we divide it further and hence second level page table 

size would be  [{ 2^(46-x) * 4 } / 2^x ] * 4B which simplifies to 2^(46-2x) * 16 B 

and another page table which is 3^rd is 2^(46-3x) * 64 B which must be equal to page size which is 2^x 

Solving this : we get  4x = 52 and x =13 and hence Page Size is 8KB