# GATE - 2013 | OS | A computer uses 46-bit virtual address, 32-bit physical address, and

## 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

## Solution:

### Option (C) is Correct.

Let the size of page is = 2p B

So the no. of entries in one page is 2p/4, where 4 is the page table entry size given in the question.

So we know that process size or virtual address space size is equal to
No. of entries × Page size

So total no. of entries for 3 level page table is,
(2p/4)×(2p/4)×(2p/4)

So, No. of entries × Page size = VAS
(2p/4)×(2p/4)×(2p/4)× (2p) = 246
24p = 252
4p = 52
p = 13

∴ Page size = 213

= 8 KB

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Architecture of physically indexed cache:

Architecture of virtual indexed physically tagged (VIPT):

VIPT cache and aliasing effect and synonym.