GATE CSE 2015 | Set 1 | Question: 1‏2

Question

Consider a system with byte-addressable memory, $32\text{-bit}$ logical addresses, $4\;\text{kilobyte}$ page size and page table entries of $4\;\text{bytes}$ each. The size of the page table in the system in $\text{megabytes}$ is___________.

Answer 1

Total no of pages $= \frac{2^{32}}{2^{12}} = 2^{20}$

We need a PTE for each page and an entry is $4$ bytes.

So, page table size  $= 4 \times 2^{20} = 2^{22}\;\textsf{B} = 4\;\textsf{MB}.$

Comments

Raju Kalagoni

Given that system is byte-addressable memory. Which means a bit can address a byte. Total 4 kilobytes = 4096 bytes, which can be addressed by 12 bits. Ie page offset is 12 bits.

Bits required to represent the number of pages = 32-12 =20 bits, and total number of pages = $2^{20}$

Now

What is mean by 32 bit logical address ? 32 bits are used to represent a word. It doesn't mean that a word is bit. A word may be a byte or 2 byte etc.. its depend on the design of the machine. Usually we represent word = 1 byte. That is why page size is in byte.

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here frame no means bits required to refer frames?

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@Neelam_$ingh_222 Yes bro . 

Answer 2

LA=32 bits

Page size or offset =4KB=>12 bits 

pages=32-12=20 bits ..so there are 2²⁰ pages .For each page there would be an entry in the page table ..and PTE=4 bytes

page table size= 2²⁰ * 4 B=> 4MB

Comments

will answer remain same if i calculate size of PT1 then size of PT2  and then add both??..

i guess this method will also give same answer

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page table size=number of pages in page table(which is 2^page number bits of virtual address)x page size.

why are we taking 2^20 bits of logical address as number  of pages.?

plz help

Answer 3