GATE CSE 2015 Set 2 | Question: 47

Question

A computer system implements $8\;\text{kilobyte}$ pages and a $32\text{-bit}$ physical address space. Each page table entry contains a valid bit, a dirty bit, three permission bits, and the translation. If the maximum size of the page table of a process is $24\;\text{megabytes}$, the length of the virtual address supported by the system is _______ bits.

Comments

@shrestha @arjun

how to identify that it is page size not number of pages ?Like in question most of the time I am assuming 8kilobutes pages means no of pages not page size.please help

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8KB means 2¹³ Bytes, How can a number have a unit (like here its has Bytes). Had it been number of pages they would have written it as 8 Kilo pages (or something like this but without any unit). We can have 10 pages 20 pages but having 10 Bytes page indicate it is the size of page.

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what is translation here. is it frame no. bits????

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8KB pages mean size of each page is 8KB

Answer 1

$8\;\textsf{KB}$ pages means $13$ offset bits.

For $32$ bit physical address, $32 - 13 = 19$ page frame bits must be there in each PTE (Page Table Entry).

We also have $1$ valid bit, $1$ dirty bit and $3$ permission bits.

So, total size of a PTE (Page Table Entry) $= 19 + 5 = 24$ bits $= 3$ bytes.

​Given in question, maximum page table size $= 24\;\textsf{MB}$

Page table size $=$ No. of PTEs $\times$ size of an entry

So, no. of PTEs $= 24\;\textsf{ MB} / 3\;\textsf{B} = 8\;\textsf{M}$

Virtual address supported $=$ No. of PTEs $\ast$ Page size (As we need a PTE for each page and assuming single-level paging)

$= 8\;\textsf{M} \ast 8\;\textsf{KB}$

$= 64\;\textsf{GB} = 2^{36}$ Bytes

So, length of virtual address supported $= 36$ bits (assuming byte addressing)

Comments

What would be the difference if we assume word addressing ? just curious.

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Hi All,  In this question what is mean by translation . Is it the frame numbers bits?

Please confirm 

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@Nandhakumar Yes bro. 

Answer 2

Ans 36 bits

 

Assume Virtual Address = x bits

entry size in page table = 19(frame bits)+5(permission bits) = 24 bits = 3 Bytes

2^(x-13)*3 = 24*2^(20)

solving we get x = 36 bits

Answer 3

let 2^x pages in pg table
2^x * (frame size + book keeping)= 24MB
2^x * (19+5) = 2^20 * 24
x=23

virtual address = page no bits + offset = 23 +13 = 36

Ans : - 36

Comments

2^x * (19+5) = 2^20 * 24
x=23
explain

Answer 4

we know page size in virtual address is equal to page size in physical address
 

so , 32bit physical address where each page is 8kb (2^13) so no.of frames in physical address is 2^32 / 2 ^13 = 2^19.

19 bits in page table entry and 1 valid bit , 1 dirty bit , 3 permission bits total 24 bits in page table entry

page table size is 24mb and each page table entry has 24 bits means 3 bytes
now Number of entries in page table = size of table/ size of each entry

24mb / 3 bytes = 800000 entries in page table

8000000 can be represented by 23bits
page number is 23 bits and page size is 8kb (2 ^ 13)

we know if a size of virtual address is 2^m has a page of size 2^n
then m-n bits of logical address represent page number and n bits offset

if m-n = 23

n = 13

m = 23 + 13

m = 36 bits of virtual address