GATE CSE 2016 Set 1 | Question: 47

Question

Consider a computer system with $40$-bit virtual addressing and page size of sixteen kilobytes. If the computer system has a one-level page table per process and each page table entry requires $48$ bits, then the size of the per-process page table is __________ megabytes.

Comments

Every process has its own page to maintain. That is what written here.

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Pardon me if my question is invalid, but they have asked “size of the per-process page table", now can we take all the 26 bits or one process? I mean total virtual memory is $2^{26}$ bits, then can we take the entire virtual memory for one process? Should not they mention something about the number of processes, then only we can calculate size of the per-process page table 

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In this kind of question where process size is not mentioned, you have to consider the worst case scenario i.e. the entire LAS will be equal to a process size. That’s how we’ll get 2^26 number of pages.

Answer 1

$\text{No. of pages} (N) =  2^{26} = \text{No. of entries in Page Table}$
$\text{Page Table Entry Size}(E) = 6\; \text{bytes}$

So, $\text{Page Table Size} = n \times e = 2^{26} \times 6 \text{ bytes} = 384 \text{ MB}$

Comments

there are 2^26 pages and for each of these page we will do entry .

And that entry is suppose to be written in some page.

And size of page is 16KB therfore 16KB/6B =x, entries in 1 page .

so no. of page is required will be 2^26/x =y.

and answer should be y*page size

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can we say it as 2^26*2^48 bytes..then how to convert it into mb??

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@mo7ammedfarooq. 

Given $40-bit$ for virtual addr and $16 KB$ page-size, no of pages is

$2^{40}$ / $2^{14}$

= $2^{26}$ 

Now, given each PTE is 48 bits in size,

Page Table size = no. of PTEs * size of PTE

= $2^{26} * 48 bits$

= $2^{26} *$ $6bytes$ ( $\because$ 48bits = 6 bytes )

= $2^{20} * 2^{6} * 6$

= $64 * 6 MB$ ( $\because 2^{20} bytes = 1 MB $ )

= $384 MB$

 

Answer 2

40 bit address system. So total Address Space =2^40 Bytes

16 Kbytes Page size=2^14bytes

Total Frames(Number of pages of page table)=(2^40)/(2^16)=2^26

As Table entry is 48 bits, so each page=6 bytes

Total=2^26*6=2^20*2^6*6=384MB

Comments

why the total Address Space =2^40 bits  is wrong ?
m  got confused most of the times whether to take bit or byte.
plz explain

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we take the basic addressible unit words..... 2^40 words by default we consider word as bytes unless word size specified

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The phrase "Total Frames" is incorrectly used here...

Answer 3

384 MB   ( 2²⁰*2⁶*6 bytes)