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A computer system implements $8$ $\text{kilobyte}$ pages and a $32-bit$ 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.
in Operating System by Veteran (105k points)
edited by | 5.9k views
+1
@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
+2

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.

0
what is translation here. is it frame no. bits????

3 Answers

+54 votes
Best answer
$8$ $KB$ 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 $= 24$ $MB$
Page table size $=$ No. of PTEs $\times$ size of an entry
So, no. of PTEs $= 24 MB / 3 B = 8 M$

Virtual address supported = No. of PTEs * Page size (As we need a PTE for each page and assuming single-level paging)
= $8 M $*$ 8 KB$
= $64 $GB$ $=  $2^{36}$ Bytes

So, length of virtual address supported = $36$ bits (assuming byte addressing)
by Veteran (425k points)
edited by
0
sir, we dont divide the page table size with the page size to get entries because of the extra bits involved? so that we wont get correct no of entries?
0
Page Table Size = #entries * Size of an entry

Why is page size relevant here?
0
sir i thought by dividing the page table size with page size/frame size  we get no of entries,my flaw in concept.
0
Yes....I got it...I was making mistake in bits and Bytes conversion...hence getting answer less by 3...thanks a lot :)
+1
@resilent What is the logic for that division? Page table is something and page is something else. Why would one divind page table size by page size?
+2
@Hrishikesh

No. 36-13 = 23 bits are for indexing to page table. So, we have $2^{23}$ possible page table entries. Each of these entry is of 24 bits (19 for indexing and 5 for other purpose as in question). So, total page table size

$= 2^{23} \times 24\text{ bits}$

$ = 2^{23} \times 3 \text{ bytes}$

$ = 24 MB$
0
Why offset bit(13) has been subtracted from physical bit(32)?
Please help
+1
@abhishek

total physical address bit=bits to point frames no + bits to point word inside  frame(offset bit)
0

@Arjun sir in the formula 

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

does No. of PTE's also mean the number of pages??

0

@srestha ma'am

how we can say that  no. of pages means offset?

i think i am missing something

+17 votes
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
by Boss (13.5k points)
0 votes
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
by (305 points)
edited by
0
2^x * (19+5) = 2^20 * 24
x=23
explain

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