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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.
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96 votes
96 votes
$8\;\textsf{KB}$ 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)
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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
1 votes
1 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
0 votes
0 votes
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
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