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A computer with a $32-bit$ address uses a two-level page table. Virtual addresses are split into a $9-bit$ top-level page table field, an $11-bit$ second-level page table field, and an offset. How large are the pages and how many are there in the address space?

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5 votes

$\underline{\mathbf {Answer:}}$

Number of bits used for virtual page numbers $=20$

Number of bits used for offset $=12$

$\therefore$ Size of Page $=\;\mathbf{4K} \;$.

$20$ bits for Virtual Page means $2^{20}$ pages.

edited by
1 votes
1 votes

Each level consists of only 1 page table, page table of 2nd level has 2^11 pages and page table of 1st level has 2^9 pages. So, total 2^20 pages.

And page size = (bits of VAS – (top level page table bits + bottom level page table bits))12 bits = 4KB.

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32 bit address,two page tables

 Virtual addresses are split into a 9−bit9−bit top-level page table field, an 11−bit     

 bit second-level page table field

32-9-11=12,12 is the offset,

if offset is equal to 12 ,page size =2^12

so max page size equal(How large are the pages ) =2^12

number of page : 2^9(in first level)*2^11(in first level)=2^20
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0 votes

Given, 32-bit address

2-level page table

virtual address: top level 9-bit, second level 11-bit

offset = 32-9-11 = 12

Offset = 12 therefore, page size = 2^12

so, virtual address = top level = 9-bit  → 2^9

                                 second level = 11-bit → 2^11

                                 offset = 12 bit → 2^12 → 4KB

No. of pages = top level * second level

                        = 2^9 * 2^11

                        = 2^20  → 1MB

 

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