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Question 1
Given:
* address space = 22 bits
* page size = 1 KiB = 1024 = 2
10
bytes
a.
No. of PTEs in onelevel page table =
±²³´µ¶´·¸¸¹³
´ º·»³
±²³´µ¶´º·¼³
= 2
22
/2
10
= 2
12
Not given in the question is the size of each PTE, usually it is of size 4 bytes (IA32 architecture).
The required space to store 2
12
PTEs = 2
12
× 4 = 2
14
bytes
The number of page for storing the onelevel page table = 2
14
/2
10
= 2
4
= 16 pages
To minimize the number of pages used to store the page tables of a twolevel table scheme, the only
solution is to have one table in the first level, which points to tables in the second level. It is known that
the second level has 2
12
entries and needs 16 pages to store these PTEs. This implies that one solution is
to have the first level table consists of 16 entries; each entry points to a small table in one single page.
Convert these information into the logical address format, we need to find v = (p, t, d).
d = no. of byte locations in a page = 1024 = 10 bits
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 Spring '10
 DR

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