Consider a computer with a 64MB of RAM and a 32- bit virtual address space. The

Question

Consider a computer with a 64MB of RAM and a 32- bit virtual address space. The offset within a page is 12 bits. a) What is the size, in bytes, of a single virtual memory page on this computer? b) Each page table entry is 4 bytes in size. How much memory must be reserved for an architecture that uses a single level page table? c) If the computer architecture is updated to use a 2-level page table scheme like the diagram below How large of a virtual address range would a single Is1 level page table entry cover?

Explanation / Answer

Given, RAM = 64MB and having address space as 32bit.
Offset with in a page = 12 bits

a) Offset --> 12 bits --> It requires 4 KBytes/page

So, page size = 4 KB/page

Number of pages = 2^32 / 4KB = 2^32 / 2^12 = 2^20
given, Physical memory being 64 MB,physical address must be 26 bits having page of size 4KB address needs 26-12 = 14 address bits.
So, each page table entry must be at least 14 bits it requires 2 bytes long.
Size of page table = (2^20 *2) = 2MB

b) Assume a single level page table having 4 bytes entry.Thus, the page number is nothing but difference of address space and offset bits as 32 - 12 = 20 bits

Memory required = 20 bits

c)

* level 1 pages = 2^10 = 1K pages.

Physical memory is 64 MB,physical address must be 26 bits having page of size 4KB address needs 26-12 = 14 address bits.So, each page table entry must be at least 14 bits it requires 2 bytes long.

* Each second level page table has 2^10 entries, we need 16 (i.e. 2^4) second level page tables.There is one first-level page table and 16 second level page tables, the total
space occupied by the page tables --> (17 * 4KB) = 68KB

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