Part 1. Consider a demand paging virtual memory system with 22 bit virtual addre
ID: 3760001 • Letter: P
Question
Part 1. Consider a demand paging virtual memory system with 22 bit virtual address space, if page size of 8KB is used when we split a logical address into page number and offset within the page, how many bits are used for the page number, and how many bits are used for the offset? Using the 22 bit virtual address space from above, if we use a page size of 2KB, when we split a logical address into page number and offset within the page, how many bits are used for the page number, and how many bits are used for the offset?Explanation / Answer
A 22 bit address can address 2^22 bytes in a byte addressable machine.
a)Since the size of a page 8K bytes= (2^13 )Bits
the number of addressable pages is 2^22 / >2^13 = 2^9 bits
Offset = log2 (page size) bits
=lod2(2^13)=13 bits
Page size = 8 KB = 2^13 B
Offset = 13 bits
b)Since the size of a page 2K bytes=
Page size = 2 KB = 2^20 B
the number of addressable pages is 2^22 / >2^20 = 2^2 bits
Offset =20 bits