A virtual memory system with 32 bit address space uses a pur

A virtual memory system with 32 bit address space uses a purge paging system. What is the page table size of the page size a) 1k b) 4k c) 1M

Solution

Total number of addresses = 2³². Assuming byte level addressing,

a) Page size 1k = 1 kilobit = 1024/8 = 256 Bytes = 2⁸ bytes.

Then, number of entries in page table = 2³² / 2⁸  = 2²⁴.

Since each address is 32 bit, the page table size = 2²⁴ * 2⁵ = 2²⁹. Each 32 bit address is mapped to 1 of the 2²⁴ pages.

b) Page size 4k = 4 kilobit = 4096/8 = 1024 Bytes = 2¹⁰ bytes.

Then, number of entries in page table = 2³² / 2¹⁰  = 2²².

Since each address is 32 bit, the page table size = 2²² * 2⁵ = 2²⁷. Each 32 bit address is mapped to 1 of the 2²² pages.

c) Page size 1M = 1024 Kilobytes = 2²⁰ bytes.

Then, number of entries in page table = 2³² / 2²⁰  = 2¹².

Since each address is 32 bit, the page table size = 2¹² * 2⁵ = 2¹⁷. Each 32 bit address is mapped to 1 of the 2¹² pages.