Virtual Memory Consider a virtual memory system with the following specification
ID: 2084882 • Letter: V
Question
Virtual Memory Consider a virtual memory system with the following specifications 32-bit physical memory addresses (physical memory is byte-addressable) 40-bit virtual memory address (virtual memory is byte-addressable) Page size = 16K bytes (a) Determine the number of entries in the page table. (b) Determine the total size of the page table. Assume that the page table includes 1 valid bit, 1 dirty bit, and 2 bits for the LRU replacement policy. Assume, also, that the disk address is NOT stored in the page table.Explanation / Answer
Each table entry has protection bit,dirty bit,valid bit , tag address
Page Offset Length = 14 (as the page size is 16KB and 214 = 16K).
Physical Page Length = Physical Address Length - Page Offset Length = 32 – 14 = 18
Virtual Page Length = Virtual Address Length - Page Offset Length = 40 – 14 = 26
Table Size = 4 + 26 + 18 = 48
Table = Table entries*Table Entry Size = 8 * 48 = 384 bits = 48B
Each Page Table Entry has:1 valid,1 bits for LRU
Page Table Entry Size =4+18=22
Page Table Entries = 226
Page Table Size = Page Table Entries * Page Table Entries Size = 22 * 226 bits = 176 MB
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