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Page Table

Consider a system with byte-addressable memory, 32-bit logical addresses, 4Kb page size and page table entries of 4 bytes each.The size of the page table in the system in mb is 

Answer

No. of page table = logical address/ page size

                           =232 / 2 * 210

                                     =220

Size of Page Table =  No. of Page Table * Page Entry Size

                               = 220 *4

                               =4Mb

2Comments
Digvijay Pandey @digvijay
2 Mar 2016 03:27 pm
Page size is 4Kb i.e. 0.5KB.
Number of pages = 2^32/0.5*2^10 = 8M

Size of page table = Number of pages * Page table entry size = 8M*4B = 32MB = 32M*8b = 256Mb

PS : there is difference between kb and kB

PREETI KAUR @preetikaur
2 Mar 2016 05:11 pm

Thanx for correction Sir and sorry to all for my mistake