Question

Assume a 128KB virtual address space, with page size 64 bytes.
a) What is the number of entries in the linear page table for these parameters?
b) If each page table entry requires 4 bytes, what is the space occupied by a page table?
c) If at any point only 1% of the actual translations in the page table are valid, does a two-
level page table improve the space consumption? Explain with detailed calculation.

Answer 1

a) What is the number of entries in the linear page table for these parameters?

128KB = 2^7 x 2^10 => 2^17 i.e 17 bits in the virtual address
Page Size. = 64 bytes => 2^6 => 17 bits in the virtual address - 6 (page size) = 11 bits in the virtual page number
Number of entries = 2^11 = 2048

b) If each page table entry requires 4 bytes, what is the space occupied by a page table?

Page table size = 2^11 x 4 bytes => 8KB

Thanks, PLEASE COMMENT if there is any concern.