Question

Given a 32 bit processor, with 2 MB of physical RAM split into 512 frames. What is the size of the single level page table, in entries, if the maximum addressable virtual memory space is 16MB

Answer 1

Size of physical memory = 2 MB = 2 x 2²⁰ Bytes = 2²¹ B

Size of virtual memory = 16 MB = 2⁴ x 2²⁰ = 2²⁴ B

Physical RAM is divided into 512 frames = 2⁹ frames

Size of one frame = (2²¹ ) / (2⁹) = 2¹² B = 4 KB

Therefore, Size of frame = size of page = 2¹² B = 4 KB

Total number of pages = size of virtual memory / page size = (2²⁴ / 2¹²) = 2¹² pages.

Each page has a entry. Since there are 2¹² pages, we have 2¹² entries. Each entry contains frame number and some other bits. Since we have 2⁹ frames we need 9 bits to store frame number. Neglecting all other bits, size of page table entry = 9 bits.

So, Size of single level page table = No. of entries x size of each entry

= 2¹² x 9 bits = 4 x 2¹⁰ x 9 bits = 36K bits = (36/8) K Bytes = 4.5 KB

Therefore, Size of single level Page Table = 4.5 KB