Consider a system with Virtual memory 8G Physical memory 3

Consider a system with Virtual memory = 8G Physical memory = 32G Page size = 2M Number of processes = 512 Analyze this system to find entries such as page table entry size: size of page table, total size of page table, size of page table entry, number of page table entries, size of the entire page table. assume 2 level paging with 3 bits for identifying the entry in level 1 page table. In other words, L1 = 3 bits.

Solution

Answer:

Answer:

We have given : Virtual address = 8GB = 2^3* 2^30 B = 2^33B

Virtual address = 33bits

Physical address = 32GB = 2^5* 2^30 B = 2^35

Physical address space = 35bits

Page size = 2M = 2 *2^20= 2 ^22 B

Therefore page offset = 22 bits

Frames = 2^13

therefore number of bytes needed to store each entry of the page table = 13 bytes

and page table size = (2^22 * 13 ) bytes.