Consider a system with Virtual memory 64G Physical memory

Consider a system with: Virtual memory = 64G Physical memory = 8G Page size = 8K Number of processes = 256 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:

We have given : Virtual address = 64GB = 2^6 * 2^30 B = 2^36B

Virtual address = 36bits

Physical address = 8GB = 2^3* 2^30 B = 2^33

Physical address space = 33bits

Page size = 8k = 2^3 * 2^10 B

Therefore page offset = 13 bits

Frames = 2^20

therefore number of bytes needed to store each entry of the page table = 20 bytes

and page table size = (2^13 * 20 ) bytes.