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: page table entry size, size of page table, total size of page table etc 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.