Question

Given a virtual memory configuration with: • 4 TB virtual memory space • 32 GB physical memory space • 16 KB page size How many bits are needed for the virtual address? Choose... How many bits are needed for the physical address? Choose... How many bits in the address correspond to page offset bits? Choose... - How many bits will the virtual page number (VPN) be? Choose... - How many bits will the physical page number (PPN) be? Choose... → Please answer all parts of the question.

Answer 1

Vrtual memory space = 4TB = 4*(2^40)= 2^42 bytes

physical memory space = 32GB = 32*(2^30) = 2^35 bytes

page size = 16KB = 16*(2^10) = 2^14 bytes

Virtual Page size =Physical Page size

Number of physical pages = physical memory size / physical page size = (2^35)/(2^14) = 2^21

physical page number requires = 21 bits

Number of virtual pages = virtual memory size / virtual page size = (2^42)/(2^14) = 2^28

virtual page number requires = 28 bits

page size = 2^14, so offset length = 14 bits

a) virtual address bits = virtual page number bits + offset bits = 28+14=42 bits

b) physical address bits = physical page number bits + offset bits = 21+14=35 bits

c) page offset bits=14

d) virtual page number = 28 bits

e) physical page number = 21 bits