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For a computer architecture with 64-bit virtural and physical addresses and a page size of 4KB.
How many page frames will there be?
How many entries are there in each page table?
What is the smallest possible size of a page table entry in BYTES, rounded up to an even number?
Expert Answer
Given Virtal address =64 bit
Physical address =64 bit
Page size = 4KB = 2^12
Number of entries in page table =
(virtual address space size)/(page size)
Therefore Number of entries= 2^64 /2^12 = 2^52
No. of bits required for Physical memory =64.
No.of page frames = physical address size/page size
= 2^64 /2^12 = 2^52 page frames
