A machine has 64bit virtual addresses and 32bit physical add

A machine has 64-bit virtual addresses and 32-bit physical addresses. Page size is 8KB. How many entries are needed for the page table?

Solution

The virtual address =64bits

physica address=32bits

page size=8KB

The physical memory of a machine is 2^32 bytes, since the physical address is 32 bits.

Since the page size is 8 KB, the number of pages is (2^32)/(2^13) i.e., 2^19.

Since each page table entry is four bytes in size and each page must

have one page table entry, the size of the page table is (2^19)*4 bytes or 2 MB