Question

Consider a system with byte-addressable memory, 32 bit logical addresses, 4 kilobyte page size and page table entries of 4 bytes each. The size of the page table in the system in megabytes is_____

Answer 1

Page table size = Number of page table entries $\times$ Entry size
= Number of pages $\times$ 4 bytes
= $\frac{2^{32}}{2^{12}}\times 4bytes=4MB$