2 votes 2 votes Consider a system with byte-addressable memory, $\text{31-bit}$ logical addresses, $4\;\text{kilobyte}$ page size and page table entries of $8\;\text{bytes}$ each. The size of the page table in the system in $\text{megabytes}$ is ________ Operating System go2025-os-2 numerical-answers memory-management paging virtual-memory + – gatecse asked Dec 7, 2020 gatecse 114 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes Number of PTEs $=$ Logical Address Space/Page size $ = 2^{31}/2^{12} = 2^{19}$ Each PTE being $8$ bytes, total size of the page table $ = 8 \times 2^{19} = 4\:\text{MB}$ gatecse answered Dec 7, 2020 • selected Nov 6, 2021 by Arjun gatecse comment Share Follow See all 0 reply Please log in or register to add a comment.
