3 3 votes Consider a computer system using a $3$-level paging scheme. The Virtual Address is $46$ bits, and the Physical Address is $32$ bits. The page size is $\mathbf{8 ~ K B}$. The page table base register (PTBR) stores the base address of the first-level table $(T_1)$, which occupies exactly one page. Each entry in $T_1$ points to a second-level table $\left(T_2\right)$, and each entry in $T_2$ points to a third-level table ( $T_3$ ). Each entry in $T_3$ is a Page Table Entry (PTE). If the size of each PTE is $\mathbf{3 2}$ bits, what is the maximum number of entries possible in a single $T_3$ table? Operating System goclasses operating-system goclasses-cs-dpp goclasses-cs-dpp-day-190 goclasses-os-practice-questions numerical-answers + – GO Classes 299 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
2 2 votes The page size is $8 \mathrm{~KB}=2^{13}$ bytes. Since the first-level table $T_1$ occupies exactly one page and each PTE (or pointer to the next level) is $32$ bits $( 4$ bytes$)$, the number of entries per page is: $$ \frac{2^{13}}{2^2}=2^{11}=2048 \text { entries } $$ This structure remains consistent for each level if they are all page-sized. Therefore, a single $T_3$ table can hold $2048$ entries. GO Classes answered Feb 3 GO Classes comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes 2048 sandeshrbhat24 answered Feb 3 sandeshrbhat24 comment Share Follow 0 reply Please log in or register to add a comment.