1 votes 1 votes A machine has $48-bit$ virtual addresses and $32-bit$ physical addresses. Pages are $8\: KB.$ How many entries are needed for a single-level linear page table? Operating System tanenbaum operating-system memory-management paging descriptive + – admin asked Oct 26, 2019 • edited Oct 26, 2019 by Lakshman Bhaiya admin 715 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Answer: $\mathbf{\;2^{35}\;entries}$ $\because$ Total number of bits in the page number field = $48-13 \;\text{bits} = 35$ $\therefore \text{Number of page table entries } = 2^{35} \;\text{entries}$ `JEET answered Oct 29, 2019 • edited Jan 29, 2020 by `JEET `JEET comment Share Follow See all 3 Comments See all 3 3 Comments reply yagnesh commented Oct 30, 2019 reply Follow Share Why 12 bits? 1 votes 1 votes Ram Swaroop commented Jan 29, 2020 reply Follow Share page size give 8KB =13 bit virtual address= 48 bit so no of entries 2^35 0 votes 0 votes `JEET commented Jan 29, 2020 reply Follow Share @yagnesh Corrected! 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes 2^35 pages will be there Anup dogrial answered Jan 26, 2020 Anup dogrial comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes page size give 8KB =13 bit virtual address= 48 bit bit for no of entries in page table at first level=48-13=35 bit so no of entries 2^35 Ram Swaroop answered Jan 29, 2020 Ram Swaroop comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes page size is 8KB total bit in page offset =13 (2^13=8KB) Virtual address space will be divided into two part i.e. Page no + Page offset Hence,Total pages = 2^(no of bits in page no) = 2^(48-13)=2^35 himanshukumarpatel answered Jan 29, 2020 himanshukumarpatel comment Share Follow See all 0 reply Please log in or register to add a comment.