2 2 votes Operating System operating-system paging + – ashish pal 1.9k views answer comment Share Follow Print See all 9 Comments 9 9 Comments reply Show 6 previous comments Karan Saini commented Dec 21, 2017 reply Follow flag A 64-bit processor. -> every 8 bytes are physically addressed. 0 0 replyShare Vishal Rana commented Jan 16, 2018 reply Follow flag Please help me out in this. Our page offset bits = 16( in this question). And we also know that the outermost page table should fit in one frame. So in worst case we can have 2^16 addressable locations in a page, so how come the outermost page table is indexed by 32 bits (which implies that the page of outermost page table has 2^32 addressable locations). Thanks in advance. 0 0 replyShare sachin486 commented Sep 9, 2020 reply Follow flag quite a ambiguous question.there has to be way to find PTE size by the way u can read arjun sir comments on this question here https://gateoverflow.in/72799/ace-test-series-os 0 0 replyShare Please log in or register to add a comment.
0 0 votes The total entries possible in outer page table are 2^32 each one of size 8 bytes.. i.e, 2^32 * 8 bytes 4 GB * 8 = 32 GB Answer = option D. Kshitij Mhatre answered Dec 26, 2017 Kshitij Mhatre comment Share Follow See all 2 Comments 2 2 Comments reply ashish pal commented Dec 26, 2017 reply Follow flag How the entry size is 8 bytes ? 0 0 replyShare Osheen commented Jan 3, 2019 reply Follow flag I think the answer is 8GB...2^32 * 2=8GB 0 0 replyShare Please log in or register to add a comment.
