0 votes 0 votes Gate Fever asked Nov 9, 2018 Gate Fever 310 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments Gate Fever commented Nov 10, 2018 reply Follow Share THIS IS WHAT THEY HAVE GIVEN!!I DIDN'T GET IT 0 votes 0 votes Gate Fever commented Nov 10, 2018 reply Follow Share @Hemanth_13 i tried many times but am not getting your solution, pls explain your approach!! 0 votes 0 votes Hemanth_13 commented Nov 10, 2018 reply Follow Share Virtual address spaces is 2^32 B Page size is 2^10 B no of pages = 2^32/2^10=> 2^22, each will get on entry in page table Given page table entry is 4 B => Page table size is 2^22 * 4=> 2^24 as it greater than one page dividing further. no of pages=>2^24/2^10=2^14 Page size of 2nd level page table is 2^14* 4 => 2^16 B no of pages=>2^16/ 2^10==> 2^6 Page size for 3rd level page table is 2^6*4 => 2^8 as it can fit in one page (1)stopping here Total no of pages required is 1+2^14+2^6==>16449 0 votes 0 votes Please log in or register to add a comment.
