0 votes 0 votes The virtual memory system uses the demand paging for its implementation. The probability of getting page faults is $0.25$, the normal memory access time is $200$ nanoseconds. If it takes $2$ millseconds to service a page fault, then what is effective memory access time? $500000$ ns $500075$ ns $500150$ ns $500250$ ns Operating System tbb-os-2 + – Bikram asked Dec 26, 2016 edited Aug 27, 2019 by go_editor Bikram 537 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 0 votes 0 votes E.M.A.T. = (1 - p) * ma + p * page fault service time , where p = page fault rate EMAT = [ 0.75 * 200 + 0.25 * 2000000 ]ns = 500150 n.s Bikram answered Dec 26, 2016 selected Jan 12, 2017 by Bikram Bikram comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments Bikram commented Jul 12, 2017 reply Follow Share @ hem chandra joshi yes, we consider memory access time . in this formula Effective Memory Access Time = (1 - p) * ma + p * page fault service time we use memory access time as ma . 0 votes 0 votes Jai Kataria commented Sep 17, 2017 reply Follow Share Sir, We should use hierarchical approach as page fault service time to move page from disk to memory and update page table, after that again memory reference takes place. 0 votes 0 votes slow_but_detemined commented Jan 30, 2020 reply Follow Share For those who are confused about which formula to use. https://gateoverflow.in/41917/confusion-in-effective-memory-access 1 votes 1 votes Please log in or register to add a comment.