1 votes 1 votes In a paged memory, the page hit ratio is 0.35. The page fault service time takes 100 ns. The time required to access a page in primary memory is 10 ns. The effective memory access time required to access a page is ____ ns Operating System test-series operating-system effective-memory-access page-fault + – LRU asked Oct 20, 2021 LRU 660 views answer comment Share Follow See 1 comment See all 1 1 comment reply Awe111 commented Oct 20, 2021 reply Follow Share @LRU Avg memory access time =0.35*10+0.65*100 =68.5 ns .. 1. https://gateoverflow.in/33767/what-is-effective-memory-access-time 1 votes 1 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes EMAT = hit ratio * MMAT + (1 – hit ratio) * (PFST) = 0.35 * 10 + (1 - 0.35) * (100) = 3.5+0.65*100 = 3.5+ 65 = 68.5 ns Vishal_kumar98 answered Oct 20, 2021 selected Oct 26, 2021 by LRU Vishal_kumar98 comment Share Follow See 1 comment See all 1 1 comment reply NIKHIL ADHIKARI commented Nov 23, 2021 reply Follow Share Why aren't we accessing MM 2 TIMES because if it is hit 1st we access page table that take 10ns then we fetch frame from MM that take again 10ns?? EMAT= HIT * (2MM) + MISS*(PFST) ??????? 1 votes 1 votes Please log in or register to add a comment.