0 votes 0 votes Consider a two-level memory hierarchy, L1 (cache) has an accessing time of 5 ns and main memory has an accessing time of 100 ns. Writing or updating contents takes 20 ns and 200 ns for L1 and main memory respectively. Assume L1 gives misses 20% of the time with 60% of the instructions being read-only instructions. What is the average access time for the system (in ns) if it uses the WRITETHROUGH technique? CO and Architecture co-and-architecture cache-memory multilevel-cache numerical-answers + – Shivangi Parashar 2 asked Nov 25, 2018 • edited Aug 2, 2022 by Shubham Sharma 2 Shivangi Parashar 2 1.3k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments VIDYADHAR SHELKE 1 commented Jan 26, 2020 reply Follow Share @srestha what is correct ans ma'am 0 votes 0 votes srestha commented Jan 26, 2020 reply Follow Share what should be?? Can u give the correct logic? " 60% of the instructions are read only instructions. " right?? 0 votes 0 votes VIDYADHAR SHELKE 1 commented Feb 5, 2020 reply Follow Share @srestha ma'am avg time ={(0.8*5+0.2*(100+5)}*0.60 + {(0.8*20+0.2*(200+20)}*0.40 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes read:0.8(5)+0.2(5+100)=25 ns write:0.8* max(20,200)+0.2(200)=200 ns total time=0.6*25+0.4*200 =95 ns arun yadav answered Sep 10, 2020 arun yadav comment Share Follow See all 0 reply Please log in or register to add a comment.
