1 votes 1 votes Actually there is solution also but not able to take screen shot. My solution is : TA1 = 0.95*30+0.05*300 TA2=0.95*30+0.05*270 TA1-TA2=3ns Am I correct? CO and Architecture co-and-architecture effective-memory-access + – Sumit1311 asked Jan 21, 2016 • retagged Nov 13, 2017 by Arjun Sumit1311 982 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 2 votes 2 votes TA1 = 0.95 * 30(HIT) + 0.05 * 330( CACHE ACCESS TIME + MEMORY ACCESS) TA2 = 0.95 * 30(HIT) + 0.05 * 300(MAIN MEMORY ACCESS as searching is done simultaneously) Therefore TA1 - TA2 = 1.5 ns Tushar Jain answered Jan 21, 2016 • selected Jan 21, 2016 by Arjun Tushar Jain comment Share Follow See all 4 Comments See all 4 4 Comments reply Sumit1311 commented Jan 21, 2016 reply Follow Share What does simulteneous access mean here then? I was assuming it as for first 30 secs also the memory access will be happening so I have taken it as 270. Is it not like that? Am I missing something? 1 votes 1 votes shivanisrivarshini commented Jan 21, 2016 reply Follow Share Could you plz explain why u have added 30 in case of miss 0 votes 0 votes shivanisrivarshini commented Jan 21, 2016 reply Follow Share Even untill hit takes place both main memory and cache is accessed na ?? then why u took only 30 incase of hit 0 votes 0 votes Adityasharma17 commented Oct 4, 2019 reply Follow Share While calculating TA2 why cache time is added there should be main memory access time? As it is write though cache. Ta2= 0.95×300+0.05×270 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes T1 = .95(30)+0.05(30+300) = 28.5 +16.5 = 45 T2 = .95(30)+0.05(30+270) = 28.5 + 15.0 = 43.5 T1 - T2 = 1.5 ns mystylecse answered Jun 23, 2017 mystylecse comment Share Follow See all 0 reply Please log in or register to add a comment.
