2 2 votes Consider a two level memory hierarchy L1 (cache) has an accessing time of 10 nsec and main memory has accessing time 100 nsec. Assume the hit ratio read operation is 0.75 and 40% references are for write operation. The average access time for system (in nsec) if it uses write through technique ________. (Upto 1 decimal places) Ans. 67.6 CO & Architecture effective-memory-access co-and-architecture + – Na462 4.0k views answer comment Share Follow Print See all 9 Comments 9 9 Comments reply Na462 commented Jul 24, 2018 reply Follow flag Please Help me around here:- Please tell me where i am wrong and why i am wrong:- Teff = 0.75 * (60 % * 10 + 40 % * 100 ) + 0.25 * ( 60 % * 110 + 40 % *100) = 61. 60 % -----> If its a read and 40% if its a write 0 0 replyShare Na462 commented Jul 24, 2018 reply Follow flag Made Easy Solution:- 0 0 replyShare MiNiPanda commented Jul 24, 2018 reply Follow flag Solution provided is wrong. You can see they have done (1-0.75)=0.35 instead of 0.25 for which the answer they obtained = 67.6 1 1 replyShare Na462 commented Jul 24, 2018 reply Follow flag Ohk i didn't saw that is my solution correct sir ? 0 0 replyShare MiNiPanda commented Jul 24, 2018 reply Follow flag Teff = 0.75 * (60 % * 10 + 40 % * 100 ) + 0.25 * ( 60 % * 110 + 40 % *100) = 61. I am getting the same answer but i am not sure how to interpret this line. It seems like you are doing like this -> 0.75 * (60 % * 10+ 40%*100) When there is a hit, Read operation takes 60% time of the L1 time to access it and write ops take 40% of the memory time. I do like -> When there is a hit, read op takes 10ns (i.e. the time to access L1) and when there is miss, it takes 110ns. Hence, 0.75(10)+ 0.25(110) For write operations, whether there is hit or miss does not matter as the effective memory access time = time to access main memory as it follows write through policy. So 100ns. Now 40% of the ops are write and 60% are reads..So, 60%( 0.75*10+ 0.25*110 ) + 40%( 100) 1 1 replyShare Na462 commented Jul 24, 2018 reply Follow flag Well sir you are calculating seperately that's ok. What i did is that here when there is hit. Now in hit we can have 40% of write and 60% of read and like wise in miss. Therefore i wrote:- 0.75 * (60 % * 10 + 40 % * 100 ) + 0.25 * ( 60 % * 110 + 40 % *100) = 61 By the way both are giving the same answer yours as well as mine i just wanna know that my approach is good or not :) 0 0 replyShare MiNiPanda commented Jul 24, 2018 reply Follow flag Yes we will be getting same answers anyway..it was just that i wasn't getting your approach.. Now it is okay :) And do you really think i am a teacher even after seeing this username? :P :P 1 1 replyShare Mk Utkarsh commented Aug 23, 2018 reply Follow flag 61 :/ 0 0 replyShare RubyGATE commented Jan 12, 2020 reply Follow flag (1- Hit%) * (cache + main memory) , why is cache memory access time considered ? 0 0 replyShare Please log in or register to add a comment.