0 votes 0 votes L1 and L2 cache access time 2ns and 5ns respt. Hit percentage 3% and 5% respt. If Average access time =2.4ns then memory access time=? utk0203 asked Mar 23, 2017 utk0203 604 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments srestha commented Mar 23, 2017 reply Follow Share yes may be some wrong data I am getting -ve ans equation will be $0.03\times 2+0.97\times 0.05(2+5)+0.97\times 0.05(2+5+m)=2.4$ 0 votes 0 votes Heisenberg commented Mar 23, 2017 reply Follow Share yes it's possible . if we add the access time of previous memory hierarchy then memory time comes out to be -ve and if you dont, you still get a value very close to access time of L1 which is impossible. 0 votes 0 votes Heisenberg commented Mar 23, 2017 reply Follow Share some books have the convention of adding L1 time to L2 and some books don't add. It is still a grey area in COA 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes solution... akash.dinkar12 answered Mar 24, 2017 akash.dinkar12 comment Share Follow See 1 comment See all 1 1 comment reply JeetRoy commented Mar 27, 2017 reply Follow Share H1T1+(1-H1)H2T2+(1-H1)(1-H2)H3T3=T .03*2 + .97*.05*5 + .97*.95*T3 = 2.4 T3 = 2.276 ns 0 votes 0 votes Please log in or register to add a comment.