0 0 votes IS THE ANSWER 1.794?? CO & Architecture co-and-architecture cache-memory hit-ratio numerical-answers made-easy-test-series + – Gate Fever 1.4k views answer comment Share Follow Print See all 11 Comments 11 11 Comments reply Shubhanshu commented Nov 3, 2018 reply Follow flag Either hit or miss ratio can't be greater than 1. 0 0 replyShare Gate Fever commented Nov 3, 2018 reply Follow flag oh yess, then what is the correct answer?? i know answer given in solution is wrong!! 0 0 replyShare Gate Fever commented Nov 3, 2018 reply Follow flag see where am i wrong?? 0 0 replyShare Gate Fever commented Nov 19, 2018 reply Follow flag @Shubhanshu ?? any idea?? am still confused!!not getting it 0 0 replyShare Gate Fever commented Nov 19, 2018 reply Follow flag is the hit ratio = 0.7983?? pls check @Shubhanshu 0 0 replyShare Shubhanshu commented Nov 19, 2018 reply Follow flag @Gate Fever check this out:- 1 1 replyShare Gate Fever commented Nov 19, 2018 reply Follow flag you calculated miss ratio blockwise not element wise , why?? i also got same no. of misses,but element wise first time total no. of misses will be 6, yes! so i did it like this :- on 42 elements there will be 6 misses; total misses =6/42 now when loop goes from 2nd iteration to 10th iteraion in each iteration we have 2 misses on 34 elements so total misses = 2/34 hence total miss rate = 6/42 + 2/34 = 0.20168 hence hit ratio = 1-0.20168 = 0.7983 whts incorrect in this???? 0 0 replyShare Shubhanshu commented Nov 19, 2018 reply Follow flag you calculated miss ratio blockwise not element wise , why?? Because transfer between Cache memory to/from Main Memory is blockwise, not wordwise. on 42 elements there will be 6 misses; total misses =6/42 Ok. now when loop goes from 2nd iteration to 10th iteraion in each iteration we have 2 misses on 34 elements so total misses = 2/34 from 2nd to 10th there are 9 iterations. And in each iteration, there are 2 misses and 2 hits. So total hit and miss will be $18$. You have to take misses for all 9 iterations, not for the single iteration. thus getting $\frac{18}{42}$ So miss ratio will be $\frac{6}{18} + \frac{18}{42} = \frac{24}{42}$ And hit ratio will be $\frac{18}{42}$ 1 1 replyShare Gate Fever commented Nov 19, 2018 reply Follow flag now when loop goes from 2nd iteration to 10th iteraion in each iteration we have 2 misses on 34 elements therefore (2*9)/(34*9) thats why 2/34 0 0 replyShare Shubhanshu commented Nov 19, 2018 reply Follow flag What formula are using for Miss ratio calculation? Btw standard formyal to calculate miss ratio is = $\frac{\text{no of misses}}{\text{Total no of memory references}}$ 0 0 replyShare Gate Fever commented Nov 19, 2018 reply Follow flag i dont know where ia m going wrong!! ill solve it with a fresh mind again & then will see!! one more thing Shubhanshu, from where can i prepare questions for I/O INTERFACE like for DMA,POLLING etc, pls tell me a good source 0 0 replyShare Please log in or register to add a comment.