0 votes 0 votes Consider the following cache A and B .let the average access times in cache A and B is $t_A$ and $t_B$ respectively.the value of $t_A+t_B$ _(in ns) Answer is given as 30.18 ns I am getting 31.2 ns .Please verify it . CO and Architecture co-and-architecture cache-memory made-easy-test-series + – Prateek Raghuvanshi asked Dec 22, 2018 • edited Mar 4, 2019 by ajaysoni1924 Prateek Raghuvanshi 1.0k views answer comment Share Follow See all 16 Comments See all 16 16 Comments reply Astitva Srivastava commented Dec 22, 2018 i moved by Shaik Masthan Dec 22, 2018 reply Follow Share 31.2 ns is correct. 1 votes 1 votes aambazinga commented Dec 22, 2018 reply Follow Share 31.2ns. 1 votes 1 votes Magma commented Dec 22, 2018 reply Follow Share i too got 31.2 1 votes 1 votes Prateek Raghuvanshi commented Dec 22, 2018 reply Follow Share thank you for verifying it.they just ignored the time access of L1 when hit in L2. 0 votes 0 votes himgta commented Dec 22, 2018 reply Follow Share @Prateek Raghuvanshi @Magma @aambazinga please explain the solution! 0 votes 0 votes Prateek Raghuvanshi commented Dec 22, 2018 reply Follow Share @ himgta , $t_A=h_1*t_1+(1-h_1)(t_1+t_m)$ $t_B=h_1*t_1+(1-h_1)h_2(t_1+t_2)+(1-h_1)(1-h_2)(t_1+t_2+t_m)$ put the value you will get the answer 0 votes 0 votes himgta commented Dec 22, 2018 reply Follow Share @Prateek Raghuvanshi there is another formula also for default access...how to identify that which formula has to be used when? 0 votes 0 votes Prateek Raghuvanshi commented Dec 22, 2018 reply Follow Share @ himgta by default cache memory is hierarchical. for more refer this -https://gateoverflow.in/108780/simultaneous-vs-hierarchical-memory 1 votes 1 votes Deepanshu commented Jan 2, 2019 reply Follow Share 21 + 9.30= 30.30 i am getting( only rounding up to 2 everything ) 0 votes 0 votes Deepanshu commented Jan 2, 2019 reply Follow Share and in nanosecond not in ms 0 votes 0 votes Nandkishor3939 commented Jan 6, 2019 reply Follow Share @Deepanshu will you please elaborate it , m not getting the answer What i did Ta=0.8*5+0.2*(80+5) Tb=0.6*3+o.4*0.15*(6+3)+o.4*0.85(80+6+3) 0 votes 0 votes Nandkishor3939 commented Jan 6, 2019 i edited by Nandkishor3939 Jan 6, 2019 reply Follow Share someone please elaborate it What i did Ta=0.8*5+0.2*(80+5) Tb=0.6*3+o.4*0.15*(6+3)+o.4*0.85(80+6+3) Edit: I took 0.15 instead of 0.85 Ta=0.8*5+0.2*(80+5)=21 Tb=0.6*3+o.4*0.85*(6+3)+o.4*0.15(80+6+3)=10.2 so answ: 31.2 ns 0 votes 0 votes MiNiPanda commented Jan 6, 2019 reply Follow Share @Nandkishor3939 The correct solution is here https://gateoverflow.in/282699/selfdoubt-me-ft 0 votes 0 votes Deepanshu commented Jan 6, 2019 reply Follow Share 15% miss rate not hit rate u r taking it as hit rate 2 votes 2 votes Nandkishor3939 commented Jan 6, 2019 reply Follow Share omg !! Thank you now the answer is correct :) one last question... if nothing is mentioned in the question then are we supposed to take hierarchical cache memory?? @Deepanshu @MiNiPanda 0 votes 0 votes MiNiPanda commented Jan 6, 2019 reply Follow Share @Nandkishor3939 yes . 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes By default memory access in cache is hierarchical. For A EMAT = $5 + 0.2 * 80 = 21 ns$ For B EMAT = $3 + 0.4( 6 + 0.15*80) = 10.2 ns$ So sum = $31.2ns$ smsubham answered Mar 23, 2020 smsubham comment Share Follow See all 0 reply Please log in or register to add a comment.