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 959 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.