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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 .

edited | 179 views
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31.2 ns is correct.
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31.2ns.
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i too got 31.2
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thank you for verifying it.they just ignored the time access of L1 when  hit in L2.
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@ @ please explain the solution!

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, $t_A=h_1*t_1+(1-h_1)(t_1+t_m)$

$t_B=h_1*t_1+(1-h_1)h_2(t_1+t_2)+(1-h_1)(1-h_2)(t_1+t_2+t_m)$

put the value you will get the answer

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@ there is another formula also for default access...how to identify that which formula has to be used when?

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by default cache memory is hierarchical.

for more refer this -https://gateoverflow.in/108780/simultaneous-vs-hierarchical-memory

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21 + 9.30= 30.30 i am getting(  only rounding up to 2 everything )
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and in nanosecond not in ms
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@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)

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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
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15% miss rate not hit rate u r taking it as hit rate
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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??

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yes .