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Ques. In a two-level memory hierarchy, let $t_1 = 10^{-7s}$ and $t_2 = 10^{-2s}$. If ta denotes the average access
time of the memory hierarchy, and if we define the access efficiency to be equal to $t_1=t_a$, then what
must the hit ratio $H$ be in order for the access efficiency to be at least $80\%$ of the maximum value?


Solution: For a $2-$level hierarchy, the average access time is given by:
$t_a = t_1 + (1 - H) * t_2$
where $H$ is the hit ratio, $t_1$ is the hit time, and $t_2$ s the miss penalty. Note that when on a cache miss,
the amount of time it takes to service the memory access request is $(t_1 + t_2)$, and hence, the miss
penalty is $t_2$.
Hence, we want to find H such that the access efficiency $\dfrac{t_1}{t_a} \geq 0.8$. That is,
 $t1 \geq 0.8 * (t_1 + (1 - H) * t_2)$
On Solving :-

$H \geq 1- t_1 / (4*t_2)$
Substituting the values for $t_1$ and $t_2$, you can get:
$H = 0.9999975$


My Doubt is when we solve such Question we Studied that

$T_{avg} = (HitRatio * HitTime + MissRatio * Miss Time)$

Why they have considered only Hit Time instead of Hit Ratio and Hit time....

and plz explain which one to use and when...
in CO and Architecture by
edited by | 165 views

1 Answer

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Best answer
your first eqn is simplified eqn of hierarchical access. your doubt part eqn is simultaneous eqn.

unless specified consider hierarchical organisation
selected by
Ohhk that means when we are doing parallel access then we apply Tavg = (HitRatio * HitTime + MissRatio * Miss Time) and in hierarchical the above one right?

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