MADE EASY -COA

Question

consider a cache memory which is 30 times faster than the main memory and uses 90% of the total time.

what is the speedup gain by the cache memory

Comments

By using Ambdahl's law we can find speed up enhancement.

Speed up enhancement=$((1-F)+F/S)^{-1}$.

Here F=fraction of time used and S= speed factor

Enhancement(speed up gain)=$((1-0.9)+0.9/30)^{-1}$

                    =30/3.9

                    =7.69

Hence speed up gain=7.69

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what should be the default case simulataneous or heirarchial access?

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Just for everyone understanding, if the word "level" or "hierarchical" is not mentioned, it should be simultaneous access.:)

Answer 1

Let, access time without cache = $M$  (Main Memory access time)

Cache is $30$ times faster than MM. So, $T_c = \frac{M}{30}$
Access time with cache $= hT_c$ + (1-h)M = $0.9 \frac{M}{30} + 0.1M$

Speed up $=\frac{Speed_{old}}{Speed_{new}}\\ =\frac{M}{ 0.9 \frac{M}{30} + 0.1M}\\ =\frac{1}{0.13}\\ =7.69$

Comments

Here if we consider hierarchical access then shouldn't formulae be:

Cache access time=hTc + (1-h)(M+T_c)

then the answer is coming 7.5.

So which one is correct?

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@Surajit you figure out this or not? 
               i mean we have to take simultaneous access always!? (as default)