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Consider the cache memory that is 30 times faster than main memory and used 90% of the total time. What is the speedup gain by the cache memory?
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Speedup will be 7.69

Direct application of Amdahl's Law -

$Speedup = (1 - 0.9 + \frac{0.9}{30})^{-1} = 7.69$

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@shashin1989 Sir, if we consider it as this: Let it takes x cycles to access memory, hence x/30 cycles to access the cache ,hence, AMAT in case of this cached organization is 0.90*(x/30)+0.10*(x+x/30) and in case of memory organization it takes x cycles only. Then by taking their ratio I got 7.50 as answer....can you tell where am I going wrong?:(


@Deterministic I don't think you need need to do (x+x/30) instead it should only be x because execution for the rest 10% will remain the same.


@Pratyush Priyam Kuan in case of a cache miss we also need to take into account the cache access we only get to know that we have a miss after we access the cache... correct me if I am wrong..

Question is asking about the speedup in a theoretical system where part of it has some enhancement. Why are you taking into account cache misses and other factors ?

If half of the road from Delhi to Agra is a new highway that lets you drive at twice the speed limit - and somebody asks you what is the time improvement in driving from Delhi to Agra, you won't respond by taking into account traffic jams, cows on the road, etc right ? Same thing here. Question is asking speedup of system with enhancement over system without enhancement. How efficient that enhancement is - is beyond the scope of the question.

Thanks @shashin got your point:)

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