Electronics and Communication Engineering - Digital Electronics - Discussion

Assume that main memory access time is 100 ns instead of ms as given here. When we try to search/ access a bit or word through memory, first the cache memory is searched for access. So it's related to hit ratio directly. The performance of cache memory is measured in terms of hit ratio. If a data item requested by CPU is found in the cache, it is called a 'HIT'.If the requested data item is not found i the cache that means it's in main memory and it is called a 'MISS'. Hit ratio is the ratio of the number of hits divided by total no. of requests.

Now given that the memory access time (MAT. cache) for cache is 10 ns and for main memory is 100 ns.

So, Average memory access time or efficiency = (Hit ratio * MAT. cache) + (1- hit ratio) * (MAT. main memory).
= (0.95* 10 ns) + (0.05* 100 ns),
= 9.5 ns + 5 ns,
= 14.5 ns.

I hope everyone understand this.

That means the access time from the main memory is 100ns or 100ms? I think time is 100ns and not 100ms.

Please anyone explain it clearly.

Here, T(avg)=h*Tc+(1-h)*M.

where,
T(avg)=Average time.

h=bit rate,(1-h)=Min Rate,
Tc=Time to access information from cache,
M=Miss Penalty(Time to main memory).

Thank you @Bikram.

Thank you @Bikram and @Pooja.

Thank you so much, @Bikram Ji. You can join as a teacher.

Thank you @Bkram and @Pooja.

You explained in a good manner. Thank you @ Bikram.

Please mention the formula and explanation.

1-0.095=0.05 according to the formula