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.

There is clearly mentioned that hit ratio is 0.95 and it is always prefer cache memory, so 10*0.95 and 0.05 is failure case it is from main memory. So 100*0.05. So answer is correct.

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).

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.

0.95 * 10ns + 0.05 * 100ms = If you are consider ms and ns, then what will be the answer?
Please explain.

Please, anyone gives the clear solution for this question with a correct formulas or logic.

@Heena.

Hit rate only they said in formula 1-0.95 is 0.05 only, that's applied there.

Hi, I want a clear explanation of the answer. Please provide me.

Formula is right, but 100 ms needs to be converted to ns.

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