Electronics and Communication Engineering - Digital Electronics - Discussion

How do we get 0.05 hit rate from main memory?

Thank you @Bikram.

Thank you all for explaining the solution.

Thank you the explanation @Bikram.

Thank you @Bikram Kesharee Nayak.

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.

Please mention the formula and explanation.

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

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

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