Electronics and Communication Engineering - Digital Electronics - Discussion
Discussion Forum : Digital Electronics - Section 1 (Q.No. 16)
16.
The access time of a word in 4 MB main memory is 100 ms. The access time of a word in a 32 kb data cache memory is 10 ns. The average data cache bit ratio is 0.95. The efficiency of memory access time is
Answer: Option
Explanation:
Access time = 0.95 x 10 + 0.05 x 100.
Discussion:
34 comments Page 4 of 4.
Kokila said:
9 years ago
I need clear explanation, please elaborate the solution.
Pradeepa said:
9 years ago
Hi, I want a clear explanation of the answer. Please provide me.
Monica said:
9 years ago
Please, anyone gives the clear solution for this question with a correct formulas or logic.
Ram said:
9 years ago
0.95 * 10ns + 0.05 * 100ms = If you are consider ms and ns, then what will be the answer?
Please explain.
Please explain.
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