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 2 of 4.
Kokila said:
9 years ago
I need clear explanation, please elaborate the solution.
Tejas said:
1 decade ago
@ Sateesh.
I could not understand, how you calculated?
I could not understand, how you calculated?
Ramesh said:
10 years ago
If anyone knows clear explanation then please explain.
Dheeraja said:
1 decade ago
How to taken 0.05 & what formula will be applied?
Ali Muhammad said:
5 years ago
You explained in a good manner. Thank you @ Bikram.
Jhklhlkhkgku said:
1 decade ago
What about the 100 ms which has converted in ns?
Baishali said:
1 decade ago
Please give a proper explanation for the result.
Heena said:
8 years ago
How do we get 0.05 hit rate from main memory?
S.ROHINI said:
9 years ago
Please mention the formula and explanation.
Surya said:
9 years ago
Thank you all for explaining the solution.
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