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 1 of 4.
Daleep said:
1 decade ago
Please mention the formula used above.
Nithya said:
1 decade ago
Can you please explain clearly?
Sateesh said:
1 decade ago
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.
Jhklhlkhkgku said:
1 decade ago
What about the 100 ms which has converted in ns?
Tejas said:
1 decade ago
@ Sateesh.
I could not understand, how you calculated?
I could not understand, how you calculated?
Sneha kulkarni said:
1 decade ago
Ya what about conversion?
CR7 said:
1 decade ago
Any alternative?
Kar said:
1 decade ago
Formula is right, but 100 ms needs to be converted to ns.
Baishali said:
1 decade ago
Please give a proper explanation for the result.
Dheeraja said:
1 decade ago
How to taken 0.05 & what formula will be applied?
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