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.
Abarna said:
8 years ago
@Heena.
Hit rate only they said in formula 1-0.95 is 0.05 only, that's applied there.
Hit rate only they said in formula 1-0.95 is 0.05 only, that's applied there.
Heena said:
8 years ago
How do we get 0.05 hit rate from main memory?
Rishi Ram Panth said:
8 years ago
Thank you @Bikram.
Surya said:
9 years ago
Thank you all for explaining the solution.
Amrita said:
9 years ago
Thank you the explanation @Bikram.
Padmaja said:
9 years ago
Thank you @Bikram Kesharee Nayak.
Amit kumar said:
4 years ago
@Bikram.
Good explanation. Thanks.
Good explanation. Thanks.
Dheeraja said:
1 decade ago
How to taken 0.05 & what formula will be applied?
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.
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