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 3 of 4.
Amrita said:
9 years ago
Thank you the explanation @Bikram.
Surya said:
9 years ago
Thank you all for explaining the solution.
Rishi Ram Panth said:
8 years ago
Thank you @Bikram.
Heena said:
8 years ago
How do we get 0.05 hit rate from main memory?
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.
Anonym said:
8 years ago
1-0.095=0.05 according to the formula
Ms Pooja Choudhari said:
7 years ago
Here, T(avg)=h*Tc+(1-h)*M.
where,
T(avg)=Average time.
h=bit rate,(1-h)=Min Rate,
Tc=Time to access information from cache,
M=Miss Penalty(Time to main memory).
where,
T(avg)=Average time.
h=bit rate,(1-h)=Min Rate,
Tc=Time to access information from cache,
M=Miss Penalty(Time to main memory).
(1)
Harshini said:
6 years ago
Thank you @Bikram.
(1)
Vanajavivek said:
6 years ago
Thank you @Bkram and @Pooja.
Pujitha said:
6 years ago
Thank you so much, @Bikram Ji. You can join as a teacher.
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