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.
Amit kumar said:
4 years ago
@Bikram.
Good explanation. Thanks.
Good explanation. Thanks.
Tshering Dorji said:
5 years ago
That means the access time from the main memory is 100ns or 100ms? I think time is 100ns and not 100ms.
Please anyone explain it clearly.
Please anyone explain it clearly.
(4)
Ali Muhammad said:
5 years ago
You explained in a good manner. Thank you @ Bikram.
Prajwal said:
6 years ago
Thank you @Bikram and @Pooja.
Pujitha said:
6 years ago
Thank you so much, @Bikram Ji. You can join as a teacher.
Vanajavivek said:
6 years ago
Thank you @Bkram and @Pooja.
Harshini said:
6 years ago
Thank you @Bikram.
(1)
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)
Anonym said:
8 years ago
1-0.095=0.05 according to the formula
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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers