• edited by
544 views
0 0 votes
Consider a single-level cache with an access time of 2.5 ns with a block size of 64 bytes. Main memory uses a block transfer capability that has a first word (4 bytes) access time of 50 ns and an access time of 5 ns for each word thereafter. If hit ratio of cache memory is 95%, then average memory access time is __________. [Upto 3 decimal places]

I'm getting 8.750. But the answer given is 8.875.

My approach is

0.95*2.5+0.05(2.5+50+15*5).

Answer explain

2.5+0.05(2.5+50+15*5)

which is correct?

I find made easy accurate with answers for all subjects except Co.

1 Answer

0 0 votes
you  are using simulatenous access and made easy used heirarchical access.that’s why there is difference in the answer
Position:
Show:

Related questions

1 1 vote
1 1 answer
1.0k
1.0k views
dragonball asked Jan 1, 2018
1,044 views
Compute the average access time for a machine with 80%cache hit ratio.The cache access time and memory access time are 20 ns and 200 ns .DOUBT:: While solving these type ...
2 2 votes
2 2 answers
2.7k
2.7k views
2 2 votes
3 answers 3 answers
407
407 views
js__ asked Nov 5, 2025
407 views