William Stalling cache memory || 4.21

Question

Consider a $\color{maroon}{\text{single-level cache}}$ with an access time of $2.5$ ns, a line size of $64$ bytes, and a hit ratio of $\large\color{maroon}{\text{H } =   0.95}$. 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.

A. What is the access time when there is a cache miss? Assume that the cache waits until the line has been fetched from main memory and then re-executes for a hit.

b. Suppose that increasing the line size to $128$ bytes increases the $\large\color{marron}{\text{H}}$ to $0.97$. Does this reduce the average memory access time?

Answer 1

A. What is the access time when there is a cache miss? Assume that the cache waits until the line has been fetched from main memory and then re-executes for a hit. 

Directly given there is miss.

Total Words=64B/4B = 16

So Total Time = 2.5ns +[ 1*50ns + 15*5ns ] + 2.5 = 130ns

For part B.

https://gateoverflow.in/11135/consider-the-following 

Comments

@papesh why are you adding 2.5 two times...

first time it will only search cache and second time it will access a cache..

according to you, cache search time = cache access time, but genrally we neglect cache search time??

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It will be added only one time.

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Why did you include 2.5 ns two times ?

Why did you not include the miss ratio or hit ratio ?