Question

A computer has a two-level cache. Suppose that 60% of the memory references hit on the first level cache, 35% hit on the second level, and 5% miss. The access times are 5 nsec, 15 nsec, and 60 nsec, respectively, where the times for the level 2 cache and memory start counting at the moment it is known that they are needed (e.g., a level 2 cache access does not even start until the level 1 cache miss occurs). What is the average access time?

Answer 1

Answer:

t=14ns

Explanation:

We make the relation between the specific access time and the memory percentage in each level, so

$60\% \Rightarrow 60/100 = 0.60\\35\% \Rightarrow 35/100 = 0.35\\05\% \Rightarrow 05/100 = 0.05$

$t= 0.6(5) + 0.35(5+15) + 0.05(5+15+60)\\t= 0.6(5) + 0.35(20) + 0.05(80)\\t= 3 + 7 + 4\\t= 14 ns$

Average Access Time is 14 nsec.