Question

Suppose the cache access time is 15ns, main memory access time is 220ns, and the cache hit rate is 95%. Assuming parallel (overlapped) access (CPU starts the data request in parallel to both cache and to main memory at the same, so if a cache missing occurs, we don't have to add this cache search time to the memory access), what is the average access time for the processor to access an item

Answer 1

Answer:

Check the explanation

Explanation:

Generally, average access time is :

hit rate * ( cache access time ) //since data found in cache

+

miss rate * ( cache access time + time to hit memory to get data from memory) // miss occurred

In this case, parallel access is happening, so we don't need to add cache access time when

a miss occurs. Therefore, in this case the formula becomes:

avg(time) = hit rate * ( cache access time )

+

miss rate * ( time to hit memory to get data from memory)

Given,

hit rate= 0.95

miss rate = 1- hit rate = 1-0.95 = 0.05

cache access time = 15 ns

memory access time = 220 ns

Put values in formula:

avg(time) =0.95 * 15 + 0.05 * 220

=25.25 ns