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

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

solution:

the right option is a

to adding   a new row to a table we   have to do process last table cell and pressing tab.

this is the process to add new row in the cell:

* to add a row below the cell, click insert below in the rows and columns group.

thus this is the required right answer.