NPTEL Assignment

Question

Assume that a read request takes 50 nsec on a cache miss and 5 nsec on a cache hit. While running a program, it is observed that 80% of the processor’s read requests result in a cache hit. The average read access time is ………….. nsec.

Answer 1

In this question, they have mentioned about the cache miss time and cache hit time directly and they are not saying “cache access time/ memory access time”, so we use the below formula:

 

Average Access Time = Hit Rate×Hit Time + Miss Rate×Miss Time

                                  = 0.8(5) + 0.2(50)

                                  = 14ns