A processor has a cache hit ratio of 90%, a cache access time of 5 ns, and a main-memory access time of 80 ns. Calculate the average memory access…

The cache is always checked first (5 ns). On a miss (10% of accesses) main memory is accessed in addition (80 ns more).

\(\text{AMAT} = t_{cache} + (1 - h) \times t_{mem} = 5 + 0.1 \times 80 = 5 + 8 = 13\ \text{ns}\)

Check, counting hits and misses separately:

\(0.9 \times 5 + 0.1 \times (5 + 80) = 4.5 + 8.5 = 13\ \text{ns}\)

Average memory access time = 13 ns.

— rhs