Topic: Electronics Engineering
A microprocessor has a cache memory with access time of 2 ns and a main memory with access time of 10 ns. If the cache miss ratio is 0.6, what is the average memory access time?
- A. 6.8 ns
- B. 6 ns
- C. 5.2 ns
- D. 12 ns
Answer: A. 6.8 ns
Explanation: $t_{avg}=h t_{c}+(1-h)t_{m}$
Where, $h =$ hit ratio = $1 - \text{miss ratio} = 1 - 0.6 = 0.4$
$t_{c} =$ cache access time = $2~ns$
$t_{m} =$ main memory access time = $10~ns$
$t_{avg}=0.4\times2+0.6\times10 = 0.8+6=6.8~ns$.