Topic: Computer Organization
If $t_c$, $h$ and $t_m$ specify the cache access time, hit ratio and main memory access time then the average access time is (Given $t_c = 160$ ns, $t_m = 960$ ns, $h = 0.99$)
- A. 160 ns
- B. 960 ns
- C. 256 ns
- D. $950 \times 0.9$ ns
Answer: A. 160 ns
Explanation: Average access time $t_{avg} = ht_c + (1 - h)(t_m + t_c)$\nWhere, $h =$ hit ratio\n$t_c =$ cache access time\n$t_m =$ main memory access time\n$t_{avg} = 0.99 \times 160 + 0.01(160 + 960) = 169.6$ ns