If average access time of CPU is 20 ns, access time of main memory is 110 ns…

Concept

For a one-level cache, let h be the probability of a cache hit; then 1 − h is the probability of a miss. Under the two-outcome timing model used here, average access time is the probability-weighted mean of the hit-path and miss-path times.

Thus, Tavg = hTc + (1 − h)Tm, where Tc is the cache-hit time and Tm is the complete main-memory miss-path time.

Application

1. Define the given values: Tavg = 20 ns, Tc = 10 ns, and Tm = 110 ns.

2. Substitute them into the weighted-average model: 20 = 10h + 110(1 − h).

3. Expand and collect the h-terms: 20 = 10h + 110 − 110h = 110 − 100h.

4. Solve: 100h = 90, so h = 0.90 = 90%.

Cross-check

At h = 0.90, the weighted time is 0.90(10 ns) + 0.10(110 ns) = 9 ns + 11 ns = 20 ns, exactly the stated average. Therefore, the hit ratio is 90%.