Suppose the time to service a page fault is on average 10 milliseconds, while…

Effective (average) memory access time (EAT) combines a fast path — a hit serviced directly from memory — and a slow path — a miss that must first pay the full page-fault service time. In general, EAT = (Hit Ratio × Memory Access Time) + (Miss Rate × Page-Fault Service Time), where Miss Rate = 1 − Hit Ratio. Because page-fault service time is typically orders of magnitude larger than a single memory access, even a very small miss rate can dominate the result, so every term must be converted to one consistent time unit before combining them.

1. Identify the given values: memory access time = 1 microsecond; page-fault service time = 10 milliseconds; hit ratio = 99.99% = 0.9999, so miss rate = 1 − 0.9999 = 0.0001 (0.01%).

2. Convert every time to the same unit (microseconds): 10 milliseconds = 10,000 microseconds.

3. Apply the EAT formula: EAT = (0.9999 × 1 microsecond) + (0.0001 × 10,000 microseconds).

4. Compute each weighted term separately: 0.9999 × 1 microsecond = 0.9999 microseconds; 0.0001 × 10,000 microseconds = 1 microsecond.

5. Sum the two weighted terms: 0.9999 microseconds + 1 microsecond = 1.9999 microseconds, which rounds to approximately 2 microseconds.

Cross-check: since the hit ratio is almost 1, the hit term alone contributes almost exactly 1 microsecond (0.9999 × 1 microsecond ≈ 1 microsecond), and the miss term also works out to exactly 1 microsecond (0.0001 × 10,000 microseconds = 1 microsecond); two terms each close to 1 microsecond must sum to close to 2 microseconds, confirming 1.9999 microseconds ≈ 2 microseconds independently of the step-by-step arithmetic above.

This shows how even a 0.01% miss rate can nearly double the effective access time when the page-fault penalty is four orders of magnitude larger than a single memory access.