Let the page fault service time be $10\text{ ms}$ in a computer with average memory access time being $20\text{ ns}$. If one page fault is generated for every $10^6$ memory accesses, what is the effective access time for the memory?

Given:

• Normal memory access time, $T_m = 20\text{ ns}$

• Page fault service time, $S = 10\text{ ms} = 10 \times 10^6\text{ ns} = 10^7\text{ ns}$

• Page fault rate, $p = \frac{1}{10^6} = 10^{-6}$

The effective access time (EAT) is given by:

$$\begin{aligned} EAT &= (1 - 10^{-6}) \times 20\text{ ns} + 10^{-6} \times 10^7\text{ ns} \\ &\approx 20\text{ ns} + 10\text{ ns} \\ &= 30\text{ ns} \end{aligned}$$

Hence, the effective access time is $30\text{ ns}$, which corresponds to option (B).