A system uses demand paging with an average page fault

Solution

Effective Access Time (EAT) = (1 - p)  Memory Access Time + p  Page Fault Service Time         Given:         EAT = 200 ns         Memory Access Time (MA) = 100 ns         Page Fault Service Time (PFST) = 10 ms = $10 \times 10^6$ ns (convert ms to ns)         200 = (1 - p)  100 + p  $10^7$         200 = 100 - 100p + $10^7$p         100 = $10^7$p - 100p         100 = p ($10^7$ - 100)         100 = p (9,999,900)         p = 100 / 9,999,900         p $\approx$ 0.0000100001         As a percentage: 0.0000100001  100% $\approx$ 0.001%         Let's recheck the calculation carefully:         $200 = (1-p) \times 100 + p \times 10,000,000$         $200 = 100 - 100p + 10,000,000p$         $100 = 9,999,900p$         $p = 100 / 9,999,900 \approx 0.0000100001$         Converting to percentage: $0.0000100001 \times 100\% = 0.00100001\%$         Looking at the options, 0.001% is the closest. Let's re-evaluate the options.         A) 0.000001%         B) 0.00001% (This is $1 \times 10^{-5}$)         C) 0.0001% (This is $1 \times 10^{-4}$)         D) 0.001% (This is $1 \times 10^{-3}$)         E) 0.01% (This is $1 \times 10^{-2}$)         calculated p is approximately $1 \times 10^{-5}$. So, 0.00001%.         Therefore, the correct option is B.