= Solution
With the given ready-order rule and A first, the <CPU> timeline in milliseconds is
$$
\begin{array}{c|rrrrrrrr}
\text{interval}&0\!:\!4&4\!:\!5&5\!:\!6&6\!:\!10&10\!:\!11&11\!:\!12&12\!:\!16&16\!:\!17\\
\text{CPU}&A&B&\text{idle}&A&B&\text{idle}&A&B
\end{array}
$$
A performs <input-output> during $[4,6)$ and $[10,12)$; B does so during $[5,7)$ and $[11,13)$. In particular, B becoming ready at time 7 does not interrupt A's ongoing nonpreemptive burst. The completion times are $C_A=16$ and $C_B=17$, giving
$$
\boxed{\text{elapsed time}=17\text{ ms},\qquad\text{mean turnaround}=16.5\text{ ms}.}
$$
There are 15 ms of useful CPU execution and 2 ms of idle time on this particular timeline.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/digital-systems-scheduling.png]
{title=CPU timelines and I/O waits for the two scheduling policies}
{description=Both processes arrive at zero. Readiness determines queue order; A wins initial ties. The displayed completion times give mean turnarounds 16.5 ms without preemption and 14 ms with a two-millisecond quantum.}
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