= Solution
Assume both processes are available at time zero and run the listed order, A followed by B, with no overlap even during <input-output>. A takes $12+4=16$ ms; B then takes $3+4=7$ ms. Thus
$$
\boxed{\text{elapsed time}=23\text{ ms},\qquad\text{mean turnaround}=\frac{16+23}{2}=19.5\text{ ms}.}
$$
The <scheduling turnaround time> includes the time waiting before a process first runs. Reversing the serial order would leave the total at 23 ms but change completion times to 7 and 23 ms, with mean 15 ms. If arrivals were instead delayed until each serial launch, that would define a different turnaround measurement; the simultaneous-arrival convention is used for comparing the schedulers below.
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