 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. ThusThe 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.