To find the Average Turnaround Time, we first track the execution using a Gantt chart:
- Time $\mathbf{0: P1}$ arrives and starts. $($Remaining: $\mathrm{P}1=6)$
- Time $\mathbf{2: P2}$ arrives (Burst $=2$ ). $\mathrm{P1}$ has $4 \mathrm{~ms}$ remaining. Since $2<4, \mathbf{P2}$ preempts $\mathbf{P1}$ .
- Time $\mathbf{3: P3}$ arrives (Burst $=5$ ). $\mathrm{P2}$ has $1 \mathrm{~ms}$ remaining. Since $1<5, \mathbf{P} 2$ continues.
- Time $\mathbf{4: P2}$ finishes. Ready queue: P1 $(4 \mathrm{~ms})$, P3 $(5 \mathrm{~ms})$. $\mathbf{P1}$ resumes as it is the shortest.
- Time $\mathbf{5: P4}$ arrives (Burst $=2$). P1 has $\mathrm{3 ~ms}$ remaining. Since $2<3, \mathbf{P4}$ preempts $\mathbf{P1}$.
- Time $\mathbf{7: P4}$ finishes. Ready queue: $\mathrm{P1 ~(3 ~ms)}$, $\mathrm{P3 ~(5 ~ms)}$. $\mathbf{P1}$ resumes.
- Time $\mathbf{10: P1}$ finishes. Ready queue: $\mathrm{P3 ~(5 ~ms)}$. $\mathbf{P3}$ starts.
- Time $\mathbf{15: P3}$ finishes.
Turnaround Time (TAT) Calculation
$T A T=$ Completion Time $-$ Arrival Time
\[
\begin{array}{c|c|c|c}
\text{Process} & \text{Completion Time} & \text{Arrival Time} & \text{Turnaround Time} \\
\hline
\text{P1} & 10 & 0 & 10 \\
\text{P2} & 4 & 2 & 2 \\
\text{P3} & 15 & 3 & 12 \\
\text{P4} & 7 & 5 & 2
\end{array}
\]
$\text { Average TAT }=\frac{10+2+12+2}{4}=\frac{26}{4}=6.5$
Correct Answer : C
