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Consider the following processes, with the arrival time and the length of the CPU burst given in milliseconds. The scheduling algorithm used is preemptive shortest remaining-time first.$$\small \begin{array}{|c|c|c|} \hline \textbf{Process} & \textbf{Arrival Time} & \textbf{Burst Time}\\\hline \text{$P_1$} & 0 & 10 \\\hline \text{$P_2$} & 3 & 6 \\\hline \text{$P_3$} & 7 & 1 \\\hline \text{$P_4$} & 8 & 3 \\\hline  \end{array}$$The average turn around time of these processes is ___________ milliseconds.
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SRTF Preemptive hence,


$$\begin{array}{c|c} \hline \text{Process} &\text{TAT = Completion time – Arrival time} \\\hline  P_{1} & 20 \\ P_{2} & 7 \\ P_{3} & 1 \\ P_{4} & 5 \\\hline \end{array}$$

Average TAT $= \frac{20+7+1+5}{4} =33/4 = 8.25$

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