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← GATE CSE 2017 Set 2

Q51

NAT2 marks

Consider the set of process with arrival time (in milliseonds), CPU burst time (in millisecods) and priority (00 is the highest priority) shown below. None of the process have I/O burst time

ProcessArrival TimeBurst TimePriorityP10112P25280P31223P42101P59164\small \begin{array}{|c|c|c|c|} \hline \textbf{Process} & \textbf{Arrival Time} & \textbf{Burst Time} & \textbf{Priority} \\\hline \text{$P_1$} & 0 & 11 & 2 \\\hline \text{$P_2$} & 5 & 28 & 0 \\\hline \text{$P_3$} & 12 & 2 & 3 \\\hline \text{$P_4$} & 2 & 10 & 1\\\hline \text{$P_5$} & 9 & 16 & 4 \\\hline \end{array}

The average waiting time (in milli seconds) of all the process using premtive priority scheduling algorithm is ______

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