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The arrival time, priority (lower value denoting higher priority), and duration of the CPU and I/O bursts for each of three processes $P_1, P_2$ and $P_3$ are given in the table below. Each process has a CPU burst followed by an I/O burst followed by another CPU burst. Assume that each process has its own I/O resource and hence do not block each other for I/O.
$$\small \begin{array}{|c|c|c|c|c|c|} \hline \textbf{Process} & \textbf{Arrival} & \textbf{Priority}& \textbf{Burst duration} & \textbf{Burst duration} & \textbf{Burst duration) }\\
  & \textbf{Time} & & \textbf{(CPU)} & \textbf{(I/O)} & \textbf{(CPU)}
\\\hline P_{1} & 0 & 2 & 3 & 7 & 5
\\\hline P_{2} & 1 & 3 & 5 & 5 & 3
\\\hline P_{3} & 3 & 1& 4 & 5 & 3
\\\hline \end{array}$$
If the multi-programmed operating system uses preemptive priority scheduling and the finish times of the processes $P_1,P_2$ and $P_3$ are $a,b$ and $c$ respectively, then $a+2b+3c = $?
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$a=18,b=28,c=15.$

So, $a+2b+3c = 18+56+45 = 119.$
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