scheduling

Question

Answer 1

1) FCFS:

$\small \begin{array} {|c|c|c|c|c|c|} \hline \textbf{Process No.} & \textbf{Arrival Time} &\textbf{Burst Time} & \textbf{Completion Time} & \textbf{Turn Around Time} & \textbf{Watting Time}\\\hline A & 0 & 3 & 3 & 3 & 0 \\ B & 3 & 4 & 7 & 4 & 0  \\ C &  7 & 3 & 10 & 3 & 0 \\ D & 9 &  5 & 15 & 6 & 1  \\ E &12 & 7 & 22 & 10 & 3 \\\hline \end{array}$

Average watting time :$\frac{0+0+0+1+3}{5}=0.8$ msec.

2) RR:

$\small \begin{array} {|c|c|c|c|c|c|} \hline \textbf{Process No.} & \textbf{Arrival Time} &\textbf{Burst Time} & \textbf{Completion Time} & \textbf{Turn Around Time} & \textbf{Watting Time}\\\hline A & 0 & 3 & 3 & 3 & 0 \\ B & 3 & 4 & 7 & 4 & 0  \\ C &  7 & 3 & 10 & 3 & 0 \\ D & 9 &  5 & 18 & 9 & 4  \\ E &12 & 7 & 22 & 10 & 3 \\\hline \end{array}$

Average watting time :$\frac{0+0+0+4+3}{5}=1.4$ msec.

3) SRTF:

$\small \begin{array} {|c|c|c|c|c|c|} \hline \textbf{Process No.} & \textbf{Arrival Time} &\textbf{Burst Time} & \textbf{Completion Time} & \textbf{Turn Around Time} & \textbf{Watting Time}\\\hline A & 0 & 3 & 3 & 3 & 0 \\ B & 3 & 4 & 7 & 4 & 0  \\ C &  7 & 3 & 10 & 3 & 0 \\ D & 9 &  5 & 18 & 6 & 1  \\ E &12 & 7 & 22 & 10 & 3 \\\hline \end{array}$

Average watting time :$\frac{0+0+0+4+3}{5}=0.8$ msec.

Comments

The BT of process E is given 7 in question and you have taken as 6 in your solution.

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 darshan-at typo, answer remain the same.

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Yes Actually you were right this 1 will not make any difference.