Consider the following process information for Shortest Remaining Time First (SRTF) scheduling:
\[
\begin{array}{|c|c|c|}
\hline
\textbf{Process} & \textbf{Arrival Time (AT)} & \textbf{Burst Time (BT)} \\
\hline
P1 & 0 & 10 \\
P2 & 1 & 13 \\
P3 & 2 & 6 \\
P4 & 8 & 9 \\
\hline
\end{array}
\]
Find the turnaround time for each process.
Consider the following process information for Shortest Remaining Time First (SRTF) scheduling:
\[ \begin{array}{|c|c|c|} \hline \textbf{Process} & \textbf{Arrival Time (AT)} & \textbf{Burst Time (BT)} \\ \hline P1 & 0 & 10 \\ P2 & 1 & 13 \\ P3 & 2 & 6 \\ P4 & 8 & 9 \\ \hline \end{array} \]Find the turnaround time for each process.
Show Hint
- 19
- 20
- 18
- 15
The Correct Option is A
Solution and Explanation
SRTF is a preemptive version of the Shortest Job First (SJF) scheduling algorithm. It executes the process with the shortest remaining burst time.
1. Initially, process \( P1 \) starts executing at time 0.
2. Process \( P3 \) arrives at time 2, and has a burst time of 6, which is shorter than the remaining burst time of \( P1 \). Thus, \( P3 \) executes next.
3. After \( P3 \) completes at time 8, \( P1 \) resumes execution as it has the shortest remaining burst time.
4. Process \( P2 \) arrives at time 1, but \( P1 \) continues since it has less remaining burst time than \( P2 \).
5. After \( P1 \) completes at time 10, process \( P2 \) executes.
6. Process \( P4 \) arrives at time 8 but starts after \( P2 \) as it has a larger burst time.
Step 2: Calculate the turnaround time.
Turnaround time is the difference between the completion time and the arrival time.
\[ \text{Turnaround Time for } P1 = 10 - 0 = 10 \] \[ \text{Turnaround Time for } P2 = 19 - 1 = 18 \] \[ \text{Turnaround Time for } P3 = 8 - 2 = 6 \] \[ \text{Turnaround Time for } P4 = 19 - 8 = 11 \]
Thus, the average turnaround time is:
\[ \frac{10 + 18 + 6 + 11}{4} = 19 \, \text{units}. \]
Thus, the correct answer is \( 19 \).
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