Consider the following processes, all arriving at time 0, with their respective burst times:

Process	Burst Time (ms)
P1	6
P2	8
P3	7
P4	3

Using Non-Preemptive Shortest Job First (SJF) scheduling, calculate the average waiting time.

Correct option is A

In Non-Preemptive Shortest Job First (SJF) scheduling, the process with the smallest burst time executes first.

Execution Order:

• P4 = 3 ms
• P1 = 6 ms
• P3 = 7 ms
• P2 = 8 ms

Waiting Times:

• P4 = 0 ms
• P1 = 3 ms
• P3 = 3 + 6 = 9 ms
• P2 = 3 + 6 + 7 = 16 ms

Average Waiting Time:

Average Waiting Time = ​$$\frac{0 + 3 + 9 + 16}{4} = \frac{28}{4} = 7.0 \text{ ms}$$​​

Therefore, the average waiting time is 7.0 ms.

Important Key Points:

1. SJF scheduling executes the process with the shortest burst time first.
2. In Non-Preemptive SJF, once a process starts execution, it cannot be interrupted.
3. Waiting time is calculated as the total time a process waits before execution.
4. SJF generally provides the minimum average waiting time among scheduling algorithms.

Knowledge Booster:

1. (b) 5.5 ms: Incorrect because the waiting times are not calculated correctly according to SJF execution order.
2. (c) 6.25 ms: Incorrect because this value results from an incorrect summation of waiting times.
3. (d) 5.0 ms: Incorrect as per the actual calculation; the correct average waiting time is 7.0 ms.
4. (e) 6.0 ms: Incorrect because the computed average waiting time does not equal 6.0 ms.