Consider a single processor system with four processes A, B, C, and D, represented as given below, where for each process the first value is its arrival time, and the second value is its CPU burst time. A (0, 10), B (2, 6), C (4, 3), and D (6, 7). Which one of the following options gives the average waiting times when preemptive Shortest Remaining Time First (SRTF) and Non-Preemptive Shortest Job First (NP-SJF) CPU scheduling algorithms are applied to the processes?

### 1. Preemptive Shortest Remaining Time First (SRTF)

Let the given processes be:

• A: Arrival Time $(\text{AT}) = 0$, Burst Time $(\text{BT}) = 10$

• B: $\text{AT} = 2$, $\text{BT} = 6$

• C: $\text{AT} = 4$, $\text{BT} = 3$

• D: $\text{AT} = 6$, $\text{BT} = 7$

Gantt Chart Execution:

• Time $0$ to $2$: Process A executes (remaining time of $\text{A} = 8$).

• Time $2$ to $4$: Process B arrives with remaining time $6 < 8$, so B preempts A and executes (remaining time of $\text{B} = 4$).

• Time $4$ to $7$: Process C arrives with remaining time $3 < 4$, so C preempts B and runs to completion (at $t = 7$).

• Time $7$ to $11$: At $t = 7$, ready queue has B (rem = 4), D (arrived at $t=6$, rem = 7), and A (rem = 8). Process B has the shortest remaining time and runs to completion (at $t = 11$).

• Time $11$ to $18$: Process D (rem = 7) runs to completion (at $t = 18$).

• Time $18$ to $26$: Process A (rem = 8) runs to completion (at $t = 26$).

Turnaround Time (TAT = Completion Time - Arrival Time) and Waiting Time (WT = TAT - Burst Time):

• Process A: $\text{CT} = 26$, $\text{TAT} = 26 - 0 = 26$, $\text{WT} = 26 - 10 = 16$

• Process B: $\text{CT} = 11$, $\text{TAT} = 11 - 2 = 9$, $\text{WT} = 9 - 6 = 3$

• Process C: $\text{CT} = 7$, $\text{TAT} = 7 - 4 = 3$, $\text{WT} = 3 - 3 = 0$

• Process D: $\text{CT} = 18$, $\text{TAT} = 18 - 6 = 12$, $\text{WT} = 12 - 7 = 5$

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### 2. Non-Preemptive Shortest Job First (NP-SJF)

Gantt Chart Execution:

• Time $0$ to $10$: Process A arrives first at $t = 0$ and executes non-preemptively until completion at $t = 10$.

• At $t = 10$, all remaining processes are ready: B (BT = 6), C (BT = 3), D (BT = 7).

• Time $10$ to $13$:C has the shortest burst time (3), so C executes until $t = 13$.

• Time $13$ to $19$:B has the next shortest burst time (6), so B executes until $t = 19$.

• Time $19$ to $26$:D executes until $t = 26$.

Turnaround Time and Waiting Time:

• Process A: $\text{CT} = 10$, $\text{TAT} = 10 - 0 = 10$, $\text{WT} = 10 - 10 = 0$

• Process B: $\text{CT} = 19$, $\text{TAT} = 19 - 2 = 17$, $\text{WT} = 17 - 6 = 11$

• Process C: $\text{CT} = 13$, $\text{TAT} = 13 - 4 = 9$, $\text{WT} = 9 - 3 = 6$

• Process D: $\text{CT} = 26$, $\text{TAT} = 26 - 6 = 20$, $\text{WT} = 20 - 7 = 13$

Therefore, $\text{SRTF} = 6$ and $\text{NP-SJF} = 7.5$, matching option (B).