For the processes listed in the following table, which of the following scheduling schemes will give the lowest average turnaround time?

Turnaround Time (TAT) is calculated as:

Let's evaluate each scheduling algorithm:

1. First Come First Serve (FCFS):

   • Execution order: $A[0-3] \to B[3-9] \to C[9-13] \to D[13-15]$

   • Completion times: $A=3, B=9, C=13, D=15$

   • TAT: $A = 3 - 0 = 3,\ B = 9 - 1 = 8,\ C = 13 - 4 = 9,\ D = 15 - 6 = 9$

   • $\text{Average TAT} = \frac{3 + 8 + 9 + 9}{4} = 7.25$

2. Non-preemptive Shortest Job First (SJF):

   • $t=0$: $A$ runs $[0-3]$.

   • $t=3$: Only $B$ is available, so $B$ runs $[3-9]$.

   • $t=9$: Both $C$ (burst 4) and $D$ (burst 2) are available. $D$ runs $[9-11]$, then $C$ runs $[11-15]$.

   • Completion times: $A=3, B=9, D=11, C=15$

   • TAT: $A = 3,\ B = 8,\ C = 15 - 4 = 11,\ D = 11 - 6 = 5$

   • $\text{Average TAT} = \frac{3 + 8 + 11 + 5}{4} = 6.75$

3. Shortest Remaining Time (SRT / Preemptive SJF):

   • $t=0$ to $3$: $A$ runs (burst 3). At $t=1$, $B$ arrives (burst 6 > remaining $A=2$), so $A$ completes at $t=3$.

   • $t=3$ to $4$: $B$ runs (remaining becomes 5).

   • $t=4$: $C$ arrives (burst 4). Since remaining of $B (5) > 4$, $C$ preempts $B$ and runs until $t=6$ (remaining $C$ becomes 2).

   • $t=6$: $D$ arrives (burst 2). Remaining times: $C=2, D=2, B=5$. $C$ finishes at $t=8$.

   • $t=8$ to $10$: $D$ runs and finishes at $t=10$.

   • $t=10$ to $15$: $B$ finishes its remaining 5 units at $t=15$.

   • Completion times: $A=3, C=8, D=10, B=15$

   • TAT: $A = 3 - 0 = 3,\ B = 15 - 1 = 14,\ C = 8 - 4 = 4,\ D = 10 - 6 = 4$

   • $\text{Average TAT} = \frac{3 + 14 + 4 + 4}{4} = 6.25$

4. Round Robin with Quantum = 2:

   • Average TAT is $7.25$.

Comparing the average turnaround times:

• FCFS: $7.25$

• Non-preemptive SJF: $6.75$

• Shortest Remaining Time (SRT): $6.25$

• Round Robin: $7.25$

Thus, Shortest Remaining Time gives the lowest average turnaround time.