Consider the following set of processes, assumed to have arrived at time 0. Consider the CPU scheduling algorithms Shortest Job First (SJF) and Round Robin (RR). For RR, assume that the processes are scheduled in the order $P_1, P_2, P_3, P_4$. If the time quantum for RR is 4 ms, then the absolute value of the difference between the average turnaround times (in ms) of SJF and RR (round off to 2 decimal places) is _______.

### 1. Shortest Job First (SJF - Non-preemptive since all arrive at $t=0$): Order of execution based on shortest burst time: $P_3 (2) \to P_4 (4) \to P_2 (7) \to P_1 (8)$

• Completion Times (CT):

  • $CT(P_3) = 2$

  • $CT(P_4) = 2 + 4 = 6$

  • $CT(P_2) = 6 + 7 = 13$

  • $CT(P_1) = 13 + 8 = 21$

• Turnaround Time ($TAT = CT - AT = CT$):

  • Average $TAT_{SJF} = \frac{2 + 6 + 13 + 21}{4} = \frac{42}{4} = 10.5\text{ ms}$

### 2. Round Robin (RR) with time quantum $q = 4\text{ ms}$: Queue order: $P_1, P_2, P_3, P_4$

• $0 - 4$: $P_1$ executes for 4 ms (remaining: 4)

• $4 - 8$: $P_2$ executes for 4 ms (remaining: 3)

• $8 - 10$: $P_3$ executes for 2 ms (completes at $t = 10$)

• $10 - 14$: $P_4$ executes for 4 ms (completes at $t = 14$)

• $14 - 18$: $P_1$ executes for 4 ms (completes at $t = 18$)

• $18 - 21$: $P_2$ executes for 3 ms (completes at $t = 21$)

Completion Times for RR:

• $CT(P_1) = 18$

• $CT(P_2) = 21$

• $CT(P_3) = 10$

• $CT(P_4) = 14$

• Average $TAT_{RR} = \frac{18 + 21 + 10 + 14}{4} = \frac{63}{4} = 15.75\text{ ms}$

### 3. Absolute Difference: