GATE CS 2020 Q50 — CPU Scheduling

Since all processes arrive at time $t = 0$, the turnaround time (TAT) for each process is equal to its completion time (CT).

1. Shortest Job First (SJF) Scheduling

Sorting the processes by burst time: $P_3\ (2)$, $P_4\ (4)$, $P_2\ (7)$, $P_1\ (8)$.

• Gantt Chart: | P3 (0-2) | P4 (2-6) | P2 (6-13) | P1 (13-21) |
• Turnaround times: $P_1 = 21$, $P_2 = 13$, $P_3 = 2$, $P_4 = 6$
• Average $\text{TAT}_{\text{SJF}} = \frac{21 + 13 + 2 + 6}{4} = \frac{42}{4} = 10.5\text{ ms}$

2. Round Robin (RR) Scheduling (Quantum = 4 ms)

Processes are scheduled in the order $P_1, P_2, P_3, P_4$:

• $0 - 4$: $P_1$ runs (remaining burst = $4$)
• $4 - 8$: $P_2$ runs (remaining burst = $3$)
• $8 - 10$: $P_3$ runs (finishes at $10$)
• $10 - 14$: $P_4$ runs (finishes at $14$)
• $14 - 18$: $P_1$ runs (finishes at $18$)
• $18 - 21$: $P_2$ runs (finishes at $21$)
• Turnaround times: $P_1 = 18$, $P_2 = 21$, $P_3 = 10$, $P_4 = 14$
• Average $\text{TAT}_{\text{RR}} = \frac{18 + 21 + 10 + 14}{4} = \frac{63}{4} = 15.75\text{ ms}$

3. Absolute Difference

$$\text{Difference} = |10.5 - 15.75| = 5.25\text{ ms}$$