CPU Scheduling Questions

Processes:

• P1: Arrival Time = 0, Burst Time = 8
• P2: Arrival Time = 1, Burst Time = 4
• P3: Arrival Time = 2, Burst Time = 9
• P4: Arrival Time = 3, Burst Time = 5

Gantt Chart and Timeline:

• Time 0: P1 arrives. No other process. P1 starts execution. Remaining Burst (RB) for P1 = 8.
• Time 1: P2 arrives. P1 has run for 1 unit (RB P1 = 7). Comparing P1 (RB=7) and P2 (BT=4). P2 has a shorter remaining time. P1 is preempted.
• Time 1 to 2: Context Switch (CS) occurs (1 unit idle time).
• Time 2: P2 starts execution. RB for P2 = 4.
• Time 3: P4 arrives. Current running process is P2 (RB=3). Ready processes: P1 (RB=7), P4 (BT=5), P3 (not yet arrived). P2 (RB=3) is still the shortest. P2 continues.
• Time 4: P3 arrives. Current running process is P2 (RB=2). Ready processes: P1 (RB=7), P3 (BT=9), P4 (BT=5). P2 (RB=2) is still the shortest. P2 continues.
• Time 6: P2 finishes its burst (executed for 4 units from T=2 to T=6). RB for P2 = 0.
• Time 6 to 7: Context Switch (CS) occurs (1 unit idle time).
• Time 7: Ready processes: P1 (RB=7), P3 (RB=9), P4 (RB=5). P4 (RB=5) is the shortest. P4 starts execution.
• Time 12: P4 finishes its burst (executed for 5 units from T=7 to T=12). RB for P4 = 0.
• Time 12 to 13: Context Switch (CS) occurs (1 unit idle time).
• Time 13: Ready processes: P1 (RB=7), P3 (RB=9). P1 (RB=7) is the shortest. P1 starts execution.
• Time 20: P1 finishes its burst (executed for 7 units from T=13 to T=20). RB for P1 = 0.
• Time 20 to 21: Context Switch (CS) occurs (1 unit idle time).
• Time 21: Ready processes: P3 (RB=9). P3 starts execution.
• Time 30: P3 finishes its burst (executed for 9 units from T=21 to T=30). RB for P3 = 0.

Completion Times (CT):

• P1: 20
• P2: 6
• P3: 30
• P4: 12

Turnaround Times (TAT = CT - Arrival Time):

• P1: 20 - 0 = 20
• P2: 6 - 1 = 5
• P3: 30 - 2 = 28
• P4: 12 - 3 = 9

Average Turnaround Time: