Computer Sciences > GATE 2026 SET-1 > CPU Scheduling
Consider a CPU that has to execute two types of processes. The first type, Actuators (A), requires a CPU burst of 6 seconds. The second type, Controllers (C), requires a CPU burst of 8 seconds. A new process of type A arrives at time t = 10, 20, 30, 40, and 50 (in seconds). Similarly, a new process of type C arrives at time t = 11, 22, 33, 44, and 55 (in seconds). The CPU scheduling policy is First Come First Serve (FCFS). The first process of type A starts running at t = 10 seconds.

The average waiting time (in seconds) for the 10 processes is ___________. (rounded off to one decimal place)

Correct : 9.5

To find the average waiting time using First Come First Serve (FCFS) scheduling, we must order the processes by their arrival time and calculate when each process starts and finishes.
Process Data & Arrival Sequence:
Process A burst time = 6s.
Process C burst time = 8s.
Merging the arrivals of A (10, 20, 30, 40, 50) and C (11, 22, 33, 44, 55) in chronological order gives us our execution sequence: A1, C1, A2, C2, A3, C3, A4, C4, A5, C5.
FCFS Execution Timeline:
Waiting Time (WT) is calculated as: WT = Start Time - Arrival Time.
1. A1 (Arrival: 10): CPU is idle. Starts at 10. Completes at 10 + 6 = 16. WT: 10 - 10 = 0
2. C1 (Arrival: 11): CPU is busy until 16. Starts at 16. Completes at 16 + 8 = 24. WT: 16 - 11 = 5
3. A2 (Arrival: 20): CPU is busy until 24. Starts at 24. Completes at 24 + 6 = 30. WT: 24 - 20 = 4
4. C2 (Arrival: 22): CPU is busy until 30. Starts at 30. Completes at 30 + 8 = 38. WT: 30 - 22 = 8
5. A3 (Arrival: 30): CPU is busy until 38. Starts at 38. Completes at 38 + 6 = 44. WT: 38 - 30 = 8
6. C3 (Arrival: 33): CPU is busy until 44. Starts at 44. Completes at 44 + 8 = 52. WT: 44 - 33 = 11
7. A4 (Arrival: 40): CPU is busy until 52. Starts at 52. Completes at 52 + 6 = 58. WT: 52 - 40 = 12
8. C4 (Arrival: 44): CPU is busy until 58. Starts at 58. Completes at 58 + 8 = 66. WT: 58 - 44 = 14
9. A5 (Arrival: 50): CPU is busy until 66. Starts at 66. Completes at 66 + 6 = 72. WT: 66 - 50 = 16
10. C5 (Arrival: 55): CPU is busy until 72. Starts at 72. Completes at 72 + 8 = 80. WT: 72 - 55 = 17
Calculating the Average:
Total Waiting Time = 0 + 5 + 4 + 8 + 8 + 11 + 12 + 14 + 16 + 17 = 95 seconds.
Average Waiting Time = 95 / 10 = 9.5 seconds.
Correct Answer: 9.5

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