GATE CS 2026 Set 1 — Question 64
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Operating System → Process Scheduling → FCFS & SJF
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Question
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 (in seconds). Similarly, a new process of type C arrives at time (in seconds). The CPU scheduling policy is First Come First Serve (FCFS). The first process of type A starts running at seconds. The average waiting time (in seconds) for the 10 processes is ___________. (rounded off to one decimal place)
Correct answer
9.5 to 9.5
Solution
The scheduling policy is First Come First Serve (FCFS). We list the processes in order of their arrival times and calculate their start, end, and waiting times.
Total Waiting Time = seconds.Average Waiting Time = seconds.
| Process | Arrival Time () | Burst Time (s) | Start Time | End Time | Waiting Time (Start - Arrival) |
|---|---|---|---|---|---|
| 10 | 6 | 10 | 16 | ||
| 11 | 8 | 16 | 24 | ||
| 20 | 6 | 24 | 30 | ||
| 22 | 8 | 30 | 38 | ||
| 30 | 6 | 38 | 44 | ||
| 33 | 8 | 44 | 52 | ||
| 40 | 6 | 52 | 58 | ||
| 44 | 8 | 58 | 66 | ||
| 50 | 6 | 66 | 72 | ||
| 55 | 8 | 72 | 80 |
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