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GATE CSE 2026 Session 1 Q64: Average FCFS waiting time for alternating A and C processes

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 tt = 10, 20, 30, 40, and 50 (in seconds). Similarly, a new process of type C arrives at time tt = 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 tt = 10 seconds. The average waiting time (in seconds) for the 10 processes is ___________. (rounded off to one decimal place)

Try it: numerical answer: type the value

Numerical answer. No negative marking.
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Answer: 9.5 (official GATE 2026 final answer key)

Concept used: FCFS: run in arrival order; waiting time = start time − arrival time.

First step: Order by arrival (A10, C11, A20, C22, A30, C33, …) and build the Gantt chart starting at 10 with bursts 6 and 8 alternating.

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Source: official GATE 2026 Session 1 CS question paper and final answer key, organised by IIT Guwahati. Question paper · Answer key

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