GATE CS 2026 Set 1 Q.64: 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)

Your answer

Official answer and solution

Official answer: 9.5
From: Official GATE 2026 answer key

Solution

Order all 10 processes by arrival time and compute the start time, completion time, and waiting time ( $\text{Start} - \text{Arrival}$ ) of each process under FCFS.

Ordering the 10 processes by arrival time ( $A_1, C_1, A_2, C_2, A_3, C_3, A_4, C_4, A_5, C_5$ ), the CPU stays continuously busy from $t = 10\text{ s}$ to $t = 80\text{ s}$ :

Process	Arrival Time (s)	Burst Time (s)	Start Time (s)	Completion Time (s)	Waiting Time (s)
$A_1$	10	6	10	16	$10 - 10 = 0$
$C_1$	11	8	16	24	$16 - 11 = 5$
$A_2$	20	6	24	30	$24 - 20 = 4$
$C_2$	22	8	30	38	$30 - 22 = 8$
$A_3$	30	6	38	44	$38 - 30 = 8$
$C_3$	33	8	44	52	$44 - 33 = 11$
$A_4$	40	6	52	58	$52 - 40 = 12$
$C_4$	44	8	58	66	$58 - 44 = 14$
$A_5$	50	6	66	72	$66 - 50 = 16$
$C_5$	55	8	72	80	$72 - 55 = 17$

Total waiting time $= 0 + 5 + 4 + 8 + 8 + 11 + 12 + 14 + 16 + 17 = 95\text{ seconds}$ .

Average waiting time $= \frac{95}{10} = 9.5\text{ seconds}$ .

Final answer

9.5

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2026 Computer Science and Information Technology - Session 1 (CS1)
Question: 64 of 65
Section: Computer Science and Information Technology

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