Official PYQ · 2026Set 1 · Q54CPU SchedulingNAT2 marks

GATE CS 2026 Set 1 Q54 — 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)

View worked solution

Answer: 9.5 : 9.5

To find the average waiting time, we trace the execution order of the 10 processes using the First Come First Serve (FCFS) scheduling policy. The CPU burst times are $6$ seconds for Actuators (A) and $8$ seconds for Controllers (C).

The chronological order of arrivals and execution is detailed below:

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

Now, we calculate the total waiting time for all 10 processes:

\text{Total Waiting Time} = 0 + 5 + 4 + 8 + 8 + 11 + 12 + 14 + 16 + 17 = 95 \text{ seconds}

The average waiting time is:

\text{Average Waiting Time} = \frac{95}{10} = 9.5 \text{ seconds}

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Published 10 Oct 2026

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