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CPU and I/O Scheduling

GATE CSE 2026 Set 1 · Q64

NAT2 marksMedium

Time

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,t = 10, 20, 30, 40, and 5050 (in seconds). Similarly, a new process of type C arrives at time t=11,22,33,44,t = 11, 22, 33, 44, and 5555 (in seconds). The CPU scheduling policy is First Come First Serve (FCFS). The first process of type A starts running at t=10t = 10 seconds. The average waiting time (in seconds) for the 10 processes is ________. (rounded off to one decimal place)

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Solution

Step-by-step verified solution

Official answer: 9.5

CPU and I/O Scheduling

We have 10 processes with arrival times (ATAT) and burst times (BTBT):

  • Actuators (AA): BT=6BT = 6 s at AT=10,20,30,40,50AT = 10, 20, 30, 40, 50

  • Controllers (CC): BT=8BT = 8 s at AT=11,22,33,44,55AT = 11, 22, 33, 44, 55

Sorting all processes by arrival time in FCFS order:

  1. A1A_1: AT=10AT = 10, BT=6BT = 6

  2. C1C_1: AT=11AT = 11, BT=8BT = 8

  3. A2A_2: AT=20AT = 20, BT=6BT = 6

  4. C2C_2: AT=22AT = 22, BT=8BT = 8

  5. A3A_3: AT=30AT = 30, BT=6BT = 6

  6. C3C_3: AT=33AT = 33, BT=8BT = 8

  7. A4A_4: AT=40AT = 40, BT=6BT = 6

  8. C4C_4: AT=44AT = 44, BT=8BT = 8

  9. A5A_5: AT=50AT = 50, BT=6BT = 6

  10. C5C_5: AT=55AT = 55, BT=8BT = 8

Now, let's schedule them with FCFS and compute Start Time (STST), Completion Time (CTCT), and Waiting Time (WT=ST−ATWT = ST - AT):

ProcessATATBTBTSTSTCTCTWT=ST−ATWT = ST - AT
A1A_1106101610−10=010 - 10 = 0
C1C_1118162416−11=516 - 11 = 5
A2A_2206243024−20=424 - 20 = 4
C2C_2228303830−22=830 - 22 = 8
A3A_3306384438−30=838 - 30 = 8
C3C_3338445244−33=1144 - 33 = 11
A4A_4406525852−40=1252 - 40 = 12
C4C_4448586658−44=1458 - 44 = 14
A5A_5506667266−50=1666 - 50 = 16
C5C_5558728072−55=1772 - 55 = 17

Total Waiting Time:

∑WT=0+5+4+8+8+11+12+14+16+17=95 seconds\sum WT = 0 + 5 + 4 + 8 + 8 + 11 + 12 + 14 + 16 + 17 = 95\text{ seconds}

Average Waiting Time:

Average WT=9510=9.5 seconds\text{Average } WT = \frac{95}{10} = 9.5\text{ seconds}

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This CPU and I/O Scheduling previous year question is Q64 of the GATE CSE 2026 Set 1 paper. Find more GATE CSE PYQs on CPU and I/O Scheduling in Operating Systems, or browse all GATE CSE PYQs.

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CPU and I/O Scheduling in GATE CSE

CPU and I/O Scheduling has 25 GATE CSE PYQs across 22 papers from 2011 to 2026. 10 of them are 2-mark NAT questions like this one, and it is the only one in GATE CSE 2026 Set 1.

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GateAI solution cardGATE CSE 2026 Set 1 · Q64 · NAT

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)

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