CPU Scheduling Algorithms Questions and Answers

Q: In multilevel feedback scheduling algorithm

In multilevel feedback scheduling, a process can move to a different classified ready queue.The multilevel feedback queue scheduler keeps several ready queues of different priorities. A process that uses too much CPU time is moved to a lower-priority queue, and one that waits too long can be moved up (ageing). This movement, the 'feedback', is what distinguishes it.In the plain multilevel queue algorithm, by contrast, processes are classified once and the classification is permanent. Processes are classified into groups in both, so the other options are wrong.

Q: Which of the following scheduling algorithms selects the process that has been waiting the longest?

The algorithm is First-Come, First-Served (FCFS).FCFS runs processes in the order they arrive in the ready queue, so the next process chosen is always the one that has waited longest.Shortest Job First picks the process with the shortest burst, priority scheduling the highest priority, and round robin gives each process a time slice in turn, so none of them chooses by waiting time alone.

Q: Which of the following scheduling algorithms is non-preemptive ?

The non-preemptive algorithm is First-In First-Out.Under FIFO (first come, first served) a process, once given the CPU, keeps it until it finishes or blocks for I/O; nothing takes the CPU away from it.Round Robin preempts a process when its time quantum ends. Multilevel queue scheduling preempts lower-priority queues when a higher-priority process arrives, and the feedback version also moves processes between queues, typically using time slices.

Q: Which of the following process scheduling algorithm may lead to starvation ?

The algorithm that may lead to starvation is Shortest Job Next.Shortest Job Next always runs the waiting job with the shortest CPU burst. If short jobs keep arriving, a long job can be passed over indefinitely, which is starvation.FIFO serves jobs in order of arrival, so every job eventually runs, and Round Robin gives every process a time slice in turn. Since one option does starve, 'None of the above' is wrong.

Q: Suppose there are 5 processes P1, P2, P3, P4 and P5 with processing time requirements 10, 6, 2, 4 and 8 respectively. What is the average completion time if SJF (Shortest Job First) policy is used ?

The average completion time is 14.SJF runs the jobs in increasing order of time: 2, 4, 6, 8, 10 (P3, P4, P2, P5, P1). Each job completes when all the jobs before it and itself have run.\[ \begin{aligned} \text{completions} &= 2, 6, 12, 20, 30 \\ \text{average} &= \frac{70}{5} = 14 \end{aligned} \]The average waiting time is \((0+2+6+12+20)/5 = 8\), which is not offered; 10 is simply the longest job, and 12 and 16 match neither measure. The question asks for completion (turnaround) time with all jobs arriving at 0.

Q: Which CPU scheduling algorithm is most likely to cause 'Starvation' for long-running processes ?

Starvation is most likely under Shortest Job First (SJF) - non preemptive.SJF always runs the shortest waiting job next. If short jobs keep arriving, a long job can wait indefinitely, because there is always a shorter job ahead of it. That indefinite postponement is starvation; ageing, which raises a job's priority the longer it waits, is the usual remedy.Round robin gives every process a time slice in turn, and FCFS serves processes in arrival order, so neither starves anyone. Multilevel feedback queues can starve low-priority processes too, but they normally include ageing, so SJF is the standard answer.

Q: Which factor most directly affects system throughput under heavy load ?

The key's answer is CPU scheduling policy.Throughput is the number of processes or jobs completed per unit time, and it is one of the criteria by which CPU scheduling algorithms are judged. Under heavy load the processor is the contended resource, and the policy that decides which job runs next, and how often it switches, most directly determines how many jobs finish.Network latency mainly lengthens response time for each request rather than the system's rate of completing work, and disk fragmentation is a secondary effect on I/O. Screen resolution has no bearing on throughput.

Q: Four Jobs to be executed on a single processor system arrive at time 0 in the order A,B, C,D. Their burst CPU time requirements are 4,1,8,1 time units, respectively. The completion time of A under round robin scheduling with time slice of one time unit is

A completes at time 9.With a time slice of 1 the processor serves the ready jobs in turn, A, B, C, D, dropping each job when it finishes:0-1 A, 1-2 B (B done at 2), 2-3 C, 3-4 D (D done at 4)4-5 A, 5-6 C, 6-7 A, 7-8 C8-9 A: A's fourth unit, so A is done at 94 is A's burst time, the completion time only if A ran alone first. 8 stops one slice short, and 10 adds a slice too many.

11.

In multilevel feedback scheduling algorithm

A.a process can move to a different classified ready queue
B.classification of ready queue is permanent
C.processes are not classified into groups
D.None of the mentioned above

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: A

In multilevel feedback scheduling, a process can move to a different classified ready queue.

The multilevel feedback queue scheduler keeps several ready queues of different priorities. A process that uses too much CPU time is moved to a lower-priority queue, and one that waits too long can be moved up (ageing). This movement, the 'feedback', is what distinguishes it.

In the plain multilevel queue algorithm, by contrast, processes are classified once and the classification is permanent. Processes are classified into groups in both, so the other options are wrong.

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12.

Which of the following scheduling algorithms selects the process that has been waiting the longest?

A.Shortest Job First (SJF)
B.Round Robin (RR)
C.First-Come, First-Served (FCFS)
D.Priority Scheduling

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: C

The algorithm is First-Come, First-Served (FCFS).

FCFS runs processes in the order they arrive in the ready queue, so the next process chosen is always the one that has waited longest.

Shortest Job First picks the process with the shortest burst, priority scheduling the highest priority, and round robin gives each process a time slice in turn, so none of them chooses by waiting time alone.

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13.

Which of the following scheduling algorithms is non-preemptive ?

A.Round Robin
B.First-In First-Out
C.Multilevel Queue Scheduling
D.Multilevel Queue Scheduling with Feedback

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: B

The non-preemptive algorithm is First-In First-Out.

Under FIFO (first come, first served) a process, once given the CPU, keeps it until it finishes or blocks for I/O; nothing takes the CPU away from it.

Round Robin preempts a process when its time quantum ends. Multilevel queue scheduling preempts lower-priority queues when a higher-priority process arrives, and the feedback version also moves processes between queues, typically using time slices.

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14.

Which of the following process scheduling algorithm may lead to starvation ?

A.FIFO
B.Round Robin
C.Shortest Job Next
D.None of the above

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: C

The algorithm that may lead to starvation is Shortest Job Next.

Shortest Job Next always runs the waiting job with the shortest CPU burst. If short jobs keep arriving, a long job can be passed over indefinitely, which is starvation.

FIFO serves jobs in order of arrival, so every job eventually runs, and Round Robin gives every process a time slice in turn. Since one option does starve, 'None of the above' is wrong.

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15.

Suppose there are 5 processes P1, P2, P3, P4 and P5 with processing time requirements 10, 6, 2, 4 and 8 respectively. What is the average completion time if SJF (Shortest Job First) policy is used ?

A.10
B.12
C.14
D.16

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: C

The average completion time is 14.

SJF runs the jobs in increasing order of time: 2, 4, 6, 8, 10 (P3, P4, P2, P5, P1). Each job completes when all the jobs before it and itself have run.

\[ \begin{aligned} \text{completions} &= 2, 6, 12, 20, 30 \\ \text{average} &= \frac{70}{5} = 14 \end{aligned} \]

The average waiting time is \((0+2+6+12+20)/5 = 8\), which is not offered; 10 is simply the longest job, and 12 and 16 match neither measure. The question asks for completion (turnaround) time with all jobs arriving at 0.

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16.

Which CPU scheduling algorithm is most likely to cause 'Starvation' for long-running processes ?

A.Shortest Job First (SJF) - non preemptive
B.Round Robin (RR)
C.First-Come, First-Served (FCFS)
D.Multilevel Feedback Queue

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

Starvation is most likely under Shortest Job First (SJF) - non preemptive.

SJF always runs the shortest waiting job next. If short jobs keep arriving, a long job can wait indefinitely, because there is always a shorter job ahead of it. That indefinite postponement is starvation; ageing, which raises a job's priority the longer it waits, is the usual remedy.

Round robin gives every process a time slice in turn, and FCFS serves processes in arrival order, so neither starves anyone. Multilevel feedback queues can starve low-priority processes too, but they normally include ageing, so SJF is the standard answer.

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17.

Which factor most directly affects system throughput under heavy load ?

A.Disk fragmentation
B.Network latency
C.Screen resolution
D.CPU scheduling policy

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: D

The key's answer is CPU scheduling policy.

Throughput is the number of processes or jobs completed per unit time, and it is one of the criteria by which CPU scheduling algorithms are judged. Under heavy load the processor is the contended resource, and the policy that decides which job runs next, and how often it switches, most directly determines how many jobs finish.

Network latency mainly lengthens response time for each request rather than the system's rate of completing work, and disk fragmentation is a secondary effect on I/O. Screen resolution has no bearing on throughput.

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18.

Four Jobs to be executed on a single processor system arrive at time 0 in the order A,B, C,D. Their burst CPU time requirements are 4,1,8,1 time units, respectively. The completion time of A under round robin scheduling with time slice of one time unit is

A.10
B.4
C.8
D.9

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: D

A completes at time 9.

With a time slice of 1 the processor serves the ready jobs in turn, A, B, C, D, dropping each job when it finishes:

0-1 A, 1-2 B (B done at 2), 2-3 C, 3-4 D (D done at 4)
4-5 A, 5-6 C, 6-7 A, 7-8 C
8-9 A: A's fourth unit, so A is done at 9

4 is A's burst time, the completion time only if A ran alone first. 8 stops one slice short, and 10 adds a slice too many.

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19.

The problem of indefinite blockage of low-priority jobs in general priority scheduling algorithm can be solved using :

A.Parity bit
B.Aging
C.Compaction
D.Timer

Exam Year(s)

UGC NET - 2012,

Answer: B

Aging solves the problem of indefinite blockage (starvation) of low-priority jobs in priority scheduling. The priority of a process which has waited a long time is raised step by step, so that it finally gets the CPU.

A parity bit is for error detection, compaction is the way to remove external fragmentation by moving memory contents together, and a timer is used to interrupt a running process in time-sharing. None of these stops a low-priority job from waiting forever.

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20.

In UNIX, which of the following command is used to set the task priority ?

A.init
B.nice
C.kill
D.PS

Exam Year(s)

UGC NET - 2012,

Answer: B

The nice command is used in UNIX to set the priority of a task (process): it starts a command with a changed 'niceness' value, which affects how much CPU time the scheduler gives it, and renice changes the value of a running process.

init is the first process that starts the system, kill sends a signal to a process, usually to end it, and PS lists the processes with their status. None of these three sets the priority of a task.

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