Q: 1 In the following table Group—I contains properties and attributes of scheduling algorithm and Group—II contains scheduling algorithms.
| Group—I | Group—II |
|---|---|
| (P) Processes has categories like foreground and background. | (a) Gand scheduling. |
| (Q) Processes announce its deadline and processes are not periodic. | (b) Multi level queue scheduling. |
| (R) Processes have producer consumer relationship. | (c) Rate monotonic scheduling. |
| (S) Processes announce its deadline and processes are periodic. | (d) Earliest deadline first scheduling. |
The matches for (P), (Q), (R), (S) respectively are
(a), (b), (c), (d)
(b), (d), (a), (c)
(d), (a), (b), (c)
(c), (d), (b), (a)
[ Option B ]
For statement (P), where processes are divided into categories like foreground and background, the correct match is Multilevel Queue Scheduling. This algorithm classifies processes into separate queues such as interactive and batch, each with its own scheduling policy, making it suitable for such categorization.
For statement (Q), processes announce their deadlines and are not periodic, which corresponds to Earliest Deadline First Scheduling. This is a dynamic real-time scheduling technique where priorities are assigned based on the nearest deadline, and it works effectively for aperiodic tasks.
In (R), processes have a producer–consumer relationship, indicating interdependent or communicating processes. This matches Gang Scheduling, where related processes are scheduled simultaneously on multiple processors to ensure proper synchronization and efficient execution.
Finally, (S) describes processes that announce deadlines and are periodic, which fits Rate Monotonic Scheduling. This is a fixed-priority scheduling algorithm designed specifically for periodic tasks, where priority is assigned based on the rate.
Q: 2 Consider the following processes for round-robin scheduling
| Process | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 4 |
| P2 | 0 | 6 |
| P3 | 0 | 8 |
| P4 | 0 | 2 |
Which process completes first, if time slice is 2.
P1
P2
P3
P4
[ Option D ]
In Round Robin Scheduling, each process is given a fixed time slice (time quantum) in a cyclic order. If a process does not finish within its time slice, it goes to the end of the Ready Queue.
Here, all processes arrive at time 0 and the time quantum = 2.
| Time | Process | Remaining Burst Time |
|---|---|---|
| 0–2 | P1 | 2 Left (4-2=2) |
| 2–4 | P2 | 4 Left (6-2=4) |
| 4–6 | P3 | 6 Left (8-2=6) |
| 6–8 | P4 | 0 (Completed) (2-2=0) |
Since P4 has burst time equal to time quantum, it completes in its first execution cycle, before others finish.
Q: 3 What is Throughput in process scheduling?
Time required to complete a process
Number of processes completed per unit time
Time required by a process to respond
Time required to abort a process
[ Option B ]
Different types of operating systems are designed for different purposes such as improving CPU utilization, providing quick response, or meeting time limits.
| Operating System Type | Description |
|---|---|
| Batch Processing System | Jobs are collected and executed in batches without user interaction to reduce CPU idle time. |
| Multiprogramming Operating System | Multiple programs are kept in memory simultaneously to improve CPU and resource utilization. |
| Time Sharing Operating System | Multiple users can use the computer system at the same time with quick response. |
| Real-Time Operating System | Processes data and provides output within a fixed time limit. |
Q: 4 CPU scheduling is the basis of –
Multiprogramming Operating System
Large Memory Sized Systems
Multiprocessor Systems
None of the mentioned
[ Option A ]
CPU Scheduling is the foundation of a Multiprogramming Operating System. In multiprogramming, several processes are kept in memory at the same time, and CPU scheduling decides which process will be given the CPU when it becomes available. This ensures efficient CPU utilization by keeping the processor busy as much as possible. CPU scheduling ensures that the CPU is never idle when there are processes ready to execute, improving throughput.
Q: 5 Which of the following is a solution for starvation in priority based scheduling in operating system?
Caching
Aging
Paging
Thrashing
[ Option B ]
In priority-based scheduling, processes with lower priority may never get CPU time if higher priority processes keep arriving. This problem is called Starvation.
To solve this, the concept of Aging is used. In aging, the priority of a waiting process is gradually increased over time, so eventually it gets a chance to execute.
Q: 6 Which of the following, best represents an event to invoke medium-term scheduler?
Clock interrupts
I/O interrupts
New process creation
Need of swapping
[ Option D ]
In an Operating System, the CPU scheduling process involves three levels of schedulers:
Q: 7 A scenario in priority-based scheduling where higher priority task waits for a low priority task to complete, when both share the same resources is called _________.
Priority Ceiling
Priority Sharing
Priority Inheritance
Priority Inversion
[ Option D ]
In priority-based scheduling, normally a higher priority task should execute before a lower priority task. However, a problem occurs when:
This situation is called Priority Inversion, because the expected priority order gets inverted, the higher priority task is forced to wait for the lower priority task.
Q: 8 A given number of tasks T1 to Tn whose duration is known, are queued in a single processor machine. Which of the following will be the best scheduling algorithm?
First Come First Serve
Shortest Job First
Round Robin Scheduling
Priority Scheduling
[ Option B ]
The key point in the question is, the duration (burst time) of all tasks is known. Remember, when burst times are known in advance, the Shortest Job First (SJF) scheduling algorithm is considered the best because it minimizes the average waiting time.
Shortest Job First selects the process with the smallest execution time first. This ensures that short tasks do not wait behind long tasks, reducing overall waiting time.
Q: 9 Which of the following scheduling algorithm is always pre-emptive?
First Come First Served
Shortest Job First
Priority Scheduling
Round Robin Scheduling
[ Option D ]
In CPU scheduling, pre-emptive algorithms are those where a running process can be interrupted and the CPU is given to another process.
Round Robin Scheduling is always pre-emptive because each process is given a fixed Time Slice or Time Quantum. After this time expires, the process is interrupted, and the CPU is assigned to the next process in the queue.
Q: 10 What is the “Convoy effect” in CPU scheduling?
A situation where multiple processes arrive at the CPU at the same time.
A condition where all I/O-bound processes are executed before CPU-bound processes.
A scenario in which one long CPU-bound process delays many shorter I/O-bound processes, reducing overall system utilization.
A technique used to prioritize processes with the shortest burst time.
[ Option C ]
The Convoy Effect is a performance problem that occurs in CPU scheduling, especially in algorithms like First-Come, First-Served (FCFS). In this situation, a long CPU-Bound process gets the CPU first. Meanwhile, several I/O-Bound processes have to wait in the Ready Queue.
Since the CPU is busy executing the long process, the shorter I/O-bound processes are delayed. As a result, the CPU stays idle when those I/O-bound processes later perform I/O operations, and overall system utilization drops.
| TYPE OF PROCESS | EXPLANATION |
|---|---|
| CPU-Bound | Uses the CPU heavily with few I/O operations. When it runs first, it keeps the CPU busy for a long time, delaying other processes waiting in the queue. |
| I/O-Bound | Performs frequent I/O operations with short CPU bursts. These processes must wait for the long CPU-bound process to finish, leading to poor CPU and I/O utilization. |
Q: 11 Which of the following is true?
Priority inversion is a solution for avoiding deadlock condition
Priority inheritance is the cause for entering into deadlock condition
Priority inversion occurs when a low priority task is holding a resource when a higher priority task is waiting for the same resource
Priority inversion occurs when a high priority task is holding a resource when a lower priority task is waiting for the same resource
[ Option C ]
In operating systems, processes are scheduled based on priority. Priority Inversion is a situation where a high-priority process is forced to wait because a low-priority process is holding a resource that it needs. This leads to an unexpected inversion of priorities.
Q: 12 Processes are assigned to the CPU in the order they requested is called
Longest Job First
Shortest Job First
Priority Scheduling
FIFS (First-In-First-Serviced)
[ Option D ]
In operating systems, CPU scheduling determines the order in which processes are executed. When processes are assigned to the CPU in the exact order they arrive or request it, this scheduling technique is called First-In-First-Served (FIFS), also commonly known as First-Come-First-Served (FCFS).
For example, if processes arrive in order P1 → P2 → P3, then execution will also happen in the same order: P1 → P2 → P3.
Q: 13 Which of the following statements are true?
I. Shortest Remaining Time First (SRTF) scheduling may cause starvation.
II. Preemptive scheduling may cause starvation.
III. Round robin is better than FCFS in terms of response time.
I only
I & III only
II and III only
I, II and III
[ Option D ]
SRTF (Shortest Remaining Time First) is a Preemptive Scheduling algorithm where the process with the shortest remaining execution time is given the CPU. A long-running process might never get to execute if there is a continuous stream of new, short-burst processes arriving in the ready queue. The long-running process would always have a longer remaining time and would be repeatedly preempted, leading to starvation.
Preemptive scheduling may cause starvation, this is a general statement that is true. Starvation is a common problem in any Preemptive Scheduling algorithm that uses a priority or time-based approach, like SRTF or a Priority-Based Scheduler where low-priority processes might never get to run if there is a constant flow of high-priority processes.
Round Robin (RR) scheduling allocates CPU time in small time slices (quantum) to each process fairly, leading to better responsiveness compared to FCFS (First-Come, First-Serve), where a process waits (even short-burst ones) until the previous one completes because its Non-Preemptive Scheduling algorithm.
Q: 14
For the following list of processes, the average turnaround time and average waiting time using FCFS algorithm is:
| Process | Arrival Time (ms) | Execution Time (ms) |
|---|---|---|
| P1 | 6 | 24 |
| P2 | 0 | 3 |
| P3 | 3 | 3 |
Average Turnaround Time 30 ms and Average Waiting Time 4.5 ms
Average Turnaround Time 13 ms and Average Waiting Time 3 ms
Average Turnaround Time 10 ms and Average Waiting Time 0 ms
Average Turnaround Time 24 ms and Average Waiting Time 0 ms
[ Option C ]
In FCFS (First Come First Serve) scheduling, processes are executed in the order of their arrival time.
Step 1: Order of Execution :
| Process | Arrival Time | Execution Time |
|---|---|---|
| P2 | 0 | 3 |
| P3 | 3 | 3 |
| P1 | 6 | 24 |
So, execution order: P2 → P3 → P1
Step 2: Gantt Chart :
| Process | Start | Finish |
|---|---|---|
| P2 | 0 | 3 |
| P3 | 3 | 6 |
| P1 | 6 | 30 |
Step 3: Turnaround Time (TAT) :
TAT = Completion Time-Arrival Time
| Process | Completion | Arrival | TAT |
|---|---|---|---|
| P2 | 3 | 0 | 3 |
| P3 | 6 | 3 | 3 |
| P1 | 30 | 6 | 24 |
Average TAT: (3+3+24)/3 = 30/3 = 10 ms
Step 4: Waiting Time (WT) :
WT = TAT-Execution Time
| Process | TAT | Burst | WT |
|---|---|---|---|
| P2 | 3 | 3 | 0 |
| P3 | 3 | 3 | 0 |
| P1 | 24 | 24 | 0 |
Average WT: (0+0+0)/3 = 0 ms
Thus:
Q: 15 Consider the following process table with the arrival time and burst time given in milliseconds
| Process Time | Arrival Time | Burst Time |
|---|---|---|
| P1 | 0 | 7 |
| P2 | 1 | 4 |
| P3 | 2 | 9 |
| P4 | 3 | 5 |
If non preemptive SJF (Shortest Job First) is followed, the average waiting time is
7
6.25
7.75
7.5
[ Option A ]
In Non-Preemptive SJF, once a process starts execution, it runs till completion, and at each decision point we choose the shortest burst time among the arrived processes.
At time 0: only P1 is available → executes from 0 to 7.
At time 1: P1 already running. At 7 P2, P3, P4 have arrived. Among P2(4), P3(9), P4(5), P2 has shortest burst (4). So, P2 from 7 to 11.
From 11 onward, remaining: P3(9), P4(5). P4 is shorter. So, P4 from 11 to 16.
Finally, P3 from 16 to 25.
| Process | Arrival Time | Burst Time | Completion Time | Turnaround Time | Waiting Time |
|---|---|---|---|---|---|
| P1 | 0 | 7 | 7 | 7 | 0 |
| P2 | 1 | 4 | 11 | 10 | 6 |
| P3 | 2 | 9 | 25 | 23 | 14 |
| P4 | 3 | 5 | 16 | 13 | 8 |
Q: 16
What is the wait time of process P2 using shortest job first scheduling?
| Process | Burst Time |
|---|---|
| P1 | 6 |
| P2 | 8 |
| P3 | 7 |
| P4 | 3 |
9
3
7
16
[ Option D ]
In Shortest Job First (SJF) scheduling, processes are executed in ascending order of burst time.
Arrange Processes (Ascending Burst Time):
| Process | Burst Time |
|---|---|
| P4 | 3 |
| P1 | 6 |
| P3 | 7 |
| P2 | 8 |
Therefore, execution order are P4 → P1 → P3 → P2.
Calculate Waiting Time of P2:
The waiting time of process P2 is calculated by adding the burst times of all processes executed before it, since P4, P1, and P3 execute before P2 with burst times 3, 6, and 7 respectively, the total waiting time of P2 is 3+6+7 = 16.
Note:
Q: 17 Match the correct description :
| Column – I | Column – II |
|---|---|
| a. Throughput | i. Time from process submission to its first response being produced. |
| b. Turnaround Time | ii. Number of processes completed per unit of time. |
| c. Waiting Time | iii. Total time a process spends in the ready queue waiting for the CPU. |
| d. Response Time | iv. Total time from process submission to its completion, doing I/O, waiting in the ready queue, and executing on the CPU. |
Choose the correct option:
a – iv, b – ii, c – iii, d – i
a – ii, b – iii, c – iv, d – i
a – ii, b – iv, c – i, d – iii
a – ii, b – iv, c – iii, d – i
[ Option D ]
In an operating system, CPU scheduling determines the order in which processes are executed. To evaluate scheduling algorithms, we use certain key performance metrics, Throughput, Turnaround Time, Waiting Time, and Response Time.
| METRIC | DEFINITION | FORMULA |
|---|---|---|
| Throughput | Number of processes completed per unit of time. It shows how efficiently the system is processing tasks. | Throughput = (Number of Processes Completed) / (Total Time Taken) |
| Turnaround Time (TAT) | Total time taken from process submission to its completion. | TAT = Completion Time – Arrival Time |
| Waiting Time (WT) | Total time a process spends waiting in the ready queue before execution. | WT = Turnaround Time – Burst Time |
| Response Time (RT) | Time from process submission to the first response (first CPU allocation). | RT = First Response Time – Arrival Time |
| Turnaround Time (Average) | Average of turnaround times of all processes. | Average TAT = Σ(TAT) / Number of Processes |
| Waiting Time (Average) | Average of waiting times of all processes. | Average WT = Σ(WT) / Number of Processes |
Q: 18
Consider the five processes with given arrival and service times. What will be the average turnaround time for Shortest Process Next (SPN) scheduling algorithm?
| Process | Arrival Time | Service Time (Ts) |
|---|---|---|
| A | 0 | 3 |
| B | 2 | 6 |
| C | 4 | 4 |
| D | 6 | 5 |
| E | 8 | 2 |
7.6
7.2
7.8
8
[ Option A ]
SPN (Shortest Process Next), also known as Shortest Job First (SJF) (non-preemptive), is a CPU scheduling algorithm that selects the ready process with the shortest service (burst) time for execution next. The goal is to minimize the average waiting and turnaround times.
Turnaround Time (TAT) = Completion Time − Arrival Time.
| PROCESS | ARRIVAL | COMPLETION | TURNAROUND TIME |
|---|---|---|---|
| A | 0 | 3 | 3 – 0 = 3 |
| B | 2 | 9 | 9 – 2 = 7 |
| E | 8 | 11 | 11 – 8 = 3 |
| C | 4 | 15 | 15 – 4 = 11 |
| D | 6 | 20 | 20 – 6 = 14 |
Total Turnaround Time = 3 + 7 + 3 + 11 + 14 = 38
Average Turnaround Time = 38 / 5 = 7.6
Q: 19 In the context of operating system process scheduling, if W represents Wait Time, R represents Run Time, T represents Turnaround Time, S represents Response Time, then which of the following is correct option with respect to given diagram?

Time1:W, Time2:R, Time3:T, Time4:S
Time1:S, Time2:W, Time3:R, Time4:T
Time1:W, Time2:R, Time3:S, Time4:T
Time1:R, Time2:S, Time3:W, Time4:T
[ Option C ]
In process scheduling, the time intervals are defined as:
| TERM | MEANING |
|---|---|
| Wait Time (W) | Total time a process spends in the ready queue waiting for CPU execution. It is the time from arrival to the start of execution. |
| Run Time (R) | The total time for which the process is executed by the CPU. It is also called Burst Time. |
| Response Time (S) | The time from arrival of the process to the first response or output. Means, when the process starts executing for the first time. |
| Turnaround Time (T) | The total time taken from arrival of the process to its completion. It includes waiting time and execution time. |
From the given diagram,
Q: 20 The maximum number of processes that can be in ready state for a computer system with 16 CPUs is?
16
256
The value is independent of number of CPUs
65536
[ Option C ]
In an Operating System (OS), the ready state contains all processes that are prepared to execute and are waiting for CPU allocation. The ready queue can hold many processes at the same time.
The number of processes in the ready state does not depend on the number of CPUs.
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