Process-Scheduling
Question 1 |
Consider the following four processes with arrival times (in milliseconds) and their length of CPU bursts (in milliseconds) as shown below:

These processes are run on a single processor using preemptive Shortest Remaining Time First scheduling algorithm. If the average waiting time of the processes is 1 millisecond, then the value of Z is _____.
2 | |
7 | |
1 | |
4 |

At this point P4 arrives with burst 'Z' & P3 is in queue with burst 3.
P1 & P2 have executed with P1 incurred delay 1unit & P2 0units.
Hence, Avg = 0+1+x/4 =1
⇒ x=3, the next delay should be 3. It would happen if assume Z=2.
It executes and completes at 6.
P3 will wait totally for 3units.
Hence, Avg=1.

Z=2
Question 2 |
Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) as given below:
If the pre-emptive shortest remaining time first scheduling algorithm is used to schedule the processes, then the average waiting time across all processes is __________ milliseconds.
3 | |
4 | |
5 | |
6 |

Question 3 |
Consider the set of processes with arrival time (in milliseconds), CPU burst time (in milliseconds), and priority (0 is the highest priority) shown below. None of the processes have I/O burst time.

The average waiting time (in milliseconds) of all the processes using preemptive priority scheduling algorithm is __________.
29 | |
30 | |
31 | |
32 |

Waiting Time = 0 + (33 - 5) + (40 - 2) + (49 - 12) + (51 - 9) = 145
Average waiting time: 145/5 = 29
Question 4 |
Consider an arbitrary set of CPU-bound processes with unequal CPU burst lengths submitted at the same time to a computer system. Which one of the following process scheduling algorithms would minimize the average waiting time in the ready queue?
Shortest remaining time first | |
Round-robin with time quantum less than the shortest CPU burst | |
Uniform random | |
Highest priority first with priority proportional to CPU burst length |
We can consider an arbitrary set of CPU-bound processes with unequal CPU burst lengths submitted at the same time to a computer system.
We have to choose the appropriate process scheduling algorithms, which would minimize the average waiting time in the ready queue.
Waiting time is the time for which process is ready to run but not executed by CPU scheduler.
In all CPU Scheduling algorithms, shortest job first is optimal.
It gives minimum turnaround time, minimum average waiting time and high throughput and the most important thing is that shortest remaining time first is the preemptive version of shortest job first.
This scheduling algorithm may lead to starvation because if the short processes are added to the CPU scheduler continuously then the currently running process will never be able to execute as they will get pre-empted but here all the processes are arrived at same time so there will be no issue such as starvation.
SRTF would be same as SJF.
So, A is the answer. Shortest remaining time first.
Question 5 |
Consider the following processes, with the arrival time and the length of the CPU burst given in milliseconds. The scheduling algorithm used is preemptive shortest remaining-time first.

The average turn around time of these processes is _________ milliseconds.
8.25 | |
8.26 | |
8.27 | |
8.28 |
To answer the question we need to design the gantt chart:

In this algorithm, the processes will be scheduled on the CPU which will be having least remaining burst time.
Turnaround Time (TAT) = Completion Time (CT) - Arrival Time (AT)
TAT for P1 = 20 - 0 = 20,
TAT for P2 = 10 - 3 = 7,
TAT for P3 = 8 - 7 = 1,
TAT for P4 = 13 - 8 = 5.
Total TAT = 20 + 7 + 1 + 5 = 33 / 4 = 8.25 (Avg. TAT)
Question 6 |
Consider a uniprocessor system executing three tasks T1, T2 and T3, each of which is composed of an infinite sequence of jobs (or instances) which arrive periodically at intervals of 3, 7 and 20 milliseconds, respectively. The priority of each task is the inverse of its period and the available tasks are scheduled in order of priority, with the highest priority task scheduled first. Each instance of T1, T2 and T3 requires an execution time of 1, 2 and 4 milliseconds, respectively. Given that all tasks initially arrive at the beginning of the 1st millisecond and task preemptions are allowed, the first instance of T3 completes its execution at the end of ______________ milliseconds.
13 | |
12 | |
15 | |
16 |

∴ At the end of 12 milliseconds, 1st instance of T3 will complete its execution.
Question 7 |
The maximum number of processes that can be in Ready state for a computer system with n CPUs is
n | |
n2 | |
2n | |
Independent of n |
Maximum number of processes that will be in ready state is independent of number of processors.
Question 8 |
For the processes listed in the following table, which of the following scheduling schemes will give the lowest average turnaround time?

First Come First Serve | |
Non-preemptive Shortest Job First | |
Shortest Remaining Time | |
Round Robin with Quantum value two |

FCFS:

TAT for A = completion time(A) - AT(A) = 3 - 0 = 3
TAT of B = 9 - 1 = 8
TAT of C = 13 - 4 = 9
TAT of D = 15 - 6 = 9
∴ Avg. TAT = 29/4
SJF:

TAT of A = 3 - 0 = 3
TAT of B = 9 - 1 = 8
TAT of C = 15 - 4 = 11
TAT of D = 11 - 6 = 5
∴ Avg. TAT = 27/4
SRTF:

TAT of A = 3 - 0 = 3
TAT of B = 15 - 1 = 14
TAT of C = 8 - 4 = 4
TAT of D = 10 - 6 = 4
∴ Avg. TAT = 25/4
RR:

TAT of A = 5 - 0 = 5
TAT of B = 15 - 1 = 14
TAT of C = 13 - 4 = 9
TAT of D = 11 - 6 = 5
∴ Avg. TAT = 33/4
∴ SRTF has lowest Avg. TAT.
Question 9 |
Consider the following set of processes that need to be scheduled on a single CPU. All the times are given in milliseconds.

Using the shortest remaining time first scheduling algorithm, the average process turnaround time (in msec) is ________.
7.2 | |
7.3 | |
7.4 | |
7.5 |

TAT(A) = 8-0 = 8, TAT(B)= 5-3=2, TAT(C)= 12-5=7, TAT(D)= 21-7= 14, TAT(E)=15-10=5
Average turnaround time = 8+2+7+14+5/ 5 = 7.2ms
Question 10 |
Three processes A, B and C each execute a loop of 100 iterations. In each iteration of the loop, a process performs a single computation that requires tc CPU milliseconds and then initiates a single I/O operation that lasts for tio milliseconds. It is assumed that the computer where the processes execute has sufficient number of I/O devices and the OS of the computer assigns different I/O devices to each process. Also, the scheduling overhead of the OS is negligible. The processes have the following characteristics:
Process id tc tio A 100 ms 500 ms B 350 ms 500 ms C 200 ms 500 ms
The processes A, B, and C are started at times 0, 5 and 10 milliseconds respectively, in a pure time sharing system (round robin scheduling) that uses a time slice of 50 milliseconds. The time in milliseconds at which process C would complete its first I/O operation is ___________.
1000 | |
1001 | |
1002 | |
1003 |

So 'C' completes its I/O at 1000 time units.
Question 11 |
Consider the 3 processes, P1, P2 and P3 shown in the table.
Process Arrival time Time Units Required P1 0 5 P2 1 7 P3 3 4
The completion order of the 3 processes under the policies FCFS and RR2 (round robin scheduling with CPU quantum of 2 time units) are
FCFS: P1, P2, P3 RR2: P1, P2, P3 | |
FCFS: P1, P3, P2 RR2: P1, P3, P2 | |
FCFS: P1, P2, P3 RR2: P1, P3, P2 | |
FCFS: P1, P3, P2 RR2: P1, P2, P3 |
FCFS is clear.
RR Queue: In RR queue time slot is of 2 units.
Processes are assigned in following order
P1, P2, P1, P3, P2, P1, P3, P2, P2
This question used ready queue concept. At t=2, P2 starts and P1 is sent to the ready queue and at t=3 P3 arrives so then the job P3 is queued in ready queue after P1. So at t=4, again P1 is executed then P3 is executed for first time at t=6.
RR2: P1, P3, P2
So option C.
Question 12 |
Consider the following table of arrival time and burst time for three processes P0, P1 and P2.

The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of processes. What is the average waiting time for the three processes?
5.0 ms | |
4.33 ms | |
6.33 ms | |
7.33 ms |

CT = Completion time
TAT = Turn Around Time
WT = Waiting Time
TAT = CT - AT
WT = TAT - BT
Gantt Chart using pre-emptive shortest job first scheduling algorithm,

Avg. WT = 4+0+11/3 = 5ns
Question 13 |
Which of the following statements are true?
- I. Shortest remaining time first 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 and III only | |
II and III only | |
I, II and III |
- Pre-emptive scheduling causes starvation (for example lower priority jobs are waiting).
- Best Response time is given by RR.
Question 14 |
In the following process state transition diagram for a uniprocessor system, assume that there are always some processes in the ready state:

Now consider the following statements:
- I. If a process makes a transition D, it would result in another process making transition A immediately.
II. A process P2 in blocked state can make transition E while another process P1 is in running state.
III. The OS uses preemptive scheduling.
IV. The OS uses non-preemptive scheduling.
Which of the above statements are TRUE?
I and II | |
I and III
| |
II and III | |
II and IV |
Statement II is true, a process can move to ready state when I/O completes irrespective of other process being in running state or not.
Statement III is true, the transition C, represents preemptive scheduling.
Statement IV is false, because it is preemptive scheduling.
Correct Answer is Option-C (II & III are true).
Question 15 |
An operating system uses Shortest Remaining Time first (SRT) process scheduling algorithm. Consider the arrival times and execution times for the following processes:
Process Execution time Arrival time P1 20 0 P2 25 15 P3 10 30 P4 15 45
What is the total waiting time for process P2?
5 | |
15 | |
40 | |
55 |

Waiting time for process P2 = TAT - Execution time
= Completion time - AT - ET
= 55 - 15 - 25
= 15
Question 16 |
Consider three CPU-intensive processes, which require 10, 20 and 30 time units and arrive at times 0, 2 and 6, respectively. How many context switches are needed if the operating system implements a shortest remaining time first scheduling algorithm? Do not count the context switches at time zero and at the end.
1 | |
2 | |
3 | |
4 |

Total no.of context switches is 2.
Question 17 |
Consider three processes (process id 0, 1, 2 respectively) with compute time bursts 2, 4 and 8 time units. All processes arrive at time zero. Consider the longest remaining time first (LRTF) scheduling algorithm. In LRTF ties are broken by giving priority to the process with the lowest process id. The average turnaround time is:
13 units | |
14 units | |
15 units | |
16 units
|

Avg TAT = 12+13+14/3 = 39/3 = 13 units
Question 18 |
Consider three processes, all arriving at time zero, with total execution time of 10, 20 and 30 units, respectively. Each process spends the first 20% of execution time doing I/O, the next 70% of time doing computation, and the last 10% of time doing I/O again. The operating system uses a shortest remaining compute time first scheduling algorithm and schedules a new process either when the running process gets blocked on I/O or when the running process finishes its compute burst. Assume that all I/O operations can be overlapped as much as possible. For what percentage of time does the CPU remain idle?
0% | |
10.6% | |
30.0% | |
89.4% |

Total time needed to complete the execution = 47
Idle time = 2+3 = 5
Percentage of Idle time = 5/47 × 100 =10.6%
Question 19 |
Consider the following set of processes, with the arrival times and the CPU-burst times given in milliseconds.
Process Arrival Time Burst Time P1 0 5 P2 1 3 P3 2 3 P4 4 1
What is the average turnaround time for these processes with the preemptive shortest remaining processing time first (SRPT) algorithm?
5.50 | |
5.75 | |
6.00 | |
6.25 |

Avg. TAT = 12+3+6+1/4 = 22/4 = 5.50
Question 20 |
A uni-processor computer system only has two processes, both of which alternate 10ms CPU bursts with 90ms I/O bursts. Both the processes were created at nearly the same time. The I/O of both processes can proceed in parallel. Which of the following scheduling strategies will result in the least CPU utilization (over a long period of time) for this system?
First come first served scheduling | |
Shortest remaining time first scheduling | |
Static priority scheduling with different priorities for the two processes | |
Round robin scheduling with a time quantum of 5 ms
|
(i) FCFS:
0-10: Process 1 can execute
10-20: Process 2 can execute
100-110: Process 1 Terminate
110-120: Process 2 Terminate
CPU utilization = 20/100 [In every 100ms it utilizes]
=20%
(ii) SRTF: can process P and Q same as FCFS
then CPU utilization = 20%
(iii) Round robin: with TQ-5
0-5: Process P1
5-10: Process P2
10-15: Process P1
15-20: Process P2
105-110: Process P1
110-115: Process P2
CPU utilization = 20/105 = 19.5
Question 21 |
Which of the following scheduling algorithms is non-preemptive?
Round Robin | |
First-In First-Out | |
Multilevel Queue Scheduling | |
Multilevel Queue Scheduling with Feedback |
Question 22 |
Consider a set of n tasks with known runtimes r1, r2, ..., rn to be run on a uniprocessor machine. Which of the following processor scheduling algorithms will result in the maximum throughput?
Round-Robin | |
Shortest-Job-First | |
Highest-Response-Ratio-Next | |
First-Come-First-Served |
Question 23 |
Consider n processes sharing the CPU in a round-robin fashion. Assuming that each process switch takes s seconds, what must be the quantum size q such that the overhead resulting from process switching is minimized but at the same time each process is guaranteed to get its turn at the CPU at least every t seconds?
q ≤ t-ns/n-1 | |
q ≥ t-ns/n-1 | |
q ≤ t-ns/n+1 | |
q ≥ t-ns/n+1 |

Question 24 |
The process state transition diagram in below figure is representative of

a batch operating system | |
an operating system with a preemptive scheduler | |
an operating system with a non-preemptive scheduler | |
a uni-programmed operating system |
So this is preemptive.
Question 25 |
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
10 | |
4 | |
8 | |
9 |

∴ Completion time of A is 9 unit.
Question 26 |
Consider the following statements about process state transitions for a system using preemptive scheduling.
- I. A running process can move to ready state.
II. A ready process can move to running state.
III. A blocked process can move to running state.
IV. A blocked process can move to ready state.
Which of the above statements are TRUE?
II and III only | |
I, II and III only
| |
I, II, III and IV
| |
I, II and IV only
|

Question 27 |
Consider the following set of processes, assumed to have arrived at time 0. Consider the CPU scheduling algorithms Shortest Job First (SJF) and Round Robin (RR). For RR, assume that the processes are scheduled in the order P1,P2,P3,P4.
If the time quantum for RR is 4 ms, then the absolute value of the difference between the average turnaround times (in ms) of SJF and RR (round off to 2 decimal places) is _____.
5.25 |

Turn Around Time = (21 – 0) + (13 – 0) + (2 – 0) + (6 – 0), Average = 42/4 = 10.50

Turn Around Time (TAT) = (18 – 0) + (21 – 0) + (10 – 0) + (14 – 0), Average = 63/4 = 15.75
Absolute difference = |10.50-15.75| = 5.25
Question 28 |
Let a memory have four free blocks of sizes 4k, 8k, 20k, 2k. These blocks are allocated following the best-fit strategy. The allocation requests are stored in a queue as shown below.

The time at which the request for J7 will be completed will be
16 | |
19 | |
20 | |
37 |

At t = 8

At t = 10

At t = 11

J7 can be finished at t = 19.
Question 29 |
Consider n jobs J1, J2,......Jn such that job Ji has execution time ti and a non-negative integer weight wi. The weighted mean completion time of the jobs is defined to be , where Ti is the completion time of job Ji. Assuming that there is only one processor available, in what order must the jobs be executed in order to minimize the weighted mean completion time of the jobs?
Non-decreasing order of ti | |
Non-increasing order of wi | |
Non-increasing order of witi | |
None-increasing order of wi/ti |
So answer is the non-increasing order of wi/ti.
Question 30 |
The process state transition diagram of an operating system is as given below.

Which of the following must be FALSE about the above operating system?
It is a multiprogrammed operating system | |
It uses preemptive scheduling | |
It uses non-preemptive scheduling | |
It is a multi-user operating system |
Question 31 |
The arrival time, priority, and duration of the CPU and I/O bursts for each of three processes P1, P2 and P3 are given in the table below. Each process has a CPU burst followed by an I/O burst followed by another CPU burst. Assume that each process has its own I/O resource.

The multi-programmed operating system uses preemptive priority scheduling. What are the finish times of the processes P1, P2 and P3 ?
11, 15, 9 | |
10, 15, 9 | |
11, 16, 10 | |
12, 17, 11 |

Hence, finish times of process.
P1 - 10
P2 - 11
P3 - 9
Question 32 |
Round Robin | |
First come first serve | |
preemptive | |
random scheduling |
● Processing time requirements (including any OS delay) are measured in tenths of seconds or shorter increments of time.
● A real time system is a time bound system which has well defined fixed time constraints. Processing must be done within the defined constraints or the system will fail. They either are event driven or time sharing.
● Event driven systems switch between tasks based on their priorities while time sharing systems switch the task based on clock interrupts. Most RTOS’s use a pre-emptive scheduling algorithm.
Question 33 |

a | |
b | |
c | |
d |
Total 4 processors:

Question 34 |
Which of the following statements is FALSE?
The long term scheduler controls the degree of multiprogramming
| |
Multiple process of a single program cannot exist | |
Ready queue of the processes resides in main memory
| |
A process can have multiple sub processes
|
→ Ready queue of the processes resides in main memory.
→ A process can have multiple sub processes.
→ Multiple process of a single program cannot exist because each program can have only one process.
Question 35 |
13 units | |
14 units | |
15 units | |
16 units |

Avg TAT = 12+13+14/3 = 39/3 = 13 units
Question 36 |
0% | |
10.6% | |
30.0% | |
89.4% |

Total time needed to complete the execution = 47
Idle time = 2+3 = 5
Percentage of Idle time = 5/47 × 100 =10.6%
Question 37 |
1 | |
2 | |
3 | |
4 |

Total no.of context switches is 2.
Question 38 |
FIFO | |
Round Robin | |
Shortest Job Next | |
None of the option |
● In the Shortest job next scheduling shortest job process will run first and later higher burst time.Here the process with higher burst time need to wait for longer time.
Question 39 |
This algorithm is equivalent to the first come first serve algorithm | |
This algorithm is equivalent to the round-robin algorithm | |
This algorithm is equivalent t the shortest-job-first algorithm | |
This algorithm is equivalent to the shortest-remaining time-first algorithm |

Consider scheduler schedule processes priority after S times units so order of execution will be P1 P2 P3 P4 P1 P2 P3 P4 (S< execution time of any process) which is exactly same as round robin so answer – B.
Question 40 |
Priority | |
Round Robin | |
FCFS | |
none of the above |
Question 41 |
State whether TRUE or FALSE
- (i) Shortest remaining time (SRT) algorithm is the preemptive version of the shortest job next(SJN) CPU scheduling algorithm
(ii) A “context switch” is the mechanism to store and restore the state or context of a CPU in PCB
(i) True, (ii) True | |
(i) False, (ii) False
| |
(i) False, (ii) True
| |
(i) True, (ii) False
|
TRUE: A “context switch” is the mechanism to store and restore the state or context of a CPU in PCB.
Question 42 |
Running | |
Ready | |
Interrupted | |
Blocked |
A program which is going to be picked up by the OS into the main memory is called a new process.
2. Ready
Whenever a process is created, it directly enters in the ready state, in which, it waits for the CPU to be assigned. The OS picks the new processes from the secondary memory and put all of them in the main memory. The processes which are ready for the execution and reside in the main memory are called ready state processes.
3. Running
One of the processes from the ready state will be chosen by the OS depending upon the scheduling algorithm.
4. Block or wait
From the Running state, a process can make the transition to the block or wait state depending upon the scheduling algorithm or the intrinsic behavior of the process.
Question 43 |

Process creation | |
Process state | |
Queueing diagram | |
Communication model |
→ All processes, upon entering into the system, are stored in the Job Queue.
→ Processes in the Ready state are placed in the Ready Queue.
→ Processes waiting for a device to become available are placed in Device Queues. There are unique device queues available for each I/O device.
Question 44 |
Priority scheduling | |
Round robin | |
FCFS | |
SJF |
→ SJF scheduling algorithm will result in the maximum throughput
Question 45 |
Swapping | |
Dirty bit | |
Aging | |
Compaction |
→ A task with a high priority is allowed to access a specific system resource before a task with a lower priority is allowed to do the same.
→ A disadvantage of this approach is that tasks assigned with a lower priority may be starved when a large number of high priority tasks are queued. Aging is used to gradually increase the priority of a task, based on its waiting time in the ready queue.
Question 46 |
1 | |
5 | |
6 | |
7 |
→ But according to their intention will cause total 6 switches. 0-2, 2-4, 4-6, 6-8, 8-10, 10-12, 12-14. We are initially started with 0. So, It will take 6 connected switches
Note: We thought they made this question wrong.
Question 47 |
Arrival times and time units required for execution are given below:

Which of the following is the correct completion order if RR scheduling algorithm is used with time quantum of 2 units?
P1,P2,P3 | |
P1,P3,P2 | |
P3,P1,P2 | |
P2,P1,P3 |

The completion of order of processes are P3,P1 and P2
Question 48 |
a. First-come-first-served
b. Round Robin
c. Priority
d. Shortest process next
e. Shortest remaining time first
a, c and e | |
c, d and e | |
b, d and e | |
b, c and d |
1. Priority: A process ready to run for CPU can wait indefinitely because of low priority
2. Shortest process next : longest process possibility to get starvation
3. Shortest remaining time first: longest process possibility to get starvation
Solution: Aging
→ FCFS and Round Robin scheduling never happen indefinite blocking.
Question 49 |
File descriptor table | |
System call state | |
Scheduling parameters | |
Kernel stack |
1. File descriptor table
2. System call state
3. Kernel stack
Scheduling parameters: Process priority, amount of CPU time consumed recently, amount of time spent sleeping recently. Together, these are used to determine which process to run next.
Question 50 |

a-3, b-4, c-2, d-1 | |
a-4, b-3, c-2, d-1 | |
a-2, b-4, c-1, d-3 | |
a-3, b-4, c-1, d-2 |
Batch processing→ FIFO
Time sharing→ Round robin with the help of time quantum.
Interrupt processing→ LIFO
Question 51 |
Shortest-Job-first | |
Round-robin | |
Priority based | |
Shortest-Job-next |
→ Round robin is purely Nonpreemptive.
→ Priority based scheduling also preemptive and Nonpreemptive.
→ Shortest-Job-next scheduling algorithm is either preemptive and Nonpreemptive.
Question 52 |
FILO | |
FIFO | |
SJF | |
Longest time first |
Question 53 |

a-iii, b-i, c-ii, d-iv | |
a-iv, b-i, c-iii, d-ii | |
a-iv, b-iii, c-i, d-ii | |
a-iii, b-iii, c-ii, d-i |
Blocked→ Ready: Request made by the process is satisfied or an event for which it was waiting occurs.
Running→ Blocked: Process wishes to wait for some action by another process.
Running→ Ready: The process is preempted.
Question 54 |

The Gantt Chart for preemptive SJF scheduling algorithm is
![]() | |
![]() | |
![]() | |
![]() |

Question 55 |

The time quantum q=1. When two processes want to join ready queue simultaneously, the process which has not executed recently is given priority. The finish time of processes P1, P2, P3 and P4 will respectively be
4, 5, 7 and 8 | |
8, 2, 7 and 5 | |
2, 5, 7 and 8 | |
8, 2, 5 and 7 |

Question 56 |
Before the CPU time slice expires | |
To allow starving processes to run | |
When it requests I/O | |
To avoid collision |
→ Preemptive scheduling means the processes is always interrupted by higher priorities.
Question 57 |
block | |
dispatch | |
wakeup | |
none of the above |
Question 58 |
3 | |
4 | |
2 | |
5 |
1. New→ The process is being created.
2. Running→ Instructions are being executed.
3. Waiting→ The process is waiting for some event to occur
4. Ready→ The process is waiting to be assigned to a processor
5. Terminated→ The process has finished execution
Question 59 |
short term scheduling | |
long term scheduling | |
medium term scheduling | |
mutual exclusion |
Medium-term scheduling:The decision to add to the number of processes that are partially or fully in main memory.
Long-term scheduling:The decision to add to the pool of processes to be executed.
Short-term scheduling:The decision as to which available process will be executed by the processor.
Question 60 |

(a)-3, (b)-4, (c)-2, (d)-1 | |
(a)-4, (b)-3, (c)-2, (d)-1 | |
(a)-2, (b)-4, (c)-1, (d)-3 | |
(a)-1, (b)-4, (c)-3, (d)-2 |
Batch Processing→ FIFO
Time Sharing→ Round-robin
Interrupt Processing→ LIFO
Question 61 |
RR | |
SSN | |
SSF | |
Priority based |
Question 62 |
J1 J2 J3 J4 | |
J4 J3 J2 J1 | |
J3 J4 J1 J2 | |
J3 J4 J2 J1 |



Question 63 |
Round Robin | |
Priority Scheduling | |
Shortest job first | |
2 level scheduling |
→ FCFS and SJF is non-preemptive based algorithms but in SJF are supporting another version is SRTF(Shortest remaining time first)
Question 64 |
Shortest job first scheduling. | |
Round robin scheduling. | |
Priority scheduling. | |
Fair share scheduling. |
→ FCFS and SJF is non-preemptive based algorithms but in SJF are supporting another version is SRTF(Shortest remaining time first)
Question 65 |

The average waiting time for preemptive SJF scheduling algorithm is __________.
6.5 | |
7.5 | |
6.75 | |
7.75 |

Question 66 |
a. Processor utilization by an individual process.
b. Weight assigned to a user or group of users.
c. Processor utilization by a user or group of processes In fair share scheduler, priority is calculated based on:
only (a) and (b) | |
only (a) and (c) | |
(a), (b) and (c) | |
only (b) and (C) |
Question 67 |
4 | |
2 | |
3 | |
1 |

Total no.of context switches is 2.
Question 68 |
Queues have different priorities | |
Each queue may have different scheduling algorithm | |
Processes are permanently assigned to a queue | |
This algorithm can be configured to match a specific system under design |
● There are more than one queues and each queue have assigned a different priority.
● Since there are more than one queue, therefore more than one scheduling algorithm is applied.
● Job starts in the highest priority queue.
● If time slice expires, drop the job by one level and if time slice does not expire push the job up by one level.
● This algorithm can be configured to match a specific system under design
Question 69 |
Match the following .

A-III, B-IV, C-I, D-II | |
A-II, B-I, C-IV, D-III | |
A-III, B-I, C-II, D-IV | |
A-II, B-IV, C-I, D-III |
Moves suspended process to secondary storage → Medium Term Scheduler
Loades the process into memory for execution → Long Term Scheduler
Moves one of the process to running time → Short Term Scheduler
Allocate CPU to a process → Dispatcher
Question 70 |
6.875 unit time | |
6.25 unit time | |
5.785 unit time | |
5.5 unit time |

Question 71 |
Considering the process details given in the following table, what will be the completion order of the processes under the policies FCFS and RR (with CPU quantum of 2 time units)?

FCFS: P1, P2, P3 and RR: P1, P2, P3 | |
FCFS: P3, P2, P1 and RR: P3, P1, P2 | |
FCFS: P3, P2, P1 and RR: P3, P2, P1 | |
FCFS: P1, P3, P2 and RR: P1, P2, P3 |

Question 72 |
Consider the process details given in the following table.

A new pre-emptive scheduling algorithm is proposed, i.e., the Longest Remaining Time Next (LRTN), wherein ties are broken by giving priority to the process with the highest priority.
Calculate the following.
i. Turnaround time for Process P2
ii. Response time for Process P3
(i)20, (ii)6 | |
(i)19, (ii)3 | |
(i)19, (ii)8 | |
(i)20, (ii)7 |

Question 73 |
Shortest job scheduling algorithm | |
Round robin scheduling algorithm | |
Priority scheduling algorithm | |
Multilevel queue scheduling algorithm |
Question 74 |
The speed of dispatching a process from running to the ready state | |
The time of dispatching a process from running to ready state and keeping the CPU idle | |
The time to stop one process and start running another one | |
None of these |
Question 75 |
Ignore the context switches at time 0 and at the end.
0 < t < 3 | |
t = 0 | |
t <= 3 | |
3 < t < 8 |
Question 76 |
Process ID | Priority level | CPU Burst time |
P1 | 2 | 10 |
P2 | 3 | 8 |
P3 | 1 | 12 |
P4 | 4 | 5 |
16 | |
24.75 | |
18 | |
25 |

Question 77 |
In processor management, round-robin method essentially uses the pre-emptive version of
FILO
| |
FIFO | |
SJF | |
longest time first
|
Question 78 |
Consider the following table of arrival time and burst time for three processes P0,P1,P2
The preemptive shortest job first scheduling algorithm is used scheduling is carried out only at arrival or completion of process. The average waiting time for three process is____.
7 Seconds | |
4 seconds
| |
6 seconds
| |
5 seconds |

∴ Avg. WT = 4+0+11/3 = 5
Question 79 |
Which scheduling algorithm gives a minimum average waiting time?
RR
| |
Priority
| |
SJF
| |
FCFS
|
Question 80 |
In which Disk scheduling algorithm starvation is possible?
FCFS | |
SCAN
| |
Shortest Seek Time First (SSTF)
| |
C-SCAN
|
Question 81 |
Which scheduling algorithm from the following is allows processes to move from one queue (different classified queues based on scheduling parameters) to another?
Multilevel Queue Scheduling
| |
Priority Based Scheduling
| |
Round Robin Scheduling
| |
Multilevel Feedback Queue Scheduling |
Question 82 |
Increases | |
Decreases | |
Remains constant | |
Varies irregularly |
Question 83 |
Executing more than one program at a time | |
Ability to accommodate multiple tasks in main memory | |
Presence of multiple processors in one system | |
All the above |
Question 84 |
80 % | |
50 % | |
60 % | |
30 % |
Service time for 1 process--------> 6sec
Service time for 5 processes------> 5*6 sec
= 30 seconds
CPU utilization = (30/60)*100
= 50%
Question 85 |
In the following process state transition diagram for a uniprocessor system, assume that there are always some processes in the ready state. Now, consider the following statements.

I. If a process makes a transition D, it would result in another process making transition A immediately
II. A process P2 in blocked state can make transition E while another process PI is in running state
III. The OS uses preemptive scheduling
IV. The OS uses non-preemptive scheduling
Which of the above statements are TRUE?
I and II | |
I and III | |
II and III | |
II and IV |
S2 is true.
S3 is true due to the transition C, because transition C states that if a process is in running state it can be pulled out to a ready state when some high priority process comes in the ready state.
S4 is false ,since S3 is true.
Question 86 |
Round-robin | |
Shortest job first | |
Priority queuing | |
Latest job first | |
None of the others |