MCQOPTIONS
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MCQOPTIONS
This section includes 104 Mcqs, each offering curated multiple-choice questions to sharpen your Operating System knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A disk has tracks from 0 to 179. Currently the head is at track 40. The R/W requests for tracks are in the following order- 80, 102, 26, 160, 35, 10, 135. What is the total head movement when SSTF algorithm is used? |
| A. | 180 |
| B. | 190 |
| C. | 258 |
| D. | 542 |
| Answer» B. 190 | |
| 2. |
Consider a disk queue with requests for I/O to blocks on cylinders 98, 183, 37, 122, 14, 124, 65, 67. Suppose SSTF disk scheduling algorithm implemented to meet the requests then the total number of head movements are ______ if the disk head is initially at 53. |
| A. | 224 |
| B. | 248 |
| C. | 236 |
| D. | 240 |
| Answer» D. 240 | |
| 3. |
Consider the following table of arrival time and burst time for three processes P0, P1 and P2. Process Arrival time Burst time P0 0 ms 9 ms P1 1 ms 4 ms P2 2ms 9 msThe 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? |
| A. | 5.0 ms |
| B. | 4.33 ms |
| C. | 6.33 ms |
| D. | 7.33 ms |
| Answer» B. 4.33 ms | |
| 4. |
At time 0, an OS with round-robin scheduling (time quantum = 5 units) has 5 processes P1, P2, P3, P4, P5 with burst times 15, 10, 20, 5, 15 respectively. What is the average turn-around time? |
| A. | 40 units |
| B. | 46 units |
| C. | 56 units |
| D. | 65 units |
| Answer» C. 56 units | |
| 5. |
Let processes are scheduled on a system using round robin scheduling; assume that there is only one process of r time units. Round robin time quantum is q time units and process switch time is s time units. If s < q < r, then process switch over head units are |
| A. | r/(r+s) |
| B. | (r/q - 1)*s |
| C. | q/(r+s) |
| D. | None of the above |
| Answer» C. q/(r+s) | |
| 6. |
Let disk request come to a disk driver for cylinders in the order 10, 22, 20, 2, 40, 6 and 38, at the time when the disk driver is reading from cylinder 20. If the seek time is 6 ms per cylinder, then the total seek time for the disk arm scheduling time algorithm (FCFS - First Come First Serve) is: |
| A. | 360 ms |
| B. | 212 ms |
| C. | 848 ms |
| D. | 876 ms |
| Answer» E. | |
| 7. |
A scheduling algorithm assigns priority proportional to the waiting time of a process. Every process starts with priority zero (the lowest priority). The scheduler re-evaluates the process priorities everyT time units and decides the next process to schedule. Which one of the following is TRUE if theprocesses have no I/O operations and all arrive at time zero? |
| A. | This algorithm is equivalent to the first-come-first-serve algorithm. |
| B. | This algorithm is equivalent to the round-robin algorithm. |
| C. | This algorithm is equivalent to the shortest-job-first algorithm. |
| D. | This algorithm is equivalent to the shortest-remaining-time-first algorithm. |
| Answer» C. This algorithm is equivalent to the shortest-job-first algorithm. | |
| 8. |
In _______ disk scheduling algorithm, the disk head moves from one end to other end of the disk, serving the requests along the way. When the head reaches the other end, it immediately returns to the beginning of the disk without serving any requests on the return trip. |
| A. | LOOK |
| B. | SCAN |
| C. | C - LOOK |
| D. | C - SCAN |
| Answer» E. | |
| 9. |
For the processes listed in the following table, which of the following scheduling schemes will give the lowest average turnaround time?ProcessArrival TimeProcessing TimeA03B16C44D62 |
| A. | First come First Serve |
| B. | Non-peemptive Shortest Job First |
| C. | Shortest Remaining Time |
| D. | Round Robin with Quantum value two |
| Answer» D. Round Robin with Quantum value two | |
| 10. |
Consider a disk sequence with 100 cylinders. The request to access the cylinder occur in the following sequence :4, 34, 10, 7, 19, 73, 2, 15, 6, 20Assuming that the head is currently at cylinder 50, what is the time taken to satisfy all requests if it takes 2 ms to move from one cylinder to adjacent one and shortest seek time first policy is |
| A. | 190 |
| B. | 238 |
| C. | 233 |
| D. | 276 |
| Answer» C. 233 | |
| 11. |
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? |
| A. | Shortest remaining time first |
| B. | Round robin with time quantum less than the shortest CPU burst |
| C. | Uniform random |
| D. | Highest priority first with priority proportional to CPU burst length |
| Answer» B. Round robin with time quantum less than the shortest CPU burst | |
| 12. |
Consider the following set of processes:ProcessP1P2P3P4P5Burst time10 ms29 ms3 ms7 ms12 ms First Come First Serve (FCFS), non-preemptive Shortest Job First (SJF) and Round Robin (RR) (quantum = 10 ms) Scheduling Algorithms for this process set would imply which of the following features?1. The SJF policy results in less than half of the average waiting time obtained with FCFS scheduling.2. The RR_ algorithm gives an intermediate value for the average waiting time.3. Deterministic modeling takes a particular predetermined workload and designs the performance of each algorithm for that workload.Select the correct answer using the code given below. |
| A. | 1, 2 and 3 |
| B. | 1 and 2 only |
| C. | 1 and 3 only |
| D. | 2 and 3 only |
| Answer» B. 1 and 2 only | |
| 13. |
Consider a disk system having 60 cylinders. Disk requests are received by a disk drive for cylinders 10, 22, 20, 2, 40, 6, and 38, in that order. Assuming the disk head is currently at cylinder 20, what is the time taken to satisfy all the requests if it takes 2 milliseconds to move from one cylinder to an adjacent one and Shortest Seek Time First (SSTF) algorithm is used ? |
| A. | 240 milliseconds |
| B. | 96 milliseconds |
| C. | 120 milliseconds |
| D. | 112 milliseconds |
| Answer» D. 112 milliseconds | |
| 14. |
Disk requests come to a disk driver for cylinders in the order 10, 22, 20, 2, 40, 6 and 38, at a time when the disk drive is reading from cylinder 20. The seek time is 6 ms/cylinder. The total seek time, if the disk arm scheduling algorithms is first-come-first-served is |
| A. | 360 ms |
| B. | 850 ms |
| C. | 900 ms |
| D. | None of the above |
| Answer» E. | |
| 15. |
Consider the following. Process IDCPU BurstArrival TimeP150P272P333 The completion order of the three processes under the policies FCFS and RR (with CPU quantum of 2 time units) are: |
| A. | FCFS: P1, P2, P3 and RR: P1, P3, P2 |
| B. | FCFS: P1, P2, P3 and RR: P1, P2, P3 |
| C. | FCFS: P1, P2, P3 and RR: P3, P2, P1 |
| D. | FCFS: P1, P3, P2, and RR: P1, P2 P3 |
| Answer» B. FCFS: P1, P2, P3 and RR: P1, P2, P3 | |
| 16. |
An operating system uses shortest remaining time (SRT) scheduling algorithm. Consider the arrival times and execution time for the following processes:ProcesArrival TimeExecution TimeP1020P21525P33010P44515What is total waiting time for process P2? |
| A. | 5 |
| B. | 15 |
| C. | 40 |
| D. | 55 |
| Answer» C. 40 | |
| 17. |
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 and 1 time units. respectively. The completion time of A under round robin scheduling with time slice of one time unit is: |
| A. | 11 |
| B. | 4 |
| C. | 8 |
| D. | 9 |
| Answer» E. | |
| 18. |
Match the following.List IList III. Stack AlgorithmP. DeadlockII. Elevator AlgorithmQ. Disk SchedulingIII. Priority Scheduling AlgorithmR. Page ReplacementIV. Havender’s AlgorithmS. CPU Scheduling |
| A. | I – R, II – Q, III – S, IV – P |
| B. | I – Q, II – R, III –S, IV – P |
| C. | I – R, II – Q, III – P, IV – S |
| D. | I – Q, II – R, III – P, IV – S |
| Answer» B. I – Q, II – R, III –S, IV – P | |
| 19. |
Consider the following table of arrival time and burst time for three processes P0, P1, P2ProcessArrival Time (Seconds)Burst Time (Second)P009P114P229 The pre-emptive shortest job first scheduling algorithm is used. Scheduling is carried out only at arrival or completion of process. The average waiting time for the three process is _____ |
| A. | 5 seconds |
| B. | 4 seconds |
| C. | 6 seconds |
| D. | 7 seconds |
| Answer» B. 4 seconds | |
| 20. |
Consider the following four processes with the arrival time and length of CPU burst given in milliseconds :ProcessArrival TimeBurst TimeP108P214P329P435 The average waiting time for preemptive SJF scheduling algorithm is |
| A. | 6.5 ms |
| B. | 7.5 ms |
| C. | 6.75 ms |
| D. | 7.75 ms |
| Answer» B. 7.5 ms | |
| 21. |
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? |
| A. | Round - robin |
| B. | Shortest job first |
| C. | FCFS |
| D. | Priority scheduling |
| Answer» C. FCFS | |
| 22. |
Consider a disk system with 100 cylinders. The request to access the cylinders occur in the following sequences4, 37, 10, 7, 19, 73, 2, 15, 6, 20Assuming the head is currently at cylinder 50, what is the time taken to satisfy all requests if it takes 1 ms to move from one cylinder to adjacent one and shortest seek time first algorithm is used |
| A. | 95 ms |
| B. | 119 ms |
| C. | 233 ms |
| D. | 276 ms |
| Answer» C. 233 ms | |
| 23. |
If the time quantum size is 2 units of time and there is only one job of 14 time units in a ready queue, the round-robin scheduling algorithm will cause ______context switches. |
| A. | 5 |
| B. | 7 |
| C. | 6 |
| D. | 8 |
| Answer» D. 8 | |
| 24. |
Consider following groups, in context of operating system:Group - IGroup - IIP. Gang Scheduling1. Guaranteed SchedulingQ. Rate monotonic Scheduling2. Real time SchedulingR. Fair share Scheduling3. Thread SchedulingCorrect match of entries in Group - I and Group - Ii is: |
| A. | P - 2, Q - 3, R - 1 |
| B. | P - 1, Q - 2, R - 3 |
| C. | P - 3, Q - 2, R - 1 |
| D. | P - 1, Q - 3, R - 2 |
| Answer» D. P - 1, Q - 3, R - 2 | |
| 25. |
Consider the following five disk access requests of the form (request id, cylinder number) that are present in the disk scheduler queue at a given time.(P, 155), (Q, 85), (R, 110), (S, 30), (T, 115)Assume the head is positioned at cylinder 100. The scheduler follows Shortest Seek Time First scheduling to service the requests.Which one of the following statements is FALSE? |
| A. | T is serviced before P. |
| B. | Q is serviced after S, but before T. |
| C. | The head reverses its direction of movement between servicing of Q and P. |
| D. | R is serviced before P. |
| Answer» C. The head reverses its direction of movement between servicing of Q and P. | |
| 26. |
One disk queue with requests for I/O to blocks on cylinders. The Request are in the following manner:98 183 37 122 14 124 65 67Considering SSTF (shortest seek time first) scheduling, the total number of head movements is, if the disk head of initially at 53 is: |
| A. | 236 |
| B. | 246 |
| C. | 220 |
| D. | 240 |
| Answer» B. 246 | |
| 27. |
Consider the following scheduling:A.Gang schedulings.Guaranteed schedulingB.Rate Monotonic schedulingt.Thread schedulingC.Fair share schedulingu.Real time schedulingMatching the table in the order A, B, C gives |
| A. | t, u, s |
| B. | s, t, u |
| C. | u, s, t |
| D. | u, t, s |
| Answer» B. s, t, u | |
| 28. |
Let RR3 means Round Robin Scheduling algorithm with time quantum of 3 units and FCFS means First Come First Serve (FCFS) scheduling algorithm. Now if 3 processes P1, P2, P3 has arrival time in order is (0, 2, 4) and execution time is (8, 5, 4) respectively then the completion order of P1, P2, P3 under RR3 and FCFS scheduling is: |
| A. | RR3 : P2, P1, P3 and FCFS : P3, P2, P1 |
| B. | RR3 : P3, P2, P1 and FCFS : P1, P2, P3 |
| C. | RR3 : P2, P1, P3 and FCFS : P1, P2, P3 |
| D. | RR3 : P1, P2, P3 and FCFS : P3, P2, P1 |
| Answer» D. RR3 : P1, P2, P3 and FCFS : P3, P2, P1 | |
| 29. |
A scheduler which selects processes from secondary storage device is called |
| A. | Short term scheduler |
| B. | Long term scheduler |
| C. | Medium term scheduler |
| D. | Process scheduler |
| Answer» D. Process scheduler | |
| 30. |
State which statement is true for Suspended process?i) The process is not immediately available for execution.ii) The process may be removed from suspended state automatically without removal order. |
| A. | i only |
| B. | ii only |
| C. | i and ii only |
| D. | None of the above |
| Answer» B. ii only | |
| 31. |
Three CPU intensive processes requires 10, 20 and 30 time units and arrive at times 0, 2 and 6 respectively. The operating system implements a shortest remaining time first scheduling algorithm. Considering that the context switches at time zero and at the end are not counted the number of context switches are needed is ______. |
| A. | 4 |
| B. | 3 |
| C. | 2 |
| D. | 1 |
| Answer» D. 1 | |
| 32. |
Program ‘preemption’ is |
| A. | forced de allocation of the CPU from a program which is executing on the CPU. |
| B. | release of CPU by the program after completing its task. |
| C. | forced allotment of CPU by a program to itself. |
| D. | a program terminating itself due to detection of an error. |
| Answer» B. release of CPU by the program after completing its task. | |
| 33. |
In process scheduling, …………………… determines when new processes are admitted to the system. |
| A. | long term scheduling |
| B. | medium term scheduling |
| C. | short term scheduling |
| D. | none of the above |
| Answer» B. medium term scheduling | |
| 34. |
The process that are residing in the main memory and are waiting to execute are kept on a list called the ............ |
| A. | job queue |
| B. | ready queue |
| C. | wait queue |
| D. | device queue |
| Answer» C. wait queue | |
| 35. |
“Throughput” of a system is |
| A. | Number of programs processed by it per unit time |
| B. | Number of times the program is invoked by the system |
| C. | Number of requests made to a program by the system |
| D. | None of the above |
| Answer» B. Number of times the program is invoked by the system | |
| 36. |
What is the non - preemptive scheduling algorithm? |
| A. | Multilevel queue scheduling with feedback |
| B. | Multilevel queue scheduling |
| C. | First - In - First - Out |
| D. | Round Robin |
| Answer» D. Round Robin | |
| 37. |
The scheduling in which CPU is allocated to the process with least CPU-burst time is called |
| A. | Priority Scheduling |
| B. | Shortest job first Scheduling |
| C. | Round Robin Scheduling |
| D. | Multilevel Queue Scheduling |
| Answer» C. Round Robin Scheduling | |
| 38. |
Match the following mechanisms for interrupting the execution of a process and their uses.i) Interrupt a) Call to an operating system functionii) Trap b) Reaction to an asynchronous external eventiii) Supervisor Call c) Handling of a error or an exception condition |
| A. | i-a, ii-b, iii-c |
| B. | i-c, ii-a, iii-b |
| C. | i-b, ii-c, iii-a |
| D. | i-a, ii-c, iii-b |
| Answer» D. i-a, ii-c, iii-b | |
| 39. |
Round robin scheduling is essentially the preemptive version of ........... |
| A. | first in first out |
| B. | shortest job first |
| C. | shortest remaining |
| D. | longest time first |
| Answer» B. shortest job first | |
| 40. |
Following is/are the reasons for process suspension. |
| A. | Swapping parent process |
| B. | Interrupt request |
| C. | Timing |
| D. | All of the above |
| Answer» E. | |
| 41. |
Interval between the time of submission and completion of the job is called |
| A. | Waiting time |
| B. | Turnaround time |
| C. | Throughput |
| D. | Response time |
| Answer» C. Throughput | |
| 42. |
___________ is a technique of improving the priority of process waiting in Queue for CPU allocation. |
| A. | Starvation |
| B. | Ageing |
| C. | Revocation |
| D. | Relocation |
| Answer» C. Revocation | |
| 43. |
Which of the following are the states of a five state process model?i) Running ii) Ready iii) New iv) Exit v) Destroy |
| A. | i, ii, iii and v only |
| B. | i, ii, iv and v only |
| C. | i, ii, iii, and iv only |
| D. | All i, ii, iii, iv and v |
| Answer» D. All i, ii, iii, iv and v | |
| 44. |
.............. refers to a situation in which a process is ready to execute but is continuously denied access to a processor in deference to other processes. |
| A. | Synchronization |
| B. | Mutual Exclusion |
| C. | Dead lock |
| D. | Starvation |
| Answer» E. | |
| 45. |
Consider the following set of processes, the length of the CPU burst time given in milliseconds :Process Burst timeP1 6P2 8P3 7P4 3i) Assuming the above process being scheduled with the SJF scheduling algorithm : |
| A. | The waiting time for process P1 is 3ms. |
| B. | The waiting time for process P1 is 0ms. |
| C. | The waiting time for process P1 is 16ms. |
| D. | The waiting time for process P1 is 9ms. |
| Answer» B. The waiting time for process P1 is 0ms. | |
| 46. |
Consider the following set of processes, the length of the CPU burst time given in milliseconds :Process Burst timeP1 6P2 8P3 7P4 3Assuming the above process being scheduled with the SJF scheduling algorithm: |
| A. | The waiting time for process P4 is 3ms. |
| B. | The waiting time for process P4 is 0ms. |
| C. | The waiting time for process P4 is 16ms. |
| D. | The waiting time for process P4 is 9ms. |
| Answer» C. The waiting time for process P4 is 16ms. | |
| 47. |
How you will describe Throughput in operating system? |
| A. | Number of times a program produces output. |
| B. | Number of times the program is running. |
| C. | Number of programs processed by it per unit time |
| D. | None of the above. |
| Answer» D. None of the above. | |
| 48. |
We have an operating system, which is capable of loading and executing a single sequential user process at a time. If the disk head scheduling algorithm which is initially First Come First Served (FCFS) is replaced by Shortest Seek Time First (SSTF) claimed by the vendor to give 50% better benchmark results then what is the expected improvement in the I / O performance of the user programs? |
| A. | 0% |
| B. | 25% |
| C. | 50% |
| D. | 75% |
| Answer» B. 25% | |
| 49. |
What is the most suitable scheduling policy for a time - shared operating systems? |
| A. | Shortest Job First |
| B. | Elevator |
| C. | First Come First Served |
| D. | Round Robin |
| Answer» E. | |
| 50. |
Consider a uniprocessor computer system that has 2 processes and both of them alternate 10ms CPU bursts with 90ms I / O bursts. Both the processes were created at nearly the same time and can proceed in parallel. What is the scheduling strategy for this system that will result in the least CPU utilization (over a long period of time) ? |
| A. | Round Robin scheduling with a time quantum of 5ms |
| B. | First Come First Served scheduling |
| C. | Shortest remaining time first |
| D. | Static priority scheduling with different priorities for the two processes |
| Answer» B. First Come First Served scheduling | |