OS OBJECTIVE QUESTIONS

Transcription

1 OS OBJECTIVE QUESTIONS Which one of the following is Little s formula Where n is the average queue length, W is the time that a process waits 1)n=Lambda*W 2)n=Lambda/W 3)n=Lambda^W 4)n=Lambda*(W-n) Answer:1 A CPU scheduling algorithm determines an order for the execution of its scheduled processes. Given n processes to be scheduled on one processor, how many possible different schedules are there? 1) n! 2) n*n 3) 1 4) log n CPU scheduling is the basis of operating systems 1) Batch operating systems 2) Real time OS 3) Multiprogramming 4) Mono programming Answer:3 The SJF algorithm 1) executes first the job that last entered the queue 2) executes first the job that first entered the queue 3) executes first the job with the least processor needs 4) executes first the job that has been in the queue already Answer:3 Which scheduling policy is most suitable for a time shared operating systems? 1) SJF 2) RR 3) FCFS 4) Elevator Answer:2 Scheduling is 1) allowing job to use the processor 2) unrelated to performance conditions 3) quite simple to implement, even on large frames 4) the same regardless of the purpose of the system Answer:1 Context Switching is 1) Part of spooling 2) part of polling 3) part of interrupt handling 4) part of interrupt servicing The dispatcher 1) actually schedules the task to the processor 2) is always very small 3) is never used on large components

2 4) should be used only on large computers 126) Race condition results for the following 1) all processes 2) co-operating processes 3) competing processes 4) none Answer:3 Which of the following is not related to disk scheduling? 1) transfer time 2) waiting time 3) seek time 4) rotational latency Answer:1 Which of the following is not related to the disk access 1) Wait Time 2) Transfer Time 3) seek time 4) Latency Time Answer:2 Consider the following reference strings {0, 2, 4, 1, 1, 4, 5, 7} {2, 3, 2, 5, 6, 3, 2, 6} The FIFO page replacement algorithm with 3 frames would give lesser number of faults for 1) Reference string I 2) Reference string II 3) Both will have the same number of faults 4) Can t determine from the given data Scheduling is 1) allows processor to use the jobs 2) Not required in uni processor system 3) Allowing jobs to use the processor 4) None. In which of the following scheduling schemes does context switching never take place. 1) Round robin 2) SJF 3) FCFS 4) Priority In multilevel queue scheduling (CPU scheduling) arrange the following in the order lowest to highest priority. a) Interactive process b) Batch process c) System process d) Student process 1) a, b, c, d 2) d, c, b, a

3 3) c, a, b, d 4) c, b, a, d Belady s Anomaly is present in 1) LIFO 2)FIFO 3) FCFS 4)SJF Answer:2 FCFS scheduling is the simplest algorithm, but it can cause short process to wait for very long processes. True/False Answer: True In which of the following scheduling schemes does context switching take place. 1) SJF 2) FCFS 3) Priority 4) both 1 and 3 Which of the following is not a CPU scheduling algorithm? 1) SCAN 2) SJF 3) FCFS 4) Round robin Process A has arrived in the system by time 0 and requires 5 time units in the CPU, process B has arrived by time 1 and requires 3 time units, process C has arrived by time 2 and requires 1 time unit. Assuming negligible processes switching time, and that all processes are CPU bound and that a new quantum starts whenever a process finishes, the average time for Round robin (Q=3) is 1) ) ) ) 4.67 Which of the scheduling algo gives the minimum average waiting time for a given set of processes? 1) FCFS 2) SJF 3) Round Robin 4) Priority 83) Forcible removal of a resource from a process is possible through 1) Scheduling 2) Preemption 3) Dispatching 4) None Which of the following gives control of the CPU to the process selected by short term scheduler? 1) Context switching 2) Thread 3) Interrupt 4) Dispatcher Which of the following is strictly preemptive? 1) FCFS 2) Priority 3) Round Robin 4) SJF The performance of round robin algorithm depends on 1) Preemption 2) Time Quantum 3) Non-preemption 4) Priority

4 The following are the processes and their burst time: P1 : 6, P2 : 8 P3 : 7 P4 : 3.. The average turn around time if SJF is used is mille seconds 1) 7 2) 13 3) 19 4) 7.6 Allocate the smallest hole that is big enough, is which type of fit in dynamic storage allocation problem. 1) First fit 2) Best fit 3) Worst fit 4) None Round robin algorithm is suitable for which type of systems? 1) Real time systems 2) Embedded 3) Multiprogramming 4) Time sharing The major problem with priority scheduling algorithm is. 1) Indefinite blocking 2) Takes large turnaround time 3) Deadlock occurs 4) All the above If the time quantum is very big in round robin CPU scheduling algorithm, then it acts as 1) SJF scheduling 2) FCFS 3) Priority 4) SJF with preemption Medium term scheduler is based on 1) Scroll in, scroll out 2) Wrap in, wrap out 3) Swap in, swap out 4) Fetch in, fetch out In which of the following context switching never takes place? 1) SJF 2) Round robin 3) Preemptive policy 4) None The hardware mechanism that enables a device to notify the CPU is called. 1) Interrupt 2) Polling 3) System call 4) None Round robin scheduling algorithm is suitable for Time sharing systems. True/ False Answer: True selects the process from secondary storage and loads them into memory for execution. 1) Medium term scheduler 2) Short term scheduler 3) Long term scheduler

5 4) None To maximize CPU utilization and throughput we should maximize turnaround time and minimize waiting time. True/ False? Answer: False Exp: we should minimize turnaround time and minimize waiting time. The function of resident monitor is Memory management. True/ False Answer: False Expl: The function of resident monitor is Automatic job scheduling. Round robin scheduling is essentially the preemptive version of. 1) FIFO 2) Shortest job first 3) Shortest remaining 4) FCFS Ans: 1 If the Disk head is located initially at 32, find the number of disk moves required with FCFS if the disk queues of I/O blocks requests are 98,37,14,124,65,67. 1) 310 2) 324 3) 315 4) 321 Ans:- (4) FIFO scheduling is. 1) Preemptive 2) Non Preemptive 3) Deadline 4) Fair share Ans: - 2 Using Priority Scheduling algorithm, find the average waiting time for the following set of processes given with their priorities in the order: Process : Burst Time : Priority respectively. P1 : 10 : 3, P2 : 1 : 1, P3 : 2 : 4, P4 : 1 : 5, P5 : 5 : 2. 1) 8 ms 2) 8.2 ms 3) 7.75 ms 4) 3 ms Ans: 2 103) Which of the following disk scheduling techniques has a drawback of starvation? 1) SCAN 2) SSTF 3) FCFS 4) LIFO Ans:- 2 Which of the following is a criterion to evaluate a scheduling algorithm? 1) CPU Utilization: Keep CPU utilization as high as possible. 2) Throughput: number of processes completed per unit time. 3) Waiting Time: Amount of time spent ready to run but not running. 4) All of the above Ans: - 4

6 Which of the following is not a disk scheduling algorithm? 1) SCAN 2) Elevator / LOOK 3) FCFS 4) Working set Ans: 4 Which of these scheduling algorithms allows a process to move up and down between queues? 1) Round robin 2) First-Come, First-Served scheduling 3) Multilevel feedback queue scheduling 4) Priority scheduling Ans:- 3 If a new process arrives with CPU burst time less than remaining CPU burst time of currentexecuting process, preempt. This scheduling algorithm is known as. 1) Shortest-Job-First 2) Priority 3) FCFS 4) None of the above Ans :- 1 Which of these is not a part of the process scheduler? 1. Context Switch 2. Long, medium and short term scheduler 3. Dispatcher 4. Program Control Ans:- 4 The ready list in an operating system contains jobs. 1) waiting for I/O 2) being executed by the processor 3) waiting for the processor to be allotted to them 4) waiting for input only Ans:- ( 3) Round robin scheduling. 1) Is quite complex to implement 2) Allows interactive tasks quicker access to the processor 3) Allows processor bound tasks more time in the processor 4) Gives each task the same chance at the processor Ans 4 Which of these scheduling algorithms allows a process to move up and down between queues? 1) Round robin 2) First-Come, First-Served scheduling 3) Multilevel feedback queue scheduling 4) Priority scheduling Ans: - ( 3 )

7 In a process scheduling, the mechanism for scheduling and policy can be set by and respectively. 1) Kernel & user process 2) User Process & threads 3) Shell & Kernel 4) Kernel & Threads Ans:- 1 scheduler selects the jobs from the pool of jobs and loads into the ready queue. 1) Long term 2) Short term 3) Medium term 4) None of the above Ans: - 1 Which one of the following is true for a CPU having a single interrupt request line and a single interrupt grant line? 1) Neither vectored interrupt nor multiple interrupting devices are possible. 2) Vectored interrupts are not possible but multiple interrupting devices are possible 3) Vectored interrupts and multiple interrupting devices are both possible. 4) Vectored interrupt is possible but multiple interrupting devices are not possible Which of the process scheduling algo do you expect to produce the smallest average waiting time 1) SJF 2) FCFS 3) round-robin 4) All the above Answer:1 A disc queue with requests for I/O to blocks on cylinders 98, 183, 37, 122, 14, 124, 65, 67 If the head is presently at 53rd cylinder, what will be the total number of cylinders traversed if we are using SCAN disc scheduling algorithm. 1) 152 2) 252 3) 352 4) 462 A disc queue with requests for I/O to blocks on cylinders. 98, 183, 37, 122, 14, 124, 65, 67 If the head is presently at 53rd cylinder, what will be the total number of cylinders traversed if we are using C-SCAN disc scheduling algorithm. 1) 182 2) 282 3) 382 4) 482 A disc queue with requests for I/O to blocks on cylinders 98, 183, 37, 122, 14, 124, 65, 67 If the head is presently at 53rd cylinder, what will be the total number of cylinders traversed if we are using C-LOOK disc scheduling algorithm. 1) 155 2) 255 3) 355 4) 455 In which disc scheduling algorithm the head traverses to the extreme end cylinders. 1) FCFS 2) SSTF 3) SCAN 4) C-LOOK

8 In which disc scheduling algorithm the head traverses to the nearest available cylinder. 1) FCFS 2) SSTF 3) SCAN 4) C-LOOK In which disc scheduling algo the head traverses to the first cylinder encountered in the given order. 1) FCFS 2) SSTF 3) SCAN 4) C-LOOK In which disc scheduling algorithm the head traverses till one end then returns directly to the other end before encountering any other cylinder. 1) FCFS 2) SSTF 3) C-SCAN 4) C-LOOK C-SCAN is also known as 1) Circular scan 2) Cylindrical scan 3) Constant scan 4) None In priority scheduling (CPU scheduling) Process Burst time Priority Arrival time P P P P P The first process to get executed is 1) P1 2) P2 3) P3 4) P4 In priority scheduling (CPU scheduling) Process Burst time Priority Arrival time P P P P P The process which will be executed at the last is 1) P5 2) P2 3) P3 4) P4 In priority scheduling (CPU scheduling) Process Burst time Priority Arrival time P P P P P The turn around time is 1) 9.6 sec 2) 10.6 sec 3) 11.6 sec 4) 12.6 sec

9 In priority scheduling (CPU scheduling) Process Burst time Priority Arrival time P P P P P The waiting time is. 1) 4.8 sec 2) 5.8 sec 3) 6.8 sec 4) 5 sec In priority scheduling (CPU scheduling) Process Burst time Priority Arrival time P P P P P The response time is 1) 3.4 sec 2) 4.4 sec 3) 5.4 sec 4) 6.4 sec In round robin scheduling ( CPU scheduling),time quantum=4 millisecs Process Burst time Arrival time P P2 3 0 P3 3 0 The first process which will get executed is 1) P1 2) P2 3) P3 4) P2 and P3 In round robin scheduling ( CPU scheduling), time quantum=4 millisecs Process Burst time Arrival time P P2 3 0 P3 3 0 The process which will get executed at last 1) P1 2) P2 3) P3 4) P2 and P3 In round robin scheduling ( CPU scheduling), time quantum=4 millisecs Process Burst time Arrival time P P2 3 0 P3 3 0

10 The waiting time for the given problem is. 1) 4 sec 2) 5 sec 3) 6 sec 4) 7 sec In round robin scheduling ( CPU scheduling), time quantum=4 millisecs Process Burst time Arrival time P P2 3 0 P3 3 0 The turn around time for the given problem is. 1) 12 sec 2) 13 sec 3) 14 sec 4) 15sec In round robin scheduling ( CPU scheduling), time quantum=4 millisecs Process Burst time Arrival time P P2 3 0 P3 3 0 The response time for the given problem is. 1) 1 sec 2) 2 sec 3) 3 sec 4) 4 sec In round robin scheduling ( CPU scheduling), time quantum=4 millisecs Process Burst time Arrival time P P2 3 0 P3 3 0 The process which will complete the last 1) P1 2) P2 3) P3 4) P4 In round robin scheduling ( CPU scheduling), time quantum=4 millisecs Process Burst time Arrival time P P2 3 0 P3 3 0 All the processes will get executed 1) At the starting of 24th sec 2) At the ending of 24th sec 3) At the ending of 25th sec 4) At the ending of 23rd sec In SJF algorithm for CPU scheduling Process Burst Time Arrival time P1 6 1 P2 8 2 P3 7 3 P4 3 4 The first process which will complete its execution

11 1) P1 2) P2 3) P3 4) P4 Checkpoint mechanism should be performed frequently when there are no failures? True/false Answer: false Exp: If no failures occur, the system must incur the cost of performing checkpoints that are Essentially unnecessary. In this situation, performing checkpoints less often will lead to better system performance. In case of a system crash checkpoint occurrence should be done more frequently? True/false? Answer: true Exp: checkpoint record means that an operation will not have to be redone during system recovery. In this situation, the more often checkpoints were performed, the faster the recovery time is from a system crash. In case of disk crash more checkpoints have to be made? True/false Answer: true Exp: The time it takes to recover from a disk crash The existence of a checkpoint record means that an operation will not have to be redone during system recovery. In this situation, the more often checkpoints were performed, the faster the recovery time is from a disk crash. what is a transaction? a) A transaction is a series of read and write operations upon some data followed by a commit operation b) a transaction is a series of read and write operation upon some data followed by a roll back operation. c) a transaction is any set of operations which may/may not be committed. d) none of the above Answer: a I If the series of operations in a transaction cannot be completed a) the transaction must be aborted and the operations that did take place must be rolled back b) the transaction must be aborted without roll-backing the operations. c) the transaction must be aborted and the operations that are not completed must be committed. d) none of the above Answer: a Exp: so that after recovery we can start a new transaction what is transaction activity? a) a transaction in which operations appear as one indivisible operation to ensure the integrity of the data being updated. b) a transaction in which most of the operations appear as a single operation. c) a transaction in which all operations are committed successfully. d) none Answer: a Some schedules are possible under the two-phase locking Protocol but not possible under the timestamp protocol

12 True/false? Answer: true Exp: both cannot occur at a time Some schedules are not possible under time-stamp protocol but not under two-phase locking protocol. True/false? Answer: false Exp: both cannot occur at a time State under which protocol the following schedule comes? a) time stamp protocol b) two-phase locking protocol c) both d) none Answer: b Exp: This schedule is not allowed in the timestamp protocol because at step7, the W-timestamp of B is 1. State under which protocol the following schedule comes? Step T0 T1 T2 1 Write(A) 2 Write(A) 3 Write(A) 4 Write(B) 5 Write(B) a) time stamp protocol b) two-phase locking protocol c) both d) none Answer: a This schedule cannot have lock instructions added to make it legal under two-phase locking protocol because T1 must unlock (A) between steps 2 and 3, and must lock (B) between steps 4 and 5. We can use Simulations to get more accurate evaluation of scheduling algorithms. True/False Answer: True Which of the following statements are false? 1) The time to completion of a CPU-bound process is largely determined by the amount of CPU time it receives. 2) The time to completion of an I/O-bound process is largely determined by the time taken to service its I/O requests. CPU time plays little part in the completion time of I/O-bound processes. 3) both 4) None Exp: the completion time is determined by cpu and I/O depending on the user needs Which of the following are true about multitasking? 1) In Cooperative multitasking, the thread specifically releases the CPU. 2) In pre-emptive multitasking, the thread has no choice.

13 3) Both 4) None. is the state of processor, when a process is waiting for some event to occur. 1) ready state 2) wait state 3) dead state 4) None Most CPUs have the following type interrupt request. 1) Non-mask able 2) Mask able 3) Both 4) None A disc queue with requests for I/O to blocks on cylinders 98, 183, 37, 122, 14, 124, 65, 67 If the head is presently at 53rd cylinder, what will be the total number of cylinders traversed if we are using FCFS disc scheduling algorithm. 1) 340 2) 440 3) 540 4) 640 A disc queue with requests for I/O to blocks on cylinders 98, 183, 37, 122, 14, 124, 65, 67 If the head is initially at 53rd cylinder, what will be the total number of cylinders traversed if we are using SSTF disc scheduling algorithm. 1) 156 2) 236 3) 366 4) 456 ) Why Solaris, Windows XP, and Linux implement multiple locking mechanisms. a) for more security b) for application developer s needs c) both a and b d) none Answer: b Exp: option b is more specific to the question rather than others why spinlocks, mutexes used for a) multiple locking mechanisms b) kernel process mapping c) process control d) none Answer:a Spinlocks are useful for multiprocessor systems True/false? Answer: true Exp: In spinlocks a thread can run in a busy-loop (for a short period of time) rather than incurring the overhead of being put in a sleep queue. what is adaptive mutexes? 1) mutex implemented with a spin lock on multiprocessor machines. 2) mutex used for synchronization 3) mutex used as a locking mechanism

14 4) none of the above Answer:1 adaptive mutexes are used in a) Win XP b) Solaris 2 c) Linux d) none Answer:b Exp: compatable with solaris 2 what are volatile devices a) devices which store data permanently, devices which cannot survive system crashes b) devices which store data temporarily c) 2,3 Exp: Volatile storage refers to main and cache memory and is very fast. However, volatile storage cannot survive system crashes or powering down the system. what are Nonvolatile storage devices? a) devices which survives system crashes b) devices which survives powered-down systems c) devices which store data permanently d) all the above Exp: Non-volatile devices survive system crashes.disks and tapes are examples of nonvolatilestorage Stable storage refers to a) storage devices in which information residing is never lost b) storage devices which survive system crashes c) storage devices which store data permanently d) none of the above Answer: a Explain the purpose of the checkpoint mechanism. a) System performance when no failure occurs b) The time it takes to recover from a system crash c)the time it takes to recover from a disk crash d) all the above Answer: d Exp:A checkpoint log record indicates that a log record and its modified data has been written to stable storage and that the transaction need not to be redone in case of a system crash. Which of the following are Scheduling algorithms: 1) multi level queue 2) shortest job first 3) Least recently used 4) both 1 and 2

15 systems are required to complete a critical task within a guaranteed amount of time. 1) Real time 2) Networking systems 3) Distributed systems 4) None of above The following are the criteria for CPU scheduling algorithm evaluation 1) Maximum CPU utilization under the constraint that maximum response time is one second 2) Maximum throughput such that turnaround time is (on an average) linearly proportional to total execution time. 3) Both 4) None The thread library schedules user level threads to run on an available Light wait process, this scheme is known as 1) Process Local Scheduling 2) Global Scheduling 3) CPU scheduling 4)None The selected thread runs on the CPU until one of the following occurs 1) it blocks 2) it uses its time slice( if its not a system thread) 3) it is preempted by a higher priority thread 4) All of the above In UNIX if multiple threads have the same priority, the scheduler uses 1) First come first serve 2) Shortest job first 3) Round robin queue 4) Multi level queue scheduling Disabling interrupts frequently could affect the system s clock. True /false? Answer: true Exp: The system clock is updated at every clock interrupt. If interrupts were disabled particularly for a long period of time it is possible the system clock could easily lose the correct time. how system clock can be maintained correct time instead of disabling the interrupts. 1) keeping the disabling time for very short periods. 2) keeping one more system clock so that if one is disabled other works. 3) do not disable the interrupt. 4) None The system clock is updated at every clock interrupt. If interrupts Were disabled particularly for a long period of time it is Possible the system clock could easily lose the correct time Under which circumstances the scheduling scheme is non preemptive.

16 1) When a process switches from running to waiting state, When a process terminates 2) When a process switches from running to the ready state. 3) When a process switches from waiting to the ready state. 4).1, Under which circumstances the scheduling scheme is preemptive. 1) When a process switches from running to waiting state 2) When a process switches from running to the ready state. 3) When a process switches from waiting to the ready state. 4) 2,3. gives control of the CPU to the process selected by the short term Scheduler 1) long term scheduler 2) medium term scheduler 3) Dispatcher 4) None Which of the following are the functions of the dispatcher? 1) swapping 2) Jumping to the proper location in the user program to restart that program 3) Scheduling 4) All the above Time taken by the dispatcher to stop one process and start another running is known as Answer : Dispatcher Latency Which of the following are CPU Scheduling Criteria? 1) CPU utilization 2) Waiting time 3) Dispatcher latency 4) 1,2 Which of the following is true for the algorithms for allocating regions of contiguous memory : 1. First-Fit 2. Next-Fit 3. Best-Fit 4. all the above answer: The First-Fit algorithm for allocating regions of contiguous memory does 1) Scan memory region list from start for first fit. 2) Scan memory region list from point of last allocation to next fit 3) Tends to leave small unusable regions, and slower due to requirement of search the entire list. 4) There is no such algorithm The Best-Fit algorithm for allocating regions of contiguous memory does 1) Scan memory region list from start for first fit 2) Pick the closest free region in the entire list. 3) There is no such algorithm 4) Find the worst fit in the entire list Question: From the given data Bytes/Sector = 500

17 Sectors/Track = 50 Tracks/Platter = 150 Platters/ Hard Disk = 10 Data Transfer rate = 10 millisecs per byte Total number of millisecs taken to transfer all the bytes in the hard disk? 1) 500 x 50 x 10 x 10 2) 500 x 50 x 150 x 10 3) 500 x 50 x 150 x 10 x10 4) 500 x 150 x 10 x10 From the given data Bytes/Sector = 500 Sectors/Track = 50 Tracks/Platter = 150 Platters/ Hard Disk = 10 Data Transfer rate = 10 millisecs per byte Total number of millisecs taken to transfer the bytes from one single platter? 1) 500 x 50 x 150 x 10 2) 500 x 50 x 10 3) 500 x 150 x 50 x 10 4) 500 x 150 x 10 x10 From the given data Bytes/Sector = 500 Sectors/Track = 50 Tracks/Platter = 150 Platters/ Hard Disk = 10 Data Transfer rate = 10 millisecs per byte Total number of ms taken to transfer all the bytes from a cylinder cylinder? 1) 500 x 50 x 10 x 10 2) 500 x 150 x 50 3) 500 x 150 x 10 x10 4) 50 x 150 x 10 x10 In a Hard disk, Bytes per sector = 512 Sectors/track = 63 Number of Magnetic disks = 5 Tracks / platter = 50 Total number of bytes in a cylinder is 13) 512X 63X 5 14) 512X 50X 5 15) 512X 63 X 50 16) 512 X 5 In a Hard disk, Bytes per sector = 512 Sectors/track = 63 Number of Magnetic disks = 5

18 Tracks / platter = 50 The total number of bytes/platter is 17) 512X 63X 5 18) 512X 50X 5 19) 512X 63 X 50 20) 512 X 5 In a Hard disk, Bytes per sector = 512 Sectors/track = 63 Number of Magnetic disks = 5 Tracks / platter = 50 The total number of cylinders in the hard disk is. 1) 63 2) 50 3) 512 4) 63 X 5 The size of single unit of allocation on disk is called. 1) Sector size 2) Page size 3) Segment size 4)File size MS-DOS supports multiprogramming fully. True / False? Answer: False Which of the following is crucial time while accessing data on the disk? 1) Seek time 2) Rotational time 3) Transmission time 4) Waiting time Ans: 1 A is software that manages the time of a microprocessor to ensure that all time critical events are processed as efficiently as possible. This software allows the system activities to be divided into multiple independent elements called tasks. 1) Kernel 2) Shell 3) Processor 4) Device Driver Ans: 1