Operating-Systems
Question 1 |
The difference between a named pipe and a regular file in Unix is that
Unlike a regular file, named pipe is a special file | |
The data in a pipe is transient, unlike the content of a regular file | |
Pipes forbid random accessing, while regular files do allow this. | |
All of the above |
Question 1 Explanation:
→ Named pipe is a special instance of a file that has no contents on the filesystem.
→ A FIFO special file (a named pipe) is similar to a pipe, except that it is accessed as part of the filesystem. It can be opened by multiple processes for reading or writing. When processes are exchanging data via the FIFO, the kernel passes all data internally without writing it to the filesystem. Thus, the FIFO special file has no contents on the filesystem; the filesystem entry merely serves as a reference point so that processes can access the pipe using a name in the filesystem.
→ The kernel maintains exactly one pipe object for each FIFO special file that is opened by at least one process. The FIFO must be opened on both ends (reading and writing) before data can be passed. Normally, opening the FIFO blocks until the other end is opened also.
→ A FIFO special file (a named pipe) is similar to a pipe, except that it is accessed as part of the filesystem. It can be opened by multiple processes for reading or writing. When processes are exchanging data via the FIFO, the kernel passes all data internally without writing it to the filesystem. Thus, the FIFO special file has no contents on the filesystem; the filesystem entry merely serves as a reference point so that processes can access the pipe using a name in the filesystem.
→ The kernel maintains exactly one pipe object for each FIFO special file that is opened by at least one process. The FIFO must be opened on both ends (reading and writing) before data can be passed. Normally, opening the FIFO blocks until the other end is opened also.
Question 2 |
Processes P1 and P2 have a producer-consumer relationship, communicating by the use of a set of shared buffers.
Increasing the number of buffers is likely to do which of the following?
I. Increase the rate at which requests are satisfied (throughput)
II. Decrease the likelihood of deadlock
III. Increase the ease of achieving a correct implementation
Increasing the number of buffers is likely to do which of the following?
I. Increase the rate at which requests are satisfied (throughput)
II. Decrease the likelihood of deadlock
III. Increase the ease of achieving a correct implementationIII Only | |
II Only | |
I Only | |
II and III Only |
Question 2 Explanation:
It only satisfied statement I. because increasing the memory size increases the rate at which requests are satisfied but can not alter the possibility of deadlock and neither does it play any role in implementation.
Question 3 |
The term ‘aging’ refers to
booting up the priority of the process in multi-level of queue without feedback | |
keeping track of the following a page has been in memory for the purpose of LRU replacement | |
letting job reside in memory for a certain amount of time so that the number of pages required can be estimated accurately | |
gradually increasing the priority of jobs that wait in the system for a long time to remedy infinite blocking | |
In Operating systems, aging is a scheduling technique used to avoid starvation. In this, the priority of the jobs that have a longer waiting time is increased as compared to the newer processes, to avoid the starvation of older processes
|
Question 4 |
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 4 Explanation:
Throughput means total number of tasks executed per unit time i.e. sum of waiting time and burst time.
Shortest job first scheduling is a scheduling policy that selects the waiting process with the smallest execution time to execute next.
Thus, in shortest job first scheduling, shortest jobs are executed first. This means CPU utilization is maximum. So, maximum number of tasks are completed.
Shortest job first scheduling is a scheduling policy that selects the waiting process with the smallest execution time to execute next.
Thus, in shortest job first scheduling, shortest jobs are executed first. This means CPU utilization is maximum. So, maximum number of tasks are completed.
Question 5 |
Consider a job scheduling problem with 4 jobs J1, J2, J3, J4 and with corresponding deadlines: ( d1, d2, d3, d4) = (4, 2, 4, 2).
Which of the following is not a feasible schedule without violating any job schedule?
Which of the following is not a feasible schedule without violating any job schedule?
J2, J4, J1, J3 | |
J4, J1, J2, J3. | |
J4, J2, J1, J3. | |
J4, J2, J3, J1 |
Question 5 Explanation:
→ Feasible schedule is completing all the jobs within deadline.
→ From the dead line, we can deduce that Job J2 & J4 will complete by time “2” whereas remaining two requires time “4”.
→ So the order of completion of Jobs are Either J2 or J4 and followed by either J1 or J3.
From the given options , Option A,C & D gives the solution because after completion of Jobs J2 and J4 then only jobs J1 and J3 is going to complete.
→ But option B , order of completing jobs is J4,J1,J2 ,J3 which is not possible and it is not feasible schedule
→ From the dead line, we can deduce that Job J2 & J4 will complete by time “2” whereas remaining two requires time “4”.
→ So the order of completion of Jobs are Either J2 or J4 and followed by either J1 or J3.
From the given options , Option A,C & D gives the solution because after completion of Jobs J2 and J4 then only jobs J1 and J3 is going to complete.
→ But option B , order of completing jobs is J4,J1,J2 ,J3 which is not possible and it is not feasible schedule
Question 6 |
Round Robin schedule is essentially the preemptive version of
FIFO | |
Shortest job first | |
Shortest remaining time | |
Longest remaining time |
Question 6 Explanation:
FIFO is when implemented in preemptive version, it acts like round-robin algorithm.
Question 7 |
What is the name of the technique in which the operating system of a computer executes several programs concurrently by switching back and forth between them?
Partitioning | |
Multi-tasking | |
Windowing | |
Paging |
Question 7 Explanation:
In a multitasking system, a computer executes several programs simultaneously by switching them back and forth to increase the user interactivity. Processes share the CPU and execute in an interleaving manner. This allows the user to run more than one program at a time.
Question 8 |
Virtual memory is
Part of Main Memory only used for swapping | |
A technique to allow a program, of size more than the size of main memory, to run | |
Part of secondary storage used in program execution | |
None of these |
Question 8 Explanation:
A computer can address more memory than the amount physically installed on the system. This extra memory is actually called virtual memory and it is a section of a hard disk that's set up to emulate the computer's RAM.
The main visible advantage of this scheme is that programs can be larger than physical memory. Virtual memory serves two purposes. First, it allows us to extend the use of physical memory by using disk. Second, it allows us to have memory protection, because each virtual address is translated to a physical address
The main visible advantage of this scheme is that programs can be larger than physical memory. Virtual memory serves two purposes. First, it allows us to extend the use of physical memory by using disk. Second, it allows us to have memory protection, because each virtual address is translated to a physical address
Question 9 |
Disk requests are received by a disk drive for cylinder 5, 25, 18, 3, 39, 8 and 35 in that order. A seek takes 5 msec per cylinder moved. How much seek time is needed to serve these requests for a Shortest Seek First (SSF) algorithm? Assume that the arm is at cylinder 20 when the last of these requests is made with none of the requests yet served
125 msec | |
295 msec | |
575 msec | |
750 msec |
Question 9 Explanation:
The arm is at cylinder 20, so the service order = 18, 25, 35, 39, 8, 5, 3.
Seek time = (20−18) + (25−18) + (35−25) + (39−35) + (39−8) + (8−5) + (5−3)
= 2 + 7 + 10 + 4 + 31 + 3 + 2 = 59
Total seek time = 59 * 5 = 29
Seek time = (20−18) + (25−18) + (35−25) + (39−35) + (39−8) + (8−5) + (5−3)
= 2 + 7 + 10 + 4 + 31 + 3 + 2 = 59
Total seek time = 59 * 5 = 29
Question 10 |
Consider a system having ‘m’ resources of the same type. The resources are shared by 3 processes A, B, C, which have peak time demands of 3, 4, 6 respectively. The minimum value of ‘m’ that ensures that deadlock will never occur is
11 | |
12 | |
13 | |
14 |
Question 10 Explanation:
Minimum resources required to avoid deadlock = (m1 – 1) + (m2 – 1) +..+ (my – 1) + 1
where m = resource required by process
y = number of processes
so, Number of resources that ensures that deadlock will never occur is = (3-1) + (4-1) + (6-1) + 1 = 11
Option (A) is correct.
where m = resource required by process
y = number of processes
so, Number of resources that ensures that deadlock will never occur is = (3-1) + (4-1) + (6-1) + 1 = 11
Option (A) is correct.
Question 11 |
A task in a blocked state
is executable | |
is running | |
must still be placed in the run queues | |
is waiting for some temporarily unavailable resources |
Question 11 Explanation:
Waiting or Blocked state is when a process has requested some input/output and is waiting for the resource.
Question 12 |
Semaphores
synchronize critical resources to prevent deadlock | |
synchronize critical resources to prevent contention | |
are used to do I/O | |
are used for memory management |
Question 12 Explanation:
Semaphore is a variable and is used to solve critical section problem and to achieve process synchronization in the multi processing environment.
Question 13 |
On a system using non-preemptive scheduling, processes with expected run times of 5, 18, 9 and 12 are in the ready queue. In what order should they be run to minimize wait time?
5, 12, 9, 18 | |
5, 9, 12, 18 | |
12, 18, 9, 5 | |
9, 12, 18, 5 |
Question 13 Explanation:
The processes should execute in SJF manner to get the lowest waiting time. So, the order should be 5, 9, 12, 18.
Question 14 |
The number of page frames that must be allocated to a running process in a virtual memory environment is determined by
the instruction set architecture | |
page size | |
number of processes in memory | |
physical memory size |
Question 14 Explanation:
There are two important tasks in virtual memory management: a page-replacement strategy and a frame-allocation strategy. Frame allocation strategy says gives the idea of minimum number of frames which should be allocated. The absolute minimum number of frames that a process must be allocated is dependent on system architecture, and corresponds to the number of pages that could be touched by a single (machine) instruction.So, it is instruction set architecture
Question 15 |
Consider a small 2-way set-associative cache memory, consisting of four blocks. For choosing the block to be replaced, use the least recently (LRU) scheme. The number of cache misses for the following sequence of block addresses is 8, 12, 0, 12, 8
2 | |
3 | |
4 | |
5 |
Question 15 Explanation:
8, 12, 0, 12, 8
Miss 8
Miss 12
Miss 0
No miss for 12
Again a miss for 8
Total misses = 4
Miss 8
Miss 12
Miss 0
No miss for 12
Again a miss for 8
Total misses = 4
Question 16 |
Which of the following RAID level provides the highest Data Transfer Rate (Read/Write)
RAID 1 | |
RAID 3 | |
RAID 4 | |
RAID 5 |
Question 16 Explanation:
Disk mirroring, also known as RAID 1, is the replication of data to two or more disks. Disk mirroring is a good choice for applications that require high performance and high availability, such as transactional applications, email and operating systems.
Question 17 |
Feedback queues
are very simple to implement | |
dispatch tasks according to execution characteristics | |
are used to favour real time tasks | |
require manual intervention to implement properly |
Question 17 Explanation:
n a multilevel queue-scheduling algorithm, processes are permanently assigned to a queue on entry to the system. Processes do not move between queues. This setup has the advantage of low scheduling overhead, but the disadvantage of being inflexible.
Multilevel feedback queue scheduling, however, allows a process to move between queues. The idea is to separate processes with different CPU-burst characteristics. If a process uses too much CPU time, it will be moved to a lower-priority queue. Similarly, a process that waits too long in a lower-priority queue may be moved to a higher-priority queue. This form of aging prevents starvation.
Multilevel feedback queue scheduling, however, allows a process to move between queues. The idea is to separate processes with different CPU-burst characteristics. If a process uses too much CPU time, it will be moved to a lower-priority queue. Similarly, a process that waits too long in a lower-priority queue may be moved to a higher-priority queue. This form of aging prevents starvation.
Question 18 |
A particular parallel program computation requires 100 sec when executed on a single processor. If 40% of this computation is inherently sequential (i.e. will not benefit from additional processors), then theoretically best possible elapsed times of this program running with 2 and 4 processors, respectively, are
20 sec and 10 sec | |
30 sec and 15 sec | |
50 sec and 25 sec | |
70 sec and 55 sec |
Question 18 Explanation:
→ The computation requires 100 seconds on a single processor implies that 40% of the
computation takes 40 seconds on any number of processors and the remaining 60%
takes 60 / (#processors) seconds on parallel computation which becomes 30 seconds on two
processors and 15 seconds on four.
→ Hence, in total, the computation takes 40+30=70 seconds on two processors and 40+15=55 seconds on four processors.
→ Hence, in total, the computation takes 40+30=70 seconds on two processors and 40+15=55 seconds on four processors.
Question 19 |
The Operating System of a computer may periodically collect all the free memory space to form contiguous block of free space. This is called:
Concatenation | |
Garbage collection | |
Collision | |
Dynamic Memory Allocation |
Question 19 Explanation:
→ The Operating System of a computer may periodically collect all the free memory space to form a contiguous block of free space. This is called garbage collection
→ We can also use compaction to minimize the probability of external fragmentation.
→ In compaction, all the free partitions are made contiguous and all the loaded partitions are brought together.
→ We can also use compaction to minimize the probability of external fragmentation.
→ In compaction, all the free partitions are made contiguous and all the loaded partitions are brought together.
Question 20 |
The following C program
If we execute this core segment, how many times the string yes will be printed ?
If we execute this core segment, how many times the string yes will be printed ?Only once | |
2 times | |
4 times | |
8 times |
Question 20 Explanation:
There is a direct formula to find the number of child processes is 2n -1 =3 times and already one ‘yes’ they are given. So, it will print 4 times ‘yes’
Question 21 |
Determine the number of page faults when references to pages occur in order – 1, 2, 4, 5, 2, 1, 2, 4. Assume that the main memory can accommodate 3 pages and the main memory already has the pages 1 and 2, with page 1 having brought earlier than page 2.(assume LRU algorithm is used)
3 | |
4 | |
5 | |
None of these |
Question 21 Explanation:
Page fault table is
The main memory already has the pages 1 and 2, with page 1 having brought earlier than page 2. So, total 6 page faults. 6-2=4
The main memory already has the pages 1 and 2, with page 1 having brought earlier than page 2. So, total 6 page faults. 6-2=4Question 22 |
Consider a system having m resources of the same type. These resources are shared by 3 processes A, B, C which have peak time demands of 3, 4, 6 respectively. The minimum value of m that ensures deadlock will never occur is
11 | |
12 | |
13 | |
14 |
Question 22 Explanation:
A requires 3, B-4, C-6;
→ If A has 2, B has 3, C has 5 then a deadlock will occur i.e., 2+3+5=10.
→ If we have one extra resource then a deadlock will not occur i.e., 10+1=11.
→ If we have equal (or) more than 11 resources then deadlock will never occur.
→ If A has 2, B has 3, C has 5 then a deadlock will occur i.e., 2+3+5=10.
→ If we have one extra resource then a deadlock will not occur i.e., 10+1=11.
→ If we have equal (or) more than 11 resources then deadlock will never occur.
Question 23 |
A computer has 1000 K of main memory. The jobs arrive and finish in the sequence
Job 1 requiring 200 K arrives
Job 2 requiring 350 K arrives
Job 3 requiring 300 K arrives
Job 1 finishes
Job 4 requiring 120 K arrives
Job 5 requiring 150 K arrives
Job 6 requiring 80 K arrives
Among the best fit and first fit, which performs better for this sequence?
First fit | |
Best fit | |
Both perform the same | |
None of the above |
Question 23 Explanation:
Main memory = 1000K
Job 1 requiring 200 K arrives
Job 2 requiring 350 K arrives
Job 3 requiring 300 K arrives and assuming continuous allocation:
Free memory = 1000 − 850(200 + 350 + 300) = 150 K (till these jobs first fit and best fit are same)
Since, job 1 finishes, Free memory = 200 K and 150 K
Case 1: First fit
Job 4 requiring 120 K arrives
Since 200 K will be the first slot, so Job 4 will acquire this slot only. Remaining memory = 200 – 120 = 80 K
Job 5 requiring 150 K arrives
It will acquire 150 K slot
Job 6 requiring 80 K arrives
It will occupy 80 K slot, so, all jobs will be allocated successfully.
Case 2: Best fit
Job 4 requiring 120 K arrives
It will occupy best fit slot which is 150 K. So, remaining memory = 150 − 120 = 30 K
Job 5 requiring 150 K arrives
It will occupy 200 K slot. So, free space = 200 − 150 = 50 K
Job 6 requiring 80 K arrives
There is no continuous 80 K memory free. So, it will not be able to allocate.
So, first fit is better.
Job 1 requiring 200 K arrives
Job 2 requiring 350 K arrives
Job 3 requiring 300 K arrives and assuming continuous allocation:
Free memory = 1000 − 850(200 + 350 + 300) = 150 K (till these jobs first fit and best fit are same)
Since, job 1 finishes, Free memory = 200 K and 150 K
Case 1: First fit
Job 4 requiring 120 K arrives
Since 200 K will be the first slot, so Job 4 will acquire this slot only. Remaining memory = 200 – 120 = 80 K
Job 5 requiring 150 K arrives
It will acquire 150 K slot
Job 6 requiring 80 K arrives
It will occupy 80 K slot, so, all jobs will be allocated successfully.
Case 2: Best fit
Job 4 requiring 120 K arrives
It will occupy best fit slot which is 150 K. So, remaining memory = 150 − 120 = 30 K
Job 5 requiring 150 K arrives
It will occupy 200 K slot. So, free space = 200 − 150 = 50 K
Job 6 requiring 80 K arrives
There is no continuous 80 K memory free. So, it will not be able to allocate.
So, first fit is better.
Question 24 |
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 is360 ms
360 ms | |
850 ms | |
900 ms | |
None of the above |
Question 24 Explanation:
FCFS
Total seek time in FCFS Scheduling when the disk drive is reading from cylinder 20 for cylinders in the order 10, 22, 20, 2, 40, 6, and 38.
Total seek time in FCFS Scheduling when the disk drive is reading from cylinder 20 for cylinders in the order 10, 22, 20, 2, 40, 6, and 38.
Question 25 |
In multiprogramming systems, it is advantageous if some programs such as editors and compilers can be shared by several users.
Which of the following must be true of multi-programmed systems in order that a single copy of a program can be shared by several users.
I. The program is a macro
II. The program is recursive
III. The program is reentrant
I only | |
II only | |
Ill only | |
I, II and III |
Question 25 Explanation:
→ Reentrant code is commonly required in operating systems and in applications intended to be shared in multi-use systems. A programmer writes a reentrant program by making sure that no instructions modify the contents of variable values in other instructions within the program.
→ Each time the program is entered for a user, a data area is obtained which keep all the variable values for that user. The data area is in another part of memory from the program itself. When the program is interrupted to give another user a turn to use the program, information about the data area associated with that user is saved.
→ When the interrupted user of the program is once again given control of the program, information in the saved data area is recovered and the program can be re-entered without concern that the previous user has changed some instruction within the program.
→ Each time the program is entered for a user, a data area is obtained which keep all the variable values for that user. The data area is in another part of memory from the program itself. When the program is interrupted to give another user a turn to use the program, information about the data area associated with that user is saved.
→ When the interrupted user of the program is once again given control of the program, information in the saved data area is recovered and the program can be re-entered without concern that the previous user has changed some instruction within the program.
Question 26 |
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 |
Question 26 Explanation:
1. All processes are arrived at time 0.
2. Algorithm used for scheduling is round robin with time quantum of one unit time.
3. The order of execution of the processes A B C D A C A C A,C,C,C,C,C
4. After 8 context switches, process A completes it execution So the completion time is 9
2. Algorithm used for scheduling is round robin with time quantum of one unit time.
3. The order of execution of the processes A B C D A C A C A,C,C,C,C,C
4. After 8 context switches, process A completes it execution So the completion time is 9
Question 27 |
What problem is solved by Dijkstra banker’s algorithm?
Cache coherence | |
Mutual exclusion | |
Deadlock recovery | |
Deadlock avoidance |
Question 27 Explanation:
Deadlock avoidance is solved by Dijkstra banker’s algorithm
Question 28 |
What is the output of the following program?
10 and 11 | |
10 | |
11 | |
11 and 11 |
Question 28 Explanation:
→ The purpose of the fork system call is to create a child process.
→ The parent process fork call will return process ID which will make if condition false then parent process will print 10
→ The child process will execute the next instruction is a++ because in the child process if the condition is not tested. Execution starts from next instruction and it will print 11
→ The parent process fork call will return process ID which will make if condition false then parent process will print 10
→ The child process will execute the next instruction is a++ because in the child process if the condition is not tested. Execution starts from next instruction and it will print 11
Question 29 |
Given the reference to the following pages by a program
0, 9, 0, 1, 8, 1, 8, 7, 8, 7, 1, 2, 8, 2, 7, 8, 2, 3, 8, 3
How many page faults will occur if the program has three-page frames available to it and uses an optimal replacement?
7 | |
8 | |
9 | |
None of these |
Question 29 Explanation:
Optimal page replacement: Replace the page that will not be used for the longest period of time.
Note: Total 7-page faults will occur.
Note: Total 7-page faults will occur.
Question 30 |
Mutual exclusion problem occurs
Between two disjoint processes that do not interact | |
Among processes that share resources | |
Among processes that do not use the same resource | |
Between two processes that uses different resources of different machine |
Question 30 Explanation:
→ Mutual exclusion is used to avoid the concurrent use of the same resources. But sometimes the problem occurred in mutual exclusion when process or program is not sharing the same resources.
Question 31 |
In a resident – OS computer, which of the following systems must reside in the main memory under all situations?
Assembler | |
Linker | |
Loader | |
Compiler |
Question 31 Explanation:
Loader is the part of an operating system that is responsible for loading programs and libraries. It is one of the essential stages in the process of starting a program, as it places programs into memory and prepares them for execution.
Loading a program involves tasks such as reading the contents of the executable file containing the program instructions into memory, and then carrying out other required preparatory tasks to prepare the executable for running.
The operating system starts the program by passing control to the loaded program code once the loading process is completed.
Loading a program involves tasks such as reading the contents of the executable file containing the program instructions into memory, and then carrying out other required preparatory tasks to prepare the executable for running.
The operating system starts the program by passing control to the loaded program code once the loading process is completed.
Question 32 |
Raid configurations of the disks are used to provide
Fault-tolerance | |
High speed | |
High data density | |
A & B |
Question 32 Explanation:
RAID stands for Redundant Array of Independent Disks. A RAID-enabled system uses two or more hard disks to improve the performance and provide some level of fault tolerance for a machine.
Question 33 |
A critical region
is a piece of code which only one process executes at a time | |
is a region prone to deadlock | |
is a piece of code which only a finite number of processes execute | |
is found only in Windows NT operating system |
Question 33 Explanation:
A critical region is a piece of code which only one process executes at a time. It cannot be executed by more than one process at a time.
Question 34 |
The Linux command “mknod myfifo b 4 16”
Will create a character device if the user is root | |
Will create a named pipe FIFO if the user is root | |
Will create a block device if the user is root | |
None of the above |
Question 34 Explanation:
The mknod command makes a directory entry and corresponding inode for a special file. The first parameter is the name of the Name entry device. Select a name that is descriptive of the device. The mknod command has two forms that have different flags.
Syntax: mknod Name { b | c } Major Minor
Syntax: mknod Name { b | c } Major Minor
Question 35 |
Which of the following need not necessarily be saved on a context switch between processes?
General purpose registers | |
Translation lookaside buffer | |
Program counter | |
All of the above |
Question 35 Explanation:
The values stored in registers, stack pointers and program counters are saved on context switch between the processes so as to resume the execution of the process.
There’s no need of saving the contents of TLB as it is invalidated after each context switch.
Question 36 |
Which of the following statement is true?
Hard real-time OS has less jitter than soft real-time OS | |
Hard real-time OS has more jitter than soft real-time OS | |
Hard real-time OS has equal jitter as a soft real-time OS | |
None of the above |
Question 36 Explanation:
→ A key characteristic of an RTOS is the level of its consistency concerning the amount of time it takes to accept and complete an application's task; the variability is jitter.
→ A hard real-time operating system has less jitter than a soft real-time operating system. The chief design goal is not high throughput, but rather a guarantee of a soft or hard performance category.
→ An RTOS that can usually or generally meet a deadline is a soft real-time OS, but if it can meet a deadline deterministically it is a hard real-time OS.
→ A hard real-time operating system has less jitter than a soft real-time operating system. The chief design goal is not high throughput, but rather a guarantee of a soft or hard performance category.
→ An RTOS that can usually or generally meet a deadline is a soft real-time OS, but if it can meet a deadline deterministically it is a hard real-time OS.
Question 37 |
The total time to prepare a disk drive mechanism for a block of data to be read from it is
seek time | |
latency | |
latency plus seek time | |
transmission time |
Question 37 Explanation:
Seek time is the time taken by the head to move to the correct track and Rotational latency is the time taken by the disk to rotate until the head is over the desired block to read
So, the total time to prepare a disk drive mechanism for a block of data to be read from is the sum of both the seek time and the latency.
So, the total time to prepare a disk drive mechanism for a block of data to be read from is the sum of both the seek time and the latency.
Question 38 |
Feedback queues
are very simple to implement | |
dispatch tasks according to execution characteristics | |
are used to favour real time tasks | |
require manual intervention to implement properly |
Question 38 Explanation:
Consider feedback queue as multilevel feedback queue. Multilevel feedback queue scheduling allows a process to move between queues. This movement is facilitated by the characteristic of the CPU burst of the process. If a process uses too much CPU time, it will be moved to a lower-priority queue.
Question 39 |
With Round-Robin CPU scheduling in a time shared system
using very large time slices (quantas) degenerates into First Come First served (FCFS) algorithm. | |
using extremely small time slices improves performance | |
using very small time slices degenerates into Last-In First-Out (LIFO) algorithm. | |
using medium sized times slices leads to shortest Request time First (SRTF) algorithm |
Question 39 Explanation:
If time quantum is very large, then scheduling happens according to FCFS
Question 40 |
Dynamic address translation
is part of the operating system paging algorithm | |
is useless when swapping is used | |
is the hardware necessary to implement paging | |
storage pages at a specific location on disk |
Question 40 Explanation:
Dynamic address translation is the hardware necessary to implement paging as it converts the logical address to physical address
Question 41 |
Thrashing
always occurs on large computers | |
is a natural consequence of virtual memory systems | |
can always be avoided by swapping | |
can be caused by poor paging algorithms |
Question 41 Explanation:
When the degree of multiprogramming increases, the CPU utilization will drastically fall down and the system will spent more time only in the page replacement and the time taken to complete the execution of the process will increase. This situation in the system is called as thrashing. It can also be a consequence of poor programming algorithms.
Question 42 |
What is the name of the operating system that reads and reacts in terms of operating system?Dynamic address translation
Batch system | |
Quick response time | |
real time system | |
Time sharing system |
Question 42 Explanation:
A real-time operating system is an operating system that guarantees to process events or data by a specific moment in time.
Question 43 |
Which of the following is an example of spooled device?
A line printer used to print the output of a number of jobs | |
A terminal used to enter input data to a running program | |
A secondary storage device in a virtual memory system | |
A graphic display device |
Question 43 Explanation:
Spooling works like a typical request queue where data, instructions and processes from multiple sources are accumulated for execution later on. Generally, it is maintained on computer’s physical memory, buffers or the I/O device-specific interrupts. The spool is processed in FIFO manner i.e. whatever first instruction is there in the queue will be popped and executed.
Example: printer
Example: printer
Question 44 |
Checkpointing a job
allows it to be completed successfully | |
allows it to continue executing later | |
prepares it for finishing | |
occurs only when there is an error in it |
Question 44 Explanation:
Checkpointing is a method of periodically saving the state of a job so that, if for some reason, the job does not complete, it can be restarted from the saved state.
Question 45 |
Overlaying
requires use of a loader | |
allows larger programs, but requires more effort | |
is most used on large computers | |
is transparent to the user | |
A,B and D |
Question 45 Explanation:
Overlaying means "the process of transferring a block of program code or other data into internal memory, replacing what is already stored".Overlaying is a programming method that allows programs to be larger than the computer's main memory.overlaying is transparent to the user.
Question 46 |
A critical section is a program segment
which should run in a certain amount of time | |
which avoids deadlocks | |
where shared resources are accessed | |
which must be enclosed by a pair of semaphore operations P and V |
Question 46 Explanation:
In concurrent programming, concurrent accesses to shared resources can lead to unexpected or erroneous behavior, so parts of the program where the shared resource is accessed are protected. This protected section is the critical section .
Question 47 |
In which of the following four necessary conditions for deadlock processes claim exclusive control of the resources they require?
no preemption | |
mutual exclusion | |
circular wait | |
hold and wait |
Question 47 Explanation:
Mutual Exclusion is a condition when one or more than one resource are non-sharable (Only one process can use at a time) i.e. processes claim exclusive control of the resources they require.
Question 48 |
Fork is
the creation of a new job | |
the dispatching of a task | |
increasing the priority of a task | |
the creation of a new process |
Question 48 Explanation:
fork() creates a new process by duplicating the calling process, The new process, referred to as child, is an exact duplicate of the calling process, referred to as parent, except for the following :
The child has its own unique process ID, and this PID does not match the ID of any existing process group.
The child’s parent process ID is the same as the parent process ID.
The child does not inherit its parent’s memory locks and semaphore
adjustments.
The child does not inherit outstanding asynchronous I/O operations from its parent nor does it inherit any asynchronous I/O contexts from its parent.
The child has its own unique process ID, and this PID does not match the ID of any existing process group.
The child’s parent process ID is the same as the parent process ID.
The child does not inherit its parent’s memory locks and semaphore
adjustments.
The child does not inherit outstanding asynchronous I/O operations from its parent nor does it inherit any asynchronous I/O contexts from its parent.
Question 49 |
Which of the following need not necessarily be saved on a Context Switch between processes?
General purpose registers | |
Translation lookaside buffer | |
Program counter | |
Stack pointer |
Question 49 Explanation:
The values stored in registers, stack pointers and program counters are saved on context switch between the processes so as to resume the execution of the process.
There’s no need of saving the contents of TLB as it is invalidated after each context switch.
There’s no need of saving the contents of TLB as it is invalidated after each context switch.
Question 50 |
At a particular time, the value of a counting semaphore is 10, it will become 7 after:
(a) 3 V operations
(b) 3 P operations
(c) 5 V operations and 2 P operations
(d) 2 V operations and 5 P operations
Which of the following option is correct?
Only (b) | |
Only(d) | |
Both (b) and (d) | |
None of these |
Question 50 Explanation:
P: Wait operation decrements the value of the counting semaphore by 1.
V: Signal operation increments the value of counting semaphore by 1.
Current value of the counting semaphore = 10
a) after 3 P operations, value of semaphore = 10-3 = 7
d) after 2 v operations, and 5 operations value of semaphore = 10 + 2 – 5 = 7
V: Signal operation increments the value of counting semaphore by 1.
Current value of the counting semaphore = 10
a) after 3 P operations, value of semaphore = 10-3 = 7
d) after 2 v operations, and 5 operations value of semaphore = 10 + 2 – 5 = 7
Question 51 |
Consider a logical address space of 8 pages of 1024 words mapped into memory of 32 frames. How many bits are there in the logical address?
13 bits | |
15 bits | |
14 bits | |
12 bits |
Question 51 Explanation:
logical address space = 8 pages of 1024 words
number of bits in logical address space = p (page bits) + d (offset bits)
number of bits = log28 + log21024 = 3 + 10 = 13 bits
number of bits in logical address space = p (page bits) + d (offset bits)
number of bits = log28 + log21024 = 3 + 10 = 13 bits
Question 52 |
The performance of Round Robin algorithm depends heavily on
size of the process | |
the I/O bursts of the process | |
the CPU bursts of the process | |
the size of the time quantum |
Question 52 Explanation:
In Round Robin algorithm, it is very important to choose the the quantum carefully as smaller time quanta leads to more context switches thereby reducing the efficiency of the CPU and the larger time quanta makes the round robin algorithm regenerate into FCFS algorithm.
Question 53 |
The performance of Round Robin algorithm depends heavily on
size of the process | |
the I/O bursts of the process | |
the CPU bursts of the process | |
the size of the time quantum |
Question 53 Explanation:
In Round Robin algorithm, it is very important to choose the the quantum carefully as smaller time quanta leads to more context switches thereby reducing the efficiency of the CPU and the larger time quanta makes the round robin algorithm regenerate into FCFS algorithm.
Question 54 |
The page replacement algorithm which gives the lowest page fault rate is
LRU | |
FIFO | |
Optimal page replacement | |
Second chance algorithm |
Question 54 Explanation:
In Optimal Page replacement algorithm, pages are replaced which are not used for the longest duration of time in the future. This page replacement algorithm ensures the lowest page fault rate.
Question 55 |
The primary purpose of an operating system is
To make most efficient use of the computer hardware | |
To allow people to use the computer | |
To keep systems programmers employed | |
To make computers easier to use |
Question 55 Explanation:
Explanation: An operating system has three main functions:
(1) manage the computer’s resources, such as the central processing unit, memory, and other input – output sources
(2) establish a user interface, and
(3) execute and provide services for applications software.
OS provides an interface between the user and the hardware and thus making the computer easy to use for the user but the primary function of OS is to manage the hardware in the most efficient way.
(1) manage the computer’s resources, such as the central processing unit, memory, and other input – output sources
(2) establish a user interface, and
(3) execute and provide services for applications software.
OS provides an interface between the user and the hardware and thus making the computer easy to use for the user but the primary function of OS is to manage the hardware in the most efficient way.
Question 56 |
Which is the correct definition of a valid process transition in an operating system?
Wake up: ready → running | |
Dispatch: ready → running | |
Block: ready → running | |
Timer runout: ready → running |
Question 56 Explanation:
From the below process state diagram, we can easily that the option (B) is correct.
Question 57 |
The correct matching of the following pairs is
(A) Disk check (1) Round robin
(B) Batch processing (2) Scan
(C) Time sharing (3) LIFO
(D) Stack operation (4) FIFO
(A) Disk check (1) Round robin
(B) Batch processing (2) Scan
(C) Time sharing (3) LIFO
(D) Stack operation (4) FIFO
A B C D 3 4 2 1 | |
A B C D 4 3 2 1 | |
A B C D 3 4 1 2 | |
A B C D 2 4 1 3 |
Question 57 Explanation:
Scan is disk scheduling algorithm
In the round robin algorithm, each process will execute for particular time quantum,So it is the example for time sharing,
Stack is last in first out(LIFO)
In the batching processing, whatever process arrives first, it execute that process. So it First in First out(FIFO)
In the round robin algorithm, each process will execute for particular time quantum,So it is the example for time sharing,
Stack is last in first out(LIFO)
In the batching processing, whatever process arrives first, it execute that process. So it First in First out(FIFO)
Question 58 |
A page fault
Occurs when a program accesses an available page on memory | |
is an error in a specific page | |
is a reference to a page belonging to another program | |
occurs when a program accesses a page not currently in memory |
Question 58 Explanation:
A page fault is a type of exception raised by computer hardware when a running program accesses a memory page that is not currently mapped by the memory management unit (MMU) into the virtual address space of a process
Question 59 |
Let the page fault service time be 10 ms in a computer with average memory access time being 20 ns. If the one-page fault is generated for every 106 memory accesses, what is the effective access time for the memory?
21.4 ns | |
29.9 ns | |
23.5 ns | |
35.1 ns |
Question 59 Explanation:
Question 60 |
Consider a system having “n” resources of same type. These resources are shared by 3 processes, A, B, C. These have peak demands of 3, 4, and 6 respectively. For what value of “n” deadlock won’t occur
15 | |
9 | |
10 | |
11 |
Question 60 Explanation:
Number of min resources required = (3-1) + (4-1) + (6-1) + 1 = 11
Question 61 |
Consider a set of 5 processes whose arrival time, CPU time needed and the priority are given below:
(smaller the number, higher the priority)
If the CPU scheduling policy is priority scheduling without preemption, the average waiting time will be
(smaller the number, higher the priority)
If the CPU scheduling policy is priority scheduling without preemption, the average waiting time will be
12.8 ms | |
11.8 ms | |
10.8 ms | |
9.8 ms |
Question 61 Explanation:
Following is the Gantt diagram:
Average Waiting Time = (30 + 3 + 3 + 18)/ 5 = 10.8
Average Waiting Time = (30 + 3 + 3 + 18)/ 5 = 10.8
Question 62 |
With single resource, deadlock occurs
If there are more than two processes competing for that resources | |
If there are only two processes competing for that resources | |
If there is a single process competing for that resources | |
None of these |
Question 62 Explanation:
Only Single resource available never occur deadlock because it violates circular wait and hold & wait condition.
Question 63 |
At a particular time of computation the value of a counting semaphore is 7. Then 20 P operations and xV operations were completed on this semaphore. If the new value of semaphore is 5 ,x will be
18 | |
22 | |
15 | |
13 |
Question 63 Explanation:
Here, 20P operations means 20 wait operations. It decrement value by 1 every time.
xV operations means x increments operations. It increment value by 1 every time.
→ New value of semaphore is 5 after performing xV operations
= -13 + xV
= 5
xV = 5 + 13
= 18
→ After applying 20P operations in semaphore value is = 7-20 = -13
xV operations means x increments operations. It increment value by 1 every time.
→ New value of semaphore is 5 after performing xV operations
= -13 + xV
= 5
xV = 5 + 13
= 18
→ After applying 20P operations in semaphore value is = 7-20 = -13
Question 64 |
A system has 3 processes sharing 4 resources. If each process needs a maximum of 2 units, then
Deadlock can never occur | |
Deadlock may occur | |
Deadlock has to occur | |
None of these |
Question 64 Explanation:
If the system is deadlocked, it implies that each process is holding one resource and is waiting for one more. Since there are 3 processes and 4 resources, one process must be able to obtain two resources. This process requires no more resources and therefore it will return its resources when done.
Question 65 |
Determine the number of page faults when references to pages occur in the following order:
1, 2, 4, 5, 2, 1, 2, 4
Assume that the main memory can accommodate 3 pages and the main memory already has the pages 1 and 2, with page one having brought earlier than page 2. (LRU page replacement algorithm is used)
3 | |
5 | |
4 | |
None of these |
Question 65 Explanation:
Here, total 6 page faults but in question, they are clearly mentioned that the main memory already has the pages 1 and 2, with page one having brought earlier than page 2. It means 6-2=4.
Question 66 |
Working Set(t,k) at an instant of time t is
the set of future references that the OS will make | |
the set of future references that the OS will make in the next unit of time | |
the set of references with high frequency | |
the set of pages that have been referenced in the last time units |
Question 66 Explanation:
→ Working set defines the amount of memory that a process requires in a given time interval.
→ It defines “the working set of information W(t,τ) of a process at time t to be the collection of information referenced by the process during the process time interval (t−τ,t)”.
→ Typically the units of information in question are considered to be memory pages. This is suggested to be an approximation of the set of pages that the process will access in the future (say during the next τ time units), and more specifically is suggested to be an indication of what pages ought to be kept in main memory to allow most progress to be made in the execution of that process.
→ It defines “the working set of information W(t,τ) of a process at time t to be the collection of information referenced by the process during the process time interval (t−τ,t)”.
→ Typically the units of information in question are considered to be memory pages. This is suggested to be an approximation of the set of pages that the process will access in the future (say during the next τ time units), and more specifically is suggested to be an indication of what pages ought to be kept in main memory to allow most progress to be made in the execution of that process.
Question 67 |
A CPU generates 32-bit virtual addresses. The page size is 4 KB. The processor has a translation lookaside buffer (TLB) which can hold a total of 128 page table entries and is 4-way set associative. The minimum size of the TLB tag is:
11 bits | |
13 bits | |
15 bits | |
20 bits |
Question 67 Explanation:
Page size = 4 KB = 4 × 210 Bytes = 212 Bytes
Virtual Address = 32 bit
No. of bits needed to address the page frame = 32 - 12 = 20
TLB can hold 128 page table entries with 4-way set associative
⇒ 128/4=32=25
→ 5 bits are needed to address a set.
→ The size of TLB tag = 20 - 5 = 15 bits
Virtual Address = 32 bit
No. of bits needed to address the page frame = 32 - 12 = 20
TLB can hold 128 page table entries with 4-way set associative
⇒ 128/4=32=25
→ 5 bits are needed to address a set.
→ The size of TLB tag = 20 - 5 = 15 bits
Question 68 |
For the real time operating system, which of the following is the most suitable scheduling scheme?
Round robin | |
Preemptive | |
First come first serve | |
Random scheduling |
Question 68 Explanation:
→ Preemption is the act of temporarily interrupting a task being carried out by a computer system, without requiring its cooperation, and with the intention of resuming the task at a later time. Such changes of the executed task are known as context switches.
→ It is normally carried out by a privileged task or part of the system known as a preemptive scheduler, which has the power to preempt, or interrupt, and later resume, other tasks in the system.
→ Preemptive scheduling is the most suitable scheduling scheme for real time operating systems as it can always preempt other processes based on priority.
→ In real time operating systems, we have different tasks to perform simultaneously and also there are a few tasks which need to be performed first at priority basis.
→ It is normally carried out by a privileged task or part of the system known as a preemptive scheduler, which has the power to preempt, or interrupt, and later resume, other tasks in the system.
→ Preemptive scheduling is the most suitable scheduling scheme for real time operating systems as it can always preempt other processes based on priority.
→ In real time operating systems, we have different tasks to perform simultaneously and also there are a few tasks which need to be performed first at priority basis.
Question 69 |
In which one of the page replacement policies, Belady’s anomaly may occur?
FIFO | |
Optimal | |
LRU | |
MRU |
Question 69 Explanation:
→ Belady's anomaly is the phenomenon in which increasing the number of page frames results in an increase in the number of page faults for certain memory access patterns.
→ This phenomenon is commonly experienced when using the first-in first-out (FIFO) page replacement algorithm.
→ In FIFO, the page fault may or may not increase as the page frames increase, but in Optimal and stack-based algorithms like LRU, as the page frames increase the page fault decreases.
→ This phenomenon is commonly experienced when using the first-in first-out (FIFO) page replacement algorithm.
→ In FIFO, the page fault may or may not increase as the page frames increase, but in Optimal and stack-based algorithms like LRU, as the page frames increase the page fault decreases.
Question 70 |
Special software to create a job queue is called a
Driver | |
Spooler | |
Interpreter | |
Linkage editor |
Question 70 Explanation:
Spooling is a specialized form of multi-programming for the purpose of copying data between different devices.
A dedicated program, the spooler, maintains an orderly sequence of jobs for the peripheral and feeds it data at its own rate
Conversely, for slow input peripherals, such as a card reader, a spooler can maintain a sequence of computational jobs waiting for data, starting each job when all of the relevant input is available
A dedicated program, the spooler, maintains an orderly sequence of jobs for the peripheral and feeds it data at its own rate
Conversely, for slow input peripherals, such as a card reader, a spooler can maintain a sequence of computational jobs waiting for data, starting each job when all of the relevant input is available
Question 71 |
Process is:
A program in high level language kept on disk | |
Contents of main memory | |
A program in execution | |
A job in secondary memory |
Question 71 Explanation:
A process is the instance of a computer program that is being executed by one or many threads. It contains the program code and its activity.
Depending on the operating system (OS), a process may be made up of multiple threads of execution that execute instructions concurrently
Depending on the operating system (OS), a process may be made up of multiple threads of execution that execute instructions concurrently
Question 72 |
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 |
Question 72 Explanation:
Question 73 |
Using a larger block size in a fixed block size file system leads to
better disk throughput but poorer disk space utilization | |
better disk throughput and better disk space utilization | |
poorer disk throughput but better disk space utilization | |
poorer disk throughput and poorer disk space utilization |
Question 73 Explanation:
→ While using a larger block size means that contains less number of blocks then that results better throughput.
→ This can be implemented in a fixed block size then the space utilization is not upto the mark. So the statement results better disk throughput but poorer disk space utilization
→ This can be implemented in a fixed block size then the space utilization is not upto the mark. So the statement results better disk throughput but poorer disk space utilization
Question 74 |
When a process is rolled back as a result of deadlock the difficulty which arises is
Starvation | |
System throughput | |
Low device utilization | |
Cycle stealing |
Question 74 Explanation:
→ When a process is rolled back as a result of deadlock the difficulty which arises is starvation.
→ Resource starvation is a problem encountered in concurrent computing where a process is perpetually denied necessary resources to process its work.
→ Starvation may be caused by errors in a scheduling or mutual exclusion algorithm, but can also be caused by resource leaks, and can be intentionally caused via a denial-of-service attack such as a fork bomb
Question 75 |
The following table shows the processes in the ready queue and time required for each process for completing its job.
If round-robin scheduling with 5 ms is used what is the average waiting time of the processes in the queue?
If round-robin scheduling with 5 ms is used what is the average waiting time of the processes in the queue?
27 ms | |
26.2 ms | |
27.5 ms | |
27.2 ms |
Question 75 Explanation:
→In the round robin algorithm, time slices (also known as time quanta) are assigned to each process in equal portions and in circular order, handling all processes without priority (also known as cyclic executive).
→Given scheduling algorithm is round robin with time quantum value is 5ms.
→In the given example, every process will execute for 5ms. process P1 will execute for two times , P2 for one time, P3 for four times , P4 for two times and P5 for three times.
→Waiting time of process P1 = 0+(25-20)=5
→Waiting time of process P2 = 5
→Waiting time of process P3 = 10 + (30-15 )+ (43-35) + (53-48)=10 + 15 + 8 + 5 = 38
→Waiting time of process P4 = 15+(35-20)=15 + 15 = 30
→Waiting time of process P5 =20+(38-25)+48-43= 20 + 13 + 5 = 38
→Average waiting time = 20 + 5 + 38 + 30 + 38 =131/5 = 26.2
→Given scheduling algorithm is round robin with time quantum value is 5ms.
→In the given example, every process will execute for 5ms. process P1 will execute for two times , P2 for one time, P3 for four times , P4 for two times and P5 for three times.
→Waiting time of process P1 = 0+(25-20)=5
→Waiting time of process P2 = 5
→Waiting time of process P3 = 10 + (30-15 )+ (43-35) + (53-48)=10 + 15 + 8 + 5 = 38
→Waiting time of process P4 = 15+(35-20)=15 + 15 = 30
→Waiting time of process P5 =20+(38-25)+48-43= 20 + 13 + 5 = 38
→Average waiting time = 20 + 5 + 38 + 30 + 38 =131/5 = 26.2
Question 76 |
Below is the precedence graph for a set of tasks to be executed on a parallel processing system S.
What is the efficiency of this precedence graph on S if each of the tasks T1, T2, T3,….T8 takes the same time and the system S has five processors?
What is the efficiency of this precedence graph on S if each of the tasks T1, T2, T3,….T8 takes the same time and the system S has five processors?
25% | |
40% | |
50%
| |
90%
|
Question 76 Explanation:
From the precedence graph, we say that the following tasks executed sequentially
I. T1 ,T2
II. T3 and T6
III. T4 and T7
IV. T5 and T8
(T3,T6),(T4,T7) and (T5,T8) will execute parallely.
So total number of processes that can be executed in 4 units time using 5 available processors = 5*4 = 20
Maximum number of tasks are 8
Efficiency = 8/20 * 100 = 40%
I. T1 ,T2
II. T3 and T6
III. T4 and T7
IV. T5 and T8
(T3,T6),(T4,T7) and (T5,T8) will execute parallely.
So total number of processes that can be executed in 4 units time using 5 available processors = 5*4 = 20
Maximum number of tasks are 8
Efficiency = 8/20 * 100 = 40%
Question 77 |
If the page size in a 32-bit machine is 4K bytes then the size of the page table is
1 M bytes | |
2 M bytes | |
4 M bytes | |
4 K bytes
|
Question 77 Explanation:
→Page size is total space taken up by page and Page table entry size is memory taken for indexing the Page in Page Table
→Size of logical address = 32 bits
→Page size = 4K =22210=212 Bytes
→Number of pages = logical address space/ size of each page = 232/ 212= 220
→Page table size = number of pages * size of a page table entry
= 220 * 22
= 222
→Size of logical address = 32 bits
→Page size = 4K =22210=212 Bytes
→Number of pages = logical address space/ size of each page = 232/ 212= 220
→Page table size = number of pages * size of a page table entry
= 220 * 22
= 222
Question 78 |
In a system using a single processor, a new process arrives at the rate of six processes per minute and each such process requires seven seconds of service time. What is the CPU utilization?
70%
| |
30%
| |
60%
| |
64%
|
Question 78 Explanation:
From the given question,
The number of new processes will arrive per minute = 6
Each process require to complete its task = 7 secs
CPU utilization time within a minute = 6*7 = 42 secs
The percentage of CPU utilization = time which is spent for utilization / total time * 100
= (42/60) * 100
= 70%
The number of new processes will arrive per minute = 6
Each process require to complete its task = 7 secs
CPU utilization time within a minute = 6*7 = 42 secs
The percentage of CPU utilization = time which is spent for utilization / total time * 100
= (42/60) * 100
= 70%
Question 79 |
What is the bitrate of a video terminal unit with 80 characters/line, 8 bits/character and a horizontal sweep time of 100 μs (including 20 μs of retrace time)?
8 Mbps | |
6.4 Mbps | |
0.8 Mbps | |
0.64 Mbps |
Question 79 Explanation:
Bit rate is the number of bits that are conveyed or processed per unit of time
Given data is video terminal unit has 80 lines and each line consists of 8 bits
Total number of bits transmitted = 80 * 8 = 640 bits
Horizontal sweep time = 100 us=100x 10-6 seconds
Bit rate = (640 * 106) / 100 =6400000=6.4 Mbps
Given data is video terminal unit has 80 lines and each line consists of 8 bits
Total number of bits transmitted = 80 * 8 = 640 bits
Horizontal sweep time = 100 us=100x 10-6 seconds
Bit rate = (640 * 106) / 100 =6400000=6.4 Mbps
Question 80 |
Belady’s anomaly means
Page fault rate is constant even on increasing the number of allocated frames | |
Page fault rate may increase on increasing the number of allocated frames | |
Page fault rate may increase on decreasing the number of allocated frames | |
Page fault rate may decrease on increasing the number of allocated frames |
Question 80 Explanation:
Bélády's anomaly is the phenomenon in which increasing the number of page frames results in an increase in the number of page faults for certain memory access patterns.
This phenomenon is commonly experienced when using the first-in first-out (FIFO) page replacement algorithm.
This phenomenon is commonly experienced when using the first-in first-out (FIFO) page replacement algorithm.
Question 81 |
Which RAID level gives block-level striping with double distributed parity?
RAID 10 | |
RAID 2 | |
RAID 6 | |
RAID 5 |
Question 81 Explanation:
RAID (Redundant Array of Inexpensive Disks or Drives, or Redundant Array of Independent Disks) is a data storage virtualization technology that combines multiple physical disk drive components into one or more logical units for the purposes of data redundancy, performance improvement, or both.
RAID levels and their corresponding functionality as shown below
RAID 0: Stripping
RAID 1: Mirroring
RAID 2: Hamming code for error detection
RAID 3: Byte-level striping with a dedicated parity disk
RAID 4: Block-level striping with block-level striping with two parity blocks parity
RAID 5: Block-level striping with distributed parity
RAID 6: Block-level striping with double distributed parity.
RAID levels and their corresponding functionality as shown below
RAID 0: Stripping
RAID 1: Mirroring
RAID 2: Hamming code for error detection
RAID 3: Byte-level striping with a dedicated parity disk
RAID 4: Block-level striping with block-level striping with two parity blocks parity
RAID 5: Block-level striping with distributed parity
RAID 6: Block-level striping with double distributed parity.
Question 82 |
Consider a 32-bit machine where four-level paging scheme is used. If the hit ratio to TLB is 98%, and it takes 20 nanosecond to search the TLB and 100 nanoseconds to access the main memory what is effective memory access time in nanoseconds?
126 | |
128 | |
122 | |
120 |
Question 82 Explanation:
Hit ratio to TLB(H) is 98%
Searching time of TLB(T) is 20ns
Access time(M) is 100ns and four level paging scheme is used.
Effective Memory access Time, EAT = H* T+ (1 - H)[ T+ 4*M] + M]
EAT = (0.98 *20) + 0.02(20 + 400) + 100
= 19.6 + 8.4 + 100 = 128 ns
Searching time of TLB(T) is 20ns
Access time(M) is 100ns and four level paging scheme is used.
Effective Memory access Time, EAT = H* T+ (1 - H)[ T+ 4*M] + M]
EAT = (0.98 *20) + 0.02(20 + 400) + 100
= 19.6 + 8.4 + 100 = 128 ns
Question 83 |
There are three processes in the ready queue. When the currently running process requests for I/O how many process switches take place?
1 | |
2 | |
3 | |
4 |
Question 83 Explanation:
Context switching is the procedure of storing the state of an active process for the CPU when it has to start executing a new one.
For example, process A with its address space and stack is currently being executed by the CPU and there is a system call to jump to a higher priority process B;
the CPU needs to remember the current state of the process A so that it can suspend its operation, begin executing the new process B and when done, return to its previously executing process A
Only one context switch process happened
For example, process A with its address space and stack is currently being executed by the CPU and there is a system call to jump to a higher priority process B;
the CPU needs to remember the current state of the process A so that it can suspend its operation, begin executing the new process B and when done, return to its previously executing process A
Only one context switch process happened
Question 84 |
A total of 9 units of a resource type available, and given the safe state shown below, which of the following sequence will be a safe state?
(P4, P1, P3, P2) | |
(P4, P2, P1, P3) | |
(P4, P2, P3, P1) | |
(P3, P1, P2, P4) |
Question 84 Explanation:
Question 85 |
A fast wide SCSI-II disk drive spins at 7200 RPM, has a sector size of 512 bytes, and holds 160 sectors per track. Estimate the sustained transfer rate of this drive
576000 Kilobytes / sec | |
9600 Kilobytes / sec | |
4800 Kilobytes / sec | |
19200 Kilobytes / sec |
Question 85 Explanation:
Given data,
Disk drive spins=7200 RPM,
Sector size=512 bytes
Disk drive holds= 160 sectors/track
Step-1: First find the number of rotations for 1 second. They given RPM for minute.
1 sec= 7200 / 60 = 120 rotations
Step-2: Disk Transfer rate =120*160*512
= 98,30,400 Bytes per second
Step-3: Equivalent in kilobytes = 9830400 / 1024
= 9600 KB
Note: The disk spins 120 times per second, and each spin transfers a track of 80 KB. Thus, the sustained transfer rate can be approximated as 9600 KB/s.
Disk drive spins=7200 RPM,
Sector size=512 bytes
Disk drive holds= 160 sectors/track
Step-1: First find the number of rotations for 1 second. They given RPM for minute.
1 sec= 7200 / 60 = 120 rotations
Step-2: Disk Transfer rate =120*160*512
= 98,30,400 Bytes per second
Step-3: Equivalent in kilobytes = 9830400 / 1024
= 9600 KB
Note: The disk spins 120 times per second, and each spin transfers a track of 80 KB. Thus, the sustained transfer rate can be approximated as 9600 KB/s.
Question 86 |
Which of the following strategy is employed for overcoming the priority inversion problem?
Temporarily raise the priority of lower priority level process | |
Have a fixed priority level scheme | |
Implement kernel preemption scheme | |
Allow lower priority process to complete its job |
Question 86 Explanation:
Priority inversion is the condition in which a high priority task needs to wait for a low priority task to release a resource between the medium priority task and a low priority task.
Question 87 |
Consider the following set of processes, with arrival times and the required CPU-burst times given in milliseconds.
What is the sequence in which the processes are completed? Assume round robin scheduling with a time quantum of 2 milliseconds
What is the sequence in which the processes are completed? Assume round robin scheduling with a time quantum of 2 milliseconds
P1, P2, P3 | |
P2, P1, P3 | |
P3, P2, P1 | |
P2, P3, P1 |
Question 87 Explanation:
Given scheduling algorithm is round robin algorithm .
To schedule processes fairly, a round-robin scheduler generally employs time-sharing, giving each job a time slot or quantum(its allowance of CPU time), and interrupting the job if it is not completed by then. The job is resumed next time a time slot is assigned to that process. If the process terminates or changes its state to waiting during its attributed time quantum, the scheduler selects the first process in the ready queue to execute.
Given time quantity is 2ms which means each process will execute 2ms and switches to next process.
This procedure will continue till completion of all processes execution.
From the given data P1 burst time 4 ms and P2 is 2 and P3 is 1.
So first P1 and P2 , P2 will complete first.
After completion of P2 . there are two processes in the queue. From the two processes P1 entered in the queue first and later P3
So P1 will complete next and finally P3
The order of completion of processes are P2,P1 and P3
To schedule processes fairly, a round-robin scheduler generally employs time-sharing, giving each job a time slot or quantum(its allowance of CPU time), and interrupting the job if it is not completed by then. The job is resumed next time a time slot is assigned to that process. If the process terminates or changes its state to waiting during its attributed time quantum, the scheduler selects the first process in the ready queue to execute.
Given time quantity is 2ms which means each process will execute 2ms and switches to next process.
This procedure will continue till completion of all processes execution.
From the given data P1 burst time 4 ms and P2 is 2 and P3 is 1.
So first P1 and P2 , P2 will complete first.
After completion of P2 . there are two processes in the queue. From the two processes P1 entered in the queue first and later P3
So P1 will complete next and finally P3
The order of completion of processes are P2,P1 and P3
Question 88 |
Suppose we have variable logical records of lengths of 5 bytes, 10 bytes, and 25 bytes while the physical block size in disk is 15 bytes. What is the maximum and minimum fragmentation seen in bytes?
25 and 5 | |
15 and 5 | |
15 and 0 | |
10 and 5 |
Question 88 Explanation:
The maximum fragmentation that is observed is in the case of record length of 25 bytes and it is of 10 bytes. This is explained as follows. Suppose we have a logical record of 25 bytes. Then, since the physical block length is only 15 bytes so we will require 2 physical blocks to store data. One block will be fully occupied and for the second block only 15 bytes will be used, resulting in unused 10 bytes.
Physical block length is 15 bytes. In case of 10 bytes record, the unused space is 5 bytes. The minimum fragmentation that is observed in this case is 5 bytes for the record length of 10 bytes
Physical block length is 15 bytes. In case of 10 bytes record, the unused space is 5 bytes. The minimum fragmentation that is observed in this case is 5 bytes for the record length of 10 bytes
Question 89 |
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?
n | |
n2 | |
n! | |
2n |
Question 89 Explanation:
For ‘n’ processes to be scheduled on one processor, there can be n! different schedules possible.
If you are making choices from n objects, then on your first pick you have n choices. On your second pick, you have n-1 choices, n-2 for your third choice and so forth.
For example 5 processes, the number of ways to schedules is 5!
If you are making choices from n objects, then on your first pick you have n choices. On your second pick, you have n-1 choices, n-2 for your third choice and so forth.
For example 5 processes, the number of ways to schedules is 5!
Question 90 |
Which of the following are the likely causes of thrashing?
Page size was very small | |
There are too many programs running in the system | |
Least recently used policy is used for page replacement | |
First in First out policy is used for page replacement |
Question 90 Explanation:
If a process is given too few frames,its faulting rate will rise dramatically.
If this occurs for many or all the processes, the resulting situation in which the system is doing very little useful work due to the high I/O requirements for all the page faults is called thrashing.
Question 91 |
Consider a logical address space of 8 pages of 1024 words each, mapped onto a physical memory of 32 frames. How many bits are there in the physical address and logical address respectively?
5, 3 | |
10, 10 | |
15, 13 | |
15, 15 |
Question 91 Explanation:
→Number of pages= 8= 23=(3 bits)
→Each page consists of 1024 words =210(10 bits)
→Logical address space consists of 8 pages of 1024 words each,
→Then the number of bits required for logical address is 3+10=13 bits.
→Total number of frames =32=25(5 bits).
→The logical memory is mapped to physical memory which means mapping should done between pages and frames.
→Physical address = 5(number of bits for frames) + 10 (number of bits for pages)= 15 bits
→Each page consists of 1024 words =210(10 bits)
→Logical address space consists of 8 pages of 1024 words each,
→Then the number of bits required for logical address is 3+10=13 bits.
→Total number of frames =32=25(5 bits).
→The logical memory is mapped to physical memory which means mapping should done between pages and frames.
→Physical address = 5(number of bits for frames) + 10 (number of bits for pages)= 15 bits
Question 92 |
In a 64-bit machine, with 2 GB RAM, and 8 KB page size, how many entries will be there in the page table if it is inverted?
218 | |
220 | |
233 | |
251 |
Question 92 Explanation:
Given data is
Memory size = 2 GB = 231
Page size = 8 KB = 213
Number of entries in inverted page table = physical address space / page size = 231/213 = 218
Memory size = 2 GB = 231
Page size = 8 KB = 213
Number of entries in inverted page table = physical address space / page size = 231/213 = 218
Question 93 |
Which of the following is not a necessary condition for deadlock?
Mutual exclusion | |
Reentrancy | |
Hold and wait | |
No pre-emption |
Question 93 Explanation:
Four Necessary and Sufficient Conditions for Deadlock are:
→Mutual exclusion
The resources involved must be unshareable; otherwise, the processes would not be prevented from using the resource when necessary.
→Hold and wait or partial allocation
The processes must hold the resources they have already been allocated while waiting for other (requested) resources. If the process had to release its resources when a new resource or resources were requested, deadlock could not occur because the process would not prevent others from using resources that it controlled.
→No pre-emption
The processes must not have resources taken away while that resource is being used. Otherwise, deadlock could not occur since the operating system could simply take enough resources from running processes to enable any process to finish.
→Resource waiting or circular wait
A circular chain of processes, with each process holding resources which are currently being requested by the next process in the chain, cannot exist. If it does, the cycle theorem (which states that "a cycle in the resource graph is necessary for deadlock to occur") indicated that deadlock would occur
→Mutual exclusion
The resources involved must be unshareable; otherwise, the processes would not be prevented from using the resource when necessary.
→Hold and wait or partial allocation
The processes must hold the resources they have already been allocated while waiting for other (requested) resources. If the process had to release its resources when a new resource or resources were requested, deadlock could not occur because the process would not prevent others from using resources that it controlled.
→No pre-emption
The processes must not have resources taken away while that resource is being used. Otherwise, deadlock could not occur since the operating system could simply take enough resources from running processes to enable any process to finish.
→Resource waiting or circular wait
A circular chain of processes, with each process holding resources which are currently being requested by the next process in the chain, cannot exist. If it does, the cycle theorem (which states that "a cycle in the resource graph is necessary for deadlock to occur") indicated that deadlock would occur
Question 94 |
Consider the following process and resource requirement of each process.
Predict the state of this system, assuming that there are a total of 5 instances of resource type 1 and 4 instances of resource type 2.
Can go to safe or unsafe state based on sequence | |
Safe state | |
Unsafe state | |
Deadlock state |
Question 94 Explanation:
Question 95 |
A starvation free job scheduling policy guarantees that no job indefinitely waits for a service. Which of the following job scheduling policies is starvation free?
Priority queuing | |
Shortest Job First | |
Youngest Job First | |
Round robin |
Question 95 Explanation:
Starvation is nothing but the indefinite postponement of a process because it requires some resource before it can run, but the resource, though available for allocation,is never allocated to this process
Round Robin is a starvation free scheduling algorithm because the process will execute up to quantum value and switches to next process.
In the round robin algorithm, All processes will execute for a given time quantum value.so indefinite postponement of process is not possible.
Round Robin is a starvation free scheduling algorithm because the process will execute up to quantum value and switches to next process.
In the round robin algorithm, All processes will execute for a given time quantum value.so indefinite postponement of process is not possible.
Question 96 |
The state of a process after it encounters an I/O instruction is
ready | |
blocked | |
idle | |
running |
Question 96 Explanation:
According to process state diagram, whenever the process encounter I/O operation , the process enters into waiting or blocked state.
Once I/O operation completed, the process will enter into ready state to execute the process.
Question 97 |
A particular parallel program computation requires 100 seconds when executed on a single CPU. If 20% of this computation is strictly sequential, then theoretically the best possible elapsed times for this program running on 2 CPUs and 4 CPUs respectively are
55 and 45 seconds | |
80 and 20 seconds | |
75 and 25 seconds | |
60 and 40 seconds |
Question 97 Explanation:
→From the given data, one CPU will require 100 seconds to complete given task.
→20% of computation is done by sequential which means 20% of 100 seconds (20 seconds)
→Remaining computation 80% of work(80 seconds) can be done by the parallel processing.
→2 Processors: 20 % of sequential work is done by processor p1 (20 sec) and remaining 80 % of work (80 seconds) can be distributed among p1 and p2 equally(each process will get 40 sec). so time required = 20 + 40 = 60 seconds
→4 Processors: 20 % of work is done by any of the processor sequentially(20 seconds) and remaining 80 % of work (80 seconds) can be divided among 4 processors(each process will get 20 sec), so maximum time required = 20 + 20 = 40 seconds
→20% of computation is done by sequential which means 20% of 100 seconds (20 seconds)
→Remaining computation 80% of work(80 seconds) can be done by the parallel processing.
→2 Processors: 20 % of sequential work is done by processor p1 (20 sec) and remaining 80 % of work (80 seconds) can be distributed among p1 and p2 equally(each process will get 40 sec). so time required = 20 + 40 = 60 seconds
→4 Processors: 20 % of work is done by any of the processor sequentially(20 seconds) and remaining 80 % of work (80 seconds) can be divided among 4 processors(each process will get 20 sec), so maximum time required = 20 + 20 = 40 seconds
Question 98 |
Consider the list of page references in the timeline as below:
9 6 2 3 4 4 4 4 3 4 4 2 5 8 6 8 5 5 3 2 3 3 9 6 2 7
What is the working set at the penultimate page reference if ∆ is 5?
{8,5,3,2,9,6} | |
{4,3,6,2,5} | |
{3,9,6,2,7} | |
{3,9,6,2,7} |
Question 98 Explanation:
Penultimate means in second last working set model.
The set of pages that a process is currently using is called its working set. If the entire working set is in memory, the process will run without causing many faults until it moves into another execution phase (e.g., the next pass of the compiler).
Many paging systems try to keep track of each process' working set and make sure that it is in memory before letting the process run. This approach is called the working set model.. It is designed to greatly reduce the page fault rate.
Working sets are as below:
9 - {9}
6 - {9,6}
2 - {9,6,2}
3 - {9,6,2,3}
4 - {9,6,2,3,4}
4 - {6,2,3,4}
4 - {2,3,4}
4 - {3,4}
3 - {3,4}
4 - {3,4}
4 - {3,4}
2 - {3,4,2}
5 - {3,4,2,5}
8 - {2,4,5,8}
6 - {2,4,5,8,6}
8 - {2,5,8,6}
5 - {5,8,6}
5 - {5,8,6}
3 - {3,5,8,6}
2 - {2,3,5,8}
3 - {2,3,5}
9 - {2,3,9}
6 - {2,3,9,6}
2 - {3,9,6,2}
7 - {3,9,6,2,7}
The set of pages that a process is currently using is called its working set. If the entire working set is in memory, the process will run without causing many faults until it moves into another execution phase (e.g., the next pass of the compiler).
Many paging systems try to keep track of each process' working set and make sure that it is in memory before letting the process run. This approach is called the working set model.. It is designed to greatly reduce the page fault rate.
Working sets are as below:
9 - {9}
6 - {9,6}
2 - {9,6,2}
3 - {9,6,2,3}
4 - {9,6,2,3,4}
4 - {6,2,3,4}
4 - {2,3,4}
4 - {3,4}
3 - {3,4}
4 - {3,4}
4 - {3,4}
2 - {3,4,2}
5 - {3,4,2,5}
8 - {2,4,5,8}
6 - {2,4,5,8,6}
8 - {2,5,8,6}
5 - {5,8,6}
5 - {5,8,6}
3 - {3,5,8,6}
2 - {2,3,5,8}
3 - {2,3,5}
9 - {2,3,9}
6 - {2,3,9,6}
2 - {3,9,6,2}
7 - {3,9,6,2,7}
Question 99 |
Consider the following segment table in the segmentation scheme:
What happens if the logical address requested is Segment ID 2 and offset 1000?
What happens if the logical address requested is Segment ID 2 and offset 1000?
Fetches the entry at the physical address 2527 for segment Id2 | |
A trap is generated | |
Deadlock | |
Fetches the entry at offset 27 in Segment Id 3 |
Question 99 Explanation:
From the question we need to find the logical address for segment id-2.
From given table,
Segment-2 has a base address = 1527 ‘
limit address = 498.
Process can access memory from the location 1527 to 2025(1527+498)
If the process tries to access the memory with offset 1000 then a segmentation fault trap will be generated.
In computing and operating systems, a trap, also known as an exception or a fault, is typically a type of synchronous interrupt caused by an exceptional condition (e.g., breakpoint, division by zero, invalid memory access)
From given table,
Segment-2 has a base address = 1527 ‘
limit address = 498.
Process can access memory from the location 1527 to 2025(1527+498)
If the process tries to access the memory with offset 1000 then a segmentation fault trap will be generated.
In computing and operating systems, a trap, also known as an exception or a fault, is typically a type of synchronous interrupt caused by an exceptional condition (e.g., breakpoint, division by zero, invalid memory access)
Question 100 |
A computer has 16 pages of virtual address space but the size of main memory is only four frames. Initially the memory is empty. A program references the virtual pages in the order 0, 2, 4, 5, 2, 4, 3, 11, 2, 10. How many page faults occur if LRU page replacement algorithm is used?
3 | |
5 | |
7 | |
8 |
Question 100 Explanation:
LRU Page Replacement Algorithm:
when a page fault occurs, throw out the page that has been unused for the longest time.
when a page fault occurs, throw out the page that has been unused for the longest time.
Question 101 |
Consider a system where each file is associated with a 16-bit number. For each file, each user should have the read and write capability. How much memory is needed to store each user’s access data?
16 KB | |
32 KB | |
64 KB | |
128 KB |
Question 101 Explanation:
The user has read and write capability to the file and there are four possible combinations to file read, write,no-read,no-write operations.
We can represent that four options we require 2 bits(22=4).
So total memory(in bits) to require to access the data is 216x2=217bits
Convert the above into bytes 2421023=16KB(210=1 KB and 23=8=1Byte)
We can represent that four options we require 2 bits(22=4).
So total memory(in bits) to require to access the data is 216x2=217bits
Convert the above into bytes 2421023=16KB(210=1 KB and 23=8=1Byte)
Question 102 |
What are the minimum number of resources required to ensure that deadlock will never occur, if there are currently three processes P1, P2 and P3 running in a system whose maximum demand for the resources of the same type are 3, 4, and 5 respectively?
3 | |
7 | |
9 | |
10 |
Question 102 Explanation:
let the resources needed by P1 is R1 =3,
the number of resources needed by P2 is R2 = 4 and so on.
. Minimum resources required to ensure that deadlock will never occur = (R1-1) + (R2-1) + (R3-1) + 1
= (3-1) + (4-1)+ (5-1) + 1 = 10.
the number of resources needed by P2 is R2 = 4 and so on.
. Minimum resources required to ensure that deadlock will never occur = (R1-1) + (R2-1) + (R3-1) + 1
= (3-1) + (4-1)+ (5-1) + 1 = 10.
Question 103 |
Dirty bit is used to indicate which of the following?
A page fault has occurred | |
A page has corrupted data | |
A page has been modified after being loaded into cache | |
An illegal access of page |
Question 103 Explanation:
→ The dirty bit allows for a performance optimization i.e., Dirty bit for a page in a page table helps to avoid unnecessary writes on a paging device
→ When a page is modified inside the cache and the changes need to be stored back in the main memory, the valid bit is set to 1 so as to maintain the record of modified pages.
→ When a page is modified inside the cache and the changes need to be stored back in the main memory, the valid bit is set to 1 so as to maintain the record of modified pages.
Question 104 |
What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid and invalid bit) and a page size of 512
211 | |
215 | |
219 | |
220 |
Question 104 Explanation:
Size of Physical Address = Paging bits + Offset bits
Paging bits = 11 – 1 = 10 (As 1 valid bit is also included)
Offset bits = log2 (page size) =log2 (512) =9
Size of Physical Address = 10 + 9 = 19 bits
Paging bits = 11 – 1 = 10 (As 1 valid bit is also included)
Offset bits = log2 (page size) =log2 (512) =9
Size of Physical Address = 10 + 9 = 19 bits
Question 105 |
Which of the following is not an optimization criterion in the design of a CPU scheduling algorithm?
Minimum CPU utilization | |
Maximum throughput | |
Minimum turnaround time | |
Minimum waiting time |
Question 105 Explanation:
Goal of an operating system is “maximum utilization of CPU”.
The following are criterion for designing CPU scheduling
● CPU utilization - Ideally the CPU would be busy 100% of the time, so as to waste 0 CPU cycles. On a real system CPU usage should range from 40% ( lightly loaded ) to 90% ( heavily loaded. )
● Throughput - Number of processes completed per unit time. May range from 10 / second to 1 / hour depending on the specific processes.
● Turnaround time - Time required for a particular process to complete, from submission time to completion. ( Wall clock time. )
● Waiting time - How much time processes spend in the ready queue waiting their turn to get on the CPU.
→ In general one wants to optimize the average value of a criteria ( Maximize CPU utilization and throughput, and minimize all the others.) However sometimes one wants to do something different, such as to minimize the maximum response time.
The following are criterion for designing CPU scheduling
● CPU utilization - Ideally the CPU would be busy 100% of the time, so as to waste 0 CPU cycles. On a real system CPU usage should range from 40% ( lightly loaded ) to 90% ( heavily loaded. )
● Throughput - Number of processes completed per unit time. May range from 10 / second to 1 / hour depending on the specific processes.
● Turnaround time - Time required for a particular process to complete, from submission time to completion. ( Wall clock time. )
● Waiting time - How much time processes spend in the ready queue waiting their turn to get on the CPU.
→ In general one wants to optimize the average value of a criteria ( Maximize CPU utilization and throughput, and minimize all the others.) However sometimes one wants to do something different, such as to minimize the maximum response time.
Question 106 |
Using the page table shown below, translate the physical address 25 to virtual address. The address length is 16 bits and page size is 2048 words while the size of the physical memory is four frames.
-
-25 | |
6169 | |
2073 | |
4121 |
Question 106 Explanation:
Given data,
Virtual address length =16 bits,
Page size=2048 words
= 211 bytes
Step-1: Total number of pages = 216/211
= 25
Step-2: The physical address is nothing but [number of frames * size of each frame] Physical address= 4*211
= 213
Step-3: Given physical address (25)10 = (0000000011001)2 in 13 bits
The 13 bits address, we are representing into
Virtual address length =16 bits,
Page size=2048 words
= 211 bytes
Step-1: Total number of pages = 216/211
= 25
Step-2: The physical address is nothing but [number of frames * size of each frame] Physical address= 4*211
= 213
Step-3: Given physical address (25)10 = (0000000011001)2 in 13 bits
The 13 bits address, we are representing into
Question 107 |
At a particular time of computation, the value of a counting semaphore is 10. Then 12 P operations and “x” V operations were performed on this semaphore. If the final value of semaphore is 7, x will be :
8 | |
9 | |
10 | |
11 |
Question 107 Explanation:
Initial value of counting semaphore is 10. And when a process enters the critical section it decreases the value of counting semaphore using P operation and when a process leaves the critical section it increases the value of counting semaphore using V operation.
So, here 12P (subtraction) and x(addition) operation are given. And after performing P(subtraction) and V(addition) the value of counting semaphore is 7.
7 = 10 -12 + x
x = 9
7 = 10 -12 + x
x = 9
Question 108 |
In a paged memory, the page hit ratio is 0.40. The time required to access a page in secondary memory is equal to 120 ns. The time required to access a page in primary memory is 15 ns. The average time required to access a page is
105 | |
68 | |
75 | |
78 |
Question 108 Explanation:
Average time to access a page = page hit ratio(time required to access a page in primary memory) + page miss ratio(time required to access a page in primary memory + time required to access a page in secondary memory)
Average time to access a page = 0.40(15) + 0.60(120)
Average time to access a page = 6 + 72
Average time to access a page = 78
Average time to access a page = 0.40(15) + 0.60(120)
Average time to access a page = 6 + 72
Average time to access a page = 78
Question 109 |
In a multi-user operating system, 30 requests are made to use a particular resource per hour, on an average. The probability that no requests are made in 40 minutes, when arrival pattern is a poisson distribution, is
e-15
| |
1 - e-15
| |
1 - e-20
| |
e-20
|
Question 109 Explanation:
→ In probability theory, a Poisson process is a stochastic process that counts the number of events and the time points at which these events occur in a given time interval.
→ The time between each pair of consecutive events has an exponential distribution with parameter λ and each of these inter-arrival times is assumed to be independent of other inter-arrival times.
→ λ = (40*30)/60
= 20
P(T > 40min) = 1 - P(T ≤ 40 min)
= 1 - (1 - e-40min/λ)
= 1 - (1 - e-40min/20)
= e-20
→ The time between each pair of consecutive events has an exponential distribution with parameter λ and each of these inter-arrival times is assumed to be independent of other inter-arrival times.
→ λ = (40*30)/60
= 20
P(T > 40min) = 1 - P(T ≤ 40 min)
= 1 - (1 - e-40min/λ)
= 1 - (1 - e-40min/20)
= e-20
Question 110 |
Normally user programs are prevented from handling I/O directly by I/O instructions in them. For CPUs having explicit I/O instructions, such I/O protection is ensured by having the I/O instructions privileged. In a CPU with memory mapped I/O, there is no explicit I/O instruction. Which one of the following is true for a CPU with memory mapped I/O ?
I/O protection is ensured by operating system routines. | |
I/O protection is ensured by a hardware trap. | |
I/O protection is ensured during system configuration.
| |
I/O protection is not possible. |
Question 110 Explanation:
→ I/O protection can be ensured by operating system. Because all the user application are not modified by user mode. Those are sent to kernel mode as a system calls.
→ Normally user programs are prevented from handling I/O directly by I/O instructions in them , For CPUs having explicit I/O instructions, such I/O protection is ensured by having the I/O instruction privileged.
→ Normally user programs are prevented from handling I/O directly by I/O instructions in them , For CPUs having explicit I/O instructions, such I/O protection is ensured by having the I/O instruction privileged.
Question 111 |
Which UNIX/Linux command is used to make all files and sub-directories in the directory “progs” executable by all users ?
chmod−R a+x progs | |
chmod−R 222 progs | |
chmod−X a+x progs
| |
chmod−X 222 progs |
Question 111 Explanation:
Here, observe 2 key points is
1. All files and their subdirectories
2. Execute by all users
Step-1: Normally any file consists of 3 categories.
1. Owner
2. Group
3. Others
Step-2: Every category is represented in 3 accesses.
1. Read(Octal value 4)
2. Write(Octal value 2)
3. Execute(Octal value 1)
Step-3: To change user permission, we are using command is chmod.
Chmod syntax: chmod permissions filename
1. All files and their subdirectories (using recursive command ‘R’)
2. Execute by all users(using ‘a+x’ subcommand. ‘a’ is nothing but all and ‘+x’ is execute permissions)
Actual command required: chmod− R a+x progs
1. All files and their subdirectories
2. Execute by all users
Step-1: Normally any file consists of 3 categories.
1. Owner
2. Group
3. Others
Step-2: Every category is represented in 3 accesses.
1. Read(Octal value 4)
2. Write(Octal value 2)
3. Execute(Octal value 1)
Step-3: To change user permission, we are using command is chmod.
Chmod syntax: chmod permissions filename
1. All files and their subdirectories (using recursive command ‘R’)
2. Execute by all users(using ‘a+x’ subcommand. ‘a’ is nothing but all and ‘+x’ is execute permissions)
Actual command required: chmod− R a+x progs
Question 112 |
Which of the following statements are true ?
- (a) External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is contiguous.
(b) Memory Fragmentation can be internal as well as external.
(c) One solution to external Fragmentation is compaction.
(a) and (b) only | |
(a) and (c) only | |
(b) and (c) only | |
(a), (b) and (c) |
Question 112 Explanation:
External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is not contiguous.
Yes, it is true that memory Fragmentation can be internal as well as external.
Yes, compaction is a solution to external Fragmentation.
Yes, it is true that memory Fragmentation can be internal as well as external.
Yes, compaction is a solution to external Fragmentation.
Question 113 |
Page information in memory is also called as Page Table. The essential contents in each entry of a page table is/are
Page Access information
| |
Virtual Page number
| |
Page Frame number
| |
Both virtual page number and Page Frame Number |
Question 113 Explanation:
→ For every page table it contains page frame number.
→ Virtual page number can represents index in the page table to get the page frame number.
→ Virtual page number can represents index in the page table to get the page frame number.
Question 114 |
Consider a virtual page reference string 1, 2, 3, 2, 4, 2, 5, 2, 3, 4. Suppose LRU page replacement algorithm is implemented with 3 page frames in main memory. Then the number of page faults are
5 | |
7 | |
9 | |
10 |
Question 114 Explanation:
So, total number of page faults are 7.
Question 115 |
Consider the following three processes with the arrival time and CPU burst time given in milliseconds :
Process Arrival Time Burst Time P1 0 7 P2 1 4 P3 2 8
The Gantt Chart for preemptive SJF scheduling algorithm is
 | |
 | |
 | |
 |
Question 115 Explanation:
Question 116 |
In which of the following scheduling criteria, context switching will never take place ?
ROUND ROBIN
| |
Preemptive SJF | |
Non-preemptive SJF | |
Preemptive priority
|
Question 116 Explanation:
ROUND ROBIN :
In this because of time quantum context switching will take place.
Preemptive SJF :
In this context switching will take place if a smaller burst time process arrives in ready queue.
Non-preemptive SJF :
Since it is non-preemptive SJF scheduling so a process will leave CPU only when it is completely executed.
Preemptive priority :
In this Context switching will take place if a process with higher priority enters the ready queue.
In this because of time quantum context switching will take place.
Preemptive SJF :
In this context switching will take place if a smaller burst time process arrives in ready queue.
Non-preemptive SJF :
Since it is non-preemptive SJF scheduling so a process will leave CPU only when it is completely executed.
Preemptive priority :
In this Context switching will take place if a process with higher priority enters the ready queue.
Question 117 |
Among all given option, ___ must reside in the main memory.
Assembler | |
Compiler | |
Linker | |
Loader |
Question 117 Explanation:
→ A loader is the part of an operating system that is responsible for loading programs and libraries.
→ It is one of the essential stages in the process of starting a program, as it places programs into memory and prepares them for execution.
→ Loading a program involves reading the contents of the executable file containing the program instructions into memory, and then carrying out other required preparatory tasks to prepare the executable for running.
→ Once loading is complete, the operating system starts the program by passing control to the loaded program code.
Note: The loader is a program that loads the object program from the secondary memory into the main memory for execution of the program
→ It is one of the essential stages in the process of starting a program, as it places programs into memory and prepares them for execution.
→ Loading a program involves reading the contents of the executable file containing the program instructions into memory, and then carrying out other required preparatory tasks to prepare the executable for running.
→ Once loading is complete, the operating system starts the program by passing control to the loaded program code.
Note: The loader is a program that loads the object program from the secondary memory into the main memory for execution of the program
Question 118 |
The process executes the following code and after execution ____ number of child process get created.
fork();
fork();
fork();
fork();
4 | |
1 | |
15 | |
16 |
Question 118 Explanation:
Step-1: The number of child processes of “n” fork() system calls are 2n-1.
Step-2: Total 4 fork() system calls. It means n=4
= 2n-1
= 24-1
= 15
Step-2: Total 4 fork() system calls. It means n=4
= 2n-1
= 24-1
= 15
Question 119 |
Find the effective access time for the memory for given data.
Page fault service time=8ms
Average memory access time=20ms
One page fault is generated for every memory access=106
29ns | |
33ns | |
28ns | |
30ns |
Question 119 Explanation:
Given data,
Page fault service time=8ms
Average memory access time=20ms
One page fault is generated for every memory access=106
P = page fault rate
Effective Memory Access = p × page fault service time + (1 – p) × Memory access time
=(1/106)*8*106+(1-1/106)*20
≅27.99998 ns
Page fault service time=8ms
Average memory access time=20ms
One page fault is generated for every memory access=106
P = page fault rate
Effective Memory Access = p × page fault service time + (1 – p) × Memory access time
=(1/106)*8*106+(1-1/106)*20
≅27.99998 ns
Question 120 |
Identify the true statement from the given statements
(1) FIFO is non preemptive
(2) Round robin is non preemptive
(3) Multilevel queue scheduling is non preemptive
(1) FIFO is non preemptive
(2) Round robin is non preemptive
(3) Multilevel queue scheduling is non preemptive
(1) | |
(1) and (2) | |
(1),(2) and (3) | |
(2) |
Question 120 Explanation:
Nonpreemptive Scheduling
A scheduling discipline is nonpreemptive if, once a process has been given the CPU, the CPU cannot be taken away from that process.
Eg: FIFO→ Non preemptive
Preemptive Scheduling
A scheduling discipline is preemptive if, once a process has been given the CPU can taken away.
Eg: Round robin is non preemptive
Multilevel queue scheduling is non preemptive
Priority,SRTF,Multilevel feedback scheduling is non preemptive
A scheduling discipline is nonpreemptive if, once a process has been given the CPU, the CPU cannot be taken away from that process.
Eg: FIFO→ Non preemptive
Preemptive Scheduling
A scheduling discipline is preemptive if, once a process has been given the CPU can taken away.
Eg: Round robin is non preemptive
Multilevel queue scheduling is non preemptive
Priority,SRTF,Multilevel feedback scheduling is non preemptive
Question 121 |
In the disk, swap space is used to _____
Save XML files | |
Save process data | |
Save drivers | |
Save HTML files |
Question 121 Explanation:
→ A swap file (or swap space or a pagefile) is a space on a hard disk used as the virtual memory extension of a computer's real memory (RAM).
→ Having a swap file allows your computer's operating system to pretend that you have more RAM than you actually do.
→ Having a swap file allows your computer's operating system to pretend that you have more RAM than you actually do.
Question 122 |
In a system, counting semaphore was initialized to 10. Then 6P(wait) operations and 4V (signal) operations were completed on this semaphore. So ___ is the final value of the semaphore.
7 | |
8 | |
13 | |
12 |
Question 122 Explanation:
Given data,
Counting semaphore(S)=10,
Wait operations(P)=6,
Signal operations(V)=4,
→ P(wait) operations means decrementing the value and V(signal) operations means incrementing the Semaphore values.
→ After completion of 6P operations and 4V operations the final value of S will be
S=10-6+4
= 8
Counting semaphore(S)=10,
Wait operations(P)=6,
Signal operations(V)=4,
→ P(wait) operations means decrementing the value and V(signal) operations means incrementing the Semaphore values.
→ After completion of 6P operations and 4V operations the final value of S will be
S=10-6+4
= 8
Question 123 |
Decreasing the RAM causes ______
Fewer page faults | |
More page faults | |
Virtual memory get increases | |
Virtual memory get decreases |
Question 123 Explanation:
A page fault is a type of exception raised by computer hardware when a running program accesses a memory page that is not currently mapped by the memory management unit (MMU) into the virtual address space of a process.
When handling a page fault, the operating system generally tries to make the required page accessible at the location in physical memory, or terminates the program in case of an illegal memory access.
So, Decrease the physical RAM on your machine could result in more page faults
When handling a page fault, the operating system generally tries to make the required page accessible at the location in physical memory, or terminates the program in case of an illegal memory access.
So, Decrease the physical RAM on your machine could result in more page faults
Question 124 |
The maximum combined length of the command-line arguments including the spaces between adjacent arguments is:
128 characters | |
256 characters | |
67 characters | |
It may very from one OS to another |
Question 124 Explanation:
Question 125 |
Which of the following is not an input device?
Mouse | |
Keyboard | |
Light Pen | |
VDU |
Question 125 Explanation:
→ Mouse, Keyboard and Light Pen are input devices.
→ Computer monitors were formerly known as visual display units ( VDU ), but this term had mostly fallen out of use by the 1990s.
→ Computer monitors were formerly known as visual display units ( VDU ), but this term had mostly fallen out of use by the 1990s.
Question 126 |
The Banker's algorithm is used:
to rectify deadlock | |
to prevent deadlock | |
to detect deadlock | |
to detect and solve deadlock |
Question 126 Explanation:
● The Banker algorithm, sometimes referred to as the detection algorithm, is a resource allocation and deadlock avoidance algorithm
● This algorithm tests for safety by simulating the allocation of predetermined maximum possible amounts of all resources, and then makes an "s-state" check to test for possible deadlock conditions for all other pending activities, before deciding whether allocation should be allowed to continue.
Question 127 |
Which of the following statements about semaphores is true?
P and V operations should be indivisible operations | |
A semaphore implementation should guarantee that threads do not suffer indefinite postponement | |
If several threads attempt a P(S) operation simultaneously, only one thread should be allowed to proceed. | |
All of the above |
Question 127 Explanation:
● A semaphore is a variable or abstract data type used to control access to a common resource by multiple processes in a concurrent system such as a multitasking operating system
● Semaphores are a useful tool in the prevention of race conditions; however, their use is by no means a guarantee that a program is free from these problems.
● Semaphores which allow an arbitrary resource count are called counting semaphores, while semaphores which are restricted to the values 0 and 1 (or locked/unlocked, unavailable/available) are called binary semaphores and are used to implement locks.
● Semaphores are a useful tool in the prevention of race conditions; however, their use is by no means a guarantee that a program is free from these problems.
● Semaphores which allow an arbitrary resource count are called counting semaphores, while semaphores which are restricted to the values 0 and 1 (or locked/unlocked, unavailable/available) are called binary semaphores and are used to implement locks.
Question 128 |
The file manager is responsible for
naming files | |
saving files | |
deleting files | |
all the above |
Question 128 Explanation:
● A file manager is a software program that helps a user manage all the files on their computer. For example, all file managers allow the user to view, edit, copy, and delete the files on their computer storage devices.
● Note: Although a file manager helps the user view and manage their files, it is the operating system that is responsible for accessing and storing the files on a storage device.
● Note: Although a file manager helps the user view and manage their files, it is the operating system that is responsible for accessing and storing the files on a storage device.
Question 129 |
In real time operating systems, which of the following is the most suitable scheduling scheme?
Round Robin | |
First come first serve | |
preemptive | |
random scheduling |
Question 129 Explanation:
● A real-time operating system (RTOS) is any operating system (OS) intended to serve real-time applications that process data as it comes in, typically without buffer delays.
● 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.
● 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 130 |
If there are 32 segments, each of size 1 K byte, then the logical address should have
13 bits | |
14 bits | |
15 bits | |
16 bits |
Question 130 Explanation:
There are 32 segments which is equal to 25
Each segment size 1K byte =2 10
Then total number of bits that logical address consists is 15 bits.
Each segment size 1K byte =2 10
Then total number of bits that logical address consists is 15 bits.
Question 131 |
Relocation bits used by relocating loader are specified by
relocating loader itself | |
Linker | |
Assembler | |
Macro Processor |
Question 131 Explanation:
● A linker or link editor is a computer utility program that takes one or more object files generated by a compiler and combines them into a single executable file, library file, or another 'object' file.
● Relocating loader in which some of the addresses in the program to be loaded are expressed relative to the start of the program rather than in absolute form.
● An assembler is a type of computer program that interprets software programs written in assembly language into machine language, code and instructions that can be executed by a computer.
● A macro processor is a program that copies a stream of text from one place to another, making a systematic set of replacements as it does so. Macro processors are often embedded in other programs, such as assemblers and compilers. Sometimes they are standalone programs that can be used to process any kind of text.
● Relocating loader in which some of the addresses in the program to be loaded are expressed relative to the start of the program rather than in absolute form.
● An assembler is a type of computer program that interprets software programs written in assembly language into machine language, code and instructions that can be executed by a computer.
● A macro processor is a program that copies a stream of text from one place to another, making a systematic set of replacements as it does so. Macro processors are often embedded in other programs, such as assemblers and compilers. Sometimes they are standalone programs that can be used to process any kind of text.
Question 132 |
The most powerful parser is
SLR | |
LALR | |
Canonical LR | |
Operator Precedence |
Question 132 Explanation:
Canonical LR is more powerful than SLR as every grammar which can be parsed by SLR parser, can also be parsed by CLR parser.
So CLR > LALR > SLR
In computer science, a canonical LR parser or LR(1) parser is an LR(k) parser for k=1, i.e. with a single lookahead terminal. The special attribute of this parser is that any LR(k) grammar with k>1 can be transformed into an LR(1) grammar.However, back-substitutions are required to reduce k and as back-substitutions increase, the grammar can quickly become large, repetitive and hard to understand. LR(k) can handle all deterministic context-free languages.
So CLR > LALR > SLR
In computer science, a canonical LR parser or LR(1) parser is an LR(k) parser for k=1, i.e. with a single lookahead terminal. The special attribute of this parser is that any LR(k) grammar with k>1 can be transformed into an LR(1) grammar.However, back-substitutions are required to reduce k and as back-substitutions increase, the grammar can quickly become large, repetitive and hard to understand. LR(k) can handle all deterministic context-free languages.
Question 133 |
Serial access memories are useful in applications where
data consists of numbers | |
short access time is required | |
each stored word is processed differently | |
data naturally needs to flow in and out in serial form |
Question 133 Explanation:
In serial-access media the access time depends on the data’s location and the position of the read-write head. The typical serial-access medium is magnetic tape.
Question 134 |
Suppose P, Q and R are co-operating processes satisfying Mutual Exclusion condition. Then if the process Q is executing in its critical section then
‘P’ executes in critical section
| |
‘R’ executes in critical section | |
Neither ‘P’ nor ‘Q’ executes in their critical section | |
Both ‘P’ and ‘R’ executes in critical section |
Question 134 Explanation:
• A mutual exclusion (mutex) is a program object that prevents simultaneous access to a shared resource.
• This concept is used in concurrent programming with a critical section, a piece of code in which processes or threads access a shared resource.
• Only one thread owns the mutex at a time, thus a mutex with a unique name is created when a program starts.
• When a thread holds a resource, it has to lock the mutex from other threads to prevent concurrent access of the resource.
• In the question, three processes are cooperating processes and satisfying mutual exclusion condition. If process Q is executing in its critical section means remaining two processes in wait stating and they won’t enter into critical section.
• This concept is used in concurrent programming with a critical section, a piece of code in which processes or threads access a shared resource.
• Only one thread owns the mutex at a time, thus a mutex with a unique name is created when a program starts.
• When a thread holds a resource, it has to lock the mutex from other threads to prevent concurrent access of the resource.
• In the question, three processes are cooperating processes and satisfying mutual exclusion condition. If process Q is executing in its critical section means remaining two processes in wait stating and they won’t enter into critical section.
Question 135 |
Consider the following set of processes and the length of CPU burst time given in milliseconds:
Assume that processes being scheduled with Round-Robin Scheduling Algorithm with Time Quantum 4ms. Then The waiting time for P4 is _________ ms.
0 | |
4 | |
12 | |
6 |
Question 135 Explanation:
Waiting time = Turnaround time - Burst time
Turnaround time
= Completion time - Arrival time
Since in question Arrival time is not given so consider it as “0” for all the processes.
= Completion time - Arrival time
Since in question Arrival time is not given so consider it as “0” for all the processes.
Question 136 |
How many wires are threaded through the cores in a coincided-current core memory?
2 | |
3 | |
4 | |
6 |
Question 136 Explanation:
The most common form of core memory, X/Y line coincident-current, used for the main memory of a computer, consists of a large number of small toroidal ferrimagnetic ceramic ferrites (cores) held together in a grid structure (organized as a "stack" of layers called planes), with wires woven through the holes in the cores' centers.
In early systems there were four wires: X, Y, Sense, and Inhibit, but later cores combined the latter two wires into one Sense/Inhibit line. Each toroid stored one bit (0 or 1).
In early systems there were four wires: X, Y, Sense, and Inhibit, but later cores combined the latter two wires into one Sense/Inhibit line. Each toroid stored one bit (0 or 1).
Question 137 |
Which access method is used for obtaining a record from cassette tape?
Direct | |
Sequential | |
Random | |
Parallel |
Question 137 Explanation:
Question 138 |
A CPU generates 32 bit virtual addresses. The page size is 4KB. The processor has a Translation Lookaside Buffer(TLB) which can hold a total of 128 page table entries and is 4-way set associative. The minimum size of the TLB tag is
11 bits | |
13 bits | |
15 bits | |
20 bits |
Question 138 Explanation:
Page size = 4 KB = 4 × 210 Bytes = 212 Bytes
Virtual Address = 32 bit
No. of bits needed to address the page frame = 32 - 12 = 20
TLB can hold 128 page table entries with 4-way set associative
⇒ 128/4=32=25
→ 5 bits are needed to address a set.
→ The size of TLB tag = 20 - 5 = 15 bits
Virtual Address = 32 bit
No. of bits needed to address the page frame = 32 - 12 = 20
TLB can hold 128 page table entries with 4-way set associative
⇒ 128/4=32=25
→ 5 bits are needed to address a set.
→ The size of TLB tag = 20 - 5 = 15 bits
Question 139 |
Computer uses 46-bit virtual address, 32 bit physical address, and a three level paged page table organization. The page table base register stores the base address of the first level table(T1), which occupies exactly one page. Each entry of T1 stores the base address of a page of the second level table (T2). Each entry of T2 stores the base address of a page of the third level table(T3). Each entry of T3 stores a page table entry(PTE). The PTE is 32 bits in size, The processor used in the computer has a 1MB 16 way set associative virtually indexed physically tagged cache. the cache block size is 64 bytes.
What is the size of a page in KB in this computer?
2 | |
4 | |
8 | |
16 |
Question 139 Explanation:
Architecture of physically indexed cache:
VIPT cache and aliasing effect and synonym.
Alias: Same physical address can be mapped to multiple virtual addresses.
Synonym: Different virtual addresses mapped to same physical address (for data sharing).
So these synonyms should be in same set to avoid write-update problems.
In our problem VA = 46bits
We are using 16 bits for indexing into cache.
To have two synonym is same set we need to have same 16 bits index for PA & VA.
Assume that physical pages are colored and each set should have pages of same color so that any synonyms are in same set.
Since page size = 8KB ⇒ 13 bits
These 13bits are not translated during VA→PA. So 13 bits are same out of 16 Index bits, 13 are same we need to make 3 bits (16-13) same now.
3 bits can produce, 23 = 8 combinations which can be mapped on the different sets, so we need 8 different colors to color our pages. >br> In physically indexed cache indexing is done via physical address bits, but in virtual indexed cache, cache is indexed from (offset + set) bits. In physical Index cache indexing is done one to one (1 index maps to one page in one block of cache). In VIPT we have more/ extra bits, so mapping is not one-one. Hence these extra bits have to be taken care, such that if two virtual address refers to same page in cache block of different sets then they have to be assumed same i.e., we say of same color and put same color page in one set to avoid write update problems.
VIPT cache and aliasing effect and synonym.
Alias: Same physical address can be mapped to multiple virtual addresses.
Synonym: Different virtual addresses mapped to same physical address (for data sharing).
So these synonyms should be in same set to avoid write-update problems.
In our problem VA = 46bits
We are using 16 bits for indexing into cache.
To have two synonym is same set we need to have same 16 bits index for PA & VA.
Assume that physical pages are colored and each set should have pages of same color so that any synonyms are in same set.
Since page size = 8KB ⇒ 13 bits
These 13bits are not translated during VA→PA. So 13 bits are same out of 16 Index bits, 13 are same we need to make 3 bits (16-13) same now.
3 bits can produce, 23 = 8 combinations which can be mapped on the different sets, so we need 8 different colors to color our pages. >br> In physically indexed cache indexing is done via physical address bits, but in virtual indexed cache, cache is indexed from (offset + set) bits. In physical Index cache indexing is done one to one (1 index maps to one page in one block of cache). In VIPT we have more/ extra bits, so mapping is not one-one. Hence these extra bits have to be taken care, such that if two virtual address refers to same page in cache block of different sets then they have to be assumed same i.e., we say of same color and put same color page in one set to avoid write update problems.
Question 140 |
Consider the following snapshot of a system running n processes. Process i is holding Xi instances of a resource R, 1<=i<=n. Currently, all instances of R are occupied. Further, for all i, process i has placed a request for an additional Yi instances while holding the Xi instances it already p and q such that Yp=Yp=0. which one of the following can serve as a necessary condition to guarantee that the system is not approaching a deadlock?
min(Xp,Xq) | |
Xp+Xq>=min(Yi)where i!=p and i!=q | |
max(Xp,Xq) >1 | |
min(Xp,Xq)>1 |
Question 140 Explanation:
Deadlock refers stops the execution of process due to non-availability of resources.
→ When two (or) more processes waiting for another process to release the resources.
→ P and Q can execute if they have sufficient resources, they don’t wait for extra resources (i.e., Xp+ Xq) required.
→ Option B can satisfies the corresponding equation i.e., Xp+ Xq >= min(Yk) where k != p and k != q.
Here we have sufficient resources.
→ When two (or) more processes waiting for another process to release the resources.
→ P and Q can execute if they have sufficient resources, they don’t wait for extra resources (i.e., Xp+ Xq) required.
→ Option B can satisfies the corresponding equation i.e., Xp+ Xq >= min(Yk) where k != p and k != q.
Here we have sufficient resources.
Question 141 |
Which of the following conditions must be met to avoid race around problem?
△t < t p < T | |
T > △t > t p | |
2t p < △t < T | |
none of these |
Question 141 Explanation:
When we are using T > △t > tp this condition, we will avoid race around condition.
Question 142 |
A system has n resources R0,..,Rn-1 and k processes P0,..Pk-1. The implementation of the resources request logic of each process Pi is a follows:
if(i%2==0)
{
if(i<n) request Ri
if(i+2 <n) request Ri+2
}
else
{
if(i<n) request Rn-i
if(i+2 < n) request Rn-i-2
}
In which one of the following situations is a deadlock possible?
if(i%2==0)
{
if(i<n) request Ri
if(i+2 <n) request Ri+2
}
else
{
if(i<n) request Rn-i
if(i+2 < n) request Rn-i-2
}
In which one of the following situations is a deadlock possible?
n=40, k=26 | |
n=21, k=12 | |
n=20, k=10 | |
n=41, k=19 |
Question 142 Explanation:
Consider the case where i = 10 & i = 11, n = 21 & k = 12
P10 requests R10 & R11
P11 requests R10 & R8
Hence P10 & P11 inorder in deadlock.
P10 requests R10 & R11
P11 requests R10 & R8
Hence P10 & P11 inorder in deadlock.
Question 143 |
A system contains three programs and each requires three tape units for its operation. The minimum number of tape units which the system must have such that deadlocks never arise is__
6 | |
7 | |
8 | |
9 |
Question 143 Explanation:
Three programs and each requires 3 tapes.
We will allocate minimum two tapes to each program then total tapes are 6 and each program requires one more tape to complete its operation.
So, we will allocate one tape to one program and that program will complete its operation.
After completion of operation of one program, all tapes allocated to that program are free. So no need of the extra tape to complete action.
Minimum tapes required are=(3*2 tape units) + 1 tape unit = 7
We will allocate minimum two tapes to each program then total tapes are 6 and each program requires one more tape to complete its operation.
So, we will allocate one tape to one program and that program will complete its operation.
After completion of operation of one program, all tapes allocated to that program are free. So no need of the extra tape to complete action.
Minimum tapes required are=(3*2 tape units) + 1 tape unit = 7
Question 144 |
The process of entering data into a storage location
causes variation in its address number | |
adds to the contents of the location | |
is called a readout operation | |
is destructive of previous contents |
Question 144 Explanation:
The process of entering data into a storage location is destructive of previous contents using memory management techniques. In paging we are using LRU for replacing existing data.
Question 145 |
For Providing large storage space the semiconductor based storage is not permitted now a days due to their:
Lack of sufficient resources | |
High cost per bit value | |
Lack of speed of operation | |
High cost of raw material |
Question 145 Explanation:
→ In the case of semiconductor based memory technology, we get speed but the increase in the integration of various devices the cost is high.
Question 146 |
Let the next generation computer systems of 64 bit has virtual address space is of the same size as the physical address space. In such a case, if we remove the virtual space completely in new systems then which one of the following is true in this scenario?
Efficient implementation of multi user supports is no longer possible | |
The processor cache organization can be made more efficient now | |
Hardware support for memory management is no longer needed | |
CPU scheduling can be made more efficient now |
Question 146 Explanation:
→ When designer decides to get rid of virtual memory entirely then hardware support is no longer needed.
→ Because special hardware support needed only for virtual memory.
→ Virtual memory is an integral part of a modern computer architecture. Implementations usually require hardware support, typically in the form of a memory management unit built into the CPU.
→ While not necessary, emulators and virtual machines can employ hardware support to increase performance of their virtual memory implementations.
→ Because special hardware support needed only for virtual memory.
→ Virtual memory is an integral part of a modern computer architecture. Implementations usually require hardware support, typically in the form of a memory management unit built into the CPU.
→ While not necessary, emulators and virtual machines can employ hardware support to increase performance of their virtual memory implementations.
Question 147 |
Consider a disk pack with 32 surfaces, 64 tracks and 512 sectors per pack. 256 bytes of data are stored in a bit serial manner in a sector. The number of bits required to specify a particular sector in the disk is
19 | |
20 | |
18 | |
22 |
Question 147 Explanation:
There are 32(25) surfaces, each surface have 64(26) tracks and each surface have 512(29) sectors.
So, to identify each sector uniquely,
5+6+9 = 20-bits are needed.
So, to identify each sector uniquely,
5+6+9 = 20-bits are needed.
Question 148 |
Dirty bit is used to show the
Page with low frequency occurrence
| |
Wrong page | |
Page with corrupted data | |
Page that is modified after being loaded into cache memory |
Question 148 Explanation:
The Dirty bit in a PCB(process control block) is set if the page is modified after being loading into cache memory.
This bit helps in maintaining updated data in the hard disk.
This bit helps in maintaining updated data in the hard disk.
Question 149 |
A linker is given object module for a set of programs that were compiled separately. What information need not be included in an object module?
Object code | |
relocation bits | |
Names and locations of all external symbols defined in the object module | |
Absolute addresses of internal symbols |
Question 149 Explanation:
● In object module it includes names and locations of all external symbols defined in the object module.
● To link to external symbols it must know the location of external symbols.
● So object module won’t consists of absolute addresses of internal symbols , Option D is correct.
● To link to external symbols it must know the location of external symbols.
● So object module won’t consists of absolute addresses of internal symbols , Option D is correct.
Question 150 |
The file structure that redefines its first record at a base of zero uses the term
relative organization | |
key fielding | |
dynamic reallocation | |
all of these |
Question 150 Explanation:
● A relative record file contains records ordered by their relative key, a record number that represents the location of the record relative to where the file begins.
● For example, the first record in a file has a relative record number of 1, the tenth record has a relative record number of 10, and so forth. The records can have fixed length or variable length.
● For example, the first record in a file has a relative record number of 1, the tenth record has a relative record number of 10, and so forth. The records can have fixed length or variable length.
Question 151 |
A long-term monitor
Should show any immediate performance problems | |
should show I/O, paging and processor activity | |
need show only the I/O and processor activity | |
usually reports only on terminal displays |
Question 151 Explanation:
● Long term monitor metrics, as the name indicates, are the metrics that are collected for a resource for a longer term monitoring perspective. They are useful for continuous ongoing monitoring of the managed system.
● Monitoring the performance of operating systems and processes is essential to debug processes and systems, effectively manage system resources, making system decisions, and evaluating and examining systems.
● Monitoring the performance of operating systems and processes is essential to debug processes and systems, effectively manage system resources, making system decisions, and evaluating and examining systems.
Question 152 |
Determine the number of page faults when references to pages occur in the following order:1,2,4,5,2,1,2,4. Assume that the main memory can accommodate 3 pages and the main memory already has the pages 1 and 2, with page 1 having been brought earlier than page 2.(LRU algorithm is used)
3 | |
5 | |
4 | |
none of these |
Question 152 Explanation:
Here, total 6 page faults but in question, they are clearly mentioned that the main memory already has the pages 1 and 2, with page one having brought earlier than page 2. It means 6-2=4.
Question 153 |
Working set(t,k) at an instant of time,t,is
The set of k future references that the operating system will make | |
The set of future references that the operating system will make in the next 'k' time units | |
The set of k references with high frequency | |
The set of pages that have been referenced in the last k time units |
Question 153 Explanation:
Working set defines the amount of memory that a process requires in a given time interval.
→ It defines “the working set of information W(t,τ) of a process at time t to be the collection of information referenced by the process during the process time interval (t−τ,t)”.
→ Typically the units of information in question are considered to be memory pages. This is suggested to be an approximation of the set of pages that the process will access in the future (say during the next τ time units), and more specifically is suggested to be an indication of what pages ought to be kept in main memory to allow most progress to be made in the execution of that process.
→ It defines “the working set of information W(t,τ) of a process at time t to be the collection of information referenced by the process during the process time interval (t−τ,t)”.
→ Typically the units of information in question are considered to be memory pages. This is suggested to be an approximation of the set of pages that the process will access in the future (say during the next τ time units), and more specifically is suggested to be an indication of what pages ought to be kept in main memory to allow most progress to be made in the execution of that process.
Question 154 |
The shell
Accepts command from the user | |
maintains directories of files | |
translates the keyboard character codes | |
none of these |
Question 154 Explanation:
● A Unix shell is a command-line interpreter or shell that provides a command line user interface for Unix-like operating systems.
● The shell is both an interactive command language and a scripting language, and is used by the operating system to control the execution of the system using shell scripts.
● The shell is both an interactive command language and a scripting language, and is used by the operating system to control the execution of the system using shell scripts.
Question 155 |
Software that measures, monitors, analyzes and controls real world events is called
System Software | |
Real time software | |
Scientific Software | |
Business Software |
Question 155 Explanation:
● System software is a type of computer program that is designed to run a computer’s hardware and application programs.
● Real-time software enables the user to execute various task and activities all at the same time, as long as the programs are kept open.Real-time software refers to programs that can perform their assigned duties and functions right when they are assigned to run their given processes, not after the program has been executed, and not on a scheduled date
● Scientific software is defined by three characteristics: (1) it is developed to answer a scientific question; (2) it relies on the close involvement of an expert in its scientific domain; and (3) it provides data to be examined by the person who will answer that question.
● Business software or a business application is any software or set of computer programs used by business users to perform various business functions. These business applications are used to increase productivity, to measure productivity and to perform other business functions accurately.
● Real-time software enables the user to execute various task and activities all at the same time, as long as the programs are kept open.Real-time software refers to programs that can perform their assigned duties and functions right when they are assigned to run their given processes, not after the program has been executed, and not on a scheduled date
● Scientific software is defined by three characteristics: (1) it is developed to answer a scientific question; (2) it relies on the close involvement of an expert in its scientific domain; and (3) it provides data to be examined by the person who will answer that question.
● Business software or a business application is any software or set of computer programs used by business users to perform various business functions. These business applications are used to increase productivity, to measure productivity and to perform other business functions accurately.
Question 156 |
When a computer is first turned on or restart, a special type of absolute loader is executed called
"Compile and GO" loader | |
Boot loader | |
Bootstrap loader | |
relating Loader |
Question 156 Explanation:
● A boot loader is a type of program that loads and starts the boot time tasks and processes of an operating system or the computer system. It enables loading the operating system with in the computer memory when a computer is started or booted up.
● A boot loader is also known as a boot manager or bootstrap loader. Bootstrap loader is a program that resides in the computer's EPROM, ROM, or other non-volatile memory. The bootstrap loader reads the hard drives boot sector to continue the process of loading the computer's operating system. The term bootstrap comes from the old phrase "Pull yourself up by your bootstraps."
● A boot loader is also known as a boot manager or bootstrap loader. Bootstrap loader is a program that resides in the computer's EPROM, ROM, or other non-volatile memory. The bootstrap loader reads the hard drives boot sector to continue the process of loading the computer's operating system. The term bootstrap comes from the old phrase "Pull yourself up by your bootstraps."
Question 157 |
Which of the following is the internal memory of the system(computer)?
CPU register | |
Cache | |
Main memory | |
All of these |
Question 157 Explanation:
Types of Internal Memory:
● RAM
● ROM
● RAM
● ROM
Question 158 |
Suppose for a process P, reference to pages in order are 1, 2, 4, 5, 2, 1, 2, 4. Assume that main memory can accommodate 3 pages and the main memory has already pages 1 and 2 in the order 1 - first, 2- second. At this moment, assume FIFO Page Replacement Algorithm is used then the number of page faults that occur to complete the execution of process P is
6 | |
4 | |
3 | |
5 |
Question 158 Explanation:
In this problem, they are mentioned that “main memory has already pages 1 and 2 in the order 1 - first, 2- second”. So, total page faults are 7-already 2 available.
= 7 - 2
= 5
Question 159 |
______________ system call creates new process in Unix.
fork | |
fork new | |
create | |
create new
|
Question 159 Explanation:
fork system call is used to creates a new process in Unix.
Question 160 |
A process residing in main memory and ready and waiting for execution, is kept on
Job Queue | |
Execution Queue | |
Wait Queue | |
Ready Queue
|
Question 160 Explanation:
A process residing in main memory and ready and waiting for execution, is kept on Ready queue.
Question 161 |
In which one of the following pages replacement policies, Belady's anomaly may occur?
FIFO | |
LRU | |
Optimal | |
MRU |
Question 161 Explanation:
● Bélády's anomaly is the phenomenon in which increasing the number of page frames results in an increase in the number of page faults for certain memory access patterns.
● This phenomenon is commonly experienced when using the first-in first-out (FIFO) page replacement algorithm.
● In FIFO, the page fault may or may not increase as the page frames increase, but in Optimal and stack-based algorithms like LRU, as the page frames increase the page fault decreases.
● This phenomenon is commonly experienced when using the first-in first-out (FIFO) page replacement algorithm.
● In FIFO, the page fault may or may not increase as the page frames increase, but in Optimal and stack-based algorithms like LRU, as the page frames increase the page fault decreases.
Question 162 |
A process that is based on IPC mechanism which executes on different systems and can communicate with other processes using message based communication is called__
Local Procedure call | |
remote procedure call | |
Inter Process Communication | |
Remote Machine Invocation |
Question 162 Explanation:
● Remote Procedure Call (RPC) is a protocol that one program can use to request a service from a program located in another computer on a network without having to understand the network's details.
● A procedure call is also sometimes known as a function call or a subroutine call.
● inter-process communication or interprocess communication (IPC) refers specifically to the mechanisms an operating system provides to allow the processes to manage shared data.
● The RMI (Remote Method Invocation) is an API that provides a mechanism to create distributed application in java. The RMI allows an object to invoke methods on an object running in another JVM.
● A procedure call is also sometimes known as a function call or a subroutine call.
● inter-process communication or interprocess communication (IPC) refers specifically to the mechanisms an operating system provides to allow the processes to manage shared data.
● The RMI (Remote Method Invocation) is an API that provides a mechanism to create distributed application in java. The RMI allows an object to invoke methods on an object running in another JVM.
Question 163 |
consider a system with m resources of same type being shared by n processes. Resources can be requested and release by processes only one at a time. The system is deadlock free if and only if:
The sum of all max needs is | |
the sum of all max needs is >m+n | |
Both of above | |
None |
Question 163 Explanation:
Suppose N = Sum of all Needi,
A = Sum of all Allocation,
M = Sum of all Maxi.
Prove:
→ Assume this system is not deadlock free. If there exists a deadlock state, then A = m because there's only one kind of resource and resources can be requested and released only one at a time.
→ From condition b, N + A = M < m + n. So we get N + m < m + n. So we get N < n. It shows that at least one process i that Needi = 0. From condition a, Pi can release at least 1 resource.
→ So there are n-1 processes sharing m resources now, condition a and b still hold. Go on the argument, no process will wait permanently, so there's no deadlock.
A = Sum of all Allocation,
M = Sum of all Maxi.
Prove:
→ Assume this system is not deadlock free. If there exists a deadlock state, then A = m because there's only one kind of resource and resources can be requested and released only one at a time.
→ From condition b, N + A = M < m + n. So we get N + m < m + n. So we get N < n. It shows that at least one process i that Needi = 0. From condition a, Pi can release at least 1 resource.
→ So there are n-1 processes sharing m resources now, condition a and b still hold. Go on the argument, no process will wait permanently, so there's no deadlock.
Question 164 |
Semaphores are used to solve the problem of
I. Race Condition
II. Process Synchronization
III. Mutual Exclusion
IV. None of the above
I. Race Condition
II. Process Synchronization
III. Mutual Exclusion
IV. None of the above
I and II | |
II and III | |
All of the above | |
None of the above |
Question 164 Explanation:
Semaphore are used to solve the problem of process synchronization and mutual exclusion.
Semaphore are 2 types.
Counting semaphore
Binary semaphore
Semaphore are 2 types.
Counting semaphore
Binary semaphore
Question 165 |
If there are 32 segments, each size 1K bytes, then the logical address should have
13 bits | |
14 bits | |
15 bits | |
16 bits |
Question 165 Explanation:
Given data, total 32 segments
Each segment size is 1K bytes.
Find the logical address=?
32 segments=25
Segment size=210
⇒ 25*210
⇒ 215
So, 15 bits are required.
Each segment size is 1K bytes.
Find the logical address=?
32 segments=25
Segment size=210
⇒ 25*210
⇒ 215
So, 15 bits are required.
Question 166 |
In a lottery scheduler with 40 tickets, how we will distribute the tickets among 4 processes and such that each process gets 10%, 5%, 60% and 25% respectively?
a | |
b | |
c | |
d |
Question 166 Explanation:
Total number of tickets =40
Total 4 processors:
Total 4 processors:
Question 167 |
Suppose a system contains n processes and system uses the round-robin algorithm for CPU scheduling then which data structure is best suited for ready queue of the process
stack | |
queue | |
circular queue | |
tree |
Question 167 Explanation:
→ This type of scheduling is one of the very basic algorithms for Operating Systems in computers which can be implemented through circular queue data structure.
→ Round-robin (RR) is one of the algorithms employed by process and network schedulers in computing.
→ As the term is generally used, time slices (also known as time quanta) are assigned to each process in equal portions and in circular order, handling all processes without priority (also known as cyclic executive).
→ Round-robin scheduling is simple, easy to implement, and starvation-free.
→ Round-robin scheduling can also be applied to other scheduling problems, such as data packet scheduling in computer networks.
→ Round-robin (RR) is one of the algorithms employed by process and network schedulers in computing.
→ As the term is generally used, time slices (also known as time quanta) are assigned to each process in equal portions and in circular order, handling all processes without priority (also known as cyclic executive).
→ Round-robin scheduling is simple, easy to implement, and starvation-free.
→ Round-robin scheduling can also be applied to other scheduling problems, such as data packet scheduling in computer networks.
Question 168 |
At a particular time of computation the value of a counting semaphore is 7. Then 20 P operations and 15 V operations were completed on this semaphore. The resulting value of the semaphore is :
42 | |
2 | |
7 | |
12 |
Question 168 Explanation:
Counting semaphore value is 7
After 20 P operations value of semaphore = 7 – 20 = -13
After 15 V operations value of semaphore = -13 + 15 = 2
After 20 P operations value of semaphore = 7 – 20 = -13
After 15 V operations value of semaphore = -13 + 15 = 2
Question 169 |
Increasing the RAM of a computer typically improves performance because:
Virtual Memory increases | |
Larger RAMs are faster | |
Fewer page faults occur | |
Fewer segmentation faults occur |
Question 169 Explanation:
→ When page frames increases, then no. of page faults decreases.
→ Such as if RAM size increases, then no. of page entries increases, then no. of page faults decreases.
→ Such as if RAM size increases, then no. of page entries increases, then no. of page faults decreases.
Question 170 |
Consider the following program.
main()
{
fork();
fork();
fork();
}
How many new processes will be created?
main()
{
fork();
fork();
fork();
}
How many new processes will be created?
8 | |
6 | |
7 | |
5 |
Question 170 Explanation:
The no. of child process created = 2n -1
= 23 -1
= 7
(where n is number of fork() statements)
= 23 -1
= 7
(where n is number of fork() statements)
Question 171 |
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
|
Question 171 Explanation:
→ Long term scheduler controls the degree of multiprogramming.
→ 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.
→ 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 172 |
Dirty bit for a page in a page table
helps avoid unnecessary writes on a paging device | |
helps maintain LRU information | |
allows only read on a page | |
None of the above |
Question 172 Explanation:
→ The dirty bit allows for a performance optimization i.e., Dirty bit for a page in a page table helps to avoid unnecessary writes on a paging device.
Question 173 |
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 |
Question 173 Explanation:
Algorithm: LRTF (Longest Remaining Time First)
Avg TAT = 12+13+14/3 = 39/3 = 13 units
Avg TAT = 12+13+14/3 = 39/3 = 13 units
Question 174 |
Banker’s algorithm is used for:
Deadlock avoidance | |
Deadlock recovery
| |
Deadlock resolution
| |
Deadlock prevention
|
Question 174 Explanation:
Banker’s algorithm is used for Deadlock avoidance.
Question 175 |
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 computations, 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 f time does the CPU remain idle?
0% | |
10.6% | |
30.0% | |
89.4% |
Question 175 Explanation:
Total time needed to complete the execution = 47
Idle time = 2+3 = 5
Percentage of Idle time = 5/47 × 100 =10.6%
Question 176 |
Consider three CPU-intensive processes which require 10,20 and 30 time units and arrive at times 0,2 and 6, respectively. How 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 |
Question 176 Explanation:
Total no.of context switches is 2.
Question 177 |
Which of the following process scheduling algorithm may lead to starvation?
FIFO | |
Round Robin | |
Shortest Job Next | |
None of the option |
Question 177 Explanation:
● Starvation is method in which process with high priorities continuously uses the resources preventing low priority process to acquire the resources
● 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.
● 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 178 |
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 every T time units and decides the next process to schedule. Which one of the following is TRUE if the processes have no I/O operations and all arrive at time zero?
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 |
Question 178 Explanation:
Let's take an example:
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.
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 179 |
Suppose two jobs, each of which needs 10 minutes of CPU time, start simultaneously. Assume 50% I/O wait time. How long will it take for both to complete, if they run sequentially?
10 | |
20 | |
30 | |
40 |
Question 179 Explanation:
Given data,
2 jobs need 10 min CPU time,
50% IO waiting time
Step-1: Here, total CPU time is IO operation time and execution time.
Step-2: We can assume, CPU time = (IO operation time*execution time) =2 times
Step-3: CPU running both the process is 10+10 mins.
1 CPU time=20 mins.
Step-4: Need= 2 times *20 min
=40mins.
Note: Actual answer key, they given Option C
2 jobs need 10 min CPU time,
50% IO waiting time
Step-1: Here, total CPU time is IO operation time and execution time.
Step-2: We can assume, CPU time = (IO operation time*execution time) =2 times
Step-3: CPU running both the process is 10+10 mins.
1 CPU time=20 mins.
Step-4: Need= 2 times *20 min
=40mins.
Note: Actual answer key, they given Option C
Question 180 |
Which of the following scheduling algorithm could result in starvation?
Priority | |
Round Robin | |
FCFS | |
none of the above |
Question 180 Explanation:
Priority scheduling algorithm could result starvation. Starvation is nothing but indefinite blocking.
Question 181 |
Consider a system having 'm' resources of the same type. these resources are shared by 3 processes A,B,C; which have peak time demands of 3,4,6 respectively. The
minimum value of 'm' that ensures that deadlock will never occur is
11 | |
12 | |
13 | |
14 |
Question 181 Explanation:
A requires 3, B-4, C-6;
→ If A have 2, B have 3, C have 5 then deadlock will occur i.e., 2+3+5=10.
→ If we have one extra resource then deadlock will not occur i.e., 10+1=11.
→ If we have equal (or) more than 11 resources then deadlock will never occur.
→ If A have 2, B have 3, C have 5 then deadlock will occur i.e., 2+3+5=10.
→ If we have one extra resource then deadlock will not occur i.e., 10+1=11.
→ If we have equal (or) more than 11 resources then deadlock will never occur.
Question 182 |
Consider the following four processes with their corresponding arrival time and burst time :
What is the average turnaround time (in ms) for these processes using FCFS scheduling algorithm ?
What is the average turnaround time (in ms) for these processes using FCFS scheduling algorithm ?
15 | |
12.8 | |
13 | |
none of the options |
Question 182 Explanation:
Given, FCFS it means first come first serve. And also it is pure preemptive scheduling.
Average Turnaround Time =(8+13.4+14.2+15.6)/4=(51.2)/4=12.8
Average Turnaround Time =(8+13.4+14.2+15.6)/4=(51.2)/4=12.8
Question 183 |
Which of the following is added to page table in order to track whether a page of cache has been modified since it was read from the memory?
Reference bit | |
Dirty bit | |
Tag Bit | |
Valid Bit |
Question 183 Explanation:
A dirty bit (or) modified bit is a bit that is associated with a block of computer memory and indicates whether or not the corresponding block of memory has been modified. The dirty bit is set when the processor writes to (modifies) this memory.
→ The bit indicates that its associated block of memory has been modified and has not been saved to storage yet. When a block of memory is to be replaced, its corresponding dirty bit is checked to see if the block needs to be written back to secondary memory before being replaced or if it can simply be removed.
→ Dirty bits are used by the CPU cache and in the page replacement algorithms of an operating system.
→ Dirty bits can also be used in Incremental computing by marking segments of data that need to be processed or have yet to be processed.
→ The bit indicates that its associated block of memory has been modified and has not been saved to storage yet. When a block of memory is to be replaced, its corresponding dirty bit is checked to see if the block needs to be written back to secondary memory before being replaced or if it can simply be removed.
→ Dirty bits are used by the CPU cache and in the page replacement algorithms of an operating system.
→ Dirty bits can also be used in Incremental computing by marking segments of data that need to be processed or have yet to be processed.
Question 184 |
The time taken to switch between user and kernel modes of execution be t1 while the time taken to switch between two processes be t2. Which of the following is TRUE?
t1>t2 | |
t1=t2 | |
t1 | |
Nothing can be said about the relation between t1 and t2
|
Question 184 Explanation:
→ A context switch (also sometimes referred to as a process switch or a task switch) is the switching of the CPU (central processing unit) from one processor thread to another.
→ Context switches can occur only in kernel mode. Kernel mode is a privileged mode of the CPU in which only the kernel runs and which provides access to all memory locations and all other system resource
→ Other programs, including applications, initially operate in user mode, but they can run portions of the kernel code via system calls
→ The existence of these two modes in Unix-like operating systems means that a similar, but simpler, operation is necessary when a system call causes the CPU to shift to kernel mode. This is referred to as a mode switch rather than a context switch, because it does not change the current process.
→ Context switch between the processes involves mode switch also.
→ Context switches can occur only in kernel mode. Kernel mode is a privileged mode of the CPU in which only the kernel runs and which provides access to all memory locations and all other system resource
→ Other programs, including applications, initially operate in user mode, but they can run portions of the kernel code via system calls
→ The existence of these two modes in Unix-like operating systems means that a similar, but simpler, operation is necessary when a system call causes the CPU to shift to kernel mode. This is referred to as a mode switch rather than a context switch, because it does not change the current process.
→ Context switch between the processes involves mode switch also.
Question 185 |
A system has 3 processes sharing 4 resources. If each process needs a maximum of 2 units then, deadlock.
Can never occur | |
Has to occur | |
May occur | |
None of the options |
Question 185 Explanation:
If the system is deadlocked, it implies that each process is holding one resource and is waiting for one more. Since there are 3 processes and 4 resources, one process must be able to obtain two resources. This process requires no more resources and therefore it will return its resources when done
Question 186 |
Which of the following is the first module of a language processing system?
Preprocessor
| |
Loader
| |
Compiler
| |
Linker
|
Question 186 Explanation:
→ Preprocessor is a program that processes its input data to produce output that is used as input to another program.
→ The output is said to be a preprocessed form of the input data, which is often used by some subsequent programs like compilers.
→ The amount and kind of processing done depends on the nature of the preprocessor.
→ Preprocessor is the first module of a language processing system.
→ The output is said to be a preprocessed form of the input data, which is often used by some subsequent programs like compilers.
→ The amount and kind of processing done depends on the nature of the preprocessor.
→ Preprocessor is the first module of a language processing system.
Question 187 |
The total number of page faults for the reference string 1,2,3,4,5,6,7,8,9,10 using FIFO page replacement policy for a process m if 3 frames are allocated to its are:
9 | |
10 | |
8 | |
11 |
Question 187 Explanation:
Total number of page faults are 10.
Question 188 |
The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory fram is recorded as follows. What is the number of page faults?
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0370
15,4 | |
6,4 | |
7,2 | |
4,6 |
Question 188 Explanation:
→ Page are fitted in frames, so first we need to determine the pages but given request are just the record request in decimal. We can assume that first page to address from 0000 to 0099 and page 2 contains records from 0100 to 0199 and so on (it is given in question that each page contains 100 records) and so on. So page request string is
01, 02, 04, 04, 05, 05, 05, 01, 02, 02, 02, 03, 03.
Clearly 7 page faults.
01, 02, 04, 04, 05, 05, 05, 01, 02, 02, 02, 03, 03.
Clearly 7 page faults.
Question 189 |
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
|
Question 189 Explanation:
TRUE: Shortest remaining time (SRT) algorithm is the preemptive version of the shortest job next(SJN) CPU scheduling algorithm.
TRUE: A “context switch” is the mechanism to store and restore the state or context of a CPU in PCB.
TRUE: A “context switch” is the mechanism to store and restore the state or context of a CPU in PCB.
Question 190 |
Operating system maintains the page table for:
each process | |
each thread | |
each instruction | |
each address |
Question 190 Explanation:
Role of the page table
→ In operating systems that use virtual memory, every process is given the impression that it is working with large, contiguous sections of memory.
→ Physically, the memory of each process may be dispersed across different areas of physical memory, or may have been moved (paged out) to another storage, typically to a hard disk drive.
→ When a process requests access to data in its memory, it is the responsibility of the operating system to map the virtual address provided by the process to the physical address of the actual memory where that data is stored.
→ The page table is where the operating system stores its mappings of virtual addresses to physical addresses, with each mapping also known as a page table entry (PTE).
→ Physically, the memory of each process may be dispersed across different areas of physical memory, or may have been moved (paged out) to another storage, typically to a hard disk drive.
→ When a process requests access to data in its memory, it is the responsibility of the operating system to map the virtual address provided by the process to the physical address of the actual memory where that data is stored.
→ The page table is where the operating system stores its mappings of virtual addresses to physical addresses, with each mapping also known as a page table entry (PTE).
Question 191 |
Non contiguous memory allocation splits program into blocks of memory called ___ that can be loaded in non adjacent holes in main memory
Pages | |
Frames | |
Partition | |
Segments |
Question 191 Explanation:
Noncontiguous memory allocation splits programs into blocks of memory called Segments that can be loaded into non adjacent “holes” in main memory.
Question 192 |
A 3.5 inch micro floppy high density disk contains he data__
720 MB | |
1.44 MB | |
720 KB | |
1.44 KB |
Question 192 Explanation:
→ The 3.5-Inch floppy diskettes have dimensions of 8.9cm in width by 9.3cm in height and are referred to as floppies because of the circular magnetic floppy within the hard shell.
→ 3.5-inch floppy diskettes come in sizes of 720 KB low-density, 1.44 MB high density capacity, and IBM even developed an Extended Density disk capable of holding 2.88 MB.
→ 3.5-inch floppy diskettes come in sizes of 720 KB low-density, 1.44 MB high density capacity, and IBM even developed an Extended Density disk capable of holding 2.88 MB.
Question 193 |
What is compaction?
a technique for overcoming internal fragmentation | |
a paging technique | |
a technique for overcoming external fragmentation | |
a technique for overcoming fatal error |
Question 193 Explanation:
→ Compaction refers to combining all the empty spaces together and processes. Compaction helps to solve the problem of fragmentation, but it requires too much of CPU time.
→ It moves all the occupied areas of store to one end and leaves one large free space for incoming jobs, instead of numerous small ones.
→ In compaction, the system also maintains relocation information and it must be performed on each new allocation of job to the memory or completion of job from memory.
→ It moves all the occupied areas of store to one end and leaves one large free space for incoming jobs, instead of numerous small ones.
→ In compaction, the system also maintains relocation information and it must be performed on each new allocation of job to the memory or completion of job from memory.
Question 194 |
Process is in a ready state____
When process is scheduled to run after some execution
| |
When process is unable to run until some task has been completed | |
When process is using the CPU | |
None of the above |
Question 194 Explanation:
When process is unable to run until some task has been completed, the process is in blocked state and if process is using the CPU, it is in running state.
Question 195 |
How many times the word "PROCESS" will be printed when executing the following program?
main()
{
printf("process");
fflush();
fork();
fork();
}
main()
{
printf("process");
fflush();
fork();
fork();
}
8 | |
4 | |
6 | |
7 |
Question 195 Explanation:
→ fflush() is typically used for output stream only. Its purpose is to clear (or flush) the output buffer and move the buffered data to console (in case of stdout) or disk (in case of file output stream).
→ The two fork() calls create 3 child processes, and hence "PROCESS" will be executed 4 times if we don't use fflush.
→ If we put a '\n' at end of printf or use fflush(stdout); only 1 printf will be done.
→ The two fork() calls create 3 child processes, and hence "PROCESS" will be executed 4 times if we don't use fflush.
→ If we put a '\n' at end of printf or use fflush(stdout); only 1 printf will be done.
Question 196 |
The open file table has a/an __ associated with each file.
file content | |
file permission | |
open count | |
close count |
Question 196 Explanation:
Open count indicates the number of processes that have the file open
Question 197 |
The process of loading the operating system into memory is called:
Booting | |
Spooling
| |
Thrashing | |
Formatting |
Question 197 Explanation:
The operating system is loaded through a bootstrapping process, more succinctly known as booting. A boot loader is a program whose task is to load a bigger program, such as the operating system.
Question 198 |
Starvation can be avoided by which of the following statement:
i.By using shortest job first resource allocation policy
ii.By using first come first serve resources allocation policy
i.By using shortest job first resource allocation policy
ii.By using first come first serve resources allocation policy
i only | |
i and ii only | |
ii only | |
None of the options |
Question 198 Explanation:
Starvation possible in SJFS allocation policy because it is preemptive and always giving priority to shortest jobs.
→ Starvation never happen into FCFS because it is purely non preemptive allocation strategy.
→ Starvation never happen into FCFS because it is purely non preemptive allocation strategy.
Question 199 |
Page fault frequency in an operating system is reduced when the:
Process tend to be I/O bound | |
Locality of reference is applicable to the process | |
size of pages is reduced | |
Process tend to be CPU bound |
Question 199 Explanation:
We can reduce page faults using some methods
1.We can increase size of main memory
2.Decreasing the degree of multiprogramming
3.Locality of reference is applicable to the process
4.Possible to increase the page size.
1.We can increase size of main memory
2.Decreasing the degree of multiprogramming
3.Locality of reference is applicable to the process
4.Possible to increase the page size.
Question 200 |
Social network analysts use ____ to access facebook data
Facebook system calls | |
Facebook APIs | |
Facebook Scripts | |
Facebook System Libraries |
Question 200 Explanation:
→ The Facebook API is a platform for building applications that are available to the members of the social network of Facebook.
→ The API allows applications to use the social connections and profile information to make applications more involving, and to publish activities to the news feed and profile pages of Facebook, subject to individual users privacy settings.
→ The API allows applications to use the social connections and profile information to make applications more involving, and to publish activities to the news feed and profile pages of Facebook, subject to individual users privacy settings.
Question 201 |
Time taken to switch between user and kernel models is___the time taken to switch between two processes.
More than | |
Independent of | |
Less than | |
Equal to |
Question 201 Explanation:
Context switch between the processes involves kernel mode.So the time taken to switch between user and kernel models is less compared to switching between the processes
Question 202 |
How many page faults occur in LRU page replacement algorithm for the given reference string, with four page frames
7,0,1,2,0,3,4,2,3,0,3,2,1,2,0,1
7,0,1,2,0,3,4,2,3,0,3,2,1,2,0,1
6 | |
8 | |
7 | |
9 |
Question 202 Explanation:
In the Least Recently Used (LRU) page replacement policy, the page that is used least recently will be replaced.
Question 203 |
Which of the following is correct?
In asymmetric multiprocessing, the processor are peers. | |
In symmetric multiplexing, the processors are placed symmetrically on the motherboard | |
Clustered systems are used for high performance computing | |
All multiprocessor systems are multicore systems |
Question 203 Explanation:
→ The clustered systems are a combination of hardware clusters and software clusters. The hardware clusters help in sharing of high performance disks between the systems. The software clusters makes all the systems work together.
→ Each node in the clustered systems contains the cluster software. This software monitors the cluster system and makes sure it is working as required. If any one of the nodes in the clustered system fail, then the rest of the nodes take control of its storage and resources and try to restart.
→ asymmetric multiprocessing using master-slave strategy but symmetric multiprocessing using peer to peer.
→ In symmetric multiplexing, the processors are not placed symmetrically on the motherboard.
→ Each node in the clustered systems contains the cluster software. This software monitors the cluster system and makes sure it is working as required. If any one of the nodes in the clustered system fail, then the rest of the nodes take control of its storage and resources and try to restart.
→ asymmetric multiprocessing using master-slave strategy but symmetric multiprocessing using peer to peer.
→ In symmetric multiplexing, the processors are not placed symmetrically on the motherboard.
Question 204 |
File operations that manipulate the ‘open-count’ maintained for each entry in open file table include:
Open,write | |
Read,write | |
Write,close | |
Open,close |
Question 204 Explanation:
The open file table also has an open count associated with each file to indicate how many processes have the file open. Each close() decreases this open count, and when open count reaches zero, the file no longer in use, and file’s entry is removed from the open file table.
Question 205 |
In a computer system, memory mapped access takes 100 nanoseconds when a page is found in TLB. In case the page is not TLB, it takes 400 nanoseconds to access. Assuming a hit ratio of 80%, the effective access time is:
120ns | |
160ns | |
200ns | |
500ns |
Question 205 Explanation:
EAT := TLB_miss_time * (1- hit_ratio) + TLB_hit_time * hit_ratio
TLB time = 100ns,
Memory time = 400ns
Hit Ratio= 80%
E.A.T. = (0.80)*(500)+0.20*900
0.8*100+.2*400=160
TLB time = 100ns,
Memory time = 400ns
Hit Ratio= 80%
E.A.T. = (0.80)*(500)+0.20*900
0.8*100+.2*400=160
Question 206 |
The producer and consumer processes share the following variables:
int n;
Semaphore M=1
Semaphore E=n
Semaphore F=0
The consumer process must execute ______ and before removing an item from buffer.
int n;
Semaphore M=1
Semaphore E=n
Semaphore F=0
The consumer process must execute ______ and before removing an item from buffer.
signal(M), signal(F) | |
signal(M), wait(F) | |
Signal(F), wait(M) | |
wait(F), wait(M) |
Question 206 Explanation:
F is given 0 then no items would be produced
→ Two standard operations, wait and signal are defined on the semaphore.
→ Entry to the critical section is controlled by the wait operation and exit from a critical region is taken care by signal operation.
→ The manipulation of semaphore (S) takes place as following:
→ The wait command P(S) decrements the semaphore value by 1. If the resulting value becomes negative then P command is delayed until the condition is satisfied.
→ The V(S) i.e. signals operation increments the semaphore value by 1.
→ The wait, signal operations are also called P and V operations.
→ Two standard operations, wait and signal are defined on the semaphore.
→ Entry to the critical section is controlled by the wait operation and exit from a critical region is taken care by signal operation.
→ The manipulation of semaphore (S) takes place as following:
→ The wait command P(S) decrements the semaphore value by 1. If the resulting value becomes negative then P command is delayed until the condition is satisfied.
→ The V(S) i.e. signals operation increments the semaphore value by 1.
→ The wait, signal operations are also called P and V operations.
Question 207 |
Disk requests come to a disk driver for cylinders in the order 10, 22, 20, 2, 40, 6 and 38 at a given time when the given disk drive is reading from cylinder 20. The seek time is 6ms per cylinder.What is the total seek time, if the disk arm scheduling algorithm FCFS is used?
900 ms | |
850 ms | |
360 ms | |
876 ms |
Question 207 Explanation:
Total seek time= 10+12+2+18+38+34+32
= 146*6 ms
= 876 ms
Question 208 |
If a processor has 32 bit virtual address, 28 bit physical address, 2 KB pages. How many bits are required for the virtual,physical page number?
17,21 | |
21,17 | |
6,10 | |
None |
Question 208 Explanation:
Number of entries for virtual page number = 2 virtual address / Pages =2 32 /2KB= 2 32 /2 11 =2 21
Number of entries for physical page number = 2 physical address / Pages =2 28 /2KB= 2 28 /2 11 =2 17
The number bits for virtual page number are 21.
The number bits for physical page number are 21.
Number of entries for physical page number = 2 physical address / Pages =2 28 /2KB= 2 28 /2 11 =2 17
The number bits for virtual page number are 21.
The number bits for physical page number are 21.
Question 209 |
The address sequence generated by tracing a particular program executing in a pure demand paging system with 100 records per page, with 1 free main memory frame is recorded as follows. 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260,
0320, 0370, What is the number of page faults?
15,4 | |
6,4 | |
7,2 | |
4,6 |
Question 209 Explanation:
Memory page numbers will be calculated by using corresponding to given address as follows
The page number is 1 from the record numbers 1-100,
The page number is 2 from the record numbers 101 -200
The page number is 3 from the record numbers 201 -300
The page number is 4 from the record numbers 301 -400
The page number is 5 from the record numbers 401 -500
The page number is 6 from the record numbers 501 -600
The page number is 7 from the record numbers 601 -700
For a given address sequence 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240,0260, 0320, 0370
The page number numbers sequence is 1,2,5,5,6,6,6,2,3,3,3,4,4
The number of available frame is 1, So we need to use only one frame for executing demand page process.
So total number of page faults are 7.
The page number is 1 from the record numbers 1-100,
The page number is 2 from the record numbers 101 -200
The page number is 3 from the record numbers 201 -300
The page number is 4 from the record numbers 301 -400
The page number is 5 from the record numbers 401 -500
The page number is 6 from the record numbers 501 -600
The page number is 7 from the record numbers 601 -700
For a given address sequence 0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240,0260, 0320, 0370
The page number numbers sequence is 1,2,5,5,6,6,6,2,3,3,3,4,4
The number of available frame is 1, So we need to use only one frame for executing demand page process.
So total number of page faults are 7.
Question 210 |
The following diagram depicts a______cell.
Storage | |
Mobile | |
Memory | |
Register |
Question 210 Explanation:
● Computer memory is the storage space in the computer, where data is to be processed and instructions required for processing are stored.
● In the computer memory , we will perform read and write operations.
● In the computer memory , we will perform read and write operations.
Question 211 |
The major functions or requirements for an I/O module fall into which of the following categories?
Control and timing | |
Processor Communication | |
Data buffering | |
All of these |
Question 211 Explanation:
● The computer systems Input/output (I/O) architecture is to interface to the outside world.
● Each I/O module interfaces to the system bus and controls one or more peripheral devices.
● The major functions of an I/O module are categorized as follows:
■ Control and
■ Processor
■ Device communication
■ Data buffering
■ Error Detection
● Each I/O module interfaces to the system bus and controls one or more peripheral devices.
● The major functions of an I/O module are categorized as follows:
■ Control and
■ Processor
■ Device communication
■ Data buffering
■ Error Detection
Question 212 |
A web page contains many small images. We can speed up the performance of the browser by
Having multiple processes executed simultaneously | |
Having multiple threads | |
Increasing the speed of the transmission line | |
Increasing the main memory of the system |
Question 212 Explanation:
We will assign one thread for each image so that all images will load faster.
Question 213 |
If a process is runnable but is temporarily stopped to let another process run, in which state is the process said to be?
Running | |
Ready | |
Interrupted | |
Blocked |
Question 213 Explanation:
1. New
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.
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 214 |
What is coalescing?
It is a second strategy for allocating kernel memory | |
The buddy system allocates memory from a fixed size segment consistency of physically contiguous pages | |
Kernel memory is often allocated from a free memory pool different from the list used to satisfy ordinary user mode processes | |
An advantage of the buddy system is how quickly adjacent buddies can be combined to form larger segments using this technique |
Question 214 Explanation:
● coalescing is the act of merging two adjacent free blocks of memory.
● When an application frees memory, gaps can fall in the memory segment that the application uses.
● Among other techniques, coalescing is used to reduce external fragmentation, but is not totally effective.
● Coalescing can be done as soon as blocks are freed, or it can be deferred until some time later (known as deferred coalescing), or it might not be done at all.
● When an application frees memory, gaps can fall in the memory segment that the application uses.
● Among other techniques, coalescing is used to reduce external fragmentation, but is not totally effective.
● Coalescing can be done as soon as blocks are freed, or it can be deferred until some time later (known as deferred coalescing), or it might not be done at all.
Question 215 |
Real time operating system always runs on
Linux | |
Unix | |
Embedded system | |
Apple’s Mac OS |
Question 215 Explanation:
→ A real-time operating system (RTOS) is any operating system (OS) intended to serve real-time applications that process data as it comes in, typically without buffer delays.
→ 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
→ A multitasking operating system like Unix is poor at real-time tasks. The scheduler gives the highest priority to jobs with the lowest demand on the computer, so there is no way to ensure that a time-critical job will have access to enough resources.
→ 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
→ A multitasking operating system like Unix is poor at real-time tasks. The scheduler gives the highest priority to jobs with the lowest demand on the computer, so there is no way to ensure that a time-critical job will have access to enough resources.
Question 216 |
The diagram given below represents
Process creation | |
Process state | |
Queueing diagram | |
Communication model |
Question 216 Explanation:
The above diagram represents queueing diagram. Various queues are
→ 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.
→ 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 217 |
This begins at the root and follows a path down to the specified file, giving the directory names on the path. This is known as
Absolute path name | |
Relative path name | |
Definite path name | |
Indefinite path name |
Question 217 Explanation:
● A path, the general form of the name of a file or directory, specifies a unique location in a file system.
● An absolute or full path points to the same location in a file system, regardless of the current working directory. To do that, it must include the root directory.
● By contrast, a relative path starts from some given working directory, avoiding the need to provide the full absolute path.
● An absolute or full path points to the same location in a file system, regardless of the current working directory. To do that, it must include the root directory.
● By contrast, a relative path starts from some given working directory, avoiding the need to provide the full absolute path.
Question 218 |
Each process is contained in a single section of memory that is contiguous to the section containing the next process is called
Contiguous memory protection | |
Contiguous path name | |
Definite path name | |
Indefinite path name |
Question 218 Explanation:
Contiguous memory allocation is a classical memory allocation model that assigns a process consecutive memory blocks (that is, memory blocks having consecutive addresses).
Question 219 |
Both the first fit and best fit strategies for memory allocation suffer from
External fragmentation | |
Internal fragmentation | |
50-percent rule | |
segmentation |
Question 219 Explanation:
→ Internal fragmentation is the wasted space within each allocated block because of rounding up from the actual requested allocation to the allocation granularity.
→ External fragmentation is the various free spaced holes that are generated in either your memory or disk space.
→ Both the first-fit and best-fit strategies for memory allocation suffer from external fragmentation. As the processes are loaded and removed from memory, the free memory space is broken into little pieces.
→ External fragmentation exists when there is enough total memory space to satisfy a request, but the available spaces are not contiguous.
→ External fragmentation is the various free spaced holes that are generated in either your memory or disk space.
→ Both the first-fit and best-fit strategies for memory allocation suffer from external fragmentation. As the processes are loaded and removed from memory, the free memory space is broken into little pieces.
→ External fragmentation exists when there is enough total memory space to satisfy a request, but the available spaces are not contiguous.
Question 220 |
The simplest, but most expensive approach to introductory redundancy is duplicate to every disk. This technique is called
Swap space | |
Mirroring | |
Page slots | |
None of these |
Question 220 Explanation:
● Mirroring copies identical data onto more than one drive.
● Striping partitions each drive's storage space into units ranging from a sector (512 bytes) up to several megabytes.
●The stripes of all the disks are interleaved and addressed in order.
● Striping partitions each drive's storage space into units ranging from a sector (512 bytes) up to several megabytes.
●The stripes of all the disks are interleaved and addressed in order.
Question 221 |
_____ is a sequence of memory read-write operations that are atomic
Critical section object | |
Adaptive mutex | |
Turnstile | |
Memory transaction |
Question 221 Explanation:
An operation (or set of operations) is atomic or uninterruptible if it appears to the rest of the system to occur instantaneously.
Question 222 |
The term____ refers to the use of electronics and software within a product, as opposed to a general purpose computer such as a laptop or desktop system
Complex system | |
Simple system | |
Fuzzy logic | |
Embedded system |
Question 222 Explanation:
● Fuzzy logic is an approach to computing based on "degrees of truth" rather than the usual "true or false" (1 or 0) Boolean logic on which the modern computer is based.
● An embedded system is a controller programmed and controlled by a real-time operating system (RTOS) with a dedicated function within a larger mechanical or electrical system, often with real-time computing constraints.
● An embedded system is a controller programmed and controlled by a real-time operating system (RTOS) with a dedicated function within a larger mechanical or electrical system, often with real-time computing constraints.
Question 223 |
Sequential, direct, random and associative are access methods and key characteristics of computer____system
Stack | |
Count | |
Memory | |
Core |
Question 223 Explanation:
● An access method is a program or a hardware mechanism that moves data between the computer and an outlying device such as a hard disk (or other form of storage) or a display terminal.
● Memory can be accessed Sequential, direct, random and associative manner.
● Memory can be accessed Sequential, direct, random and associative manner.
Question 224 |
In a monolithic kernel, operating system runs in
User mode | |
Supervisor mode | |
User/supervisor mode | |
None of these |
Question 224 Explanation:
Monolithic kernel means that the whole operating system runs in kernel mode (i.e. highly privileged by the hardware). That is, no part of the OS runs in user mode (lower privilege). Only applications on top of the OS run in user mode.
Question 225 |
One scheme for communication between user thread library and the kernel is known as
Lightweight process | |
Upcall handler | |
Scheduler activation | |
Cleanup handler |
Question 225 Explanation:
Scheduler activations are a threading mechanism that, when implemented in an operating system's process scheduler, provide kernel-level thread functionality with user-level thread flexibility and performance
Question 226 |
A classic software based solution to the critical section problem is known as
Peterson’s solution | |
Process synchronization | |
Coordination | |
Race Condition |
Question 226 Explanation:
Peterson's solution is a concurrent programming algorithm for mutual exclusion that allows two or more processes to share a single-use resource without conflict, using only shared memory for
communication
Question 227 |
Consider a set of n tasks with known runtimes R1 ,R2 ,Rn , to be run uniprocessor machine. Which of the following processor scheduling algorithm will result in the maximum throughput?
Priority scheduling | |
Round robin | |
FCFS | |
SJF |
Question 227 Explanation:
→ In shortest job first scheduling algorithm, the processor selects the process with the smallest execution time to execute next.
→ SJF scheduling algorithm will result in the maximum throughput
→ SJF scheduling algorithm will result in the maximum throughput
Question 228 |
OLE, a microsoft's component document technology, means
Overlay linking and exchange | |
Online linking and embedding | |
Open learning exchange | |
Object linking and embedding |
Question 228 Explanation:
→ Object Linking & Embedding (OLE) is a proprietary technology developed by Microsoft that allows embedding and linking to documents and other objects.
→ For developers, it brought OLE Control Extension (OCX), a way to develop and use custom user interface elements. On a technical level, an OLE object is any object that implements the IOleObject interface, possibly along with a wide range of other interfaces, depending on the object's needs.
Question 229 |
The problem of indefinite blockage of low priority jobs in general priority scheduling algorithm can be solved using
Swapping | |
Dirty bit | |
Aging | |
Compaction |
Question 229 Explanation:
→ Ageing is a scheduling technique used to avoid starvation. Fixed priority scheduling is a scheduling discipline, in which tasks queued for utilizing a system resource are assigned a priority each.
→ 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.
→ 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 230 |
A Thread is also called
a scheduler | |
a virtual process | |
a heavyweight process | |
a lightweight process |
Question 230 Explanation:
→ A light-weight process (LWP) is a means of achieving multitasking.
→ A LWP runs in user space on top of a single kernel thread and shares its address space and system resources with other LWPs within the same process. Multiple user level threads, managed by a thread library, can be placed on top of one or many LWPs - allowing multitasking to be done at the user level, which can have some performance benefits.
→ A LWP runs in user space on top of a single kernel thread and shares its address space and system resources with other LWPs within the same process. Multiple user level threads, managed by a thread library, can be placed on top of one or many LWPs - allowing multitasking to be done at the user level, which can have some performance benefits.
Question 231 |
Copying a process from memory to disk to allow space for other processes is called___
Demand paging | |
Deadlock | |
page fault | |
Swapping |
Question 231 Explanation:
→ Swapping is a mechanism in which a process can be swapped temporarily out of main memory (or move) to secondary storage (disk) and make that memory available to other processes. At some later time, the system swaps back the process from the secondary storage to main memory.
→ The performance is usually affected by swapping process but it helps in running multiple and big processes in parallel and that's the reason Swapping is also known as a technique for memory compaction.
→ The performance is usually affected by swapping process but it helps in running multiple and big processes in parallel and that's the reason Swapping is also known as a technique for memory compaction.
Question 232 |
If the time quantum size is 2 units of an there is only one job of 14 time unit in a ready queue, the round robin scheduling algorithm will cause___ connected switches.
1 | |
5 | |
6 | |
7 |
Question 232 Explanation:
→ Actually they are given single job, so when we are using round robin algorithm it won’t perform any context switch because we are executing same job.
→ 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.
→ 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 233 |
The memory which does not lose its content on failure of power supply is known as___memory.
Main memory | |
Volatile | |
Non volatile | |
ROM |
Question 233 Explanation:
Non-volatile memory (NVM) or non-volatile storage is a type of computer memory that can retrieve stored information even after having been power cycled. In contrast, volatile memory needs constant power in order to retain data. Examples of non-volatile memory include read-only memory, flash memory, ferroelectric RAM, most types of magnetic computer storage devices (e.g. hard disk drives, solid state drives, floppy disks, and magnetic tape), optical discs, and early computer storage methods such as paper tape and punched cards.
Question 234 |
The kernel keeps track of the state of each executing program by using a data structure called__
Process control block | |
User control block | |
File control block | |
memory control block |
Question 234 Explanation:
Process Control Block (PCB, also called Task Controlling Block, Entry of the Process Table, Task Struct, or Switchframe) is a data structure in the operating system kernel containing the information needed to manage the scheduling of a particular process.
Question 235 |
An operating system contains 3 user processes each requiring 2 units of resource R. The minimum number of units of R such that no deadlock will ever occur is
5 | |
2 | |
3 | |
4 |
Question 235 Explanation:
There are three processes, so there is a possibility to occur deadlock when we are taking 3 resources. If we are taking 3+1 resources means it will never occur deadlock.
Note: If there are ‘n’ processes then ‘n+1’ resources allocated, will never occur deadlock.
Note: If there are ‘n’ processes then ‘n+1’ resources allocated, will never occur deadlock.
Question 236 |
The segmentation memory management scheme suffers from:
External fragmentation | |
Internal fragmentation
| |
Starvation
| |
Ageing
|
Question 236 Explanation:
• Segmentation and paging avoids external fragmentation but still it suffers from internal fragmentation.
• Internal fragmentation is the wasted space within each allocated block because of rounding up from the actual requested allocation to the allocation granularity.
• External fragmentation is the various free spaced holes that are generated in either your memory or disk space. External fragmented blocks are available for allocation, but may be too small to be of any use.
• Resource starvation is a problem encountered in concurrent computing where a process is perpetually denied necessary resources to process its work. Starvation may be caused by errors in a scheduling or mutual exclusion algorithm.
• Ageing is a scheduling technique used to avoid starvation.
• Internal fragmentation is the wasted space within each allocated block because of rounding up from the actual requested allocation to the allocation granularity.
• External fragmentation is the various free spaced holes that are generated in either your memory or disk space. External fragmented blocks are available for allocation, but may be too small to be of any use.
• Resource starvation is a problem encountered in concurrent computing where a process is perpetually denied necessary resources to process its work. Starvation may be caused by errors in a scheduling or mutual exclusion algorithm.
• Ageing is a scheduling technique used to avoid starvation.
Question 237 |
Which of the following technique allows execution of programs larger than the size of physical memory?
Thrashing | |
DMA | |
Buffering | |
Demand Paging |
Question 237 Explanation:
Virtual memory technique allows execution of programs larger than the size of physical memory. Demand paging we are using for virtual memory technique.
Question 238 |
In page replacement, ‘adding more frames may cause more page faults’ is referred to as:
Thrashing
| |
Belady’s anomaly
| |
Banker’s anomaly
| |
Ageing |
Question 238 Explanation:
In FIFO page replacement policy sometimes behaves differently. The abnormal behaviour is nothing but Belady’s anomaly. Belady’s anomaly is nothing but adding more frames may cause more page faults.
Question 239 |
____ is an IEEE 1003.1C standard API for thread creation and synchronization in operating system.
Mac OS X
| |
Solaris
| |
POSIX
| |
Kernel |
Question 239 Explanation:
POSIX Threads, usually referred to as pthreads, is an execution model that exists independently from a language, as well as a parallel execution model. It allows a program to control multiple different flows of work that overlap in time. Each flow of work is referred to as a thread, and creation and control over these flows is achieved by making calls to the POSIX Threads API. POSIX Threads is an API defined by the standard POSIX.1c, Threads extensions (IEEE Std 1003.1c-1995).
Question 240 |
If a system has multiple instances of resources, to avoid deadlock which of the following algorithms is used?
Deadlock avoidance algorithm
| |
Aging algorithm
| |
Resource allocation graph algorithm
| |
Banker’s algorithm |
Question 240 Explanation:
Banker's Algorithm :
Banker's algo can be applied even when Resource-system have multiple instances of each resource type. It can also be applied when Resource-system have single instance of each resource type But in that case, It would be less efficient than Resource-Allocation Graph Algorithm.
Banker's algo can be applied even when Resource-system have multiple instances of each resource type. It can also be applied when Resource-system have single instance of each resource type But in that case, It would be less efficient than Resource-Allocation Graph Algorithm.
Question 241 |
Which of the following is not a CPU scheduling criteria:
Dispatch latency | |
CPU utilization
| |
Throughput | |
Turnaround time
|
Question 241 Explanation:
The term dispatch latency describes the amount of time it takes for a system to respond to a request for a process to begin operation. With a scheduler written specifically to honor application priorities, real-time applications can be developed with a bounded dispatch latency.
Note: The time taken by the dispatcher to stop one process and start another running is known as dispatch latency.
Note: The time taken by the dispatcher to stop one process and start another running is known as dispatch latency.
Question 242 |
An address in the memory is called
physical address | |
logical address | |
memory address | |
word address |
Question 242 Explanation:
A physical address (also real address, or binary address), is a memory address that is represented in the form of a binary number on the address bus circuitry in order to enable the data bus to access a particular storage cell of main memory, or a register of memory mapped I/O device.
Question 243 |
In which of the following state, the process is waiting for processor?
Running | |
New | |
Ready | |
Waiting |
Question 243 Explanation:
New→ The process is being created
Running→ Instructions are being executed
Ready→ The process is waiting to be assigned to a processor
Waiting→ The process is waiting for some event to occur(such as IO exception or reception of a signal)
Terminated: The process has finished execution.
Running→ Instructions are being executed
Ready→ The process is waiting to be assigned to a processor
Waiting→ The process is waiting for some event to occur(such as IO exception or reception of a signal)
Terminated: The process has finished execution.
Question 244 |
The mechanism that brings a page memory only when it is needed in___
page replacement | |
segmentation | |
fragmentation | |
demand paging |
Question 244 Explanation:
→ Demand paging follows that pages should only be brought into memory if the executing process demands them. This is often referred to as lazy evaluation as only those pages demanded by the process are swapped from secondary storage to main memory. Contrast this to pure swapping, where all memory for a process is swapped from secondary storage to main memory during the process startup.
→ Demand paging is a method of virtual memory management. In a system that uses demand paging, the operating system copies a disk page into physical memory only if an attempt is made to access it and that page is not already in memory (i.e., if a page fault occurs).
→ It follows that a process begins execution with none of its pages in physical memory, and many page faults will occur until most of a process working set of pages are located in physical memory. This is an example of a lazy loading technique.
→ Demand paging is a method of virtual memory management. In a system that uses demand paging, the operating system copies a disk page into physical memory only if an attempt is made to access it and that page is not already in memory (i.e., if a page fault occurs).
→ It follows that a process begins execution with none of its pages in physical memory, and many page faults will occur until most of a process working set of pages are located in physical memory. This is an example of a lazy loading technique.
Question 245 |
Semaphores are used to solve the problem of
mutual exclusion | |
race condition | |
Process synchronization | |
The belady problem |
Question 245 Explanation:
Semaphores are used to solve the problem of process synchronization. There are two types of semaphores
1. Counting Semaphore
2. Binary semaphore
1. Counting Semaphore
2. Binary semaphore
Question 246 |
Consider the following statements
S1: a small page size causes large page tables
S2: Internal fragmentation is increase with small pages
S3: I/O transfers are more efficient with large pages
Which of the following is true?
S1: a small page size causes large page tables
S2: Internal fragmentation is increase with small pages
S3: I/O transfers are more efficient with large pages
Which of the following is true?
S1 is true and S3 is false | |
S1 and S2 are true | |
S2 and S3 are true | |
S1 is true and S2 is false |
Question 246 Explanation:
S1: True:A small page size causes large page tables
S2: True:Internal fragmentation is increase with small pages. Actually Internal fragmentation is decrease with small pages
S3:True:I/O transfers are more efficient with large pages
S2: True:Internal fragmentation is increase with small pages. Actually Internal fragmentation is decrease with small pages
S3:True:I/O transfers are more efficient with large pages
Question 247 |
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.
LOOK | |
SCAN | |
C-LOOK | |
C-SCAN |
Question 247 Explanation:
Circular scanning works just like the elevator to some extent. It begins its scan toward the nearest end and works it way all the way to the end of the system. Once it hits the bottom or top it jumps to the other end and moves in the same direction. Keep in mind that the huge jump doesn't count as a head movement.
Question 248 |
Suppose there are six files F1, F2, F3, F4, F5, F6 with corresponding sizes 150 KB, 225KB,75 KB, 60 KB, 275 KB and 65 KB respectively. The files are to be stored on a sequential device in such a way that optimizes access time. In what order should the files be stored ?
F5, F2, F1, F3, F6, F4 | |
F4, F6, F3, F1, F2, F5 | |
F1, F2, F3, F4, F5, F6 | |
F6, F5, F4, F3, F2, F1 |
Question 248 Explanation:
Here, clearly mentioned that files are stored in sequential device. It is like optimal merge pattern. We want to get optimum access time, first sort all the files according to their sizes. For sorting it will take worst case O(nlogn) but will give optimal result. So, Option D is correct order.
Note: Merging files required m+n-1 comprisions
Note: Merging files required m+n-1 comprisions
Question 249 |
Which module gives control of the CPU to the process selected by the short - term scheduler ?
Dispatcher | |
Interrupt | |
Schedular | |
Threading |
Question 249 Explanation:
The CPU-scheduling function is the dispatcher, which is the module that gives control of the CPU to the process selected by the short-term scheduler. It receives control in kernel mode as the result of an interrupt or system call. The functions of a dispatcher involve the following:
1. Context switches, in which the dispatcher saves the state (also known as context) of the process or thread that was previously running; the dispatcher then loads the initial or previously saved state of the new process.
2. Switching to user mode.
3. Jumping to the proper location in the user program to restart that program indicated by its new state.
The dispatcher should be as fast as possible, since it is invoked during every process switch. During the context switches, the processor is virtually idle for a fraction of time, thus unnecessary context switches should be avoided. The time it takes for the dispatcher to stop one process and start another is known as the dispatch latency.
1. Context switches, in which the dispatcher saves the state (also known as context) of the process or thread that was previously running; the dispatcher then loads the initial or previously saved state of the new process.
2. Switching to user mode.
3. Jumping to the proper location in the user program to restart that program indicated by its new state.
The dispatcher should be as fast as possible, since it is invoked during every process switch. During the context switches, the processor is virtually idle for a fraction of time, thus unnecessary context switches should be avoided. The time it takes for the dispatcher to stop one process and start another is known as the dispatch latency.
Question 250 |
Two atomic operations permissible on Semaphores are __________ and __________.
wait, stop | |
wait, hold | |
hold, signal | |
wait, signal |
Question 250 Explanation:
A semaphore is a variable or abstract data type used to control access to a common resource by multiple processes in a concurrent system such as a multitasking operating system. Two atomic operations are
1. wait: If the value of semaphore variable is not negative, decrement it by 1. If the semaphore variable is now negative, the process executing wait is blocked (i.e., added to the semaphore's queue) until the value is greater or equal to 1. Otherwise, the process continues execution, having used a unit of the resource.
2. signal: Increments the value of semaphore variable by 1. After the increment, if the pre-increment value was negative (meaning there are processes waiting for a resource), it transfers a blocked process from the semaphore's waiting queue to the ready queue.
1. wait: If the value of semaphore variable is not negative, decrement it by 1. If the semaphore variable is now negative, the process executing wait is blocked (i.e., added to the semaphore's queue) until the value is greater or equal to 1. Otherwise, the process continues execution, having used a unit of the resource.
2. signal: Increments the value of semaphore variable by 1. After the increment, if the pre-increment value was negative (meaning there are processes waiting for a resource), it transfers a blocked process from the semaphore's waiting queue to the ready queue.
Question 251 |
Consider the following statements:
a) DBA uses ‘grant’ statement to confer authentication
b) Privileges granted to ‘public’ are granted only to current users.
c) Privileges granted to a user cannot be passed on to other users
Which of the following is the correct option?
a) DBA uses ‘grant’ statement to confer authentication
b) Privileges granted to ‘public’ are granted only to current users.
c) Privileges granted to a user cannot be passed on to other users
Which of the following is the correct option?
Only a is correct | |
Both a and b are correct | |
Both a and c are correct | |
All a,b,c are correct |
Question 251 Explanation:
Authorization includes primarily two processes:
→Permitting only certain users to access, process, or alter data
→Applying varying limitations on user access or actions
A privilege is a right to execute a particular type of SQL statement or to access another user's object. Some examples of privileges include the right to:
→Connect to the database (create a session)
→Create a table
→Select rows from another user's table
→Execute another user's stored procedure
You grant privileges to users so these users can accomplish tasks required for their jobs. You should grant a privilege only to a user who requires that privilege to accomplish the necessary work
System Privileges
A system privilege is the right to perform a particular action, or to perform an action on any schema objects of a particular type
→Permitting only certain users to access, process, or alter data
→Applying varying limitations on user access or actions
A privilege is a right to execute a particular type of SQL statement or to access another user's object. Some examples of privileges include the right to:
→Connect to the database (create a session)
→Create a table
→Select rows from another user's table
→Execute another user's stored procedure
You grant privileges to users so these users can accomplish tasks required for their jobs. You should grant a privilege only to a user who requires that privilege to accomplish the necessary work
System Privileges
A system privilege is the right to perform a particular action, or to perform an action on any schema objects of a particular type
Question 252 |
An application programmer prefers using application programming interfaces rather than invoking system calls. Which of the following is NOT the correct reason for his preference?
APIs execute faster than system calls | |
APIs are easy to use | |
Use of APIs makes the program portable | |
APIs can potentially improve readability of the code |
Question 252 Explanation:
Question 253 |
Three processes are scheduled for execution.
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?
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 |
Question 253 Explanation:
The gantt chart of the above RR scheduling algorithm is
The completion of order of processes are P3,P1 and P2
The completion of order of processes are P3,P1 and P2
Question 254 |
Consider a pair of producer consumer process, sharing the following variables:
int n;
Semaphore M=1;
Semaphore E=n; /*size of buffer*/
Semaphore F=0;
The consumer process must execute ____ and _____ after removing an item from the buffer.
int n;
Semaphore M=1;
Semaphore E=n; /*size of buffer*/
Semaphore F=0;
The consumer process must execute ____ and _____ after removing an item from the buffer.
wait(M), signal(E) | |
signal(E),signal(M) | |
signal(M),signal(E) | |
signal(E),wait(M) |
Question 254 Explanation:
→Two standard operations, wait and signal are defined on the semaphore.
→Entry to the critical section is controlled by the wait operation and exit from a critical region is taken care by signal operation.
→The manipulation of semaphore (S) takes place as following:
→The wait command P(S) decrements the semaphore value by 1. If the resulting value becomes negative then P command is delayed until the condition is satisfied.
→The V(S) i.e. signals operation increments the semaphore value by 1.
→The wait, signal operations are also called P and V operations.
→Entry to the critical section is controlled by the wait operation and exit from a critical region is taken care by signal operation.
→The manipulation of semaphore (S) takes place as following:
→The wait command P(S) decrements the semaphore value by 1. If the resulting value becomes negative then P command is delayed until the condition is satisfied.
→The V(S) i.e. signals operation increments the semaphore value by 1.
→The wait, signal operations are also called P and V operations.
Question 255 |
First fit and best fit strategies for memory allocation suffer from ____ and _____ fragmentation, respectively.
Internal,internal | |
Internal,external | |
External,external | |
external,internal |
Question 255 Explanation:
→Internal fragmentation is the wasted space within each allocated block because of rounding up from the actual requested allocation to the allocation granularity.
→External fragmentation is the various free spaced holes that are generated in either your memory or disk space.
→Both the first-fit and best-fit strategies for memory allocation suffer from external fragmentation. As the processes are loaded and removed from memory, the free memory space is broken into little pieces.
→ External fragmentation exists when there is enough total memory space to satisfy a request, but the available spaces are not contiguous.
→External fragmentation is the various free spaced holes that are generated in either your memory or disk space.
→Both the first-fit and best-fit strategies for memory allocation suffer from external fragmentation. As the processes are loaded and removed from memory, the free memory space is broken into little pieces.
→ External fragmentation exists when there is enough total memory space to satisfy a request, but the available spaces are not contiguous.
Question 256 |
Consider a hard disk of capacity 1 TB, and block size 1 KB. If the free space list is maintained as a bitmap, then the size of the bit vector is
1012 bits | |
109 bits | |
106 bits | |
103 bits |
Question 256 Explanation:
1 Byte = 8 bits
1 KB=1024B(103)
1MB=1024KB(106)
1GB=1024MB(109)
1TB=1024GB(1012)
Size of the bit vector is 1 TB/1 kKB=1012/ 103=109bits
1 KB=1024B(103)
1MB=1024KB(106)
1GB=1024MB(109)
1TB=1024GB(1012)
Size of the bit vector is 1 TB/1 kKB=1012/ 103=109bits
Question 257 |
Degree of multiprogramming is controlled by
Medium term scheduler | |
Long term scheduler | |
Number of schedulers in process management module | |
CPU scheduler |
Question 257 Explanation:
→Long Term Scheduler. It is also called a jobscheduler.
→A long-term scheduler determines which programs are admitted to the system for processing. It selects processes from the queue and loads them into memory for execution.
→The primary objective of the job scheduler is to provide a balanced mix of jobs, such as I/O bound and processor bound. It also controls the degree of multiprogramming.
→A long-term scheduler determines which programs are admitted to the system for processing. It selects processes from the queue and loads them into memory for execution.
→The primary objective of the job scheduler is to provide a balanced mix of jobs, such as I/O bound and processor bound. It also controls the degree of multiprogramming.
Question 258 |
Which of the following statement(s) regarding a linker software is/are true ?
I A function of a linker is to combine several object modules into a single load module.
II A function of a linker is to replace absolute references in an object module by symbolic references to locations in other modules.
I A function of a linker is to combine several object modules into a single load module.
II A function of a linker is to replace absolute references in an object module by symbolic references to locations in other modules.
Only I | |
Only II | |
Both I and II | |
Neither I nor II |
Question 258 Explanation:
TRUE: A function of a linker is to combine several object modules into a single load module.
→ A linker combines one or more object files and possible some library code into either some executable, some library or a list of error messages.
FALSE: A function of a linker is to replace absolute references in an object module by symbolic references to locations in other modules.
FALSE: A function of a linker is to replace absolute references in an object module by symbolic references to locations in other modules.
Question 259 |
There are three processes P 1 , P 2 and P 3 sharing a semaphore for synchronizing a variable. Initial value of semaphore is one. Assume that negative value of semaphore tells us how many processes are waiting in queue. Processes access the semaphore in following order :
(a) P 2 needs to access
(b) P 1 needs to access
(c) P 3 needs to access
(d) P 2 exits critical section
(e) P 1 exits critical section
The final value of semaphore will be :
(a) P 2 needs to access
(b) P 1 needs to access
(c) P 3 needs to access
(d) P 2 exits critical section
(e) P 1 exits critical section
The final value of semaphore will be :
0 | |
1 | |
-1 | |
-2 |
Question 259 Explanation:
There are 3 processes P 1 , P 2 and P 3 . Given, Initial value of semaphore S is 1.
Step-1: P 2 needs to access, we are allocating critical section to P 2 . S value become 0.
Step-2: P 1 needs to access but processor not available. So, S value become -1. Step-3: P 3 needs to access but processor not available. So, S value become -2. Step-4: P 2 exits critical section. So, we are allocating processor to P 1 then S value is -1. Step-5: P 1 exits critical section. So, we are allocating processor to P 3 then S value is 0.
Step-1: P 2 needs to access, we are allocating critical section to P 2 . S value become 0.
Step-2: P 1 needs to access but processor not available. So, S value become -1. Step-3: P 3 needs to access but processor not available. So, S value become -2. Step-4: P 2 exits critical section. So, we are allocating processor to P 1 then S value is -1. Step-5: P 1 exits critical section. So, we are allocating processor to P 3 then S value is 0.
Question 260 |
In a paging system, it takes 30 ns to search translation Lookaside Buffer (TLB) and 90 ns to access the main memory. If the TLB hit ratio is 70%, the effective memory access time is :
48ns | |
147ns | |
120ns | |
84ns |
Question 260 Explanation:
Effective memory access(EMA)=Hit ratio*(TLB access time + Main memory access time) +(1–hit ratio) * (TLB access time + 2 * main memory time)
EAM=0.7*(30+90)+0.3(30+(2*90))
=0.7*120 + 0.3(30+(180))
=0.7*120 + 0.3*210
= 84 + 63
= 147
EAM=0.7*(30+90)+0.3(30+(2*90))
=0.7*120 + 0.3(30+(180))
=0.7*120 + 0.3*210
= 84 + 63
= 147
Question 261 |
Which of the following scheduling algorithms may cause starvation ?
a. First-come-first-served
b. Round Robin
c. Priority
d. Shortest process next
e. Shortest remaining time first
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 |
Question 261 Explanation:
→ Starvation is the name given to the indefinite postponement of a process because it requires some resource before it can run, but the resource, though available for allocation, is never allocated to this process.
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.
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 262 |
Distributed operating systems consist of:
Loosely coupled O.S. software on a loosely coupled hardware. | |
Loosely coupled O.S. software on a tightly coupled hardware. | |
Tightly coupled O.S. software on a loosely coupled hardware. | |
Tightly coupled O.S. software on a tightly coupled hardware. |
Question 262 Explanation:
→ Distributed System is a collection of independent computers which can cooperate, but which appear to users of the system as a uniprocessor computer.
1. Tightly coupled: Short delay in communication between computers, high data rate (e.g., Parallel computers working on related computations)
Eg: Tightly coupled operating system is software
2. Loosely coupled: Large delay in communications, Low data rate (Distributed Systems working on unrelated computations)
Eg: Loosely coupled operating system is hardware.
1. Tightly coupled: Short delay in communication between computers, high data rate (e.g., Parallel computers working on related computations)
Eg: Tightly coupled operating system is software
2. Loosely coupled: Large delay in communications, Low data rate (Distributed Systems working on unrelated computations)
Eg: Loosely coupled operating system is hardware.
Question 263 |
Consider a system with seven processes A through G and six resources R through W.
Resource ownership is as follows :
process A holds R and wants T
process B holds nothing but wants T
process C holds nothing but wants S
process D holds U and wants S & T
process E holds T and wants V
process F holds W and wants S
process G holds V and wants U
Is the system deadlocked ?
If yes, ______ processes are deadlocked.
No | |
Yes, A, B, C | |
Yes, D, E, G | |
Yes, A, B, F |
Question 263 Explanation:
Visual inspection shows that D, E, and G are deadlocked.
Question 264 |
Suppose that the virtual Address space has eight pages and physical memory with four page frames. If LRU page replacement algorithm is used, _____ number of page faults occur with the reference string.
0 2 1 3 5 4 6 3 7 4 7 3 3 5 5 3 1 1 1 7 2 3 4 1
11 | |
12 | |
10 | |
9 | |
None of the above |
Question 264 Explanation:
Total number of page frames are 13.
None: Actually they given options are wrong. Excluded for evaluation.
Question 265 |
Consider a system having ‘m’ resources of the same type. These resources are shared by three processes P 1 , P 2 and P 3 which have peak demands of 2, 5 and 7 resources respectively. For what value of ‘m’ deadlock will not occur ?
70 | |
14 | |
13 | |
7 | |
None of the above |
Question 265 Explanation:
A requires 3, B-4, C-7;
→ If A have 2, B have 3, C have 6 then deadlock will occur i.e., 2+3+6=11.
→ If we have one extra resource then deadlock will not occur i.e., 11+1=12.
→ If we have equal (or) more than 12 resources then deadlock will never occur.
Note: Actual answer given option-B but it satisfies option-A,B,C.
→ If A have 2, B have 3, C have 6 then deadlock will occur i.e., 2+3+6=11.
→ If we have one extra resource then deadlock will not occur i.e., 11+1=12.
→ If we have equal (or) more than 12 resources then deadlock will never occur.
Note: Actual answer given option-B but it satisfies option-A,B,C.
Question 266 |
Five jobs A, B, C, D and E are waiting in Ready Queue. Their expected runtimes are 9, 6, 3,5 and x respectively. All jobs entered in Ready queue at time zero. They must run in _____order to minimize average response time if 3 < x < 5.
B, A, D, E, C | |
C, E, D, B, A | |
E, D, C, B, A | |
C, B, A, E, D |
Question 266 Explanation:
Shortest job first is the way to minimize average response time.
0 < X ≤ 3: X, 3, 5, 6, 9
3 < X ≤ 5: 3, X, 5, 6, 9
5 < X ≤ 6: 3, 5, X, 6, 9
6 < X ≤ 9: 3, 5, 6, X, 9
X > 9: 3, 5, 6, 9, X
C, E, D, B, A
So, option B is correct.
0 < X ≤ 3: X, 3, 5, 6, 9
3 < X ≤ 5: 3, X, 5, 6, 9
5 < X ≤ 6: 3, 5, X, 6, 9
6 < X ≤ 9: 3, 5, 6, X, 9
X > 9: 3, 5, 6, 9, X
C, E, D, B, A
So, option B is correct.
Question 267 |
Consider three CPU intensive processes P1 , P2 , P3 which require 20, 10 and 30 units of time, arrive at times 1, 3 and 7 respectively. Suppose operating system is implementing Shortest Remaining Time first (preemptive scheduling) algorithm, then _____ context switches are required (suppose context switch at the beginning of Ready queue and at the end of Ready queue are not counted).
3 | |
2 | |
4 | |
5 |
Question 267 Explanation:
P 1 → P 2 is one switchP 2 → P 1 is 2nd switch
P 3 → P 3 is 3rd switch
Question 268 |
Suppose there are four processes in execution with 12 instances of a Resource R in a system.
The maximum need of each process and current allocation are given below:
With reference to current allocation, is system safe ? If so, what is the safe sequence ?
With reference to current allocation, is system safe ? If so, what is the safe sequence ?No | |
Yes, P 1 P 2 P 3 P 4 | |
Yes, P 4 P 3 P 1 P 2 | |
Yes, P 2 P 1 P 3 P 4 |
Question 268 Explanation:
Step-1: Current allocation for all processors are 3+4+2+1=10.
Step-2: Given total number of resources is 12. But we already allocated 12-10=2.
It means 2 resources are free. we can call M=2
Step-3: P4 requires 2 free resources. Total free resources are 2. So, we can allocate for P4.
Then M become M=3
Step-4: P3 requires 2 free resources. Total free resources are 2. So, we can allocate for P3.
Then M become M=5
Step-5: P1 requires 5 free resources. Total free resources are 5. So, we can allocate for P1.
Then M become M=8
Step-5: P2 requires 5 free resources. Total free resources are 8. So, we can allocate for P2.
Then M become M=12
Note: We have alternative possibility is P4,P3,P2,P1.
Step-2: Given total number of resources is 12. But we already allocated 12-10=2.
It means 2 resources are free. we can call M=2
Step-3: P4 requires 2 free resources. Total free resources are 2. So, we can allocate for P4.
Then M become M=3
Step-4: P3 requires 2 free resources. Total free resources are 2. So, we can allocate for P3.
Then M become M=5
Step-5: P1 requires 5 free resources. Total free resources are 5. So, we can allocate for P1.
Then M become M=8
Step-5: P2 requires 5 free resources. Total free resources are 8. So, we can allocate for P2.
Then M become M=12
Note: We have alternative possibility is P4,P3,P2,P1.
Question 269 |
If the Disk head is located initially at track 32, find the number of disk moves required with FCFS scheduling criteria if the disk queue of I/O blocks requests are:
98, 37, 14, 124, 65, 67
320 | |
322 | |
321 | |
319 |
Question 269 Explanation:
Question 270 |
A scheduling Algorithm assigns priority proportional to the waiting time of a process. Every process starts with priority zero (lowest priority). The scheduler reevaluates the process priority for every ‘T’ time units and decides next process to be scheduled. If the process have no I/O operations and all arrive at time zero, then the scheduler implements _________ criteria.
Priority scheduling | |
Round Robin Scheduling | |
Shortest Job First | |
FCFS |
Question 270 Explanation:
→ A scheduling Algorithm assigns priority proportional to the waiting time of a process. Every process starts with priority zero (lowest priority). The scheduler reevaluates the process priority for
every ‘T’ time units and decides next process to be scheduled. If the process have no I/O operations and all arrive at time zero, then the scheduler implements Round Robin criteria.
Round Robin Scheduling: → To schedule processes fairly, a round-robin scheduler generally employs time-sharing, giving each job a time slot or quantum(its allowance of CPU time), and interrupting the job if it is not completed by then. The job is resumed next time a time slot is assigned to that process. If the process terminates or changes its state to waiting during its attributed time quantum, the scheduler selects the first process in the ready queue to execute.
→ Round-robin algorithm is a pre-emptive algorithm as the scheduler forces the process out of the CPU once the time quota expires.
Round Robin Scheduling: → To schedule processes fairly, a round-robin scheduler generally employs time-sharing, giving each job a time slot or quantum(its allowance of CPU time), and interrupting the job if it is not completed by then. The job is resumed next time a time slot is assigned to that process. If the process terminates or changes its state to waiting during its attributed time quantum, the scheduler selects the first process in the ready queue to execute.
→ Round-robin algorithm is a pre-emptive algorithm as the scheduler forces the process out of the CPU once the time quota expires.
Question 271 |
A virtual memory has a page size of 1K words. There are eight pages and four blocks. The associative memory page table contains the following entries:
Which of the following list of virtual addresses (in decimal) will not cause any page fault if referenced by the CPU ?
Which of the following list of virtual addresses (in decimal) will not cause any page fault if referenced by the CPU ?1024, 3072, 4096, 6144 | |
1234, 4012, 5000, 6200 | |
1020, 3012, 6120, 8100 | |
2021, 4050, 5112, 7100 |
Question 271 Explanation:
Step-1: Total number of blocks=4
Total number of pages=8
Virtual memory has a page size of 1K(1024) words.
Step-2: Page ranges
Total number of pages=8
Virtual memory has a page size of 1K(1024) words.
Step-2: Page ranges
Question 272 |
Suppose that the number of instructions executed between page fault is directly proportional to the number of page frames allocated to a program. If the available memory is doubled, the mean interval between page faults is also doubled. Further,
consider that a normal instruction takes one microsecond, but if a page fault occurs, it takes 2001 microseconds. If a program takes 60 sec to run, during which time it gets 15,000 page faults, how long would it take to run if twice as much memory were
available?
60 sec | |
30 sec | |
45 sec | |
10 sec |
Question 272 Explanation:
Normal instruction takes 1 microsecond = 10-6 sec
→ Instruction with page fault takes 2001 microseconds so page fault takes additional 2000 microsecond
→ The given program takes 60 sec to run and in that program there were 15000 p.f
→ Since time taken by 1 page fault is 2000 microseconds, time taken by all 15000 page faults = 15000*2000 microseconds
=30 sec
→ Out of overall 60 sec of the program time 30 sec are taken for page faults, remaining 30 sec are consumed by program execution.
→ If memory is doubled we will have more memory available to hold the pages and hence page faults also reduce. It is given in the question that "if memory is doubled then the mean interval between the page faults is also doubled" this is nothing but "number of page faults will be reduced by half".
→ Earlier 30 sec were needed by program instructions and 30 sec for page faults.
→ Now program execution takes 30 sec and page fault will take only 15 sec.
So new total time =30+15
=45 sec
→ Instruction with page fault takes 2001 microseconds so page fault takes additional 2000 microsecond
→ The given program takes 60 sec to run and in that program there were 15000 p.f
→ Since time taken by 1 page fault is 2000 microseconds, time taken by all 15000 page faults = 15000*2000 microseconds
=30 sec
→ Out of overall 60 sec of the program time 30 sec are taken for page faults, remaining 30 sec are consumed by program execution.
→ If memory is doubled we will have more memory available to hold the pages and hence page faults also reduce. It is given in the question that "if memory is doubled then the mean interval between the page faults is also doubled" this is nothing but "number of page faults will be reduced by half".
→ Earlier 30 sec were needed by program instructions and 30 sec for page faults.
→ Now program execution takes 30 sec and page fault will take only 15 sec.
So new total time =30+15
=45 sec
Question 273 |
Consider a disk with 16384 bytes per track having a rotation time of 16 msec and average seek time of 40 msec. What is the time in msec to read a block of 1024 bytes from this disk?
57 sec | |
49 sec | |
48 sec | |
17 sec |
Question 273 Explanation:
Given data,
-- Bytes per track=16384
-- Rotational time=16 msec.
-- Average seek time= 40 msec
-- What is the time in msec to read a block of 1024 bytes from this disk?
Step-1: Total time= Seek Time + Rotational Latency + Transfer Time = ? + ? + 40 msec
Step-2: Transfer time = ((1024 bytes *16 ms)/ (16384 bytes )) = 1 msec
Step-3: Rotational latency = 16 msec / 2
= 8 msec
Step-4: Total time= Seek Time + Rotational Latency + Transfer Time
= 40+8+1 msec
= 49 msec
-- Bytes per track=16384
-- Rotational time=16 msec.
-- Average seek time= 40 msec
-- What is the time in msec to read a block of 1024 bytes from this disk?
Step-1: Total time= Seek Time + Rotational Latency + Transfer Time = ? + ? + 40 msec
Step-2: Transfer time = ((1024 bytes *16 ms)/ (16384 bytes )) = 1 msec
Step-3: Rotational latency = 16 msec / 2
= 8 msec
Step-4: Total time= Seek Time + Rotational Latency + Transfer Time
= 40+8+1 msec
= 49 msec
Question 274 |
A system has four processes and five allocatable resources. The current allocation and maximum needs are as follows:
The smallest value of ‘x’ for which the above system in safe state is __________.
The smallest value of ‘x’ for which the above system in safe state is __________.1 | |
3 | |
2 | |
Not safe for any value of x. |
Question 274 Explanation:
Case 1: Let’s consider x=1
→ Now need of Process D can be fulfilled. After completing its execution Process D leaves its allocated resources.
→ Now Need of any either Process can’t be fulfilled. So the system is not in safe state with x=1.
Case 2: x=2
Available = 0 0 2 1 1
→ Need of Process D can be fulfilled and after its execution it release its allocated resources.
Available
= 0 0 2 1 1
D
= 1 1 1 1 0
---------------------------------- New available = 1 1 2 2 1
→ Now Need of Process ‘C’ can be fulfilled.
Available
= 1 1 3 2 1
C
= 1 1 0 1 0
----------------------------------
New available = 2 2 3 3 1
→ Now Need of Process ‘B’ can be fulfilled. So new available resources are
Available
= 2 2 3 3 1
C
= 2 0 1 1 0
----------------------------------
New available = 4 2 4 4 1
→ But now need of Process ‘A’ can’t be fulfilled with above available resources. So the system is not in safe state with x=2.
Case 3: x=3
Available = 0 0 3 1 1
→ Now Need of Process ‘D’ can be fulfilled and after its execution it will release its allocated resources.
Available
= 0 0 3 1 1
D
= 1 1 1 1 0
----------------------------------
New available = 1 1 4 2 1
→ Now Need of Process ‘C’ can be fulfilled
Available
= 1 1 4 2 1
C
= 1 1 0 1 0
----------------------------------
New available = 2 2 4 3 1
→ Now Need of Process ‘B’ can be fulfilled
Available
= 2 2 4 3 1
B
= 2 0 1 1 0
----------------------------------
New available = 4 2 5 4 1
→ But with available resource need of Process ‘A’ can’t be fulfilled. So the system is not in safe state with x=3.
→ Hence option (D) is the correct answer.
Question 275 |
A data cube C, has n dimensions, and each dimension has exactly p distinct values in the base cuboid. Assume that there are no concept hierarchies associated with the dimensions. What is the maximum number of cells possible in the data cube, C?
p n | |
p | |
(2 n - 1)p+1 | |
(p + 1) n |
Question 275 Explanation:
Option-A: This is the maximum number of distinct tuples that you can form with p distinct values per dimensions.
Option-B: You need at least p tuples to contain p distinct values per dimension. In this case no tuple shares any value on any dimension.
Option-C: The minimum number of cells is when each cuboid contains only p cells, except for the apex, which contains a single cell.
Option-D: We have p+1 because in addition to the p distinct values of each dimension we can also choose ∗.
Option-B: You need at least p tuples to contain p distinct values per dimension. In this case no tuple shares any value on any dimension.
Option-C: The minimum number of cells is when each cuboid contains only p cells, except for the apex, which contains a single cell.
Option-D: We have p+1 because in addition to the p distinct values of each dimension we can also choose ∗.
Question 276 |
System calls are usually invoked by using :
A privileged instruction | |
An indirect jump | |
A software interrupt | |
Polling |
Question 276 Explanation:
→ System calls are usually invoked by using a software interrupt.
→ Software interrupts are implementing device drivers (or) transitions between protected mode of operations, such as system calls.
→ Software interrupts are implementing device drivers (or) transitions between protected mode of operations, such as system calls.
Question 277 |
A disk drive has 100 cyclinders, numbered 0 to 99. Disk requests come to the disk driver for cyclinders 12, 26, 24, 4, 42, 8 and 50 in that order. The driver is currently serving a request at cyclinder 24. A seek takes 6 msec per cyclinder moved. How much seek time is needed for shortest seek time first (SSTF) algorithm?
0.984 sec | |
0.396 sec | |
0.738 sec | |
0.42 sec |
Question 277 Explanation:
Step-1:
Step-2: Total seek time= 420 msec.
Step-3: =420/1000
=0.42
Note: 1000 msec = 1 sec
Step-2: Total seek time= 420 msec.
Step-3: =420/1000
=0.42
Note: 1000 msec = 1 sec
Question 278 |
Let P i and P j be two processes, R be the set of variables read from memory, and W be the set of variables written to memory. For the concurrent execution of two processes P i and P j which of the following conditions is not true?
R(P i ) ∩ W(P j ) = Ф | |
W(P i ) ∩ R(P j ) = Ф | |
R(P i ) ∩ R(P j ) = Ф | |
W(P i ) ∩ W(P j ) = Ф |
Question 278 Explanation:
Option-A: It performs Read and Write. So, it violates concurrent execution of two processes Pi and Pj
Option-B: It performs write and Read. So, it violates concurrent execution of two processes Pi and Pj
Option-C: It performs Read and Write. So, it is not violates concurrent execution of two processes Pi and Pj
Option-D: It performs Write and Write. So, it violates concurrent execution of two processes Pi and Pj
Option-B: It performs write and Read. So, it violates concurrent execution of two processes Pi and Pj
Option-C: It performs Read and Write. So, it is not violates concurrent execution of two processes Pi and Pj
Option-D: It performs Write and Write. So, it violates concurrent execution of two processes Pi and Pj
Question 279 |
A LRU page replacement is used with four page frames and eight pages. How many page faults will occur with the reference string 0172327103 if the four frames are initially empty?
6 | |
7 | |
8 | |
5 |
Question 279 Explanation:
Question 280 |
The Unix Kernel maintains two key data structures related to processes, the process table and the user structure. Which of the following information is not the part of user structure?
File descriptor table | |
System call state | |
Scheduling parameters | |
Kernel stack |
Question 280 Explanation:
The Unix Kernel maintains two key data structures related to processes, the process table and the user structure.
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.
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 281 |
Semaphores are used to :
Synchronise critical resources to prevent deadlock | |
Synchronise critical resources to prevent contention | |
Do I/o | |
Facilitate memory management |
Question 281 Explanation:
→ Semaphores are implemented to prevent deadlocks using wait and signal operation.
→ A semaphore is a variable or abstract data type used to control access to a common resource by multiple processes in a concurrent system such as a multitasking operating system.
→ A semaphore is simply a variable. This variable is used to solve critical section problems and to achieve process synchronization in the multi processing environment.
→ A semaphore is a variable or abstract data type used to control access to a common resource by multiple processes in a concurrent system such as a multitasking operating system.
→ A semaphore is simply a variable. This variable is used to solve critical section problems and to achieve process synchronization in the multi processing environment.
Question 282 |
In which of the following storage replacement strategies, is a program placed in the largest available hole in the memory ?
Best fit | |
First fit | |
Worst fit | |
Buddy |
Question 282 Explanation:
First fit: Allocate the first hole that is big enough. Searching can start either at the beginning of the set of holes or at the location where the previous first-fit search ended. We can stop searching as soon as we find a free hole that is large enough.
Best fit: Allocate the smallest hole that is big enough. We must search the entire list, unless the list is ordered by size. This strategy produces the smallest leftover hole.
Worst fit: Allocate the largest hole. Again, we must search the entire list, unless it is sorted by size. This strategy produces the largest leftover hole, which may be more useful than the smaller leftover hole from a best-fit approach.
Best fit: Allocate the smallest hole that is big enough. We must search the entire list, unless the list is ordered by size. This strategy produces the smallest leftover hole.
Worst fit: Allocate the largest hole. Again, we must search the entire list, unless it is sorted by size. This strategy produces the largest leftover hole, which may be more useful than the smaller leftover hole from a best-fit approach.
Question 283 |
Remote computing system involves the use of time sharing systems and :
Real time processing | |
Batch processing | |
Multiprocessing | |
All of the above |
Question 283 Explanation:
→ Remote computing system involves the use of time sharing systems and Batch processing.
→ Batch processing in which an input device is located at a distance from the main installation and has access to a computer through a communication link.
→ Batch processing in which an input device is located at a distance from the main installation and has access to a computer through a communication link.
Question 284 |
Match the following
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 |
Question 284 Explanation:
Disk scheduling→ SCAN,C-SCAN,LOOK,C-LOOK
Batch processing→ FIFO
Time sharing→ Round robin with the help of time quantum.
Interrupt processing→ LIFO
Batch processing→ FIFO
Time sharing→ Round robin with the help of time quantum.
Interrupt processing→ LIFO
Question 285 |
Producer consumer problem can be solved using :
semaphores | |
event counters | |
monitors | |
all the above |
Question 285 Explanation:
→ Producer consumer problem can be solved using semaphores.
→ A semaphore is a variable or abstract data type used to control access to a common resource by multiple processes in a concurrent system such as a multitasking operating system. Semaphore basically two types.
1. Counting semaphores
2. Binary semaphores
→ A semaphore is a variable or abstract data type used to control access to a common resource by multiple processes in a concurrent system such as a multitasking operating system. Semaphore basically two types.
1. Counting semaphores
2. Binary semaphores
Question 286 |
If you want to execute more than one program at a time, the systems software that are used must be capable of :