OPERATING SYSTEM MCQ

 

1. Which of the following is an example of a spooled device?
   a. A line printer used to print the output of a number of jobs.
   b. The terminal used to enter the input data for a Fortran program being executed.
   c. The secondary memory device in a virtual memory system.
   d. The swapping area on a disk used by the swapper.

In spooling (Simultaneous Peripheral Operation On Line), a buffer is interposed between a running program and a slow-speed device involved with the program for input/output. For example, instead of writing lines directly to the line printer, the lines are written (temporarily) onto a disk. Thus, the program is allowed to run to completion without waiting for the (slow) printer to finish. When the printer becomes free, the lines can be printed from the disk file.

2. Which of the following is not normally contained in the directory entry of a file?
a. Creation date.
b. Access control list.
c. A count of the number of free blocks in the disk.
d. Filename and extension.

The directory entry generally contains information pertaining to a file that includes the file name and possibly some or all of the following: file extension, date of creation, date of last modification, size (may be in kilobytes), access rights, etc. The number of free blocks on the disk is information related to the whole disk and is not kept as part of a file.

3. Disk scheduling involves:
a. Allocating disk space to users in a fair manner.
b. Validating the file control information stored in the file.
c. Examining pending disk requests to determine the most efficient way to service the requests.
d. Reorganizing disk requests to maximize seek time.

In multiprogramming systems, many processes may be generating requests for reading and writing disk records. Because these processes often make requests faster than they can be retrieved by moving head disks, queues build-up for such devices. Some systems use the first-come-first-serve (FCFS) policy to service the disk queues. But FCFS exhibits a seek pattern in which successive requests can cause time-consuming to seek delays. To minimize the time spent in seeking records, it is reasonable to order the request queue. Such a method of ordering the disk requests is called disk scheduling. Disk scheduling generally tries to minimize the head-movement time.

4. In capability-based systems, which of the following statements is true?
a. The unique name of a capability is reused after every time slice.
b. The unique name of a capability is never reused.
c. The unique name of a capability can be reused if there are no references to it.
d. None of these

The unique name in capabilities can be reused, but only if it can be guaranteed that no references remain to the old use of the name.

5. In operating systems, a capability is:
a. A facility that provides global access to all data in the system.
b. A unique and nonforgeable name identifying an object in the system together with access information.
c. A user-maintained list of access privileges of the objects in the system.
d. A table of the available operating system resources.

A capability is a unique nonforgeable name directly identifying an object in a system, together with access information. Users cannot manufacture or modify capabilities. Possession of a capability gives the right to access the object, governed by the access information.

6. An advantage of dynamic linking is that:
a. The segments that are not used in a run need not be linked into the process address space.
b. It reduces execution time overhead.
c. Debugging is simplified because programs are modular.
d. The linker need not construct the known segment table.

Dynamic linking allows procedures to be linked and loaded into memory when they are first called. Hence, procedure segments that are not required for a particular execution need not be in memory, thereby reducing the amount of memory required.

7. With reference to Multics, which of the following statements is false?
a. Every process must have a separate linkage segment for every shared segment.
b. The linkage segment need not be used to resolve internal references of a segment.
c. When control is transferred from one segment to another, the linkage pointer must be changed to point to the new linkage segment.
d. The linkage segment is constructed at the time of linking.

A linkage segment is only required when there are external references from a shared segment. Hence, if some objects are shared but have no external references, then there will be no linkage segment.

8. One of the ways of sharing segments in a segmented system is by:
a. Maintaining a common segment table containing the information about shared segments.
b. Dividing the program into procedures and data and allowing only the procedures to be shared.
c. Dividing the shared segment into a set of pages and allowing only certain pages to be shared.
d. None of the above, as segments are larger than pages, and hence, cannot be shared.

The sharing of segments in a segmented system can be done by having a common segment table. All virtual addresses above a certain segment number can be translated using the information in the common segment table rather than the entries in the local segment table. This method will avoid the duplication of the entries in the shared segments in each local segment table.

9. Sharing in a paged memory system is done by:
a. Giving a copy of the shared pages to each process.
b. Dividing the program into procedures and data and allowing only the procedures to be shared.
c. Several page table entries pointing to the same frame in the main memory.

Sharing pages can be achieved either by providing a copy of the page or by having the same entry in the page tables that point to the shared page. The latter does not have a duplicate copy of the page, and hence, the memory-space requirement and the overhead involved in maintaining the
consistency of shared pages is reduced. There need be no restriction that only procedures and not data can be shared in a paged system.

10. In a paged segmented scheme of memory management, the segment table points to a page table because:
a. The segment table may occasionally be too large to fit in one page.
b. each segment may be spread over a number of pages.
c. Page size is usually larger than the segment size.
d. The page table may be too large to fit into a single segment.

Segments are variable-sized memory blocks. In a paged segmented system, each segment is divided into one or more fixed-size pages. Hence, an entry for a segment points to a page table that contains the addresses of the pages corresponding to that segment.

11. It is advantageous for the page size to be small because:
a. Less unreferenced(not comparable) data will be loaded into memory.
b. Virtual addresses will be smaller.
c. Page tables will be smaller.
d. Large programs can be run.

By reducing the page size, the amount of unreferenced data held in memory will be reduced. However, the size of the page table will increase.

12. It is advantageous for the page size to be large because:
a. Less unreferenced data will be loaded into memory.
b. Virtual addresses will be smaller.
c. Page tables will be smaller.
d. Large programs can be run.

Increasing the size of the page will decrease the number of pages of the addressing space, and hence, the number of entries in the page will be smaller.

13. Which of the following statements is false?
a. With the Least Recently Used (LRU) page-replacement policy, when the page size is halved(divide into two parts of equal or roughly equal size.), the number of page faults can be more than double the original number of page faults.
b. The working set size is a monotonically nondecreasing function of the working set parameter.
c. When the working set policy is used, main memory may contain some pages which do not belong to the working set of any program.

For a given pattern of references and a given page size, the number of page faults is governed by the number of different pages that are accessed. If the page size is halved, then no more than the original number of full-sized pages can be accessed, and thus, no more than twice the original
a number of page faults can occur.

14. The FIFO page-replacement policy:
a. Is based on program locality.
b. Sometimes can cause more page faults when memory size is increased.
c. Is not easy to implement, and hence, most systems use an approximation of FIFO.

It would seem reasonable to expect that if more pages are allocated to a process, then it should experience fewer page faults. But under the FIFO page-replacement strategy, certain page-reference patterns actually cause more page faults when the number of pages allocated to the process is increased. This phenomenon is known as Belady's anomaly. For more information refer to Belady, L.A., Nelson, R.A., and Shedler, G.S., "An Anomaly in Space-Time Characteristics of Certain Programs Running in a Paging Environment"

15. In a variable partition-memory management scheme, internal fragmentation occurs when:
a. Sufficient memory is available to run a program, but it is scattered between existing partitions.
b. Insufficient memory is available to run a program.
c. The partition allocated to a program is larger than the memory required by the program.
d. A program is larger than the size of memory on the computer.

Internal fragmentation occurs when a partition allocated to a program is larger than that required. The difference between these two sizes is the amount of memory that is wasted. Choice (a) refers to external fragmentation.

16. Which of the following statements is true?
a. When the best-fit method of allocating segments is used, it is preferable to order the list of free blocks according to increasing memory addresses.
b. The best-fit method chooses the largest free block in which the given segment can fit.
c. In general, the first-fit allocation algorithm will be faster than the best-fit algorithm.
d. The tagged method of deallocating segments is fast when the list of free blocks is ordered according to increasing memory addresses.

In the first-fit allocation algorithm, the free blocks are searched sequentially until a free block whose size is greater than or equal to the requested memory size is found. The best-fit algorithm tries to find and allocate the free block (if it exists) that is the closest (but at least as large) to the required size. The empirical study shows that a first-fit allocation is generally faster than a best-fit allocation although there are circumstances where either one can be better than the other.

17. Which of the following statements is true for the Banker's algorithm?
a. It cannot be used for a system with many resources, each of which is unique with no multiple copies.
b. It is used to detect deadlock.
c. It is not applicable when a resource is shared simultaneously by many users.
d. An unsafe situation will always lead to a deadlock.

The Banker's algorithm (initially described by Dijkstra in 1965) is an algorithm that will detect whether a given allocation of resources could result in a deadlock. It does not detect the existence of a deadlock. The algorithm can handle single or multiple equivalent or independent resources. The algorithm is pessimistic in that it flags situations as being potential deadlocks when, in fact, a deadlock may not occur. Simultaneously shared resources can never be in a deadlock situation, and thus, the algorithm is inapplicable to these cases.

18. One solution to the Dining Philosophers problem which avoids deadlock is:
a. Non-preemptive scheduling.
b. Ensuring that all philosophers pick up their left fork before they pick up their right fork.
c. Ensuring that all philosophers pick up their right fork before they pick up their left fork.
d. Ensuring that odd philosophers pick up their left fork before they pick up their right fork and even philosophers pick up their right fork before they pick up their left fork.

The scheduling algorithm given (d) will guarantee that there will be no deadlock. Both (b) and (c) allow the possibility of all philosophers picking up one fork resulting in deadlock. There are other solutions to the Dining Philosophers problem.

19. Which of the following is not a necessary condition for a deadlock?
a. Mutually exclusive use of a resource by processes.
b. Partial allocation of resources to a process.
c. Preemptive scheduling of resources.
d. Circular waiting by processes.

In a deadlock, processes never finish executing, and systems resources are tied up. This prevents other jobs from ever starting. The following are the necessary conditions that must be satisfied for a deadlock to exist.

• Processes claim exclusive control of the resources they require mutual exclusion condition
• Process hold resources already allocated to them while waiting for additional resources (hold and wait condition)
• Resources cannot be removed from the processes holding them until the resources are used to completion (no preemption condition)
• A circular chain of processes exists in which each process holds one or more resources that are requested by the next process in the chain (circular-wait condition).

20. Which of the following statements if false?
a. I/O-bound processes should be given priority in scheduling over CPU-bound processes to ensure good turnaround time.
b. Users can exploit a multilevel feedback-scheduling policy by breaking a long job into several small jobs.
c. The processor scheduler normally classifies a process as being a CPU-bound process if it uses most of the previous time slice allocated to it.
d. The round-robin-scheduling policy allocates a time slice to a process depending on the number of time slices it has already used.

I/O-bound processes use little CPU time in general, and hence, in order to have better turnaround time (defined as the number of processes completing execution per unit of time), I/O processes should be given more priority over CPU-bound processes. In the case of a multilevel feedback system, multiple wait queues are maintained. The queue on which a process waits for CPU allocation is determined by its previous queue and the amount of CPU time used in the previous allocated CPU time slice. If a process uses its entire time slice (i.e., a CPU-bound process) it is moved down in the hierarchy and will thus have a lower priority in the next CPU-time-slice allocation. Hence, response to a long job can be improved by dividing it into several smaller jobs.

21. Which of the following statements is true?
a. A multiprogrammed system gives a better average turnaround than a non-multiprogrammed system.
b. When a job-scheduling policy is changed, it is possible for the average turnaround time to decrease while the average priority-weighted turnaround time increases.
c. There is no job scheduler in a time-sharing system.
d. Indefinite postponement of a job is possible if the First-Come-First-Serve-job-scheduling policy is used.

Multiprogramming systems do not always perform better than single program systems because of the overhead of switching between processes. It is possible for the average turnaround to improve because of the change in the job scheduling policy, but at the same time, it may result in a longer turnaround for high-priority jobs. This could happen when the priority of the job is given less importance in the new scheduling algorithm. Time-sharing systems sometimes have a job scheduler to limit the number of logged-on users to a level that the system can cope with. Under the first-come-first-serve algorithm, it is not possible for a job to be indefinitely postponed because jobs are executed in the order of their arrival.

22. The pure-round-robin-scheduling policy:
a. Responds poorly to short processes if the time slice is small.
b. Does not use any prior information about the service times of processes.
c. Becomes equivalent to the Shortest-Job-First Policy when the time slice is made infinitely large.
d. Ensures that the ready queue is always the same size.

With the Round-Robin-scheduling policy, processes are dispatched in FIFO order but are given a limited amount of CPU time, called the time slice or quantum. If a process does not complete before the time slice expires, it is preempted, and the CPU is given to the next waiting process in the ready queue. The preempted process is placed at the back of the ready queue.
Hence no a priori knowledge is used to determine the next process to schedule.

23. Non-preemptive-process-scheduling policies:
a. Are indispensable for interactive systems.
b. Allocate the processor to a process for a fixed time period.
c. Always use a ready queue sorted in order of decreasing priority.
d. Make short jobs wait for long jobs.

In general, a non-preemptive process-scheduling policy can make short jobs wait longer in the ready queue, thereby giving poor response to those processes. This is particularly true when the system is heavily loaded with long jobs.

24. Which of the following actions may result in a process becoming blocked?
a. A process executes a P (wait) operation on a semaphore.
b. A process executes a V (signal) operation on a semaphore.
c. A process exits from a critical section.
d. A process within a critical section changes the value of a shared variable.

V operations can never cause a process to become blocked. Processes can be blocked by executing a P operation.

25. The basic principle of a monitor is that:
a. Several resources can only be controlled by a monitor.
b. Several processes may concurrently execute a procedure of a given monitor.
c. Only one process may execute a procedure of a given monitor at any given time.
d. It schedules the execution of processes in a multiprocessor operating system.

The monitor is a synchronization mechanism based on data abstraction. The shared data is encapsulated in routines, and automatic mutual exclusion is provided on these routines; only one process may execute a procedure of a monitor at any time. Monitors also have a queuing mechanism for waiting on conditions.

26. Which of the following statements is false?
a. Disjoint processes need not use critical sections.
b. Programs with critical sections can never be used simultaneously by more than one process.
c. A process wanting to enter a critical section currently in use must wait for the process utilizing the critical section to terminate.
d. Two different critical sections may be executed concurrently if they do not use the same shared variables.

Disjoint processes that do not share any data need not require critical sections. A critical section contains shared data, and hence, any process wanting to enter the critical section which is in use by another process must wait for the process utilizing the critical section to terminate. Hence, programs with critical sections can be shared by many processes as long as only one process is accessing the data in the critical section at any time.

27. A wait operation on a semaphore should not occur within a critical section controlled by that semaphore because:
a. A deadlock will occur.
b. A semaphore is not a shared variable.
c. Another process may then enter the critical section violating the mutual exclusion constraint.
d. A signal on a semaphore is always given from outside the critical section.

When a semaphore is used to implement mutual exclusion it is initialized to 1. In order to enter the critical region, a process executes a wait on the semaphore which decrements it to zero. The next process attempting to enter will then wait. On exiting the critical region, the process executes a signal which-activates the next process (if there is one waiting) or increments the semaphore. If a process executes a wait on the semaphore while in the critical region it will be suspended, and no other process will be able to enter. Thus the processes waiting for the semaphore will be deadlocked.

28. The main difference between binary semaphores and counting semaphores is that:
a. Binary semaphores can only take the values 0 and 1, while counting semaphores can take any non-negative integer values.
b. Binary semaphores can only be used to solve problems involving up to two processes sharing the same resource, while counting semaphores can be used to solve problems involving more than two processes sharing the same resource.
c. Binary semaphores cannot solve all the problems that can be solved by counting semaphores.
d. Counting semaphores must be controlled by a monitor, while binary semaphores are called directly by user processes.

The power of counting semaphores and binary semaphores is the same. The counting semaphore is particularly useful when a resource is to be allocated from a pool of identical resources. The counting semaphore for such a resource is initialized to the number of identical resources in the pool. Each P operation decrements the value of the semaphore indicating that another resource has been allocated, and each V operation increments the semaphore value by 1 indicating that a process has returned the resource to the pool. If a process attempts a P operation on the semaphore whose value is zero, then the process has to wait until a resource is returned to the pool.

29. A counting semaphore was initialized to 9. Then 27 P (wait) operations and 23 V (signal) operations were completed on this semaphore. The resulting value of the semaphore is:
a. 5
b. 0
c. 7
d. 13

Each P (wait) operating would decrement the semaphore value by one while each V (signal) operation would incremental the semaphore value by one. Thus, for an initial value of 9, 27 waits and 23 signals would result in a semaphore value of 9 - 27 + 23 = 5.

30. A user process can become blocked only if it is:
a. In the ready state
b. In the running state.
c. In the blocked (or waiting) state.
d. Waiting for a resource.

A process can become blocked if it is either ready or in the running state. A process can become blocked while running if it requests a resource in use by another process. It is also possible for a process to become blocked while ready. For example, in a preemptive resource-allocation scheme, a low-priority process could be ready to run. While that process is waiting in the ready queue, if a higher-priority process requests a resource that is held by a ready process, the operating system can preempt the resource and allocate it to the higher-priority process, thereby blocking the process which was in the ready state.

31. The main distinction between a multiprocessor system and a
multiprogrammed system is that in a multiprocessor system:
a. The main storage is shared by several programs.
b. The input is accepted in batches of many jobs.
c. Processor time is shared among several processes.
d. Many processors may be active simultaneously.

A multiprocessor system has more than one processor. Each processor can have its own memory or share a common memory. The former belongs to the class of loosely coupled systems and the latter to the tightly coupled systems. Each processor executes a separate process, and many processors may be active simultaneously.

32. A multiprogramming system may be defined as one in which:
a. Programs are divided into pages.
b. Input is accepted in batches of many jobs.
c. Several programs can reside in memory at the same time.
d. Many processes may share the same program residing in main memory.

Multiprogramming is, by definition, having several programs in memory at the same time. In time-sharing systems, for example, a process currently requiring an input/output operation will yield the CPU to another process which is ready to run. Such switching is also called CPU scheduling. A time-sharing system generally has multiprogramming too.

33. A processor in the context of computing is:
a. A set of instructions to be executed on a computer.
b. A program in execution.
c. A piece of hardware that executes a set of instructions.
d. The main procedure of a program.

A processor is any piece of hardware that executes a set of
instructions.

34. A process may be defined as:
a. A set of instructions to be executed by a computer.
b. A program in execution.
c. A piece of hardware that executes a set of instructions.
d. The main procedure of a program.

The term process was borrowed from chemical engineering by the designers of the Multics system in the 1960s and is used interchangeably with the term task. The generally accepted definition of a process is a program in execution.