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301. |
Which is the process of invoking the wait |
A. | transaction |
B. | operation |
C. | function |
D. | all of the mentioned |
Answer» A. transaction |
302. |
Write ahead logging is a way |
A. | to ensure atomicity |
B. | to keep data consistent |
C. | that records data on stable storage |
D. | all of the mentioned |
Answer» D. all of the mentioned |
303. |
The system periodically performs checkpoints that consists of the following operation(s) |
A. | putting all the log records currently in main memory onto stable storage |
B. | putting all modified data residing in main memory onto stable storage |
C. | putting a log record onto stable storage |
D. | all of the mentioned |
Answer» D. all of the mentioned |
304. |
A locking protocol is one that |
A. | governs how locks are acquired |
B. | governs how locks are released |
C. | governs how locks are acquired and released |
D. | none of the mentioned |
Answer» C. governs how locks are acquired and released |
305. |
The two phase locking protocol consists of |
A. | growing & shrinking phase |
B. | shrinking & creation phase |
C. | creation & growing phase |
D. | destruction & creation phase |
Answer» A. growing & shrinking phase |
306. |
The growing phase is a phase in which? |
A. | a transaction may obtain locks, but does not release any |
B. | a transaction may obtain locks, and releases a few or all of them |
C. | a transaction may release locks, but does not obtain any new locks |
D. | a transaction may release locks, and does obtain new locks |
Answer» A. a transaction may obtain locks, but does not release any |
307. |
What is a reusable resource? |
A. | that can be used by one process at a time and is not depleted by that use |
B. | that can be used by more than one process at a time |
C. | that can be shared between various threads |
D. | none of the mentioned |
Answer» A. that can be used by one process at a time and is not depleted by that use |
308. |
Which of the following condition is required for a deadlock to be possible? |
A. | mutual exclusion |
B. | a process may hold allocated resources while awaiting assignment of other resources |
C. | no resource can be forcibly removed from a process holding it |
D. | all of the mentioned |
Answer» D. all of the mentioned |
309. |
Which of the following concurrency control protocols ensure both conflict serializability and freedom from deadlock? |
A. | the system can allocate resources to each process in some order and still avoid a deadlock |
B. | there exist a safe sequence |
C. | all of the mentioned |
D. | none of the mentioned |
Answer» A. the system can allocate resources to each process in some order and still avoid a deadlock |
310. |
The circular wait condition can be prevented by |
A. | defining a linear ordering of resource types |
B. | using thread |
C. | using pipes |
D. | all of the mentioned |
Answer» A. defining a linear ordering of resource types |
311. |
Which one of the following is the deadlock avoidance algorithm? |
A. | banker’s algorithm |
B. | round-robin algorithm |
C. | elevator algorithm |
D. | karn’s algorithm |
Answer» A. banker’s algorithm |
312. |
What is the drawback of banker’s algorithm? |
A. | in advance processes rarely know how much resource they will need |
B. | the number of processes changes as time progresses |
C. | resource once available can disappear |
D. | all of the mentioned |
Answer» D. all of the mentioned |
313. |
For an effective operating system, when to check for deadlock? |
A. | every time a resource request is made |
B. | at fixed time intervals |
C. | every time a resource request is made at fixed time intervals |
D. | none of the mentioned |
Answer» C. every time a resource request is made at fixed time intervals |
314. |
A problem encountered in multitasking when a process is perpetually denied necessary resources is called |
A. | deadlock |
B. | starvation |
C. | inversion |
D. | aging |
Answer» B. starvation |
315. |
Which one of the following is a visual ( mathematical ) way to determine the deadlock occurrence? |
A. | resource allocation graph |
B. | starvation graph |
C. | inversion graph |
D. | none of the mentioned |
Answer» A. resource allocation graph |
316. |
To avoid deadlock |
A. | there must be a fixed number of resources to allocate |
B. | resource allocation must be done only once |
C. | all deadlocked processes must be aborted |
D. | inversion technique can be used |
Answer» A. there must be a fixed number of resources to allocate |
317. |
The number of resources requested by a process |
A. | must always be less than the total number of resources available in the system |
B. | must always be equal to the total number of resources available in the system |
C. | must not exceed the total number of resources available in the system |
D. | must exceed the total number of resources available in the system |
Answer» C. must not exceed the total number of resources available in the system |
318. |
The request and release of resources are |
A. | command line statements |
B. | interrupts |
C. | system calls |
D. | special programs |
Answer» C. system calls |
319. |
For a deadlock to arise, which of the following conditions must hold simultaneously? |
A. | mutual exclusion |
B. | no preemption |
C. | hold and wait |
D. | all of the mentioned |
Answer» D. all of the mentioned |
320. |
For non sharable resources like a printer, mutual exclusion |
A. | must exist |
B. | must not exist |
C. | may exist |
D. | none of the mentioned |
Answer» A. must exist |
321. |
The disadvantage of a process being allocated all its resources before beginning its execution is |
A. | low cpu utilization |
B. | low resource utilization |
C. | very high resource utilization |
D. | none of the mentioned |
Answer» B. low resource utilization |
322. |
Each request requires that the system consider the to decide whether the current request can be satisfied or must wait to avoid a future possible deadlock. |
A. | resources currently available |
B. | processes that have previously been in the system |
C. | resources currently allocated to each process |
D. | future requests and releases of each process |
Answer» A. resources currently available |
323. |
A deadlock avoidance algorithm dynamically examines the to ensure that a circular wait condition can never exist. |
A. | resource allocation state |
B. | system storage state |
C. | operating system |
D. | resources |
Answer» A. resource allocation state |
324. |
A state is safe, if |
A. | the system does not crash due to deadlock occurrence |
B. | the system can allocate resources to each process in some order and still avoid a deadlock |
C. | the state keeps the system protected and safe |
D. | all of the mentioned |
Answer» B. the system can allocate resources to each process in some order and still avoid a deadlock |
325. |
A system is in a safe state only if there exists a |
A. | safe allocation |
B. | safe resource |
C. | safe sequence |
D. | all of the mentioned |
Answer» C. safe sequence |
326. |
All unsafe states are |
A. | deadlocks |
B. | not deadlocks |
C. | fatal |
D. | none of the mentioned |
Answer» B. not deadlocks |
327. |
If no cycle exists in the resource allocation graph |
A. | then the system will not be in a safe state |
B. | then the system will be in a safe state |
C. | all of the mentioned |
D. | none of the mentioned |
Answer» B. then the system will be in a safe state |
328. |
The resource allocation graph is not applicable to a resource allocation system |
A. | with multiple instances of each resource type |
B. | with a single instance of each resource type |
C. | single & multiple instances of each resource type |
D. | none of the mentioned |
Answer» A. with multiple instances of each resource type |
329. |
The data structures available in the Banker’s algorithm are |
A. | available |
B. | need |
C. | allocation |
D. | all of the mentioned |
Answer» D. all of the mentioned |
330. |
The content of the matrix Need is |
A. | allocation – available |
B. | max – available |
C. | max – allocation |
D. | allocation – max |
Answer» C. max – allocation |
331. |
The wait-for graph is a deadlock detection algorithm that is applicable when |
A. | all resources have a single instance |
B. | all resources have multiple instances |
C. | all resources have a single 7 multiple instances |
D. | all of the mentioned |
Answer» A. all resources have a single instance |
332. |
An edge from process Pi to Pj in a wait for graph indicates that |
A. | pi is waiting for pj to release a resource that pi needs |
B. | pj is waiting for pi to release a resource that pj needs |
C. | pi is waiting for pj to leave the system |
D. | pj is waiting for pi to leave the system |
Answer» A. pi is waiting for pj to release a resource that pi needs |
333. |
If the wait for graph contains a cycle |
A. | then a deadlock does not exist |
B. | then a deadlock exists |
C. | then the system is in a safe state |
D. | either deadlock exists or system is in a safe state |
Answer» B. then a deadlock exists |
334. |
If deadlocks occur frequently, the detection algorithm must be invoked |
A. | rarely |
B. | frequently |
C. | rarely & frequently |
D. | none of the mentioned |
Answer» B. frequently |
335. |
What is the disadvantage of invoking the detection algorithm for every request? |
A. | overhead of the detection algorithm due to consumption of memory |
B. | excessive time consumed in the request to be allocated memory |
C. | considerable overhead in computation time |
D. | all of the mentioned |
Answer» C. considerable overhead in computation time |
336. |
A deadlock eventually cripples system throughput and will cause the CPU utilization to |
A. | increase |
B. | drop |
C. | stay still |
D. | none of the mentioned |
Answer» B. drop |
337. |
Every time a request for allocation cannot be granted immediately, the detection algorithm is invoked. This will help identify |
A. | the set of processes that have been deadlocked |
B. | the set of processes in the deadlock queue |
C. | the specific process that caused the deadlock |
D. | all of the mentioned |
Answer» A. the set of processes that have been deadlocked |
338. |
A computer system has 6 tape drives, with ‘n’ processes competing for them. Each process may need 3 tape drives. The maximum value of ‘n’ for which the system is guaranteed to be deadlock free is? |
A. | 2 |
B. | 3 |
C. | 4 |
D. | 1 |
Answer» A. 2 |
339. |
A system has 3 processes sharing 4 resources. If each process needs a maximum of 2 units then, deadlock |
A. | can never occur |
B. | may occur |
C. | has to occur |
D. | none of the mentioned |
Answer» A. can never occur |
340. |
‘m’ processes share ‘n’ resources of the same type. The maximum need of each process doesn’t exceed ‘n’ and the sum of all their maximum needs is always less than m+n. In this setup, deadlock |
A. | can never occur |
B. | may occur |
C. | has to occur |
D. | none of the mentioned |
Answer» A. can never occur |
341. |
A deadlock can be broken by |
A. | abort one or more processes to break the circular wait |
B. | abort all the process in the system |
C. | preempt all resources from all processes |
D. | none of the mentioned |
Answer» A. abort one or more processes to break the circular wait |
342. |
The two ways of aborting processes and eliminating deadlocks are |
A. | abort all deadlocked processes |
B. | abort all processes |
C. | abort one process at a time until the deadlock cycle is eliminated |
D. | all of the mentioned |
Answer» C. abort one process at a time until the deadlock cycle is eliminated |
343. |
Those processes should be aborted on occurrence of a deadlock, the termination of which? |
A. | is more time consuming |
B. | incurs minimum cost |
C. | safety is not hampered |
D. | all of the mentioned |
Answer» B. incurs minimum cost |
344. |
The process to be aborted is chosen on the basis of the following factors? |
A. | priority of the process |
B. | process is interactive or batch |
C. | how long the process has computed |
D. | all of the mentioned |
Answer» D. all of the mentioned |
345. |
Cost factors for process termination include |
A. | number of resources the deadlock process is not holding |
B. | cpu utilization at the time of deadlock |
C. | amount of time a deadlocked process has thus far consumed during its execution |
D. | all of the mentioned |
Answer» C. amount of time a deadlocked process has thus far consumed during its execution |
346. |
If we preempt a resource from a process, the process cannot continue with its normal execution and it must be |
A. | aborted |
B. | rolled back |
C. | terminated |
D. | queued |
Answer» B. rolled back |
347. |
To to a safe state, the system needs to keep more information about the states of processes. |
A. | abort the process |
B. | roll back the process |
C. | queue the process |
D. | none of the mentioned |
Answer» B. roll back the process |
348. |
If the resources are always preempted from the same process can occur. |
A. | deadlock |
B. | system crash |
C. | aging |
D. | starvation |
Answer» D. starvation |
349. |
What is the solution to starvation? |
A. | the number of rollbacks must be included in the cost factor |
B. | the number of resources must be included in resource preemption |
C. | resource preemption be done instead |
D. | all of the mentioned |
Answer» A. the number of rollbacks must be included in the cost factor |
350. |
What is Address Binding? |
A. | going to an address in memory |
B. | locating an address with the help of another address |
C. | binding two addresses together to form a new address in a different memory space |
D. | a mapping from one address space to another |
Answer» D. a mapping from one address space to another |
351. |
Binding of instructions and data to memory addresses can be done at |
A. | compile time |
B. | load time |
C. | execution time |
D. | all of the mentioned |
Answer» D. all of the mentioned |
352. |
If the process can be moved during its execution from one memory segment to another, then binding must be |
A. | delayed until run time |
B. | preponed to compile time |
C. | preponed to load time |
D. | none of the mentioned |
Answer» A. delayed until run time |
353. |
What is Dynamic loading? |
A. | loading multiple routines dynamically |
B. | loading a routine only when it is called |
C. | loading multiple routines randomly |
D. | none of the mentioned |
Answer» B. loading a routine only when it is called |
354. |
What is the advantage of dynamic loading? |
A. | a used routine is used multiple times |
B. | an unused routine is never loaded |
C. | cpu utilization increases |
D. | all of the mentioned |
Answer» B. an unused routine is never loaded |
355. |
The idea of overlays is to |
A. | data that are needed at any given time |
B. | enable a process to be larger than the amount of memory allocated to it |
C. | keep in memory only those instructions |
D. | all of the mentioned |
Answer» D. all of the mentioned |
356. |
The must design and program the overlay structure. |
A. | programmer |
B. | system architect |
C. | system designer |
D. | none of the mentioned |
Answer» A. programmer |
357. |
The swaps processes in and out of the memory. |
A. | memory manager |
B. | cpu |
C. | cpu manager |
D. | user |
Answer» A. memory manager |
358. |
If binding is done at assembly or load time, then the process be moved to different locations after being swapped out and in again. |
A. | can |
B. | must |
C. | can never |
D. | may |
Answer» C. can never |
359. |
In a system that does not support swapping |
A. | the compiler normally binds symbolic addresses (variables) to relocatable addresses |
B. | the compiler normally binds symbolic addresses to physical addresses |
C. | the loader binds relocatable addresses to physical addresses |
D. | binding of symbolic addresses to physical addresses normally takes place during execution |
Answer» A. the compiler normally binds symbolic addresses (variables) to relocatable addresses |
360. |
The address generated by the CPU is referred to as |
A. | physical address |
B. | logical address |
C. | neither physical nor logical |
D. | none of the mentioned |
Answer» B. logical address |
361. |
The address loaded into the memory address register of the memory is referred to as |
A. | physical address |
B. | logical address |
C. | neither physical nor logical |
D. | none of the mentioned |
Answer» A. physical address |
362. |
The run time mapping from virtual to physical addresses is done by a hardware device called the |
A. | virtual to physical mapper |
B. | memory management unit |
C. | memory mapping unit |
D. | none of the mentioned |
Answer» B. memory management unit |
363. |
The size of a process is limited to the size of |
A. | physical memory |
B. | external storage |
C. | secondary storage |
D. | none of the mentioned |
Answer» A. physical memory |
364. |
If execution time binding is being used, then a process be swapped to a different memory space. |
A. | has to be |
B. | can never |
C. | must |
D. | may |
Answer» D. may |
365. |
Swapping requires a |
A. | motherboard |
B. | keyboard |
C. | monitor |
D. | backing store |
Answer» D. backing store |
366. |
The backing store is generally a |
A. | fast disk |
B. | disk large enough to accommodate copies of all memory images for all users |
C. | disk to provide direct access to the memory images |
D. | all of the mentioned |
Answer» D. all of the mentioned |
367. |
The consists of all processes whose memory images are in the backing store or in memory and are ready to run. |
A. | wait queue |
B. | ready queue |
C. | cpu |
D. | secondary storage |
Answer» B. ready queue |
368. |
The time in a swap out of a running process and swap in of a new process into the memory is very high. |
A. | context – switch |
B. | waiting |
C. | execution |
D. | all of the mentioned |
Answer» A. context – switch |
369. |
Swapping be done when a process has pending I/O, or has to execute I/O operations only into operating system buffers. |
A. | must |
B. | can |
C. | must never |
D. | maybe |
Answer» C. must never |
370. |
Swap space is allocated |
A. | as a chunk of disk |
B. | separate from a file system |
C. | into a file system |
D. | all of the mentioned |
Answer» A. as a chunk of disk |
371. |
CPU fetches the instruction from memory according to the value of |
A. | program counter |
B. | status register |
C. | instruction register |
D. | program status word |
Answer» A. program counter |
372. |
A memory buffer used to accommodate a speed differential is called |
A. | stack pointer |
B. | cache |
C. | accumulator |
D. | disk buffer |
Answer» B. cache |
373. |
Which one of the following is the address generated by CPU? |
A. | physical address |
B. | absolute address |
C. | logical address |
D. | none of the mentioned |
Answer» C. logical address |
374. |
Run time mapping from virtual to physical address is done by |
A. | memory management unit |
B. | cpu |
C. | pci |
D. | none of the mentioned |
Answer» A. memory management unit |
375. |
Memory management technique in which system stores and retrieves data from secondary storage for use in main memory is called? |
A. | fragmentation |
B. | paging |
C. | mapping |
D. | none of the mentioned |
Answer» B. paging |
376. |
The address of a page table in memory is pointed by |
A. | stack pointer |
B. | page table base register |
C. | page register |
D. | program counter |
Answer» B. page table base register |
377. |
Program always deals with |
A. | logical address |
B. | absolute address |
C. | physical address |
D. | relative address |
Answer» A. logical address |
378. |
The page table contains |
A. | base address of each page in physical memory |
B. | page offset |
C. | page size |
D. | none of the mentioned |
Answer» A. base address of each page in physical memory |
379. |
What is compaction? |
A. | a technique for overcoming internal fragmentation |
B. | a paging technique |
C. | a technique for overcoming external fragmentation |
D. | a technique for overcoming fatal error |
Answer» C. a technique for overcoming external fragmentation |
380. |
Operating System maintains the page table for |
A. | each process |
B. | each thread |
C. | each instruction |
D. | each address |
Answer» A. each process |
381. |
The main memory accommodates |
A. | operating system |
B. | cpu |
C. | user processes |
D. | all of the mentioned |
Answer» A. operating system |
382. |
What is the operating system? |
A. | in the low memory |
B. | in the high memory |
C. | either low or high memory (depending on the location of interrupt vector) |
D. | none of the mentioned |
Answer» C. either low or high memory (depending on the location of interrupt vector) |
383. |
In contiguous memory allocation |
A. | each process is contained in a single contiguous section of memory |
B. | all processes are contained in a single contiguous section of memory |
C. | the memory space is contiguous |
D. | none of the mentioned |
Answer» A. each process is contained in a single contiguous section of memory |
384. |
The relocation register helps in |
A. | providing more address space to processes |
B. | a different address space to processes |
C. | to protect the address spaces of processes |
D. | none of the mentioned |
Answer» C. to protect the address spaces of processes |
385. |
The operating system and the other processes are protected from being modified by an already running process because |
A. | they are in different memory spaces |
B. | they are in different logical addresses |
C. | they have a protection algorithm |
D. | every address generated by the cpu is being checked against the relocation and limit registers |
Answer» D. every address generated by the cpu is being checked against the relocation and limit registers |
386. |
Transient operating system code is code that |
A. | is not easily accessible |
B. | comes and goes as needed |
C. | stays in the memory always |
D. | never enters the memory space |
Answer» B. comes and goes as needed |
387. |
Using transient code, the size of the operating system during program execution. |
A. | increases |
B. | decreases |
C. | changes |
D. | maintains |
Answer» C. changes |
388. |
In fixed size partition, the degree of multiprogramming is bounded by |
A. | the number of partitions |
B. | the cpu utilization |
C. | the memory size |
D. | all of the mentioned |
Answer» A. the number of partitions |
389. |
The first fit, best fit and worst fit are strategies to select a |
A. | process from a queue to put in memory |
B. | processor to run the next process |
C. | free hole from a set of available holes |
D. | all of the mentioned |
Answer» C. free hole from a set of available holes |
390. |
In internal fragmentation, memory is internal to a partition and |
A. | is being used |
B. | is not being used |
C. | is always used |
D. | none of the mentioned |
Answer» B. is not being used |
391. |
A solution to the problem of external fragmentation is |
A. | compaction |
B. | larger memory space |
C. | smaller memory space |
D. | none of the mentioned |
Answer» A. compaction |
392. |
Another solution to the problem of external fragmentation problem is to |
A. | permit the logical address space of a process to be noncontiguous |
B. | permit smaller processes to be allocated memory at last |
C. | permit larger processes to be allocated memory at last |
D. | all of the mentioned |
Answer» A. permit the logical address space of a process to be noncontiguous |
393. |
If relocation is static and is done at assembly or load time, compaction |
A. | cannot be done |
B. | must be done |
C. | must not be done |
D. | can be done |
Answer» A. cannot be done |
394. |
The disadvantage of moving all process to one end of memory and all holes to the other direction, producing one large hole of available memory is |
A. | the cost incurred |
B. | the memory used |
C. | the cpu used |
D. | all of the mentioned |
Answer» A. the cost incurred |
395. |
External fragmentation will not occur when? |
A. | first fit is used |
B. | best fit is used |
C. | worst fit is used |
D. | no matter which algorithm is used, it will always occur |
Answer» D. no matter which algorithm is used, it will always occur |
396. |
Sometimes the overhead of keeping track of a hole might be |
A. | larger than the memory |
B. | larger than the hole itself |
C. | very small |
D. | all of the mentioned |
Answer» B. larger than the hole itself |
397. |
When the memory allocated to a process is slightly larger than the process, then |
A. | internal fragmentation occurs |
B. | external fragmentation occurs |
C. | both internal and external fragmentation occurs |
D. | neither internal nor external fragmentation occurs |
Answer» A. internal fragmentation occurs |
398. |
Physical memory is broken into fixed-sized blocks called |
A. | frames |
B. | pages |
C. | backing store |
D. | none of the mentioned |
Answer» A. frames |
399. |
Logical memory is broken into blocks of the same size called |
A. | frames |
B. | pages |
C. | backing store |
D. | none of the mentioned |
Answer» B. pages |
400. |
Every address generated by the CPU is divided into two parts. They are |
A. | frame bit & page number |
B. | page number & page offset |
C. | page offset & frame bit |
D. | frame offset & page offset |
Answer» B. page number & page offset |
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