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)
A. | 3 |
B. | 5 |
C. | 4 |
D. | none of these |
Option: C Explanation : Click on Discuss to view users comments. |
Fence register is used for
A. | CPU protection |
B. | memory protection |
C. | file protection |
D. | all of these |
Option: B Explanation : Click on Discuss to view users comments. |
Distributed systems should
A. | meet prescribed time constraints |
B. | aim better resource sharing |
C. | aim better system utilization |
D. | aim low system overhead |
Option: B Explanation : Click on Discuss to view users comments. Gokulnath said: (1:35pm on Monday 1st May 2017)
Answer B:There are four major reasons for building distributed systems: resourcesharing, computation speedup, reliability, and communication.
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Working set (t, k) at an instant of time, t, is
A. | the set of k future references that the operating system will make |
B. | the set of future references that the operating system will make in the next 'k' time units |
C. | the set of k reference with high frequency |
D. | the set of pages that have been referenced in the last k time units |
Option: D Explanation : The working set model states that a process can be in RAM if and only if all of the pages that it is currently using (often approximated by the most recently used pages) can be in RAM. The model is an all or nothing model, meaning if the pages it needs to use increases, and there is no room in RAM, the process is swapped out of memory to free the memory for other processes to use.
The main hurdle in implementing the working set model is keeping track of the working set. The working set window is a moving window. At each memory reference a new reference appears at one end and the oldest reference drops off the other end. A page is in the working set if it is referenced in the working set window.
To avoid the overhead of keeping a list of the last k referenced pages, the working set is often implemented by keeping track of the time t of the last reference, and considering the working set to be all pages referenced within a certain period of time. Click on Discuss to view users comments. |
In a paged memory, the page hit ratio is 0.35. The time required to access a page in secondary memory is equal to 100 ns. The time required to access a page in primary memory is 10 ns. The average time required to access a page is
A. | 3.0 ns |
B. | 68.0 ns |
C. | 68.5 ns |
D. | 78.5 ns |
Option: C Explanation : 0.35 x 10 + (1 - 0.35) x 100 = 68.5 ns Click on Discuss to view users comments. Shivani said: (10:16pm on Saturday 2nd June 2018)
Why not 0.35x10 (1-0.35)x(10 100) as it will first search in main memory?
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