Test: File System & Device Management- 2 - Computer Science Engineering (CSE) MCQ

1, 2, 4, 5, 2, 1 , 2 , 4
Since 1 and 2 are already in the memory which can accomodate 3 pages

Correct Answer :- d

Explanation : Working set (t, k) at an instant of time t, is the k set of pages that have been referenced in the last t time units.

Disk scheduling manages the disk access and order of request to be fulfilled. It does not have the bounds in determining the physical location, where actually the files are stored.

In Acyclic Graph directory file can be accessed using different paths from the same root directory this can be done with the helps of soft links.

The first 3 reference A, B, C fills the internal storage with A, B,C in 3 page transfers. Now the next reference D results in a page fault. So page A is downloaded and D takes its place after a page transfer. So, the internal store has D, Fand C. The next reference is A - results in a page fault. So, a page transfer takes place and swaps B and A. Continuing this way, we find totally 9 page transfers are necessary.

We find the required number of page transfer is 10. So, increasing the number of pages need not necessarily reduce the number of page faults. It is the actual sequences of references that decides.

The disk drive has to traverse totally 146 cylinders (verify). So, seek time is 6 x 146 = 876 ms.

Total seek time:
[2 + 12 + 4 + 4 + 36 + 2] x 6 msec = 360 msec

To cover 2400 x 62500 bits, 60 s are needed. Average latency time is the time needed to traverse 100 tracks i.e. 100 x 62500 bits, which is 2.5 s.

When it tries to access 0100, it results in a page fault as the memory is empty right now. So, it loads the second page (which has the addresses 100 - 199). Trying to access 200 will result in a page fault, as it is not in memory right now. So the third page with the addresses from 200 to 299 will replace the second page in memory. Trying to access 430 will result in another page fault. Proceeding this way, we find trying to access the addresses 0510,0120,0220 and 0320 will all result in page faults. So, altogether 7 page faults.

Since the page is used heavily then it can not be removed if LFU used, since it is in constant used, hence it can’t be removed if LRU is used. Using FIFO only it can be removed effectively.

If 3 page frames are used

Page hit : 0, 1, 4.
∴ Number of page fault = 9
If 4 pages frames are used

Page faults ; 0, 1, 2, 3, 4, 0, 1,2, 3, 4 Page hit : 0, 1
∴ Number of page faults = 10

Page sequence 0, 9, 0, 1,8, 1,8, 7, 8, 7, 1,2, 3, 2, 7, 8, 2, 3, 8, 3 
Given 3 page frames are available and optimal replacement policy.

• FIFO suffers from BELADY’s anomaly, in which sometimes page faults increase with increase in page frames. Hence, true.
• Locality of reference depends on demand of process to access consecutive data and hence locality of reference may not always be satisfied. Hence, true

∴ Total distance covered by the head
= (345 -123) + (123 -105) + (105 - 0) + (999 - 0) 
+ (999-874) + (874-692) + (692-475) + (475 -376)
= 222 + 18 + 105 + 999 + 125 + 182 + 217 + 99 
= 1967