UGC NET COMPUTER SCIENCE SOLVED PAPERS 2014-16 - UGC NET Computer Science Paper 2 December 2015

26. 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:

       PageBlock
        03
        21
        5 2
        70

Which of the following list of virtual addresses (in decimal) will not cause any page fault if referenced by the CPU ?

  • Option : C
  • Explanation : We have 8 pages in the memory and 4 blocks: 0-1023 (0) 1024-2047 (1) 2048-3071 (2) 3072-4095 (3) 4096-5119 (4) 5120-6143 (5) 6144-7167 (6) 7168-8191 (7) option (A) will string 1,3,4,6 so page fault will occur. option (B) will string 1,3,4,6 so page fault will occur. option (C) will string 0,2,5,7 so page fault will occur. option (D) will string 1,3,4,6 so page fault will occur. So, option (C) is correct.
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27. Suppose that the number of instructions executed between page faults 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 micro second, but if a page fault occurs, it takes 2001 micro seconds. If a program takes 60 sec to run, during which time it gets 15000 page faults, how long would it take to run if twice as much memory were available?

  • Option : C
  • Explanation :
    T = Ninstr x 1µs + 15,000 x 2,000 µs = 60s
    Ninstr x 1µs = 60,000,000 µs – 15,000,000 µs = 30,000,000 µs
    Ninstr = 30,000,000
    The number of instruction between two page faults is
    Ninstr /NPageFaults = 30,000,000/15,000 = 2,000
    If the mean interval between page faults is doubled, the number of instruction between
    two page faults is also doubled and is 4,000. Now the number of page faults is
    30,000,000/4,000 = 7,500
    T’ = 30,000,000 µs + 7,500 x 2,000 µs
    = 30,000,000 µs + 15,000,000 µs = 45,000,000 µs = 45s
    Doubling the memory, doesn’t mean that the program runs twice as fast as on the first system. Here, the performance increase is of 25%.
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28. 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?

  • Option : B
  • Explanation :
    Time in msec to read a block of 1024 bytes (Access time or Disk Latency) = seek time +average rotational delay + transfer time
    If there are 16384 bytes per track there are 1024/16384 tracks to be read for this block. Seek time = 40 msec
    Rotational delay = 16 msec
    Transfer time = (sectors_read/sectors per rev.) x rotational delay = (1024/16384) x 16 = 1
    average rotational delay = rotational delay/2 = 16/2 = 8
    access time = 40 + 8 + 1 = 49
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29. A system has four processes and five allocatable resources. The current allocation and maximum needs are as follows:

  Allocated Maximum Available
 Process A 1 0 2 1 1 1 1 2 1 3 0 0 x 1 1
 Process B 2 0 1 1 0 2 2 2 1 0 
 Process C 1 1 0 1 0 2 1 3 1 0 
 Process D 1 1 1 1 0 1 1 2 2 1

The smallest value of x for which the above system in safe state is .............

  • Option : D
  • Explanation :
    If Process A’s Maximum need is 1 1 2 1 2 instead of 1 1 2 1 3, then answer will be x=1
    The needs matrix is as follows:
    0 1 0 0 1
    0 2 1 0 0
    1 0 3 0 0
    0 0 1 1 1
    If x is 0, available vector will be 0 0 0 1 1, we have a deadlock immediately.
    If x is 1, available vector will be 0 0 1 1 1, now, process D can run to completion. When it is finished, the available vector is 1 1 2 2 1.
    Now A can run to complete, the available vector then becomes 2 1 4 3 2.
    Then C can run and finish, return the available vector as 3 2 4 4 2.
    Then B can run to complete. Safe sequence D A C B.
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30. In Unix, the login prompt can be changed by changing the contents of the file ...............

  • Option : C
  • Explanation :
  • In Unix, the login prompt can be changed by changing the contents of the file gettydefs.
  • There is nothing like contrab in unix.
  • The inittab file controls what happens whenever a Unix system is rebooted or forced to change run levels.
  • init (short for initialization) is the first process started during booting of the computer system.
  • So, option (C) is correct.
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