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

>>>>>>>>UGC NET Computer Science Paper 2 December 2015

 Page Block 0 3 2 1 5 2 7 0
• 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.

• 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%.

• 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

 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
• 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.

• 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.