Q3. UGC NET Dec 2018
To overcome difficulties in Readers-Writers problem, which of the following statement/s is/are true?
Showing posts with label UGC NET CS. Show all posts
Showing posts with label UGC NET CS. Show all posts
Suppose a system has 12 instances | UGC NET Dec 2018
Q2. UGC NET Dec 2018 :
Suppose a system has 12 instances of some resource with n processes competing for that resource. Each process may require 4 instances of the resources. The maximum value of n for which the system never enters into deadlock is
Data warehouse | UGC NET Dec 2018
Q1. UGC NET Dec 2018
Data warehouse contains _____ data that is never found in operational environment.
UGC NET CS Paper 2 June 2012
UGC NET CS Paper 2 June 2012
1. The postfix expression AB + CD - * can be evaluated using a
Net 52
CBSE NET
Q. Convert the following infix expression into its equivalent post fix expression (A + B^ D) / (E – F) + G
(A) ABD^ + EF – / G+
(B) ABD + ^EF – / G+
(C) ABD + ^EF / – G+
(D) ABD^ + EF / – G+
Ans: (A)
Net 51
CBSE NET JUNE 2015 Paper II
Q. The in-order
and pre-order Traversal of binary Tree are dbeafcg and abdecfg respectively.
The post-order Traversal is______.
A)dbefacg
B)debfagc
C)dbefcga
D)debfgca
Ans: (D)
Explanation:
Inorder (Left, Root, Right),Preorder (Root, Left, Right),Postorder (Left, Right, Root)
Given, in-order=dbeafcg, pre-order=abdecfg.
Than, binary tree from in-order and
pre-order become-
Post-order from above binary tree=debfgca
Net 50
CBSE NET JUNE 2015 Paper II
Q. A
disk drive has 100 cylinders,numbered 0 to 99. Disk request come to the disk
driver for cylinders 12,26,24,4,42,8,50 in that order. The driver is currently
serving a request at cylinder 24. A seek takes 6msec per cylinder moved. How
much seek time is needed for the shortest seek time first (sstf) algorithm?
A)0.984 sec
B)0.396 sec
C)0.738 sec
D)0.42 sec
Ans: (D)
Explanation:
SSTF, Shortest time first algorithm
Total cylinder head movements= 02+14+04+04+38+08=70.
Seek time needed=70 x 6 = 420=0.42 sec.
Net 49
CBSE NET
Q. A
LRU page replacement is used with four page frames and eight pages. How many
page faults will occur with the reference string 0172327103. if the four frames
are initially empty.
A) 6
B) 7
C) 5
D) 8.
Ans: (B)
Explanation: LRU (Least recently used) algorithm. Here, F shows page faults and H
shows page hits.
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Net 48
CBSE NET
Q. Consider an array A[20,
10], assume 4 words per memory cell and the base address of array A is 100.
What is the address of A[11, 5] ? Assume row major storage.
(A) 560
(B) 565
(C) 570
(D) 575
Ans: (A)
Explanation:
Formula for calculating address of element in row major and column major
order is as shown in below image.
Where,
B=Base address, w=element
size, Nc=number of column, Nr=number of row, Lc=column lower bound, Lr=row lower
bound.
Using formula of row major,
Address of (11,5) = 100+4(10(11-0)+(5-0))= 100+4(115)=560.
Net 47
CBSE NET
Q. An
analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many
data elements are carried by each signal element ?
(A) 0.336 bits/baud
(B) 3 bits/baud
(C) 120,00,000 bits/baud
(D) None of the above
Ans: (B)
Explanation: A same
signal has bit rate of 6000bp and baud rate of 2000 baud. So 6000 = 2000 baud. Hence a signal carried 3bits of data.
Net 46
CBSE NET
Q. In a
demand paging memory system, page table is held in registers. The time taken to
service a page fault is 8 m.sec. if an empty frame is available or if the
replaced page is not modified, and it takes 20 m.secs., if the replaced page is
modified. What is the average access time to service a page fault assuming that
the page to be replaced is modified 70% of the time ?
(A) 11.6 m.sec.
(B) 16.4 m.sec.
(C) 28 m.sec.
(D) 14 m.sec
Ans: (B)
Explanation: Let
number of page faults be 100. In which 30 page faults are without modification
so time is 30x8=240 m.sec. And rest
70 page faults are with modification so time is 20x70=1400 m.sec.
Average access time of servicing a page fault is (240+1400)/100=28 m.sec.
Net 45
CBSE NET DECEMBER 2012 PAPER II
Q. Given an empty stack, after performing push (1), push (2), Pop, push (3), push (4), Pop, Pop, push (5), pop, what is the value of the top of the stack?
(A) 4
(B) 3
(C) 2
(D) 1
Ans: (D)
Explanation: See the image below:
Net 44
CBSE NET JUNE 2012 PAPER II
Q. Consider the following page trace :
4,3, 2, 1, 4, 3, 5, 4, 3, 2, 1, 5
Percentage of page fault that would occur if FIFO page replacement algorithm is used with number of frames for the JOB m = 4 will be
(A) 8
(B) 9
(C) 10
(D) 12
Ans: (C)
Explanation: F shows page fault using FIFO page replacement algorithm.
In FIFO algorithm we see the elements which enter first in the memory, have to out first.
In FIFO algorithm we see the elements which enter first in the memory, have to out first.
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