UGCNET-Dec2012 - UPSCFEVER

Building a Computer System

Building Internet Market

Building Offline Market

Building Market

Ans .

B

Explanation :
THE DIGITAL ECOSYSTEM OF A BUSINESS IS THE COMBINATION OF ALL RELEVANT DIGITAL TOUCHPOINTS, THE PEOPLE THAT INTERACT WITH THEM, AND THE BUSINESS PROCESSES AND TECHNOLOGY ENVIRONMENT THAT SUPPORT BOTH.Thus it can be said that Eco system is a Frame work for building Internet Market

Q.2

The efficiency (E) and speed up (S) for Multiprocessor with p processors satisfies :

E ≤ p and S_p ≤ p

E ≤ 1 and S_p≤ P

E ≤ p and S_p ≤ 1

E ≤ 1 and S_p ≤ 1

Ans .

B

Explanation :
Efficiency E can never be more than 1 or 100% even 1 is just theortical

Speed up for multiprocessor Sp can never exceed (P) no of processors

hence E≤1 and S_p≤p. Hence,The efficiency (E) and speed up (S) for Multiprocessor with p processors satisfies : E ≤ 1 and S_p ≤ p

Q.3

Match the following:

List – I List – II

a. Critical region 1. Hoares Monitor

b. Wait/signal 2. Mutual exclusion

c. Working set 3. Principal of locality

d. Dead lock 4. Circular wait

2 1 3 4

1 2 4 3

2 3 1 4

1 3 2 4

Ans .

B

Explanation :

List-I List-II

a. Critical region 2. Mutual exclusion

b. Wait/signal 1. Hoares Monitor

c. Working set 3. Principal of locality

d. Dead lock 4. Circular wait

Q.4

The technique of temporarily delaying outgoing acknowledgements so that they can be hooked onto the next outgoing data frame is known as

Bit stuffing

Piggy backing

Pipelining

Broadcasting

Ans .

B

Explanation :
Piggybacking of acknowledments:The ACK for the last received packet need not be sent as a new packet, but gets a free ride on the next outgoing data frame(using the ACK field in the frame header). The technique is temporarily delaying outgoing ACKs so that they can be hooked on the next outgoing data frame is known as piggybacking. But ACK can't be delayed for a long time if receiver(of the packet to be acknowledged) does not have any data to send.

Q.5

_______ is process of extracting previously non known valid and actionable information from large data to make crucial business and strategic decisions.

Data Management

Data base

Data Mining

Meta Data

Ans .

C

Explanation :
It is the computational process of discovering patterns in large data sets involving methods at the intersection of artificial intelligence, machine learning, statistics, and database systems..Data Mining is process of extracting previously non known valid and actionable information from large data to make crucial business and strategic decisions.

Q.6

The aspect ratio of an image is defined as

The ratio of width to its height measured in unit length.

The ratio of height to width measured in number of pixels.

The ratio of depth to width measured in unit length.

The ratio of width to depth measured in number of pixels.

Ans .

A

Explanation :
The aspect ratio of an image describes the proportional relationship between its width and its height. It is commonly expressed as two numbers separated by a colon, as in 16:9..The aspect ratio of an image is defined as : the ratio of width to its height measured in unit length.

Q.7

Which of the following features will characterize an OS as multi programmed OS?

(a) More than one program may be loaded into main memory at the same time.

(b) If a program waits for certain event another program is immediately scheduled.

(c) If the execution of a program terminates, another program is immediately scheduled.

(a) only

(a) and (b) only

(a) and (c) only

(a), (b) and (c) only

Ans .

D

Explanation :
More than one program loaded into main memory at the same time. If a program waits for certain event another program is immediately scheduled If the execution of a program terminates, another program is immediately scheduled In a multi-programming system there are one or more programs loaded in main memory which are ready to execute. Only one program at a time is able to get the CPU for executing its instructions (i.e., there is at most one process running on the system) while all the others are waiting their turn. The main idea of multi-programming is to maximize the use of CPU time. Indeed, suppose the currently running process is performing an I/O task (which, by definition, does not need the CPU to be accomplished). Then, the OS may interrupt that process and give the control to one of the other in-main-memory programs that are ready to execute (i.e. process context switching). In this way, no CPU time is wasted by the system waiting for the I/O task to be completed, and a running process keeps executing until either it voluntarily releases the CPU or when it blocks for an I/O operation. Therefore, the ultimate goal of multi-programming is to keep the CPU busy as long as there are processes ready to execute.

Q.8

Using RSA algorithm, what is the value of cipher text C, if the plain text M = 5 and p = 3, q = 11 & d = 7?

33

5

25

26

Ans .

D

Explanation :
Given p=3,q=11. n=(p X q) = (3 X 11)=33

m=(p-1)X(q-1) = (2 X 10)=20

Find a small odd integer e, that is relatively prime to m.

If e=3, then GCD(3,20)=1. e should be small and prime and so we let e=3.

d is given, d=7.

Public key = (e,n). (Values of e and n are known).

To encrypt a message, we apply the public key to the function

E(s) = se(mod n)

where s is the given message and e and n represent the public key integer pair. In the above question, the plain text M = 5. Plain text needs to be encrypted using the above formula.

=53(mod 33)

= 125 (mod 33)

= 26

As a result, the encrypted message E(s) = 26. This is what gets transmitted.

Q.9

You are given an OR problem and a XOR problem to solve. Then, which one of the following statements is true?

Both OR and XOR problems can be solved using single layer perception.

OR problem can be solved using single layer perception and XOR problem can be solved using self-organizing maps.

OR problem can be solved using radial basis function and XOR problem can be solved using single layer perception.

OR problem can be solved using single layer perception and XOR problem can be solved using radial basis function.

Ans .

D

Explanation :
OR can be solved using single layer perception and XOR problem can be solved using radial basis function.An OR problem and a XOR problem to solve. Then,OR problem can be solved using single layer perception and XOR problem can be solved using radial basis function.

Q.10

Match the following:

List – I List – II

a. Application layer 1. TCP

b. Transport layer 2. HDLC

c. Network layer 3. HTTP

d. Data link layer 4. BGP

2 1 4 3

3 4 1 2

3 1 4 2

2 4 1 3

Ans .

C

Explanation :HTTP is a Transport layer protocol,TCP is an Application layer protocol,BGP ia a Network layer protocol and HDLC is Data Link Layer protocol

Q.11

The time complexities of some standard graph algorithms are given. Match each algorithm with its time complexity? (n and m are no. of nodes and edges respectively)

List – I List – II

a. Bellman Ford algorithm 1. O (m log n)

b. Kruskals algorithm 2. O (n3)

c. Floyd Warshall algorithm 3. O(mn)

d. Topological sorting 4. O(n + m)

3 1 2 4

2 4 3 1

3 4 1 2

2 1 3 4

Ans .

A

Explanation :The time complexities of - Bellman Ford algorithm = O(mn),Kruskals algorithm = O (m log n),Floyd Warshall algorithm = O (n3) and Topological sorting = O(n + m)

Q.12

Let V1 = 2I – J + K and V2 = I + J – K, then the angle between V1 & V2 and a vector perpendicular to both V1 & V2 shall be:

90০ and (–2I + J – 3K)

60০ and (2I + J + 3K)

90০ and (2I + J – 3K)

90০ and (–2I – J + 3K)

Ans .

D

Explanation :
If V1 = 2I – J + K and V2 = I + J – K, then the angle between V1 & V2 and a vector perpendicular to both V1 & V2 shall be: 90০ and (–2I – J + 3K)

Q.13

Consider a fuzzy set A defined on the interval X=[0, 10]of integers by the membership Junction μA(x)=x/(x+2). Then the α cut corresponding to $\alpha =0.5 will be

{0,1,2,3,4,5,6,7,8,9,10}

{1,2,3,4,5,6,7,8,9,10}

{2,3,4,5,6,7,8,9,10}

{}

Ans .

C

Explanation :
putting value of X from 0 to 10 in membership function =x/x+2 we get
0/2 , 1/3 , 2/4 ,3/5 ,4/6 ,5/7 , 6 /8 , 7/9 , 8/10 , 9/11 , 10/12 means they are belonging with degree 0,0.33,0.5,0.6 .........0.83

here alpha cut =0.5 so first two elements will not be included in the result as their degree of belongingness <0.5

so ans will be C

Q.14

Let T(n) be a function defined by T(n)=1 and T(n)=2T(n/2)+√n, which of the following is true?

T(n)=O(√n)

T(n)=O(log₂n)

T(n)=O(n)

T(n)=O(n²)

Ans .

D

Explanation :
By applying master theorem, we can get the answer

Q.15

In classful addressing, an IP address 123.23.156.4 belongs to ______ class format.

A

B

C

D

Ans .

A

Explanation :
0 - 126 = Class A

128 - 191 = Class B

192 - 223 = Class C

224 - 255 = Class D

So ans is A.

Q.16

The Mandelbrot set used for the construction of beautiful images is based on the following transformation:xn + 1= x2n+ z

Here,

Both x & z are real numbers.

Both x & z are complex numbers.

x is real & z is complex.

x is complex & z is real.

Ans .

B

Explanation :
ans is B both are complex numbers.The Mandelbrot set is the set of complex numbers c for which the function does not diverge when iterated fromz=0,

Q.17

Which of the following permutations can be obtained in the output using a stack of size 3 elements assuming that input, sequence is 1, 2, 3, 4, 5?

Here,

3, 2, 1, 5, 4

5, 4, 3, 2, 1

3, 4, 5, 2, 1

3, 4, 5, 1, 2

Ans .

A

Explanation :

Hence Ans is A

Q.18

In a Linear Programming Problem, suppose there are 3 basic variables and 2 non-basic variables, then the possible numbers of basic solutions are

6

8

10

12

Ans .

C

Explanation :
Total number of basic solutions are given by the eqn n!/m! * (n-m)! , where m=3 no of basic variables and n=3+2 =5 total no of variables. hence total soln =5!/3!2!=5x4/2=10. Ans is C

Q.19

Identify the following activation function:

Φ(V)=Z+(1/(1+exp(−x∗V+Y))), Z, X, Y are parameters.

Step function

Ramp function

Sigmoid function

Gaussian function

Ans .

C

Explanation :

Q.20

The number of ways to distribute n distinguishable objects into k distinguishable boxes, so that ni objects are placed into box i, i=1,2,…k equals which of the following?

n!/(n1!+n2!+...+nk!)

(n1!+n2!+....+nk!)/(n1!n2!…nk!)

n₁!/(n₁!n₂!…n_k!)

(n₁!n₂!…n_k!)/(n₁!−n_2/!–n₃!⋯−n_k!)