UGC Net Computer Science December 13 Paper 3 Page 1 Solved

UGC Net Computer Science December 13 Paper 3 Page 1 Solved

1. If the primal Linear Programming problem has unbounded solution, then it’s dual problem will have

(A) feasible solution

(B) alternative solution

(C) no feasible solution at all

(D) no bounded solution at all
Answer C
Explanation :-
If the Primal problem is feasible, but unbounded in the direction of optimisation, then the dual has no feasible solution. Otherwise, if the Primal problem has an optimal solution, then the dual has also an optimal solution.

So the answer for your question is that feasibility of the Primal problem does not imply optimality for the Dual problem. It just excludes the possibility that the Dual will be unbounded in the direction of optimisation.

2. Given the problem to maximize

f(x), X = (x1, x2,......xn)

subject to m number of inequality

constraints.

gi(x) ≤ bi, i = 1, 2......m

including the non-negativity

constraints x ≥ 0.

Which of the following conditions is a Kuhn-Tucker necessary condition for a local maxima at –x ?

(A)

(B) –λ i [gi (–X) – bi]= 0, i = 1, 2 ….m

(C) gi (–X) ≤ bi, i = 1, 2 ….m

(D) All of these
Answer D
Explanation :-

Karush Kuhn Tucker algorithm helps to minimize or maximize point of mathematical optimization.

3. The following Linear Programming problem has :

Max Z = x1 + x2

Subject to x1 – x2 ≥ 0

                3x1 – x2 ≤ –3

and           x1, x2 ≥ 0

(A) Feasible solution

(B) No feasible solution

(C) Unbounded solution

(D) Single point as solution
Answer B
Explanation :-

This is not feasible solution because a feasible solution contains all the simplex set of variable in minimum optimization of LP. Here if constraint

         X1-x2>=0

And 3x1-x2=<-3

Not possible because 3x1 is greater than x1 which proof that it cannot be negative and as we know a feasible solution only comprise non-negative value.

4. Given a flow graph with 10 nodes,

13 edges and one connected

components, the number of regions

and the number of predicate (decision)

nodes in the flow graph will be

(A) 4, 5 (B) 5, 4

(C) 3, 1 (D) 13, 8

Answer B
Explanation :-

Predicate (decision) nodes are those nodes which has out degree atleast 2.
Region =13-10+2 = 5 [i.e. edges – vertex + (p+1)] since p= 1]
predicate nodes(P)=
= region = P+1 = 5
P = 4
so (B) is ans.

5. Function points can be calculated by

(A) UFP ∗ CAF

(B) UFP ∗ FAC

(C) UFP ∗ Cost

(D) UFP ∗ Productivity
Answer A
Explanation :-
Each application maintains information internally for performing its functions. Each logical group of data or control information that is generated, used and maintained by the application is counted as a logical internal file type. A logical internal file is simple if it contains a few record type, complex if is has many type, and average if it is in between.

Once the counts for all five different types are known for all three different complexity classes, the raw or unadjusted function point can be computed as a weighted sum as follows: -

     i=5     j=3

UFP  =          ? ?      w_ij C_ij

     i=1     j=1

 Where i reflects the row and j reflects the column in Table, wij is the entry in the ith row and jth column of the table (i.e. it represents the contribution of an element of the type i and complexity j); and Cij is the count of the number of elements of type i that have been classified as having the complexity corresponding to column j.

Once the UFP is obtained, it is adjusted for the environment complexity. For this 14 different characteristics of the system are given. These are data communications, distributed processing, performance objective, operation configuration load, transaction rate, on line data entry, end user efficiency, on-line update, complex processing logic, re-usability, installation ease, operational ease, multiple sites, and desire to facilitate change. The degree of influence of each of these factors is taken to be from 0 to 5,

Function type	Simple	Average	Complex
External input	3	4	6
External output	4	5	7
Logical internal file	7	10	15
External interface file	5	7	10
External inquiry	3	4	6

representing  the six different levels : not present (0), insignificant influence (1), modern influence 92), average influence (3), significant influence (4), and strong influence (5). The 14 degrees of influence for the system are then summed, giving total N (N ranges from 0 to 14*5 = 70). This N is used to obtain a complexity adjustment factor (CAF) as follows:

CAF = 0.65 + 0.01 N.

With this equation, the value of CAF ranges between 0.65 and 1.35. The delivered function points (DFP) are simply computed by multiplying the UFP by CAF. That is,

Delivered Function Points = CAF * Unadjusted Function Points.

6. Match the following :

List – I                             List – II

a. Data                       i. Module A and Module B have shared data

coupling

b. Stamp                    ii. Dependency between modules is based on

   coupling                      the fact they communicate by only passing of data 

c. Common                iii. When complete data structure is passed from one module to

     coupling                      another

d. Content                  iv. When the control is passed from one module to the middle

coupling                         of another

Codes :

       a  b c d

(A) iii ii i iv

(B) ii iii i iv

(C) ii iii iv i

(D) iii ii iv i

Answer B
Explanation :-

Coupling 

• Coupling is the indication of the relationships between modules.
• Coupling shows the relative independence among the modules.
• Coupling is a degree to which a component / module is connected to the other modules.
• While designing you should strive for low coupling i.e. dependency between modules should be less.
• Making private fields, private methods and non public classes provides loose coupling.
• Coupling is Inter -Module Concept.

Types of Coupling

 module here refers to a subroutine of any kind, i.e. a set of one or more statements having a name and preferably its own set of variable names.

Content coupling (high)

Content coupling (also known as Pathological coupling) occurs when one module modifies or relies on the internal workings of another module (e.g., accessing local data of another module). In this situation, a change in the way the second module produces data (location, type, timing) might also require a change in the dependent module.

Common coupling

Common coupling (also known as Global coupling) occurs when two modules share the same global data (e.g., a global variable). Changing the shared resource might imply changing all the modules using it.

int global=10;

void fun1()

{

     global++;

     printf("\ni am Updating Global Data");

}

void fun2()

{

     global++;

     printf("\ni am also Updating Global Data");

}

void main()

{

    fun1();

    fun2();

}

External coupling

External coupling occurs when two modules share an externally imposed data format, communication protocol, or device interface. This is basically related to the communication to external tools and devices.

extern external=10;

void fun1()

{

     external++;

     printf("\ni am External  Global Data");

}

Control coupling

Control coupling is one module controlling the flow of another, by passing it information on what to do (e.g., passing a what-to-do flag).

int findflag(int s,int ar[3])

{

     int i;

     for(i=0;i<3;i++)

     {

           if(ar[i]==s)

                break;

     }

     if(i==3)

           return 0; /* flag zero indicate value not found */

     else

           return 1; /* flag one indicate value found */

}

void search(int search,int ar[3])

{

     if(findflag(search,ar)==1)

           printf("Found");

     else

           printf("Not Found");

}

void main()

{

    int ar[]={1,2,3};

    search(2,ar);

}

Stamp coupling (Data-structured coupling)

Stamp coupling occurs when modules share a composite data structure and use only parts of it, possibly different parts (e.g., passing a whole record to a function that only needs one field of it).

In this situation, a modification in a field that a module does not need may lead to changing the way the module reads the record.

struct student

{

     int roll;

     char name[10];

     char city[10];

};

void show(struct student s)

{

     printf("\nRoll=%d",s.roll);

     printf("\nName=%s",s.name);

     printf("\nCity=%d",s.city);

}

void main()

{

    struct student rec={101,"sona","Vns"};

    show(rec);

}

Data coupling

Data coupling occurs when modules share data through, for example, parameters. Each datum is an elementary piece, and these are the only data shared (e.g., passing an integer to a function that computes a square root).

int sqrt(int n)

{

     return n*n;

}

void main()

{

    printf("SQRT of 2 is %d",sqrt(2));

}

Message coupling (low)

This is the loosest type of coupling. It can be achieved by state decentralization (as in objects) and component communication is done via parameters or message passing.

No coupling

Modules do not communicate at all with one another.

7. A process which defines a series of tasks that have the following four primary objectives is known as

1. to identify all items that collectively define the software configuration.

2. to manage changes to one or more of these items.

3. to facilitate the construction of different versions of an application.

4. to ensure that software quality is maintained as the configuration evolves over time.

(A) Software Quality Management Process

(B) Software Configuration Management Process

(C) Software Version Management Process

(D) Software Change Management Process

Answer B
Explanation :-

Software configuration management (SCM) is the discipline for systematically controlling the changes that take place during development. Software configuration management is a process independent of the development process largely because most development models cannot accommodate change at any time during development. SCM can be considered as having three major components:

•     Software configuration identification
•     Change control
•     Status accounting and auditing

8. One weakness of boundary value analysis and equivalence partitioning is

(A) they are not effective.

(B) they do not explore combinations of input circumstances.

(C) they explore combinations of input circumstances.

(D) none of the above
Answer B
Explanation :-

Boundary value analysis:

It is based on testing on and around the boundaries between partitions. If you have done “range checking”, you were probably using the boundary value analysis technique, even if you weren’t aware of it. Note that we have both valid boundaries (in the valid partitions) and invalid boundaries (in the invalid partitions).

for example we take a month is a input column in a application...
Equivalence Partitioning testing is

        .... -2 -1  0 1 .............. 12 13  14  15 .....
      --------------|-------------------|---------------------
invalid partition 1     valid partition    invalid partition 2

In boundary value testing is
  if (month > 0 && month < 13)
we check only boundary values...

AS WE CAN SEE they do not explore combinations of input circumstances. COnsider only two number

9. Which once of the following is not a software myth ?

(A) Once we write the program and get it to work, our job is done.

(B) Project requirements continually change, but change can be easily accommodated because

software is flexible.

(C) If we get behind schedule, we can add more programmers and catch up.

(D) If an organization does not understand how to control software projects internally, it

will invariably struggle when it outsources software projects.
Answer B
Explanation :-

Myth : if i decide to outsource project to a third party , i can just relax and let that firm built it.

Reality : If an organization does not understand how to control software projects internally, it

will invariably struggle when it outsources software projects.

10. Match the following with respect to relationship between objects and classes :

List – I                                    List – II

a. State                                i.Useful for both abstract modelling and for designing actual program

diagram

b. Object diagram               ii. Describes object classes

c. Class diagram                 iii. Useful for documenting test cases

d. Instance diagram            iv. Describing the behaviour of a single class of objects.

Codes :

        a b c d

(A) iv i ii iii

(B) ii iii iv i

(C) iii iv ii i

(D) ii iv i iii

Answer A
Explanation :-

state diagram is a type of diagram used in computer science and related fields to describe the behavior of systems. State diagrams require that the system described is composed of a finite number of states.

Object diagrams are derived from class diagrams so object diagrams are dependent upon class diagrams.Object diagrams represent an instance of a class diagram. The basic concepts are similar for class diagrams and object diagrams. Object diagrams also represent the static view of a system but this static view is a snapshot of the system at a particular moment.Object diagrams are used to render a set of objects and their relationships as an instance.

In software engineering, a class diagram in the Unified Modeling Language (UML) is a type of static structure diagram that describes the structure of a system by showing the system’s classes, their attributes, operations (or methods), and the relationships among objects.

An InstanceDiagram is a part of the UnifiedModelingLanguage that one does not see mentioned too often. The basic idea is to make a static snap shot of instances (not classes) in your system or subsystem. Make it show exactly who points to whom.

Download PDF

Next Page