## UGC NET CS 2008-june-Paper-2

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Question 1 |

Which of the following does not define a tree ?

A tree is a connected acyclic graph. | |

A tree is a connected graph with n-1 edges where ‘n’ is the number of vertices in the graph. | |

A tree is an acyclic graph with n-1 edges where ‘n’ is the number of vertices in the graph. | |

A tree is a graph with no cycles. |

Question 1 Explanation:

→ An undirected graph is

→ An acyclic graph is a graph having no graph cycles. Acyclic graphs are bipartite.

→ A connected acyclic graph is known as a tree, and a possibly disconnected acyclic graph is known as a forest (i.e., a collection of trees).

→ All first 3 options are correctly defining tree without any loss of information, whereas "A tree is a graph with no cycles." is not always correct as it can be bipartite.

Hence D does not define a tree.

**connected**when there is a path between every pair of vertices. In a connected graph, there are no**unreachable**vertices. A graph that is not connected is**disconnected**. An undirected graph G is said to be**disconnected**if there exist two nodes in G such that no path in G has those nodes as endpoints.→ An acyclic graph is a graph having no graph cycles. Acyclic graphs are bipartite.

→ A connected acyclic graph is known as a tree, and a possibly disconnected acyclic graph is known as a forest (i.e., a collection of trees).

→ All first 3 options are correctly defining tree without any loss of information, whereas "A tree is a graph with no cycles." is not always correct as it can be bipartite.

Hence D does not define a tree.

Question 2 |

The complexity of Kruskal’s minimum spanning tree algorithm on a graph with ‘n’ nodes and ‘e ’ edges is :

O (n) | |

O (n log n) | |

O (e log n) | |

O (e) |

Question 2 Explanation:

MST-KRUSKAL(G,w)

1. A=∅

2. for each vertex v∈ G.V

3. MAKE-SET(v)

4. sort the edges of G.E into nondecreasing order by weight w

5. for each edge (u,v)∈G.E, taken in nondecreasing order by weight

6. if FIND-SET(u)≠FIND-SET(v)

7. A=A∪{(u,v)}

8. UNION(u,v)

9. return A

Analysis:

The running time of Kruskal’s algorithm for a graph G=(V,E) depends on how we implement the disjoint-set data structure.

→ Initializing the set A in line 1 takes O(1) time, and the time to sort the edges in line 4 is O(ElgE). (We will account for the cost of the |V| MAKE-SET operations in the for loop of lines 2–3 in a moment.)

→ The for loop of lines 5–8 performs O(E) FIND-SET and UNION operations on the disjoint-set forest.

→ Along with the |V| MAKE-SET operations, these take a total of O((V+E)α(V)) time, where α is the very slowly growing function.

→ Because we assume that G is connected, we have |E|>=|V|-1, and so the disjoint-set operations take O(Eα(V)) time.

→ Moreover, since α |V|=O(lg V)=O(lg E), the total running time of Kruskal’s algorithm is O(ElgE)

→ Observing that |E|<|V|

1. A=∅

2. for each vertex v∈ G.V

3. MAKE-SET(v)

4. sort the edges of G.E into nondecreasing order by weight w

5. for each edge (u,v)∈G.E, taken in nondecreasing order by weight

6. if FIND-SET(u)≠FIND-SET(v)

7. A=A∪{(u,v)}

8. UNION(u,v)

9. return A

Analysis:

The running time of Kruskal’s algorithm for a graph G=(V,E) depends on how we implement the disjoint-set data structure.

→ Initializing the set A in line 1 takes O(1) time, and the time to sort the edges in line 4 is O(ElgE). (We will account for the cost of the |V| MAKE-SET operations in the for loop of lines 2–3 in a moment.)

→ The for loop of lines 5–8 performs O(E) FIND-SET and UNION operations on the disjoint-set forest.

→ Along with the |V| MAKE-SET operations, these take a total of O((V+E)α(V)) time, where α is the very slowly growing function.

→ Because we assume that G is connected, we have |E|>=|V|-1, and so the disjoint-set operations take O(Eα(V)) time.

→ Moreover, since α |V|=O(lg V)=O(lg E), the total running time of Kruskal’s algorithm is O(ElgE)

→ Observing that |E|<|V|

^{2}, we have lg |E|=O(lg V), and so we can restate the running time of Kruskal’s algorithm as O(E lg V)Question 3 |

If a code is t-error correcting, the minimum Hamming distance is equal to :

2t+1 | |

2t | |

2t-1 | |

t-1 |

Question 3 Explanation:

Question 4 |

The set of positive integers under the operation of ordinary multiplication is :

not a monoid | |

not a group | |

a group | |

an Abelian group |

Question 4 Explanation:

The set of positive integers under the operation of ordinary multiplication is a monoid but not a group.

Question 5 |

In a set of 8 positive integers, there always exists a pair of numbers having the same remainder when divided by :

7 | |

11 | |

13 | |

15 |

Question 5 Explanation:

Given data,

-- Set of positive integers=8

-- A pair of numbers having the same remainder when divided by=?

Step-1: This problem, we can use pigeonhole principle.

These numbers are less than 8 only.

divided by 11, 13, 15 is greater than 8

-- Set of positive integers=8

-- A pair of numbers having the same remainder when divided by=?

Step-1: This problem, we can use pigeonhole principle.

__Pigeonhole Principle__: If there are ‘X’ positive integers to be placed in ‘Y’ possible remainder where Y < X, then there is at least one possible remainder which receives more than one positive integer.__Option-A__: Total number of possible remainders, divided by 7 is 0,1,2,3,4,5,6.These numbers are less than 8 only.

__Option-B,C and D__: Total number of possible remainders,divided by 11, 13, 15 is greater than 8

Question 6 |

An example of a tautology is :

x v y | |

x v (~y) | |

x v (~x) | |

(x=>y) v (x<=y) |

Question 6 Explanation:

Question 7 |

Among the logic families RTL, TTL, ECL and CMOS, the fastest family is :

ECL | |

CMOS | |

TTL | |

RTL |

Question 7 Explanation:

__Emitter Coupled Logic (ECL):__

The storage time is eliminated as the transistors are used in difference amplifier mode and are never driven into saturation.

1. Fastest among all logic families

2. Lowest propagation delay.

__Complementary metal oxide semiconductor(CMOS):__

The power dissipation is usually 10nW per gate depending upon the power supply voltage, output load etc.

1. Lowest power dissipation

2. Excellent noise immunity

3. High packing density

4. Wide range of supply voltage

5. Highest fan out among all logic families

__Transistor Transistor Logic__:

→ TTL devices consume substantially more power than equivalent CMOS devices at rest, but power consumption does not increase with clock speed as rapidly as for CMOS devices.

Question 8 |

The octal equivalent of the hexadecimal number FF is :

100 | |

150 | |

377 | |

737 |

Question 8 Explanation:

Step-1: Convert FF into binary number

(FF)

Step-2: Divide binary number into 3 segments from LSB(Least significant bit).

(FF)

_{16}= (1111 1111)_{2}Step-2: Divide binary number into 3 segments from LSB(Least significant bit).

Question 9 |

The characteristic equation of a T flip flop is given by :

Q _{N+1} = TQ_{N} | |

Q _{N+1} = T+Q_{N} | |

Q _{N+1} = TQ_{N} | |

Q _{N+1}=T'Q_{N} |

Question 9 Explanation:

__T-Flip flop Truth Table__:

Question 10 |

The idempotent law in Boolean algebra says that :

~(~x)=x | |

x+x=x | |

x+xy=x | |

x(x+y)=x |

Question 10 Explanation:

→

The idempotence in the context of elements of algebras that remain invariant when raised to a positive integer power, and literally means "(the quality of having) the same power", from idem + potence (same + power).

(i). X + X=X

(ii). X ∧ X=X

According to boolean algebra

__Idempotent law__:The idempotence in the context of elements of algebras that remain invariant when raised to a positive integer power, and literally means "(the quality of having) the same power", from idem + potence (same + power).

(i). X + X=X

(ii). X ∧ X=X

According to boolean algebra

Question 11 |

What is the effect of the following C code ?

for(int i=1; i≤5; i=i+½)

printf(“%d,”,i);

for(int i=1; i≤5; i=i+½)

printf(“%d,”,i);

It prints 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5, and stops | |

It prints 1, 2, 3, 4, 5, and stops | |

It prints 1, 2, 3, 4, 5, and repeats forever | |

It prints 1, 1, 1, 1, 1, and repeats forever |

Question 11 Explanation:

According to integer datatype, fraction values we are not considering. So,it will always become

1. Iteration-1: for(int i=1; i≤5; i=i+½) /*condition TRUE, then we are printing value but incrementing by 0. It means, i=1+0 */

So, ‘i’ value always become 1 only. The given condition always TRUE and will print infinite times.

1. Iteration-1: for(int i=1; i≤5; i=i+½) /*condition TRUE, then we are printing value but incrementing by 0. It means, i=1+0 */

So, ‘i’ value always become 1 only. The given condition always TRUE and will print infinite times.

Question 12 |

Consider the following declaration in C :

char a[ ];

char *p;

Which of the following statement is not a valid statement ?

char a[ ];

char *p;

Which of the following statement is not a valid statement ?

p=a; | |

p=a+2; | |

a=p; | |

p=&a[2]; |

Question 12 Explanation:

According to pointer value assigning declarations,

p=a; /*It is correct declaration */

p=p+2; /*It is correct declaration */

p=&a[2]; /*It is correct declaration */

a=p; /* It is incorrect declaration */

p=a; /*It is correct declaration */

p=p+2; /*It is correct declaration */

p=&a[2]; /*It is correct declaration */

a=p; /* It is incorrect declaration */

Question 13 |

Consider the following C code :

{

int a=5, b=9;

float r;

r=b/a;

}

What is the value of r ?

{

int a=5, b=9;

float r;

r=b/a;

}

What is the value of r ?

1.8 | |

1.0 | |

2.0 | |

0.0 |

Question 13 Explanation:

Given ‘a’ and ‘b’ variables are integer data types.

‘r’ is floating point data type.

r=9/5; /* 9/5=1 but not 1.8 */

According to integer data type values will return only integers but not fraction.

So, ‘r’ is floating point value, it will assign fraction part 6 digits. it will print 1.000000

‘r’ is floating point data type.

r=9/5; /* 9/5=1 but not 1.8 */

According to integer data type values will return only integers but not fraction.

So, ‘r’ is floating point value, it will assign fraction part 6 digits. it will print 1.000000

Question 14 |

Function overloading is a concept in which :

a function is used to implement lots of tasks at the same time. | |

a function is called too many number of times by another function. | |

a function provides common interface to the user to carry out possibly different functions in each call. | |

a function is computationally too expensive for the system to handle. |

Question 14 Explanation:

Function overloading is a concept in which a function provides common interface to the user to carry out possibly different functions in each call.

Question 15 |

Which of the following is true ?

A “static” member of a class cannot be inherited by its derived class. | |

A “static” member of a class can be initialized only within the class it is a member of. | |

A “static” member of a class can be initialized before an object of that class is created. | |

Since “static” member of a class is actually a global element, it does not require a class/object qualifier to access it independently of class/object. |

Question 15 Explanation:

FLASE: A “static” member of a class cannot be inherited by its derived class.

FALSE: A “static” member of a class can be initialized only within the class it is a member of.

TRUE: A “static” member of a class can be initialized before an object of that class is created.

FALSE: Since “static” member of a class is actually a global element, it does not require a class/object qualifier to access it independently of class/object.

FALSE: A “static” member of a class can be initialized only within the class it is a member of.

TRUE: A “static” member of a class can be initialized before an object of that class is created.

FALSE: Since “static” member of a class is actually a global element, it does not require a class/object qualifier to access it independently of class/object.

Question 16 |

A superkey for an entity consists of :

one attribute only | |

at least two attributes | |

at most two attributes | |

one or more attributes |

Question 16 Explanation:

A superkey for an entity consists of one or more attributes.

Question 17 |

Which of the following set of keywords constitutes a mapping in SQL ?

SELECT, FROM, TABLE | |

SELECT, FROM, WHERE | |

CONNECT, TABLE, CREATE | |

SELECT, TABLE, INSERT |

Question 17 Explanation:

In given question they are asking that using which SQL keywords a SQL query can be made to retrieve the data from database.

Option(A): In SQL there is no keyword exist with name "TABLE". So using these keywords no query can be made.

Option(B): This combination of keywords can be used to retrieve data from a table in database.

Syntax: SELECT FROM WHERE

Option(C): In SQL there is no keyword exist with name "TABLE". So using these keywords no query can be made.

Option(D): In SQL there is no keyword exist with name "TABLE". So using these keywords no query can be made.

Option(A): In SQL there is no keyword exist with name "TABLE". So using these keywords no query can be made.

Option(B): This combination of keywords can be used to retrieve data from a table in database.

Syntax: SELECT

Option(C): In SQL there is no keyword exist with name "TABLE". So using these keywords no query can be made.

Option(D): In SQL there is no keyword exist with name "TABLE". So using these keywords no query can be made.

Question 18 |

If a relation is in 2NF then :

every candidate key is a primary key | |

every non-prime attribute is fully functionally dependent on each relation key | |

every attribute is functionally independent | |

every relational key is a primary key |

Question 18 Explanation:

→ If a relation is in 2NF then every non-prime attribute is fully functionally dependent on each relation key.

→ 2NF ensures that the relation contains no partial dependency. A relation is in 2NF if it is in 1NF and have no partial dependency.

→ 2NF ensures that the relation contains no partial dependency. A relation is in 2NF if it is in 1NF and have no partial dependency.

Question 19 |

Which of the following is true ?

A relation in 3NF is always in BCNF | |

A relation in BCNF is always in 3NF | |

BCNF and 3NF are totally different | |

A relation in BCNF is in 2NF but not in 3NF |

Question 19 Explanation:

Question 20 |

Consider the query : SELECT student_name FROM student_data WHERE rollno (SELECT rollno FROM student_marks WHERE SEM1_MARK5SEM2_MARK);

Which of the following is true ?

Which of the following is true ?

It gives the name of the student whose marks in semester 1 and semester 2 are same. | |

It gives all the names and roll nos of those students whose marks in semester 1 and semester 2 are same. | |

It gives the names of all the students whose marks in semester 1 and semester 2 are same. | |

It gives roll numbers of all students whose marks in semester 1 and semester 2 are same. |

Question 20 Explanation:

Since both the queries are not related so you can solve inner query first and then can solve outer query.

Question 21 |

Which of the following data structures is most efficient in terms of both space and time to reverse a string of characters ?

Linked list | |

Stack | |

Array | |

Tree |

Question 21 Explanation:

To print values in reverse order we are preferring stack data structure only. Stack using Last in first out(LIFO) procedure.

Question 22 |

Which of the following can be the sequence of nodes examined in a binary search tree while searching for key 98 ?

100,50,75, 60, 98
| |

100, 120, 90, 95, 98 | |

200,70,100, 95, 98 | |

75, 150, 90, 80, 98 |

Question 22 Explanation:

Binary search tree will follow some rules

1. Left subtree is smaller than his root node

2. Right subtree is greater than or equal to his root

Based on above rules, the search operation will perform

1. Left subtree is smaller than his root node

2. Right subtree is greater than or equal to his root

Based on above rules, the search operation will perform

Question 23 |

Which of the following is true for a sorted list with ‘n’ elements ?

Insertion in a sorted array takes constant time. | |

Insertion in a sorted linear linked list takes constant time. | |

Searching for a key in a sorted array can be done in O(log n) time. | |

Searching for a key in a sorted linear linked list can be done in O(log n) time. |

Question 23 Explanation:

TRUE: Searching for a key in a sorted array can be done in O(log n) time.

Because elements are in sorted order, we are using binary search tree. it will take worst case time complexity O(logn) only.

Because elements are in sorted order, we are using binary search tree. it will take worst case time complexity O(logn) only.

Question 24 |

Files that are related to input/output and are used to model serial I/O devices such as terminals, printers and networks are called :

regular files | |

character special files | |

directories | |

block special files |

Question 24 Explanation:

Files that are related to input/output and are used to model serial I/O devices such as terminals, printers and networks are called block special files

Question 25 |

An example of a possible file attribute is :

minimum size | |

permanent flag | |

archive flag | |

EBCDIC flag |

Question 25 Explanation:

→ An example of a possible file attribute is archive flag.

Question 26 |

The ATM cells are _________ bytes long.

48 | |

53 | |

64 | |

69 |

Question 26 Explanation:

→ The ATM cell is 53 bytes long.

→ ATM for carriage of a complete range of user traffic, including voice, data, and video signals.

→ ATM provides functionality that is similar to both circuit switching and packet switching networks.

→ ATM uses asynchronous time-division multiplexing, and encodes data into small, fixed-sized packets (ISO-OSI frames) called cells.

→ ATM uses a connection-oriented model in which a virtual circuit must be established between two endpoints before the actual data exchange begins.

→ ATM for carriage of a complete range of user traffic, including voice, data, and video signals.

→ ATM provides functionality that is similar to both circuit switching and packet switching networks.

→ ATM uses asynchronous time-division multiplexing, and encodes data into small, fixed-sized packets (ISO-OSI frames) called cells.

→ ATM uses a connection-oriented model in which a virtual circuit must be established between two endpoints before the actual data exchange begins.

Question 27 |

For slotted ALOHA, the maximum channel utilization is :

100% | |

50% | |

36% | |

18% |

Question 27 Explanation:

Question 28 |

For a channel of 3 KHz bandwidth and signal to noise ratio of 30 dB, the maximum data rate is :

3000 bps | |

6000 bps | |

15000 bps | |

30000 bps |

Question 28 Explanation:

→ Shannon Capacity (Noisy Channel) In presence of Gaussian band-limited white noise, Shannon-Hartley theorem gives the maximum data rate capacity.

C=B log

where S and N are the signal and noise power, respectively, at the output of the channel.

B=3 KHz = 3000 Hz

Signal to noise ratio=30 dB = 1000

C=3000*log

=3000*9.967226258835992

= 29901.67 bps (Approximately 30000 bps)

C=B log

_{2}(1+S/N).where S and N are the signal and noise power, respectively, at the output of the channel.

B=3 KHz = 3000 Hz

Signal to noise ratio=30 dB = 1000

C=3000*log

_{2}(1+1000)=3000*9.967226258835992

= 29901.67 bps (Approximately 30000 bps)

Question 29 |

An example of a public key encryption algorithm is :

Caesar cipher algorithm | |

DES algorithm | |

AES algorithm | |

Knapsack algorithm |

Question 29 Explanation:

→ Private key algorithms are DES,AES,Caesar cipher,etc.., algorithms.

→ Public key algorithms are Knapsack algorithm, RSA, Diffie Hellman,ECC,etc..,

→ Public key algorithms are Knapsack algorithm, RSA, Diffie Hellman,ECC,etc..,

Question 30 |

With reference to hierarchical routing, the optimum number of levels for an m router subnet is :

m ^{2} | |

m | |

ln m | |

√m |

Question 30 Explanation:

With reference to hierarchical routing, the optimum number of levels for an m router subnet is ln m.

Question 31 |

Assembler program is :

dependent on the operating system | |

dependent on the compiler | |

dependent on the hardware | |

independent of the hardware |

Question 31 Explanation:

→ Assembler program is dependent on the hardware.

→ An assembler program creates object code by translating combinations of mnemonics and syntax for operations and addressing modes into their numerical equivalents.

→ An assembler program creates object code by translating combinations of mnemonics and syntax for operations and addressing modes into their numerical equivalents.

Question 32 |

In the indirect addressing scheme, the second part of an instruction contains :

the operand in decimal form | |

the address of the location where the value of the operand is stored
| |

the address of the location where the address of the operand is stored
| |

the operand in an encoded form |

Question 32 Explanation:

→ In the indirect addressing scheme, the second part of an instruction contains the address of the location where the address of the operand is stored.

→ Indirect addressing is a scheme in which the address specifies which memory word or register contains not the operand but the address of the operand.

→ Indirect addressing is a scheme in which the address specifies which memory word or register contains not the operand but the address of the operand.

__Example__: LOAD R1, @R2 Load the content of the memory address stored at register R2 to register R1.Question 33 |

At the end of parsing,

tokens are identified. | |

set of instructions are identified.
| |

the syntactic groups are identified. | |

machine instructions are identified |

Question 33 Explanation:

At end of parsing whether it is syntactically correct or not is identified
So at end of parsing

1. tokens are identified

2. whether the given code is syntactically correct or not is identified.

1. tokens are identified

2. whether the given code is syntactically correct or not is identified.

Question 34 |

Dead-code elimination in machine code optimization refers to :

removal of all labels. | |

removal of values that never get used. | |

removal of function which are not involved. | |

removal of a module after its use. |

Question 34 Explanation:

→ Dead code elimination in machine code optimization refers to removal of values that never get used.

→ Dead code includes code that can never be executed (unreachable code), and code that only affects dead variables (written to, but never read again), that is, irrelevant to the program.

→ Dead code includes code that can never be executed (unreachable code), and code that only affects dead variables (written to, but never read again), that is, irrelevant to the program.

Question 35 |

A parse tree is an annotated parse tree if :

it shows attribute values at each node. | |

there are no inherited attributes. | |

it has synthesized nodes as terminal nodes. | |

every non-terminal nodes is an inherited attribute. |

Question 35 Explanation:

A parse tree is an annotated parse tree if it shows attribute values at each node.

1. High level specification

2. Hides implementation details

3. Explicit order of evaluation is not specified

__Features__:1. High level specification

2. Hides implementation details

3. Explicit order of evaluation is not specified

Question 36 |

An example of a non-preemptive CPU scheduling algorithm is :

Shortest job first scheduling. | |

Round robin scheduling. | |

Priority scheduling. | |

Fair share scheduling. |

Question 36 Explanation:

→ Round robin and priority scheduling algorithms are preemptive based algorithms.

→ FCFS and SJF is non-preemptive based algorithms but in SJF are supporting another version is SRTF(Shortest remaining time first)

→ FCFS and SJF is non-preemptive based algorithms but in SJF are supporting another version is SRTF(Shortest remaining time first)

Question 37 |

There are ‘n’ processes in memory. A process spends a fraction ‘p’ of its time waiting for I/O to complete. The CPU utilization is given by :

p ^{n} | |

1-p ^{n} | |

(1-p) ^{n} | |

1-n p |

Question 37 Explanation:

CPU Utilization=1-p

Where,

n = number processes in memory

p = Waiting time for I/O to complete.

^{n}Where,

n = number processes in memory

p = Waiting time for I/O to complete.

Question 38 |

An example of a memory management system call in UNIX is :

fork. | |

mmap. | |

sigaction. | |

execve. |

Question 38 Explanation:

Fork() → It will create child process of already created process.

mmap() → It is memory management system call and implements demand paging.

sigaction() → Examine and change a signal action

execve() → Executes the program pointed to by filename.

mmap() → It is memory management system call and implements demand paging.

sigaction() → Examine and change a signal action

execve() → Executes the program pointed to by filename.

Question 39 |

With 64 bit virtual addresses, a 4KB page and 256 MB of RAM, an inverted page table requires :

8192 entries.
| |

16384 entries. | |

32768 entries. | |

65536 entries. |

Question 39 Explanation:

Given data,

-- Virtual addresses size= 64 bit

-- Page size = 4 KB

-- RAM size = 256 MB

-- Inverted page table requires = ?

-- Virtual addresses size= 64 bit

-- Page size = 4 KB

-- RAM size = 256 MB

-- Inverted page table requires = ?

Question 40 |

A computer has 6 tape drives with ‘n’ processes competing for them. Each process may need two drives. For which values of ‘n’ is the system deadlock free ?

1 | |

2 | |

3 | |

6 | |

None of the above |

Question 40 Explanation:

→ Each process needs 2 drives. So for deadlock just give each process onedrive. So total 6 process can be given 1 drive each and can cause deadlock. So to break deadlock just reduce 1 process.

So, maximum number of process for the system to be deadlock free is 5.

So, maximum number of process for the system to be deadlock free is 5.

Question 41 |

Waterfall model for software development is :

a top down approach. | |

a bottom up approach. | |

a sequential approach. | |

a consequential approach. |

Question 41 Explanation:

Waterfall model for software development is a linear or sequential approach.

1. Feasibility study

2. Analysis

3. Design

4. Coding & Unit testing

5. Integration & System Testing

6. Maintenance

__Phases__:1. Feasibility study

2. Analysis

3. Design

4. Coding & Unit testing

5. Integration & System Testing

6. Maintenance

Question 42 |

In software development, value adjustment factors include the following among others :

the criticality of the performance and reusability of the code. | |

number of lines of code in the software. | |

number of technical manpower and hardware costs. | |

time period available and the level of user friendliness. |

Question 42 Explanation:

→ Value adjustment factors include the criticality of the performance and reusability of the code.

→ 14 Value adjustment factors are using in functional points maximum.

→ 14 Value adjustment factors are using in functional points maximum.

Question 43 |

While designing the user interface, one should :

use as many short cuts as possible. | |

use as many defaults as possible. | |

use as many visual layouts as possible. | |

reduce the demand on short-term memory. |

Question 43 Explanation:

→ While designing the user interface, one should reduce the demand on short-term memory.

Question 44 |

In software cost estimation, base estimation is related to :

cost of similar projects already completed. | |

cost of the base model of the present project.
| |

cost of the project with the base minimum profit. | |

cost of the project under ideal situations. |

Question 44 Explanation:

In software cost estimation, base estimation is related to cost of similar projects already completed.

Question 45 |

In cleanroom software engineering :

only eco-friendly hardware is used. | |

only hired facilities are used for development. | |

correctness of the code is verified before testing. | |

implementation is done only after ensuring correctness. |

Question 45 Explanation:

In cleanroom software engineering implementation is done only after ensuring correctness.

Cleanroom software engineering (CSE) is a process model that removes defects before they can precipitate serious hazards. It is a team-oriented, theory based software, which is developed using the formal methods, correctness verification and Statistical Quality Assurance (SQA).

→ Clean room management is based on the incremental model of software development, which accumulates into the final product. The approach combines mathematical-based methods of software specification, design and correctness verification with statistical, usage-based testing to certify software fitness for use.

→ The main goal of clean room engineering is to produce zero error-based software by allowing correct designs, which avoid rework.

__Clean-room Strategy__:Cleanroom software engineering (CSE) is a process model that removes defects before they can precipitate serious hazards. It is a team-oriented, theory based software, which is developed using the formal methods, correctness verification and Statistical Quality Assurance (SQA).

→ Clean room management is based on the incremental model of software development, which accumulates into the final product. The approach combines mathematical-based methods of software specification, design and correctness verification with statistical, usage-based testing to certify software fitness for use.

→ The main goal of clean room engineering is to produce zero error-based software by allowing correct designs, which avoid rework.

Question 46 |

Amdahl’s law states that the maximum speedup S achievable by a parallel computer with ‘p’ processors is given by :

S≤f+(1-f)/p | |

S≤f/p+(1-f) | |

S≤1/[f+(1-f)/p] | |

S≤1/[1-f+f/p] |

Question 46 Explanation:

Amdahl’s law can be formulated in the following way:

where

→S

→s is the speedup of the part of the task that benefits from improved system resources;

→p is the proportion of execution time that the part benefiting from improved resources originally occupied.

Furthermore,

shows that the theoretical speedup of the execution of the whole task increases with the improvement of the resources of the system and that regardless of the magnitude of the improvement, the theoretical speedup is always limited by the part of the task that cannot benefit from the improvement.

→ Amdahl's law applies only to the cases where the problem size is fixed. In practice, as more computing resources become available, they tend to get used on larger problems (larger datasets), and the time spent in the parallelizable part often grows much faster than the inherently serial work. In this case, Gustafson's law gives a less pessimistic and more realistic assessment of the parallel performance.

where

→S

_{latency}is the theoretical speedup of the execution of the whole task;→s is the speedup of the part of the task that benefits from improved system resources;

→p is the proportion of execution time that the part benefiting from improved resources originally occupied.

Furthermore,

shows that the theoretical speedup of the execution of the whole task increases with the improvement of the resources of the system and that regardless of the magnitude of the improvement, the theoretical speedup is always limited by the part of the task that cannot benefit from the improvement.

→ Amdahl's law applies only to the cases where the problem size is fixed. In practice, as more computing resources become available, they tend to get used on larger problems (larger datasets), and the time spent in the parallelizable part often grows much faster than the inherently serial work. In this case, Gustafson's law gives a less pessimistic and more realistic assessment of the parallel performance.

Question 47 |

With reference to cluster analysis in data mining, a distance measure that is NOT used is :

Euclidean distance. | |

Manhattan distance. | |

Chebyshev's distance. | |

Lee distance. |

Question 47 Explanation:

→

→ Euclidean, Manhattan and Chebyshev's distance measure that is used is with reference to cluster analysis in data mining.

__Lee distance__is applied for phase modulation while the Hamming distance is used in case of orthogonal modulation.→ Euclidean, Manhattan and Chebyshev's distance measure that is used is with reference to cluster analysis in data mining.

Question 48 |

In a mobile communication system, a geographic region is divided into cells. For each frequency set, there is a buffer __________ wide where that frequency is not used.

one-cell | |

two-cells | |

three-cells | |

four-cells |

Question 48 Explanation:

In a mobile communication system, a geographic region is divided into cells. For each frequency set, there is a buffer two cells wide where that frequency is not used.

Question 49 |

Identify the incorrect statement :

The overall strategy drives the e-commerce data warehousing strategy. | |

Data warehousing in an e-commerce environment should be done in a classical manner. | |

E-commerce opens up an entirely new world of web servers. | |

E-commerce security threats can be grouped into three major categories. |

Question 49 Explanation:

TRUE: The overall strategy drives the E-Commerce data warehousing strategy.

TRUE: Data warehousing in an E-Commerce environment should be done in a classical manner.

TRUE: E-Commerce opens up an entirely new world of web server.

FALSE: E-Commerce security threats can be grouped into more than three major categories.

TRUE: Data warehousing in an E-Commerce environment should be done in a classical manner.

TRUE: E-Commerce opens up an entirely new world of web server.

FALSE: E-Commerce security threats can be grouped into more than three major categories.

Question 50 |

Identify the incorrect statement :

The ATM adaptation layer is not service dependent. | |

Logical connections in ATM are referred to as virtual channel connections. | |

ATM is a streamlined protocol with minimal error and flow control capabilities. | |

ATM is also known as cell relay. |

Question 50 Explanation:

FALSE: The ATM adaptation layer is service dependent.

TRUE: Logical connections in ATM are referred to as virtual channel connections.

TRUE: ATM is streamlined protocol with minimal error and flow control capabilities

TRUE: ATM is also known as cell delays.

TRUE: Logical connections in ATM are referred to as virtual channel connections.

TRUE: ATM is streamlined protocol with minimal error and flow control capabilities

TRUE: ATM is also known as cell delays.

There are 50 questions to complete.