# SOFTWARE ENGINEERING CBSE UGC NET SUBJECT WISE SOLVED

Cyclomatic complexity has a foundation in graph theory and is computed in one of three ways.
1. The number of regions correspond to the cyclomatic complexity.
2. Cyclomatic complexity V(G) for a flow graph G, is defined as,
V(G)=E-N+2

where E=Number of flow graph edges
N=Number of flow graph nodes
3.Cyclomatic complexity,V(G) for a flow graph G, is defined as,
V(G)=P+1
where P=Number of predicate nodes contained in flow graph G.

The above three formulas are very important and there are a lot of NET questions based on this topic. I am solving a few of them here.
DECEMBER 2013 – PAPER III Q.NO 4
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
Ans:-B
Explanation:-
N=10,E=13
V(G)=E-N+2
V(G)=13-10+2
=3+2=5
Therefore, No of regions = 5. (According to rule 1.)
According to rule 3, V(G)=P+1
V(G) refers to the cyclomatic complexity which is equal to the no of regions.
5=P+1
P=5-1=4
So, the number of predicate nodes in the flow graph is 4.
So, the number of regions and the number of predicate nodes in the flow graph will be 5,4
Therefore, the correct answer is B.

JUNE 2013 – PAPER II Q.No 5
5. Cyclomatic complexity of a flow graph G with n vertices and e edges is
A)V(G)=e+n-2
B)V(G)=e-n+2
C)V(G)=e+n+2
D)V(G)=e-n-2
Ans:-B
Explanation:-
According to rule 2, the formula for Cyclomatic Complexity V(G)=e-n+2 where e is no of edges, n is no of vertices.

JUNE 2012 – PAPER III Q.No 33
33. Which one of the following statements is incorrect ?
(A) The number of regions corresponds to the cyclomatic complexity.
(B) Cyclometric complexity for a flow graph G is V(G) = N – E + 2, where E is the number of edges and N is the number of nodes in the flow graph.
(C) Cyclometric complexity for a flow graph G is V(G) = E – N + 2, where E is the number of edges & N is the number of nodes in the flow graph.
(D) Cyclometric complexity for a flow graph G is V(G) = P + 1, where P is the number of predicate nodes contained in the flow graph G.
Ans:- B
Explanation:-
According to the formulas given in (1),(2) and (3), option B is incorrect, because V(G)=E-N+2 and not N-E+2.

DECEMBER 2011 – PAPER II Q.No 15
15. McCabe’s cyclomatic metric V(G) of a graph with n vertices, e edges and p connected component is
A)e
B)n
C)e-n+p
D)e-n+2p
Ans:-C
Explanation:-
According to McCabe’s cyclomatic metric of a graph V(G)=E-N+P where P refers to connected components.

JUNE 2006 – PAPER II Q.No 42
42. In Software Metrics, McCABE’s cyclomatic number is given by following formula :
(A) c=e-n+2p
(B) c=e-n-2p
(C) c=e+n+2p
(D) c=e-n*2p
Ans:-A
I hope all of you can solve problems based on cyclomatic complexity of a graph, with confidence and ease, after going through the above article.REFER TO MY OTHER POSTS ON SOFTWARE ENGINEERING

SOFTWARE ENGINEERING

SOFTWARE ENGINEERING – SOLVED PROBLEMS

# JUNE 2014 – PAPER III

17. Assume that a program will experience 200 failures in infinite time. It has now experienced 100 failures. The initial failure intensity was 20 failures/CPU hr. Then the current failure intensity will be
(A) 5 failures/CPU hr
(B) 10 failures/CPU hr.
(C) 20 failures/CPU hr
(D) 40 failures /CPU hr

Ans:- B
Explanation:– The formula for Current Failure Intensity =
Initial Failure intensity X [ 1 – Experienced failures/Failures in infinite time ]
= 20 X [ 1 – 100/200 ]
=20 X (100/200)
=10 failures /CPU hr

18. Consider a project with the following functional units :
Number of user inputs = 50
Number of user outputs = 40
Number of user enquiries = 35
Number of user files = 06
Number of external interfaces = 04
Assuming all complexity adjustment factors and weighing factors as average, the function points for the project will be
(A)135
(B)722
(C)675
(D)672

Ans:-
Explanation:-

Measurement parameter Count Weighting Factor(Average) Total
Number of user inputs 50 4 =200
Number of user outputs 40 5 =200
Number of user enquiries 35 4 =140
Number of user files 6 10 =60
Number of external interfaces 4 7 =28

Functon point (FP) = count total X (0.65 + 0.01 X S(fi))
= 628 X (0.65 + .01 X 30)
= 596.6
There is no option like that given. So I am not choosing anything.

The weighting factors of Simple project are 3,4,3,7,5

The weighting factors of Complex project are 6,7,6,15,10

Formula for Function point (FP for Simple project)= count total X (0.65+0.01 X 22)

Formula for Function point (FP for Complex project)= count total X (0.65+0.01 X 44)

TRY OUT THE FOLLOWING PROBLEMS

1. Compute the function point value for a project with the following information domain characteristics
Number of user inputs : 23
Number of user output : 60
Number of user enquiries : 24
Number of files : 10
Number of external interface : 05
Assume that all complexity adjustment values are average.
Ans : – 591.85

2. Compute the function point value for a project with the following information domain characteristics.
Number of user inputs : 28
Number of user outputs : 50
Number of user enquiries : 42
Number of files : 07
Number of external interface : 03
Assume that all complexity adjustment values are simple.
Ans:- 412.38

Problems are taken from the book on Software Engineering by Bharat Bhushan Agarwal and Sumit Prakash Tayal

# CAPABILITY MATURITY MODEL

CMM stands for Capability Maturity Model. This was developed by Software EngineeringInstitute(SEI). This is a well known benchmark for judging the quality level of the processes in the organization. CMM defines five levels determined on the basis of organisation’s support for certain “key” process areas known as KPAs. The five levels are:

 CMM – LEVEL 1 Initial Undefined, chaotic, no process in place CMM – LEVEL 2 Repeatable Basic project management processes are in place. CMM – LEVEL 3 Defined In addition to level 2 activities, organization has defined processes and management activities are documented. CMM – LEVEL 4 Managed Standards are built for organization. Managed through setting standards CMM – LEVEL 5 Optimising Leveraging on knowledge and innovative ideas. Continous process improvement and preventing the occurrence of defects.

JUNE 2014 – PAPER II
11. KPA in CMM stands for
A) Key Process Area
B) Key Product Area
C) Key Principal Area
D) Key Performance Area
Ans:- A

JUNE 2014 – PAPER III
21. Which one of the following is not a key process area in CMM level 5 ?
(A) Defect prevention
(B) Process change management
(C) Software product engineering
(D) Technology change management
Ans:- C

DECEMBER 2012 – PAPER II
19. The maturity levels used to measure a process are
(A) Initial, Repeatable, Defined, Managed, Optimized.
(B) Primary, Secondary, Defined,Managed, Optimized.
(C) Initial, Stating, Defined, Managed, Optimized.
(D) None of the above
Ans:- A

JUNE 2012 – PAPER II
21. Key process areas of CMM level 4 are also classified by a process which is
(A) CMM level 2
(B) CMM level 3
(C) CMM level 5
(D) All of the above
Ans:- B

JUNE 2009 – PAPER II
39. Capability Maturity Model is meant for:
A) Product
B) Process
C) Product and process
D) None of the above
Ans:- B

DECEMBER 2008 – PAPER II
44. Which level is called as “defined” in capability maturity model?
A) level 0
B) level 3
C) level 4
D) level 1
Ans:- B

JUNE 2005 – PAPER II
41. The capability maturity model defines 5 levels :
a) level 1     i) Managed
b) level 2     ii) Defined
c) Level 3     iii) Repeatable
d) Level 4     iv) Initial
e) Level 5     v) Optimised
Correct matching is:
(a)   (b)   (c)   (d)   (e)
A)   i   ii   iii   iv   v
B)   iv   iii   ii   i   v
C)   v   i   iii   ii   iv
D)   v   ii   i   iii   iv
Ans:- B