recategorized by
18,850 views
1 votes
1 votes

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

  1. 135 
  2. 722
  3. 675
  4. 672 
recategorized by

1 Answer

1 votes
1 votes

This one is long 

          FP(function point)=UFP*CAF        where CAF=complex adjustment factor    and= [0.65+0.01 x  ΣFi ]   Fi i=1 to 14 are the degree of influence              UFP=unadusted function points =  Σ Wij Zij(i=1 to  5) 

                        

computing function points  0---no infuence  1---incidental   2----moderate   3----average  4-------significant  5-----essential

based on 14 questions

                    function points components and weights as below

s/w components                               weighting factors

                                             simple            average      complex

No. of user inputs                       3                   4                 6

No. of user outputs                     4                   5                 7

No. of user inquires                    3                   4                 6

No. of files                                 7                 10                15

No. of external interfaces            5                  7                 10 

now coming to question

UFP=ΣWijZij=50x4+ 40x5+ 35x4+ 6x10 +4x7=628      as all weighting factors are avg 

CAF= (0.65+0.01(14x3)=1.07

FP=UFP*CAF=628X1.07=672 APPROX. hence ans is D

Answer:

Related questions

0 votes
0 votes
2 answers
3
makhdoom ghaya asked Jul 9, 2016
2,896 views
Which one of the following is not a key process area in CMM level $5$ ?Defect preventionProcess change managementSoftware product engineeringTechnology change management