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A simple stand – alone software utility is to be developed in ‘C’ programming by a team of software experts for a computer running Linux and the overall size of this software is estimated to be $20,000$ lines of code. Considering $(a,b)=2.4,1.05)$ as multiplicative and exponention factor for the basic COCOMO effort estimation equation and $(c,d)=(2.5, 0.38)$ as multiplicative and exponention factor for the basic COCOMO development time estimation equation, approximately how long does the software project take to complete?

- $10.52$ months
- $11.52$ months
- $12.52$ months
- $14.52$ months

1 vote

Basic COCOMO Model

The basic COCOMO model gives an approximate estimate of the project parameters. The basic COCOMO estimation model is given by the following expressions:

$20,000 LOC = 20 KLOC$

$Effort = a * {(KLOC)}^b $PM

$T_{dev} = c * {(Effort)}^d $Months

$Effort = 2.4 * {(20)}^{1.05} PM$

$= 55.756$

$T_{dev} = 2.5 * {(55.756)}^{0.38} Months$

$= 11.52$ Months

The basic COCOMO model gives an approximate estimate of the project parameters. The basic COCOMO estimation model is given by the following expressions:

$20,000 LOC = 20 KLOC$

$Effort = a * {(KLOC)}^b $PM

$T_{dev} = c * {(Effort)}^d $Months

$Effort = 2.4 * {(20)}^{1.05} PM$

$= 55.756$

$T_{dev} = 2.5 * {(55.756)}^{0.38} Months$

$= 11.52$ Months

1 vote

20,000 LOC (Lines of Code) = 20K LOC (Given)

(a,b) = (2.4, 1.05) (Given)

(c,d) = (2.5,0.38) (Given)

The basic COCOMO model gives an approximate estimate of the project parameters. The basic COCOMO estimation model is given by the following expression:

**Effort = a * (KLOC) ^{b} PM (PM = Person Months)**

**T _{dev }= c * (Effort)^{d }months (T _{dev} = Development TIme_{ })**

Effort = 2.4 * (20)^{1.05 }PM

= 2.4 * 23.231 = 55.756 PM

T _{dev }= 2.5 * (55.756)^{0.38 }months

= 2.5 * 4.608 = 11.52 months

Therefore, time taken for completion of the software project is 11.52 months