UGC NET CSE | November 2017 | Part 3 | Question: 47

Question

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?

• A. $10.52$ months
• B. $11.52$ months
• C. $12.52$ months
• D. $14.52$ months

Answer 1

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

Comments

means use of binomial expansion

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maybe, I never used binomial expansion for such questions.

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then how u can do approximation here  like 20 raised to power 1.05  or 55.756 raised to power .38

Answer 2

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

Comments

Thanks roma_nagpal but my question is how to solve exponential part as calculator is not allowed in NET exam.

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Well there is not an easy way for this but may be binomial expansion theorem and some log tables could help a little bit