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A company needs to develop a digital signal processing software for one of its newest inventions. The software is expected to have 20000 lines of code. The company needs to determine the effort in person-months needed to develop this software using the basic COCOMO model. The multiplicative factor for this model is given as 2.2 for the software development on embedded systems, while the exponentiation factor is given as 1.5. What is the estimated effort in person-months?

  1. 196.77
  2. 206.56
  3. 199.56
  4. 210.68
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2 Answers

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7 votes
Best answer

kLOC = 20000/1000 = 20

Effort Applied (E)=$a_{b}\times \left ( KLOC\right )^{b_{b}}$

Here given,multiplicative factor $\left ( a_{b} \right )$=2.2

exponentiation factor $\left ( b_{b} \right )$=1.5

So Effort Applied (E) =$2.2\times \left ( 20 \right )^{1.5}$=196.77

So option A is correct.

Reference:COCOMO

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4 Comments

(20)^1.2

= (32-12)^6/5

taking 32 common

=[32(1-12/32)]^6/5

now 32 will 2^5*6/5 = 64

now for other part we will apply binomial formula (1-x) ^ n one

(1-12/32)^6/5

=(1- 12/32*6/5 + 6/5 * 1/5 * 12/32 * 12/32 *1/2!)

= 1-72/160 + 27/25*64 = .566875

 

so, 64* .566875 = 36.28(which is the answer)

 

Please try it in copy once as I have explained and let me know if you have any doubts

@rishu_darkshadow
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sure.. :)  and thanku .. :) @ Regina Phalange 

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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
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4 votes
4 votes

For basic cocomo model, effort is formulated by , E = a(KLOC)

where , a =  multiplicative factor for basic model is given as 2.2

b = exponentiation factor is given as 1.5 , and 

software is expected to have 20000 lines of code = 20KLOC 

hence , E = 2.2 (20) 1.5 = 196.77

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