ISRO CSE 2016 | Question: 65

Question

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?

• A. 196.77
• B. 206.56
• C. 199.56
• D. 210.68

Answer 1

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

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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also have a look at

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 ?
(1) 10.52 months (2) 11.52 months (3) 12.52 months (4) 14.52 months

Answer 2

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

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

Answer 3

a