Engineering Optimization: Theory and Practice, Fourth Edition

712 Modern Methods of Optimization

C(i)=(ci)α (13.27)

H(X)=

∑m

j= 1

{

φ[gj( [X)]qj(X)]γ[qi(X)]

}

(13.28)

φ[qj(X)]=a

(

1 −

1

eqj(X)

)

+b (13.29)

qj( X)=max

{

0 , gj(X)

}

;j= 1 , 2 ,... , m (13.30)

wherec,α,a, andbare constants. Note that the functionqj( X)denotes the magnitude

of violation of thejth constraint,φ[qj( X)]indicates a continuous assignment function,

assumed to be of exponential form, as shown in Eq. (13.29), andγ[qi( X)]represents

the power of the violated function. The values ofc= 0 .5,α=2,a=150, andb= 10

along with

γ[qj(X)]=

{

1 if qj(X)≤ 1

2 if qj(X)> 1

(13.31)

were used by Liu and Lin [13.35].

Example 13.4 Find the maximum of the function

f (x)= −x^2 + 2 x+ 11

in the range− 2 ≤x≤2 using the PSO method. Use 4 particles (N=4) with the initial

positionsx 1 = − 1. 5 ,x 2 = 0. 0 ,x 3 = 0. 5 , andx 4 = 1. 2 5. Show the detailed computa-

tions for iterations 1 and 2.

SOLUTION

1.Choose the number of particlesNas 4.

2.The initial population, chosen randomly (given as data), can be represented

asx 1 ( 0 )=− 1. 5 , x 2 ( 0 )= 0. 0 , x 3 ( 0 )= 0. 5 , andx 4 ( 0 )= 1. 2 5. Evaluate the

objective function values at currentxj( 0 ), j= 1 , 2 , 3 ,4 asf 1 =f[x 1 ( 0 )]=

f(− 1. 5 )= 5. 75 , f 2 =f[x 2 ( 0 )]=f( 0. 0 )= 11. 0 , f 3 =f[x 3 ( 0 )]=f( 0. 5 )=

11 .75, andf 4 =f[x 4 ( 0 )]=f( 1. 25 )= 11 .9375.

3.Set the initial velocities of each particle to zero:

v 1 ( 0 )=v 2 ( 0 )=v 3 ( 0 )=v 4 ( 0 )= 0

Set the iteration number asi=1 and go to step 4.

4.(a) FindPbest, 1 = − 1. 5 ,Pbest, 2 = 0. 0 ,Pbest, 3 = 0. 5 ,Pbest, 4 = 1 .25,andGbest=

1. 2 5.

(b) Find the velocities of the particles as (by assumingc 1 =c 2 = and using 1

the random numbers in the range (0, 1) asr 1 = 0. 3 294 andr 2 = 0. 9 542):

vj(i)=vj(i− 1 )+r 1 [Pbest ,j−xj(i− 1 )]

+r 2 [Gbest−xj(i− 1 )];j= 1 , 2 , 3 , 4