Engineering Optimization: Theory and Practice, Fourth Edition

524 Geometric Programming

they can each be approximated by a single-term posynomial with the help of Eq. (8.78)

as

̃

Q 1 (X,

̃

X)=( 1 + 4

̃

x^21 )

(

x 1

̃

x 2

) 8

̃

x^21 /( 1 + 4

̃

x^21 )

̃

Q 2 (X,

̃

X)=

(

1 + ̃

x 2

̃

x 1

)(

x 1

̃

x 1

)−

̃

x 2 /(

̃

x 1 +

̃

x 2 )(x

2

̃

x 1

)

̃

x 2 /(

̃

x 1 +

̃

x 2 )

Let us start the iterative process from the pointX(^1 )=

{ 1

1

}

,which can be seen to be

feasible. By taking

̃

X=X(^1 ), we obtain

̃

Q 1 (X,X(^1 ))= 5 x

8 / 5

1

̃

Q 2 (X,X(^1 ))= 2 x 1 −^1 /^2 x 21 /^2

and we formulate the first ordinary geometric programming problem (OGP 1 ) as

Minimizex 1

subject to

4

5 x

− 8 / 5

1 x^2 ≤^1

1

2 x

− 1 / 2

1 x

− 1 / 2

2 ≤^1

x 1 > 0

x 2 > 0

Since this (OGP 1 ) is a geometric programming problem with zero degree of difficulty,

its solution can be found by solving a square system of linear equations, namely

λ 1 = 1

λ 1 −^85 λ 2 −^12 λ 3 = 0

λ 2 −^12 λ 3 = 0

The solution isλ∗ 1 = 1 ,λ∗ 2 = 135 , λ∗ 3 =^1013. By substituting this solution into the dual

objective function, we obtain

v(λ∗)=(^45 )^5   3 /^1 (^12 )^10    3 /^1 ≃ 0. 5385

From the duality relations, we get

x 1 ≃ 0. 5 385 and x 2 =^54 (x 1 )^8   5 /^1 ≃ 0. 4643

Thus the optimal solution of OGP 1 is given by

X(o^1 pt)=

{

0. 5385

0. 4643

}