Engineering Optimization: Theory and Practice, Fourth Edition

494 Geometric Programming

By multiplying Eq. (8.4) byxk, we can rewrite it as

xk

∂f

∂xk

=

∑N

j= 1

akj(cj x

a 1 j

1 x

a 2 j

2 · · ·x

ak − 1 ,j

k− 1 x

akj

k x

ak + 1 ,j

k+ 1 · · ·x

anj

n )

=

∑N

j= 1

akjUj(X)= 0 , k= 1 , 2 ,... , n (8.5)

To find the minimizing vector

X∗=











x 1 ∗

x 2 ∗

..

.

xn∗











we have to solve thenequations given by Eqs. (8.4), simultaneously. To ensure that the

pointX∗corresponds to the minimum off (but not to the maximum or the stationary

point ofX), the sufficiency condition must be satisfied. This condition states that the

Hessian matrix off is evaluated atX∗:

JX∗=

[

∂^2 f

∂xk∂xl

]

X∗

must be positive definite. We will see this condition at a latter stage. Since the vector

X∗satisfies Eqs. (8.5), we have

∑N

j= 1

akjUj(X∗) = 0 , k= 1 , 2 ,... , n (8.6)

After dividing by the minimum value of the objective functionf∗, Eq. (8.6) becomes

∑N

j= 1

∗jakj= 0 , k= 1 , 2 ,... , n (8.7)

where the quantities∗jare defined as

∗j=

Uj(X∗)

f∗

=

Uj∗

f∗

(8.8)

and denote the relative contribution ofjth term to the optimal objective function. From

Eq. (8.8), we obtain

∑N

j= 1

∗j=∗ 1 +∗ 2 + · · · +∗N

=

1

f∗

(U 1 ∗+U 2 ∗+ · · · +UN∗)= 1 (8.9)