8.11 Complementary Geometric Programming 521
whereAk( ,X) Bk( ,X) Ck( , andX) Dk( are posynomials inX) Xand possibly some of
them may be absent. We assume thatR 0 ( X)> 0 for all feasibleX. This assumption
can always be satisfied by adding, if necessary, a sufficiently large constant toR 0 ( .X)
To solve the problem stated in Eq. (8.66), we introduce a new variablex 0 > 0 ,
constrained to satisfy the relationx 0 ≥R 0 ( [i.e.,X) R 0 ( /X)x 0 ≤ ], so that the problem 1
can be restated as
Minimizex 0 (8.67)
subjectto
Ak(X)−Bk(X)
Ck(X)−Dk(X)
≤ 1 , k= 0 , 1 , 2 ,... , m (8.68)
where
A 0 (X)=R 0 (X), C 0 (X)=x 0 , B 0 (X)= 0 , and D 0 (X)= 0
It is to be noted that the constraints have meaning only ifCk(X)−Dk( X)has a constant
sign throughout the feasible region. Thus ifCk(X)−Dk( s positive for some feasibleX)i
X, it must be positive for all other feasibleX. Depending on the positive or negative
nature of the termCk(X)−Dk( X),Eq. (8.68) can be rewritten as
Ak(X)+Dk(X)
Bk(X)+Ck(X)
≤ 1
or (8.69)
Bk(X)+Ck(X)
Ak(X)+Dk(X)
≤ 1
Thus any complementary geometric programming problem (CGP) can be stated in
standard form as
Minimizex 0 (8.70)
subjectto
Pk(X)
Qk(X)
≤ 1 , k= 1 , 2 ,... , m (8.71)
X=
x 0
x 1
x 2
..
.
xn
> 0 (8.72)
wherePk( andX) Qk( are posynomials of the formX)
Pk(X)=
∑
j
ckj
∏n
i= 0
(xi)akij=
∑
j
pkj(X) (8.73)
Qk(X)=
∑
j
dkj
∏n
i= 0
(xi)bkij=
∑
j
qkj(X) (8.74)