8.9 Primal and Dual Programs in the Case of Less-Than Inequalities 511gk( X)≤ 1 , the signum functionsσk are all equal to+ 1 , and the objective function
g 0 ( X)will be a strictly convex function of the transformed variablesw 1 , w 2 ,... , wn,
where
xi=ewi, i= 0 , 1 , 2 ,... , n (8.64)In this case, the following primal–dual relationship can be shown to be valid:
f (X)≥f∗≡v∗≥ v(λ) (8.65)Table 8.2 gives the primal and the corresponding dual programs. The following char-
acteristics can be noted from this table:
1.The factorsckjappearing in the dual functionv(λ)are the coefficients of the
posynomialsgk( ,X) k= 0 , 1 , 2 ,... , m.
2.The number of components in the vectorλis equal to the number of terms
involved in the posynomialsg 0 , g 1 , g 2 ,... , gm. Associated with every term in
gk( X),there is a correspondingkj.3 .Each factor(∑
Nk
l= 1 λkl)λkj
ofv(λ)comes from an inequality constraintgk(X)≤
1. No such factor appears from the primal functiong 0 ( X)as the normality
condition forces∑N 0
j= 1 λ^0 jto be unity.
4 .The coefficient matrix [akij] appearing in the orthogonality condition is same
as the exponent matrix appearing in the posynomials of the primal program.
The following examples are considered to illustrate the method of solving geometric
programming problems with less-than inequality constraints.
Example 8.3 Zero-degree-of-difficulty ProblemSuppose that the problem considered
in Example 8.1 is restated in the following manner. Minimize the cost of constructing
the open rectangular box subject to the constraint that a maximum of 10 trips only are
allowed for transporting the 80 m^3 of grain.
SOLUTION The optimization problem can be stated as
FindX=
x 1
x 2
x 3
so as to minimizef (X)= 20 x 1 x 2 + 04 x 2 x 3 + 08 x 1 x 2subject to
80
x 1 x 2 x 3
≤ 0 or 18
x 1 x 2 x 3≤ 1
Sincen=3 andN=4, this problem has zero degree of difficulty. AsN 0 = 3 ,N 1 = , 1
andm=1, the dual problem can be stated as follows:
Findλ=
λ 01
λ 02
λ 03
λ 11
to maximize