
233Geometric Programming
subject to
g x x x x
1 1 3 1 2
0 5 0 25 1( ) . .x = + ≤
The degree of difculty of this problem is 3 − (2 + 1) = 0. Writing the
minimization problem in dual form as
Maximize
f
w w w w
w w w
( )
. .
w =
30 30 0 5 0 25
1 2 3
1 2 3
44
3 4
4
3 4
+
+
w
w w
w w( )
subject to orthogonality and normality conditions
−
−
−
1 0 1 1
1 1 0 1
1 1 1 0
1 1 0 0
1
2
3
4
w
w
w
w
=
0
0
0
1
Solving the above equation gives
−
−
−
1 0 1 1
1 1 0 1
1 1 1 0
1 1 0 0
1
2
3
4
w
w
w
w
=
−
−
−
−
1 0 1 1
1 1 0 1
1 1 1 0
1 1 0 0
0
0
0
1
1
=
2 3
1 3
1 3
1 3
/
/
/
/
This optimum value of the object ...