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Transcript
Introduction to Geometric
Programming
Basic Idea
The Geometric Mean
(1) 12 U1  12 U 2  U1 U 2
(2) 14 U1  14 U 2  14 U 3  14 U 4  U1 U 2 U 3 U 4
(3) 1 U  3 U  U U
1
2
1
2
1
4
4
1
4
2
1
4
3
4
1
2
1
4
1
4
1
4
Posynomial Form
g  u1  u 2  ...  u n
ui  c t t ...t
ai 1 ai 2
i 1 2
ci  0
aij  R
ti  0
aim
m
Solution Approaches
Primal problem:
min g 0 ( t )
s.t. : t1  0 , t 2  0 , ... , t m  0
g1 (t )  1 , g 2 (t )  1 , ... , g p (t )  1
g k (t )   ci t 2 i1 t 2ai 2 ...t maim , k  0, 1, ..., p,
a
iJ [ k ]



aij : arbitrary real numbers
ci : positive
gk(t) : posynomials
(IP)
(1)
(2)
Solution Approaches (cont’d)
Dual problem
n
max v ( )  [(
i 1
ci
i
p
) i ]  k ( ) k ( )
k 1
s.t. :  1  0,  2  0, ... n  0
 i  1
 aij i  0
i 1
k ( )    i ,
iJ [ k ]
Positivity condition
Normality condition
jJ [ 0 ]
n
(IP)
j  1,2,..., p
k  1,2,..., p
Orthogonality condition
The Duality Theory of
Geometric Programming
Theorem 1. Suppose that primal program A
is superconsistent and that the primal
function g0(t) attains its constrained
minimum value at a point that satisfies the
primal constraints. Then
i.
ii.
The corresponding dual program B is consistent
and the dual function v(δ) attains its constrained
maximum value at a point which satisfies the
dual constraints.
The constrained maximum value of the dual
function is equal to the constrained minimum
value of the primal function.
Theorem 1(cont.)
iii.
If t’ is a minimizing point for primal
program A, there are nonnegative
Lagrange multipliers μk’, k=1,2,…,p,
such that the Lagrange function
p
L(t ,  )  g0 (t )    k [ g k (t )  1]
has the property
k 1
L(t ' ,  )  g0 (t ' )  L(t ' ,  ' )  L(t ,  ' )
For arbitrary tj>0 and arbitrary μk>=0.
Theorem 2.
If primal program A is consistent and
there is a point δ* with positive
components which satisfies the
constraints of dual program B, the
primal function g0(t) attains its
constrained minimum value at a point t’
which satisfies the constraints of primal
program A.
Example 1
Problem 1: Suppose that 400 cubic yards of
gravel must be ferried across a river. Suppose
the gravel is to be shipped in an open box of
length t1, width t2, and height t3. The sides
and bottom of the box cost $10 per square
yard and the ends of the box cost $20 per
square yard. The box will have no salvage
value and each round trip of the box on the
ferry will cost 10 cents. What is the minimum
total cost of transporting the 400 cubic yards
of gravel?
Solution 1
40
 40t 2t3  20t1t3  10t1t 2
t1t 2t3
Total cost in dollars, g=
Dual function v= ( 40 ) ( 40 ) ( 20 ) ( 10 )
Orthogonality condition: D1  1   3   4  0
D2   1   2   4  0
D3   1   2   3  0
Normality condition: 1   2   3   4  1
Solution:  1'  52 ,  2'  15 ,  3'  15 ,  4'  15
Min(g)=Max(v)=100
1
1
3
2
2
3
4
4
Example 2: constrained
problem
This is the same as Problem 1, but it is
required to make the sides and bottom
of the box from scrap material. Only
four square yards of the scrap material
are available.
Solution 2
40
 40t 2t3
t1t 2t3
Total cost g0=
t1t3 t1t 2
Constraint g1= 2t1t3  t1t2  4
  1
2
4 ( 
40  40 
2 
1 
Dual function v= (  ) (  ) (  ) (  ) ( 3   4 )
Orthogonality condition: D1  1   3   4  0
D2   1   2   4  0
D3   1   2   3  0
1   2  1
Normality condition:
'
'
'
'
2
1
1







Solution: 1 3 , 2 3 , 3 3 , 4  13
Min(g)=Max(v)=60
1
1
3
2
2
3
4
4
3
4)
Degree of Difficulty
Degree=no. terms – no. variables –1
Problem 1: 4-3-1=0
Problem 2: 4-3-1=0
When degree of difficulty is k, the problem is
reduced to a maximizing problem with k
variables.
In some practical problems, there are several
constraints and the degree of difficulty can be
large.
Conclusion
For some nonlinear and nonconvex
problems, Geometric Programming
provides a systematic method to solve.
By converting, GP always produces the
global optimal(minimum).


The maximum of the dual = The minimum
of the primal
The maximum sequence for dual is also a
minimizing sequence for primal.