Introduction to Geometric Programming

Basic Idea

The Geometric Mean(1)(2)(3) 4

341

41

41

41

41

21

21

21243

141

4321441

341

241

141

21221

121

UUUU

UUUUUUUU

UUUU

Posynomial Form

0

0

...

...21

21

21

i

ij

i

am

aaii

n

t

Ra

c

tttcu

uuugimii

Solution ApproachesPrimal problem:

aij : arbitrary real numbers ci : positive gk(t) : posynomials

)(min 0 tg ( I P )

0,...,0,0:.. 21 mtttts ( 1 )

1)(,...,1)(,1)( 21 tgtgtg p ( 2 )

,...,,1,0,...)( 21

22][

pktttctg imii am

aai

kJik

Solution Approaches (cont’d)

Dual problem)(

1i

i

1)(])

c([)(max i

k

k

p

k

n

iv

( I P )

0...,0,0:.. 21 nts P o s i t i v i t y c o n d i t i o n

1]0[

iJj

N o r m a l i t y c o n d i t i o n

pja iij

n

i,...,2,1 0

1

O r t h o g o n a l i t y c o n d i t i o n

pkikJi

k ,...,2,1 ,)(][

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. 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.

ii. 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

has the property

For arbitrary tj>0 and arbitrary μk>=0.

p

kkk tgtgtL

10 ]1)([)(),(

)',()','()'(),'( 0 tLtLtgtL

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

Total cost in dollars, g=Dual function v=Orthogonality condition:

Normality condition:Solution: , , , Min(g)=Max(v)=100

213132321

10204040

ttttttttt

4

4

3

3

2

2

1

1)()()()( 10204040

0

0

0

3213

4212

4311

D

D

D

14321 52'

1 51'

2 51'

3 51'

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

Total cost g0=Constraint g1= Dual function v=Orthogonality condition:

Normality condition:Solution: , , , Min(g)=Max(v)=60

32321

4040

ttttt

42 2131 tttt 142

2131 tttt

)(43

124040 434

4

3

3

2

2

1

1)()()()()(

121 0

0

0

3213

4212

4311

D

D

D

32'

1 31'

2 31'

4 31'

3

Degree of DifficultyDegree=no. terms – no. variables –1Problem 1: 4-3-1=0Problem 2: 4-3-1=0When 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 minimu

m of the primal The maximum sequence for dual is also a

minimizing sequence for primal.