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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 iJ [ 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 jJ [ 0 ] n a ij i 0 i 1 k ( ) i , iJ [ k ] j 1,2,..., p k 1,2,..., p (IP) Positivity condition Normality condition 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 2 4 1 ( 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.