Unconstrained Geometric Programming

Transcription

1 Jim Lambers MAT 49/59 Summer Session 20-2 Lecture 8 Notes These notes correspond to Section 2.5 in the text. Unconstrained Geometric Programming Previously, we learned how to use the A-G Inequality to solve an unconstrained minimization problem. We now formalize the procedure for solving such problems, in the case where the obective function to be minimized has the following particular form. Definition Let D R m be the convex) subset of R m defined by A function g : D R m of the form D = {t, t 2,..., t m ) R m t > 0, =, 2,..., m}. gt) = m c i t α i, where c i > 0 for i =, 2,..., n and α i R for i =, 2,..., n, =, 2,..., m, is called a posynomial. We now investigate how the A-G Inequality can be used to find the minimum of a given posynomial gt) on D, if one exists. That is, we will be solving the primal geometric program GP) We denote the ith term of gt) by Minimize gt) = n c i subect to t, t 2,..., t m > 0. g i t) = c i m m tα i t α i, x i = g it), i =, 2,..., n, where δ, δ 2,..., δ n > 0 and Then, the A-G Inequality yields gt) = =. x i

2 x i δ i δ i ) δi n m ) δi n m ) δi m t t α i t α i n α i. Because this is an unconstrained minimization problem, we need the quantity on the low side of the A-G Inequality to be a constant. It follows that the exponents must satisfy α i = 0, =, 2,..., m. If a vector δ = δ, δ 2,..., δ n ) can be found that satisfies this condition, as well as the previous conditions we have imposed on the s, then we have a candidate for a solution to the dual geometric program DGP) Maximize vδ) = ) δi n subect to δ, δ 2,..., δ n > 0 Positivity Condition) n = Normality Condition) n α i = 0, =, 2,..., m Orthogonality Condition). A vector δ that satisfies all three of the above conditions is said to be a feasible vector of the DGP. By the A-G Inequality, we have, for each feasible vector δ, gt) vδ), t D. This inequality is known as the Primal-Dual Inequality. If, in addition, δ is a global maximizer of vδ), and is therefore a solution of the DGP, then vδ ) is at least a lower bound for the minimum value of gt) on D. It can be shown using the criterion gt) = 0 that if t is a global minimizer of gt) on D, and therefore is a solution of the GP, then gt ) = vδ ) for some feasible vector δ. That is, the Primal-Dual Inequality actually becomes an equality. 2

3 Therefore, if δ is a solution of the DGP, then, by the A-G Inequality, the solution to the GP t = t, t 2,..., t m) can be found from the relations or x = x 2 = = x n = vδ ), g i t ) = vδ )δ i, i =, 2,..., n. Note that the s indicate the relative contributions of each term g i t ) of gt ) to the minimum. By taking the natural logarithm of both sides of these equations, we obtain a system of linear equations for the unknowns z = ln t, =, 2,..., m. Spefically, we can solve m [ vδ )δ ] i α i z = ln, i =, 2,..., n. c i Exponentiating the z s yields the components t, =, 2,..., m, of the minimizer t. This leads to the following method for solving the GP, known as Unconstrained Geometric Programming:. Find all feasible vectors δ for the corresponding DGP. 2. If no feasible vectors can be found, then the DGP, and therefore the GP, have no solution. 3. Compute the value of vδ) for each feasible vector δ. Each vector δ that maximizes the value of vδ) is a solution to the DGP. 4. To obtain the solution t to the GP, solve the system of equations g i t ) = vδ )δ i, i =, 2,..., n for t, t 2,..., t m, which can be reduced to a system of linear equations as described above. Example We will solve the GP Minimize gt) = t t 2 t 3 + 2t 2 t 3 + 3t t 3 + 4t t 2 subect to t, t 2, t 3 > 0. This leads to the DGP Maximize ) δ ) δ2 ) δ3 ) δ4 vδ) = δ δ2 δ3 δ4 subect to δ, δ 2, δ 3, δ 4 > 0 Positivity Condition) δ + δ 2 + δ 3 + δ 4 = Normality Condition) δ + δ 3 + δ 4 = 0, δ + δ 2 + δ 4 = 0, Orthogonality Condition) δ + δ 2 + δ 3 = 0 3

4 The Normality Condition and Orthogonality Condition, together, form a system of 4 equations with 4 unknowns whose coeffient matrix is nonsingular, so the system has the unique solution 2 δ = 5, 5, 5, ), 5 which also satisfies the Positivity Condition, so it is feasible. As it is the only feasible vector for the DGP, it is also the solution to the DGP. It follows that the maximum value of vδ), which is also the minimum value of gt), is ) 5 2/5 vδ ) = 0 /5 5 /5 20 / To find the minimizer t, we can solve the equations t t 2 t 3 From these equations, we obtain the relations which yields the solutions = δ vδ ) = 2 5 vδ ) 2t 2t 3 = δ 2 vδ ) = 5 vδ ) 3t t 3 = δ 3 vδ ) = 5 vδ ) 4t t 2 = δ 4 vδ ) = 5 vδ ). 3t 3 = 4t 2, 2t 3 = 4t, 2t 2 = 3t, t 5 = 3 6vδ ), t 2 = vδ ), t 5 3 = 2 3 6vδ ). 3t )3 = 2 5 vδ ) Substituting these values into gt) yields the value of vδ ), as expected. Example We now consider the GP Minimize gt) = 2 t t 2 + t t 2 + t subect to t, t 2 > 0. 4

5 This leads to the DGP Maximize ) δ ) δ2 ) δ3 vδ) = 2 δ δ2 δ3 subect to δ, δ 2, δ 3 > 0 Positivity Condition) δ + δ 2 + δ 3 = Normality Condition) δ + δ 2 + δ 3 = 0, δ + δ 2 = 0 Orthogonality Condition) Unfortunately, the only values of δ, δ 2, δ 3 that satisfy the Normality Condition and the Orthogonality Condition are δ = 2, δ 2 = 2, δ 3 = 0. These values do not satisfy the Positivity Condition, so there are no feasible vectors for the DGP. We conclude that there is no solution to the GP. Exerses. Chapter 2, Exerse 8 2. Chapter 2, Exerse 2 3. Chapter 2, Exerse Chapter 2, Exerse 26 5