hw2sol - EGM6365 Homework #2 1. Consider the following...

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Unformatted text preview: EGM6365 Homework #2 1. Consider the following design optimization problem: Minimize 2 f (x) = x12 + x2 − 4 x1 + 4 Subject to g1 (x) = − x1 ≤ 0 g 2 ( x) = − x2 ≤ 0 g3 (x) = x2 − (1 − x1 )3 ≤ 0 (i) (ii) Find the optimum point graphically Show that the optimum point does not satisfy K-T condition. Explain (i) As shown in the figure, (1, 0) is the optimum point and f = 1 at the optimum point. x2 f=4 ∇g3(x*) x* ∇f(x*) 1.0 f=1 x1 ∇g2(x*) (ii) The Lagrangian function for the problem can be defined as 2 2 2 L = x12 + x2 − 4 x1 + 4 + λ1 (− x1 + s12 ) + λ2 (− x2 + s2 ) + λ3 [ x2 − (1 − x1 )3 + s3 ] The K-T condition is 2 x1 − 4 − λ1 + 3λ3 (1 − x1 )3 = 0 2 x2 − λ2 + λ3 =0 − x1 + s12 =0 − x2 + s =0 2 2 2 x2 − (1 − x1 )3 + s3 =0 λi si = 0, i = 1,2,3 At x = (1, 0) since g1 is inactive, and g2 and g3 are active, the slack variables should be λ1 = 0, s2 = 0, s3 = 0 The first equation in the K-T condition can’t be satisfied by substituting into these values. As is clear from the figure, the gradients of two active constraints are not independent: [0, -1] and [0, 1]. In the mathematical term, x* is not a regular point of the feasible domain. The K-T condition-based optimality conditions assume that the feasible domain is regular, which means the domain doesn’t have a singular point. 2. An engineering design problem is formulated as: 2 Minimize f (x) = x12 + 2 x2 − 5 x1 − 2 x2 + 10 Subject to the constraints h1 = x1 + 2 x2 − 3 = 0 g1 = 3x1 + 2 x2 − 6 ≤ 0 In all of the following questions, justify your answers. (i) Write K-T necessary conditions The Lagrangian function is 2 L = x12 + 2 x2 − 5 x1 − 2 x2 + 10 + λ1 ( x1 + 2 x2 − 3) + λ2 (3 x1 + 2 x2 − 6 + s 2 ) The K-T condition is 2 x1 − 5 + λ1 + 3λ2 = 0 4 x2 − 2 + 2λ1 + 2λ2 = 0 x1 + 2 x2 − 3 = 0 3x1 + 2 x2 − 6 + s 2 = 0 λ2 s = 0 Since the second constraint is inequality, its Lagrange multiplier must be greater than or equal to zero. (ii) How many cases are there to be considered? Identify those cases. Two case. Case 1: s = 0 Case 2: λ2 = 0 (iii) Find the solution for the case where g1 is active. Is this acceptable case? Case s = 0 From third and fourth relation, we have x1 = 1.5 and x2 = 0.75. From first and second relation, we have λ1 = -1.75 and λ2 = 1.25. Since λ2 is greater than zero, the solution is acceptable. (iv) Regardless of the solution you obtained in (iii), suppose the Lagrange multiplier for the constraint h1 = 0 is λ1 = -2 and the Lagrange multiplier for the constraint g1 ≤ 0 is λ2 = 1. If the equality and inequality constraints are simultaneously changed to h1 = x1 + 2x2 – 3.2 = 0 and g1 = 3x1 + 2x2 -6.2 ≤ 0, what will be the new optimum cost? For the perturbed constraint, we can write h1 = x1 + 2 x2 − 3 = 0.2 g1 = 3x1 + 2 x2 − 6 ≤ 0.2 From the assumption, two constraints are active. Thus, ∆f = −0.2λ1 − 0.2λ2 = −0.2 × −2 − 0.2 × 1 = 0.2 and f NEW = f ORG + ∆f = 4.575 3. A design problem is formulated as an unconstrained optimization problem to minimize 2 2 f ( x1 , x2 , x3 ) = x13 + 2 x2 + 2 x3 + 4 x1 x3 + 2 x2 x3 (i) Calculate the gradient of the cost function at (1, 1, 1) 3x12 + 4 x3 7 ∇f = 4 x2 + 2 x3 = 6 4 x3 + 4 x1 + 2 x2 10 (ii) Calculate Hessian at the point (2, 1, 1) 6 x1 0 4 12 0 4 H = 0 4 2 = 0 4 2 4 2 4 4 2 4 (iii) Is the cost function f(x) a convex function? Why or Why not? In order to be a convex function, the Hessian matrix must be positive definite. The determinants of three principal minor matrices are 6 x1 , 24 x1 , and 72 x1 − 64 . These determinants can be negative or positive depending on the value of x1. Thus, f(x) is not a convex function. (iv) Is the cost function f(x) convex for the region x1 > 1? Why or Why not? For x1 > 1, the determinants of three principal minor matrices become positive. Thus, f(x) is a convex function. (v) Show that (0, 0, 0) is a stationary point. Is this a minimum, maximum, or inflection point? Why? At (0, 0, 0), ∇f = 0, but H is neither positive definite nor negative definite. Thus, (0, 0, 0) is a stationary point. But, this point is neither maximum nor minimum. In fact, it is an inflection point. ...
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This note was uploaded on 06/07/2011 for the course EGM 6365 taught by Professor Staff during the Spring '08 term at University of Florida.

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