final_2006Ans - Fall2006 ARE211 Final Exam - Answer key...

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Unformatted text preview: Fall2006 ARE211 Final Exam - Answer key Problem 1 [32 points] A) Let f ( x ) = ( x- 2)( x- 1)( x + 1)( x + 2). This function can be rewritten as: f ( x ) = ( x 2- 1)( x 2- 4) or f ( x ) = x 4- 5 x 2 + 4 Consider the NPP min x R 1 f ( x ) s.t. braceleftbigg x - 2 x . 5 (a) [ 2 points ] Convert the problem to the standard format for an NPP that we have been using in this course. (b) [ 5 points ] Is the constraint qualification (CQ) satisfied at all the points in the con- straint set? (c) [ 5 points ] Find the set of all points that satisfy the KKT conditions. (d) [ 5 points ] At what point is the minimum attained? All points satisfy the CQ. The KKT conditions satisfied at {- radicalbig 5 / 2 , , . 5 } . The min is attained at x =- radicalbig 5 / 2 B) Now consider the problem max x R 2 h ( x ) s.t x 2 ( x 1- 2)( x 1- 1)( x 1 + 1)( x 1 + 2) x 1 - 2 x 1 1 where h ( x ) = x 1 + x 2 . Do not solve this optimization problem! (a) [ 7 points ] Carefully apply KKT (including checking the CQ) for x = (0 , 4) and x = (1 , 0). (b) [ 8 points ] Using only KKT, what can we conclude about x = (0 , 4) and x = (1 , 0) as potential solutions to the optimization problem? For x = (0 , 4) , only the first constraint is satisfied with equality. g 1 ( x ) = (0 , 1) . The CQ holds. h ( x ) = (1 , 1) . KKT necessary conditions not satisfied. For x = (1 , 0) , the first and third constraints are satisfied with equality. g 1 ( x ) = (0 , 1) . g 3 ( x ) = (6 , 1) . The CQ holds. h ( x ) = (1 , 1) . KKT necessary conditions are not satisfied. Using only the KKT, we can rule out both x and x as possible maximizers. 2 Problem 2 [32 points] Consider the system of equations: F ( x , ) = S x + G ( ) = 0 where S is an invertible 2x2 matrix, G is a continuously differentiable function, and R 1 . A) [ 5 points ] Write down the domain and range of F . f : R 3 R 2 B) [ 9 points ] Treating as a parameter, write down the Jacobian of F . Taking partials of F w.r.t. x , the Jacobian of F , treating as a parameter, is simply S . C) [ 9 points ] Given any , is the solution to the system of equations unique? If so, why? If not, give a counterexample. Solution is unique: S has full rank, so there exists a unique vector x such that x =- S- 1 G ( ) D) [ 9 points ] Given G ( ) = bracketleftbigg 2 3 + ( - 2) 2 bracketrightbigg and S = bracketleftbigg 1 2 0 4 bracketrightbigg , let x star ( ) denote the solution to the equation system. Set = 0. Can you compute parenleftBig x star 1 (0) , x star 2 (0) parenrightBig using the implicit function theorem? If so, do so; If not, explain which condition(s) of the theorem is violated....
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final_2006Ans - Fall2006 ARE211 Final Exam - Answer key...

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