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quiz6 and ans

# quiz6 and ans - x 1-2 x 2 = 4 and the vector[2 1 T is a...

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AMSC/CMSC 660 Quiz 6 , Fall 2006 1. (10) Let f ( x ) = 1 2 x T Hx - x T b , where H and b are constant, independent of x , and H is symmetric positive definite. Given vectors x (0) and p (0) , find the value of the scalar α that minimizes f ( x (0) + α p (0) ). Answer: Dropping superscripts for brevity, and taking advantage of sym- metry of H , we obtain f ( x (0) + α p (0) ) = 1 2 ( x + α p ) T H ( x + α p ) - ( x + α p ) T b = 1 2 x T Hx - x T b + α p T Hx + 1 2 α 2 p T Hp - α p T b . Differentiating with respect to α we obtain p T Hx + α p T Hp - p T b = 0 , so α = p T b - p T Hx p T Hp = p T r p T Hp , where r = b - Hx . If we differentiate a second time, we find that the second derivative of f with respect to α is p T Hp > 0 (when p 6 = 0 ), so we have found a minimizer. Note: This is the formula for the step in the linear conjugate gradient algorithm. 1

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2. (10) Consider the problem min x 5 x 4 1 + x 1 x 2 + 6 x 2 2 subject to the constraints x 0 and x 1 - 2 x 2 = 4. Formulate this problem as an unconstrained optimization problem using feasible directions and a barrier function. Answer: The vector [6 , 1] T is a particular solution to
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Unformatted text preview: x 1-2 x 2 = 4, and the vector [2 , 1] T is a basis for the nullspace of the matrix A = [1 ,-2]. (These choices are not unique, so there are many correct answers) Using our choices, any solution to the equality constraint can be expressed as x = • 6 1 ‚ + • 2 1 ‚ v = • 6 + 2 v 1 + v ‚ . Therefore, our problem is equivalent to min v 5(6 + 2 v ) 4 + (6 + 2 v )(1 + v ) + 6(1 + v ) 2 subject to 6 + 2 v ≥ , 1 + v ≥ . Using a log barrier function for these constraints, we obtain the unconstrained problem min v B μ ( v ) where B μ ( v ) = 5(6+2 v ) 4 +(6+2 v )(1+ v )+6(1+ v ) 2-μ log(6+2 v )-μ log(1+ v ) . Notice that if 1 + v ≥ 0, then 6 + 2 v ≥ 0. Therefore, the ±rst log term can be dropped from B μ ( v ). 2...
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