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lecture10

# lecture10 - Last Time Conjugate Gradient Solving Ax = b...

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lehigh-logo IE417: Nonlinear Programming: Lecture 10 Jeff Linderoth Department of Industrial and Systems Engineering Lehigh University 16th February 2006 Jeff Linderoth IE417:Lecture 10 lehigh-logo Last Time: Conjugate Gradient Solving Ax = b Or min φ ( x ) def = 1 / 2 x T Ax - b T x, with A ∈ S n × n ++ Conjugate Gradient Algorithm 1 Choose x 0 .r 0 = Ax 0 - b, d 0 = - d 0 , k = 0 2 α k = - r T k d k d T k Ad k 3 x k +1 = x k + α k d k 4 β k +1 = r T k +1 Ad k d T k Ad k 5 d k +1 = - r k +1 + β T k +1 d k 6 If r k = 0 , stop. Else k = k + 1 , Go to 2. Jeff Linderoth IE417:Lecture 10 lehigh-logo CG for Unconstrained Optimization min x R n f ( x ) Fletcher-Reeves Conjugate Gradient 1 Given x 0 . d 0 = -∇ f ( x 0 ) , k = 0 2 Compute α k . x k +1 = x k + α k d k 3 Compute f ( x k +1 ) , If f ( x k +1 ) = 0 , stop . Else β k +1 = f ( x k +1 ) T f ( x k +1 ) f ( x k ) T f ( x k ) 4 Compute new direction: d k +1 = -∇ f ( x k +1 ) + β k +1 d k Fletcher Reeves method is globally convergence as long as your α k satisfies the strong Wolfe conditions (with c 2 < 1 / 2 ) Jeff Linderoth IE417:Lecture 10 lehigh-logo Today: Quasi-Newton Recall: m k ( d ) = f ( x k ) + f ( x k ) T d k + 1 / 2 d T B k d Minimizer of this quadratic function is d k = - B - 1 k f ( x k ) Step: x k +1 = x k + α k d k The Question What can B do for you? Given the gradient information that you have recently seen, how would you like your model to behave? Make m k +1 ( · ) match f ( · ) at the last two points x k +1 = x k + αd k , so m k +1 (0) = f ( x k +1 ) m k +1 ( - α k d k ) = f ( x k ) Jeff Linderoth IE417:Lecture 10

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lehigh-logo The Secant Condition If B k +1 is to be symmetric, we have m k +1 ( d ) = f ( x k +1 ) + B k +1 d So m k +1 (0) = f ( x k +1 )
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lecture10 - Last Time Conjugate Gradient Solving Ax = b...

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