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You can rewrite the kkt conditions for this problem

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Unformatted text preview: verges if at each iteration k , the next iteration k + 1 results in an estimate that is closer to x ∗ = 0. In other words, we want to make sure that ￿xk +1 − x ∗ ￿2 < ￿xk − x ∗ ￿ 42 / 45 HW5 Q2 - Geometric Interpretation (a) This is a convex problem, so all you need to show for optimality is that you can find lagrange multipliers such that the point given satisfies all KKT conditions (b) For the shape of the objective function, it’s sufficient to describe the shape of an arbitrary level set of the objective function (that is, pick some value s and describe the shape of the function f (x , y ) ≤ s where f (x , y ) is the objective function) (c) Actually calculate both of the gradients, evaluate them at the optimal value, and then plot them. 43 / 45 HW5 Q3 - More Newton’s Method (a) Necessary conditions = KKT conditions (b) Remember that we can use Newton’s method to find the point at which a system of equations all equal 0 (general Newton’s method on g (x) where we take the Jacobian ∇g (x)). You can rewrite the KKT conditions for this problem so that solving the KKT conditions is equivalent to finding the point at w...
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