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
(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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This note was uploaded on 12/10/2013 for the course MS&E 211 taught by Professor Yinyuye during the Fall '07 term at Stanford.
- Fall '07