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# Recall that if the primal problem is additionally a

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Unformatted text preview: general primal and dual are just the KKT conditions. Recall that if the primal problem is additionally a convex optimization problem, then the KKT conditions are also suﬃcient for optimality. 5 / 45 Nonlinear Dual Example Find the dual of: minimize s.t. x1 + 2 x 2 2 2 x1 + x2 ≥ 1 6 / 45 Nonlinear Dual Example This gives us: 2 2 L(x , λ) = x1 + 2x2 − (x1 + x2 − 1)∗ To ﬁnd the inﬁmum, solve ∇x L = 0 for x1 and x2 ￿ ￿ 1 − 2x 1 λ ∇x L = =0 2 − 2x 2 λ This yields x1 = 1/(2λ) and x2 = 1/λ. Plug these into L to get φ φ(λ) = λ + 5/(4λ) 7 / 45 Nonlinear Dual Example So we get that the dual of: minimize s.t. x1 + 2 x 2 2 2 x1 + x2 ≥ 1 maximize s.t. λ + 5/(4λ) λ≥0 is: 8 / 45 Bisection Method We want to ﬁnd the minimum of a continuously diﬀerentiable function f : R → R given that the minimizer x ∗ falls in some interval [a, b ] such that f ￿ (a)f ￿ (b ) &lt; 0 ￿ Looking for point where f ￿ (x ) = 0 ￿ Search by repeatedly bisecting the search interval ￿ At each step, check midpoint and keep half of interval which contains x ∗ End when either: ￿ ￿ ￿ we ﬁnd an x ∗ such that f ￿ (x ∗) = 0, or the leng...
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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.

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