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Unformatted text preview: y. In many problems the constraints include variable bounds, as in minimize f0 (x) subject to fi (x) ≤ 0, i = 1, . . . , m (13) li ≤ xi ≤ ui , i = 1, . . . , n. Let µ ∈ Rn be the Lagrange multipliers associated with the constraints xi ≤ ui , and let ν ∈ Rn + + be the Lagrange multipliers associated with the constraints li ≥ xi . Thus the Lagrangian is m L(x, λ, µ, ν ) = f0 (x) + i=1 λ i fi ( x ) + µT ( x − u ) + ν T ( l − x ) . n (a) Show that for any x ∈ R and any λ, we can choose µ 0 and ν 0 so that x minimizes L(x, λ, µ, ν ). In particular, it is very easy to find dual feasible points. (b) Construct a dual feasible point (λ, µ, ν ) by applying the method you found in part (a) with x = (l + u)/2 and λ = 0. From this dual feasible point you get a lower bound on f ⋆ . Show that this lower bound can be expressed as f ⋆ ≥ f0 ((l + u)/2) − ((u − l)/2)T |∇f0 ((l + u)/2)| where | · | means componentwise. Can you prove this bound directly? 4.13 Deducing costs from samples of optimal decision. A system (such as a firm or an organism...
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This note was uploaded on 09/10/2013 for the course C 231 taught by Professor F.borrelli during the Fall '13 term at University of California, Berkeley.

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