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Unformatted text preview: y. In many problems the constraints include variable bounds, as
minimize f0 (x)
subject to fi (x) ≤ 0, i = 1, . . . , m
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.
- Fall '13
- The Aeneid