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6.252 NONLINEAR PROGRAMMING LECTURE 13: INEQUALITY CONSTRAINTS LECTURE OUTLINE • Inequality Constrained Problems • Necessary Conditions • Sufﬁciency Conditions • Linear Constraints Inequality constrained problem minimize f (x) subject to h(x) = 0, g(x) ≤ 0 where f : n → , h : n → m , g : n → r are continuously differentiable. Here h = (h1 , ..., hm ), g = (g1 , ..., gr ). TREATING INEQUALITIES AS EQUATIONS • Consider the set of active inequality constraints A(x) = j | gj (x) = 0 . • If x∗ is a local minimum: − The active inequality constraints at x∗ can be treated as equations − The inactive constraints at x∗ don’t matter Assuming regularity of x∗ and assigning zero Lagrange multipliers to inactive constraints, • m ∇f (x ∗ ) + r λ ∗i ∇hi (x∗ ) + i=1 µ ∗j = 0, µ ∗j ∇gj (x∗ ) = 0, j=1 ∀j∈ / A(x∗ ). Extra property: µ∗j ≥ 0 for all j . Intuitive reason: Relax j th constraint, gj (x) ≤ uj , • µ ∗ j = −(∆cost due to uj )/uj BASIC RESULTS Kuhn-Tucker Necessary Conditions: Let x∗ be a local minimum and a regular point. Then there exist unique Lagrange mult. vectors λ∗ = (λ∗1 , . . . , λ∗m ), µ ∗ = (µ∗1 , . . . , µ∗r ), such that ∇x L(x ∗ , λ∗ , µ∗ ) = 0, µ ∗j ≥ 0, j = 1, . . . , r, µ ∗j = 0, ∀j∈ / A(x∗ ). If f , h, and g are twice cont. differentiable, y ∇2xx L(x∗ , λ∗ , µ∗ )y ≥ 0, for all y ∈ V (x ∗ ), where ∗ ∗ ∗ ∗ V (x ) = y | ∇h(x ) y = 0, ∇gj (x ) y = 0, j ∈ A(x ) . • Similar sufﬁciency conditions and sensitivity results. They require strict complementarity, i.e., µ ∗j > 0, ∀ j ∈ A(x∗ ). PROOF OF KUHN-TUCKER CONDITIONS Use equality-constraints result to obtain all the conditions except for µ∗j ≥ 0 for j ∈ A(x∗ ). Introduce the penalty functions gj+ (x) = max 0, gj (x) , j = 1, . . . , r, and for k = 1, 2, . . ., let xk minimize r k k f (x) + ||h(x)||2 + 2 2 j=1 2 gj+ (x) + 1 ||x − x∗ ||2 2 over a closed sphere of x such that f (x∗ ) ≤ f (x). Using the same argument as for equality constraints, λ∗i = lim khi (xk ), i = 1, . . . , m, µ∗j = lim kgj+ (xk ), j = 1, . . . , r. k→∞ k→∞ Since gj+ (xk ) ≥ 0, we obtain µ∗j ≥ 0 for all j . LINEAR CONSTRAINTS • Consider the problem minaj x≤bj , j=1,...,r f (x). • Remarkable property: No need for regularity. • Proposition: If x∗ is a local minimum, there exist µ∗1 , . . . , µ∗r with µ∗j ≥ 0, j = 1, . . . , r , such that ∇f (x∗ ) + r µ∗j aj = 0, µj∗ = 0, ∀j∈ / A(x∗ ). j=1 • C Proof uses Farkas Lemma: Consider the cones and C ⊥ a2 C ⊥ = {y | aj'y ≤ 0, j=1,...,r} 0 { r C= x|x= µjaj, µj ≥ 0 } Σ j=1 a1 x∈C iff x y ≤ 0, ∀ y ∈ C ⊥ . PROOF OF FARKAS LEMMA x∈C iff x y ≤ 0, ∀ y ∈ C ⊥ . a2 C ⊥ r { C= x|x= 0 = {y | aj'y ≤ 0, j=1,...,r} µjaj, µj ≥ 0 } Σ j=1 ^x x - x^ a1 x Proof: First show that C is closed (nontrivial). Then, let x be such that x y ≤ 0, ∀ y ∈ C ⊥ , and consider its projection x̂ on C . We have x (x − x) ˆ = x − x ˆ 2, (x − x̂) aj ≤ 0, (∗) ∀ j. Hence, (x − x̂) ∈ C ⊥ , and using the hypothesis, x (x − x̂) ≤ 0. From (∗) and (∗∗), we obtain x = x̂, so x ∈ C . (∗∗) PROOF OF LAGRANGE MULTIPLIER RESULT a2 Cone generated by aj, j ∈ A(x* ) r { C= x|x= µjaj, µj ≥ 0 } Σ j=1 − ∇f(x* ) x* a1 Constraint set {x | aj'x ≤ bj, j = 1,...,r} The local min x∗ of the original problem is also a local min for the problem mina x≤bj , j∈A(x∗ ) f (x). Hence j ∇f (x∗ ) (x − x∗ ) ≥ 0, ∀ x with aj x ≤ bj , j ∈ A(x∗ ). Since a constraint aj x ≤ bj , j ∈ A(x∗ ) can also be expressed as aj (x − x∗ ) ≤ 0, we have ∇f (x∗ ) y ≥ 0, ∀ y with aj y ≤ 0, j ∈ A(x∗ ). From Farkas’ lemma, −∇f (x∗ ) has the form µ∗j aj , for some µ∗j ≥ 0, j ∈ A(x∗ ). j∈A(x∗ ) Let µ∗j = 0 for j ∈ / A(x∗ ). CONVEX COST AND LINEAR CONSTRAINTS Let f : n → be convex and cont. differentiable, and let J be a subset of the index set {1, . . . , r}. Then x∗ is a global minimum for the problem minimize f (x) subject to aj x ≤ bj , j = 1, . . . , r, if and only if x∗ is feasible and there exist scalars µ ∗j , j ∈ J , such that µ ∗j ≥ 0, µ ∗j = 0, ∗ x = arg min a x≤bj j j ∈J / ∀j∈J j ∈ J, with j ∈/ A(x∗ ), f (x) + µ ∗j (a j x − bj ) . j∈J • Proof is immediate if J = {1, . . . , r}. • Example: Simplex Constraint.