Complementarity problem

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Since a very large class of nonlinear programming problems can be reduced to [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problems]], the importance of [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problems]] on polyhedral cones is obvious both from theoretical and practical point of view.
Since a very large class of nonlinear programming problems can be reduced to [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problems]], the importance of [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problems]] on polyhedral cones is obvious both from theoretical and practical point of view.
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== Mixed complementarity problems ==
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Let <math>\mathbb{I}</math> and <math>\mathbb{J}</math> be two Hilbert spaces and <math>\mathcal{K}=\mathbb{I}\times K</math>, where <math>K</math> is an arbitrary closed convex cone in <math>\mathbb{J}</math>. Let <math>G:\,\mathbb{I}\times\mathbb{J}\rightarrow\mathbb{I}</math>, <math>H:\,\mathbb{I}\times\mathbb{J}\to\mathbb{J}</math> and <math>F=(G,H):\,\mathbb{I}\times\mathbb{J}\to\mathbb{I}\times\mathbb{J}</math>. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problem]] <math>NCP(F,\mathcal{K})</math> is equivalent to the mixed complementarity problem <math>MiCP(G,H,K)</math> defined by
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<center>
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<math>
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G(x,y)=0,\textrm{ }K\ni y\perp H(x,y)\in K^*.
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</math>
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</center>
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=== Proof ===
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Indeed, <math>MiCP(G,H,K)</math> is a simple equivalent reformulation of the [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problem]] <math>NCP(F,\mathcal{K})</math>, by noting that <math>\mathcal{K}^*=\{0\}\times K^*</math>.
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== Conic optimization problems ==
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Let <math>f:\mathbb{R}^q\to\mathbb{R}</math> be a function and <math>K\subset\mathbb{R}^q</math> be a closed convex cone, <math>A\in\mathbb{R}^{p\times q}</math> a full row rank matrix and <math>b\in\mathbb{R}^p</math>. Then, the '''conic optimization problem''' defined by <math>f</math>, <math>A</math>, <math>b</math>, <math>K</math> is the problem
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<center>
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<math>
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CO(f,A,b,K):\,\left\{
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\begin{array}{ll}
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\textrm{Minimize} & f(x)\\
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\textrm{Subject to} & Ax=b\\
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& x\in K
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\end{array}
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\right.
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</math>
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</center>
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== Reformulation of a conic optimization problem into a mixed complementarity problem ==
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Let <math>f:\mathbb{R}^q\to\mathbb{R}</math> be a differentiable convex function, <math>K\subset\mathbb{R}^q</math> be a closed convex cone, <math>A\in\mathbb{R}^{p\times q}</math> a full row rank matrix and <math>b\in\mathbb{R}^p</math>. Then, <math>x</math> is a solution of the [[Complementarity_problem#Conic_optimization_problems|conic optimization problem]] <math>CO(f,A,b,K)</math> if and only if <math>(y,x)</math> is a solution of
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the [[Complementarity_problem#Mixed_complementarity_problems|mixed complementarity problem]] <math>MiCP(G,H,K)</math>, where <math>G(y,x)=b-Ax</math>, <math>H(y,x)=\nabla f(x)-A^\top y</math>, which can be written explicitly as
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<center>
 +
<math>
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Ax=b,\textrm{ }K\ni x\perp \nabla f(x)-A^\top y\in K^*.
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</math>
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</center>
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=== Proof ===
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Consider the auxiliary problems
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<center>
 +
<math>
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(1)\,\,\,\left\{
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\begin{array}{ll}
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\textrm{Minimize}_{\,y,x} & y^\top(b-Ax)+f(x)\\
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\textrm{Subject to} & Ax=b,\\
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& x\in K
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\end{array}
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\right.
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</math>
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</center>
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and
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<center>
 +
<math>
 +
(2)\,\,\,\left\{
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\begin{array}{ll}
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\textrm{Minimize}_{\,y,x} & y^\top(b-Ax)+f(x)\\
 +
\textrm{Subject to} & x\in K,
 +
\end{array}
 +
\right.
 +
</math>
 +
</center>
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Then, obviously <math>(y,x)</math> is a solution of (1) if and only if <math>x</math> is a solution of the [[Complementarity_problem#Conic_optimization_problems|conic optimization problem]] <math>CO(f,A,b,K)</math>. Let <math>(y,x)</math> be a solution of (1) and <math>(\widetilde{y},\widetilde{x})</math> be a solution of (2). Denote <math>g(y,x)=y^\top(b-Ax)+f(x)</math> the common objective function of (1) and (2).
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Then,
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<center>
 +
<math>
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(3)\,\,\,g(\widetilde{y},\widetilde{x})\le g(y,x),
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</math>
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</center>
 +
because the feasible set of (1) is contained in the feasible set of (2).
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 +
Since [[Complementarity_problem#A_convex_optimization_problem_on_a_closed_convex_cone_is_equivalent_to_a_nonlinear_complementarity_problem|a convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem]], we obtain that <math>(\widetilde{y},\widetilde{x})</math> is a solution of the [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problem]] <math>NCP(\nabla g,\mathbb{R}^p\times K)</math>,
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[[Complementarity_problem#Mixed_complementarity_problems|which is equivalent to the mixed complementarity problem]] <math>MiCP(G,H,K)</math>, where <math>G(y,x)=b-Ax</math>, <math>H(y,x)=\nabla f(x)-A^\top y</math>. Hence, <math>A\widetilde{x}=b</math> and therefore <math>(\widetilde{y},\widetilde{x})</math> is in the feasible set of (1). Thus,
 +
<center>
 +
<math>
 +
(4)\,\,\,g(y,x)\le g(\widetilde{y},\widetilde{x}).
 +
</math>
 +
</center>
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From equations (3) and (4) it follows that <math>g(y,x)=g(\widetilde{y},\widetilde{x})</math>. This together with <math>(\widetilde{y},\widetilde{x})</math> being in the feasible set of (1) implies that (1) and (2) are equivalent. Since [[Complementarity_problem#A_convex_optimization_problem_on_a_closed_convex_cone_is_equivalent_to_a_nonlinear_complementarity_problem|a convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem]] problem (2) is equivalent to the [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problem]] <math>NCP(\nabla g,\mathbb{R}^p\times K)</math>, [[Complementarity_problem#Mixed_complementarity_problems|which is equivalent to the mixed complementarity problem]] <math>MiCP(G,H,K)</math>. Thus, (1) is equivalent to the [[Complementarity_problem#Mixed_complementarity_problems|mixed complementarity problem]] <math>MiCP(G,H,K)</math>. But we have already remarked that <math>(y,x)</math> is a solution of (1) if and only if <math>x</math> is a solution of the [[Complementarity_problem#Conic_optimization_problems|conic optimization problem]] <math>CO(f,A,b,K)</math>. Therefore, <math>x</math> is a solution of the [[Complementarity_problem#Conic_optimization_problems|conic optimization problem]] <math>CO(f,A,b,K)</math> if and only if <math>(y,x)</math> is a solution of the [[Complementarity_problem#Mixed_complementarity_problems|mixed complementarity problem]] <math>MiCP(G,H,K)</math>.

Revision as of 04:26, 29 June 2015

LaTeX: - Sándor Zoltán Németh

(In particular, we can have LaTeX: \mathbb{H}=\mathbb{R}^n everywhere in this page.)

Contents

Fixed point problems

Let LaTeX: \mathcal{A} be a set and LaTeX: T:\,\mathcal{A}\to\mathcal{A} a mapping. The fixed point problem defined by LaTeX: T\, is the problem

LaTeX: Fix(T):\,\left\{ \begin{array}{l} \textrm{Find }x\in\mathcal{A}\textrm{ such that}\\\\\\ x=T(x). \end{array} \right.

Nonlinear complementarity problems

Let LaTeX: \mathcal{K} be a closed convex cone in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: F:\,\mathbb{H}\to\mathbb{H} a mapping. Recall that the dual cone of LaTeX: \mathcal{K} is the closed convex cone LaTeX: \mathcal{K}^*=-\mathcal{K}^\circ where LaTeX: \mathcal{K}^\circ is the polar of LaTeX: \mathcal{K}. The nonlinear complementarity problem defined by LaTeX: \mathcal{K} and LaTeX: f\, is the problem

LaTeX: NCP(F,\mathcal{K}):\,\left\{ \begin{array}{l} \textrm{Find }x\in\mathcal{K}\textrm{ such that}\\\\\\ F(x)\in\mathcal{K}^*\textrm{ and }\langle x,F(x)\rangle=0. \end{array} \right.

Every nonlinear complementarity problem is equivalent to a fixed point problem

Let LaTeX: \mathcal{K} be a closed convex cone in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: F:\,\mathbb{H}\to\mathbb{H} a mapping. Then the nonlinear complementarity problem LaTeX: NCP(F,\mathcal{K}) is equivalent to the fixed point problem LaTeX: Fix(P_{\mathcal{K}}\circ(I-F)) where LaTeX: I:\,\mathbb{H}\to\mathbb{H} is the identity mapping defined by LaTeX: I(x)=x\, and LaTeX: P_{\mathcal{K}} is the projection onto LaTeX: \mathcal{K}.

Proof

For all LaTeX: x\in\mathbb{H} denote LaTeX: z=x-F(x)\, and LaTeX: y=-F(x).\, Then LaTeX: z=x+y.\,

Suppose that LaTeX: x\, is a solution of LaTeX: NCP(F,\mathcal{K}). Then LaTeX: z=x+y\, with LaTeX: x\in\mathcal{K}, LaTeX: y\in\mathcal{K}^\circ, and LaTeX: \langle x,y\rangle=0. Hence, via Moreau's theorem, we get LaTeX: x=P_{\mathcal{K}}z. Therefore LaTeX: x\, is a solution of LaTeX: Fix(P_{\mathcal{K}}\circ(I-F)).

Conversely, suppose that LaTeX: x\, is a solution of LaTeX: Fix(P_{\mathcal{K}}\circ(I-F)). Then LaTeX: x\in\mathcal{K} and via Moreau's theorem

LaTeX: z=P_{\mathcal{K}}(z)+P_{\mathcal{K}^\circ}(z)=x+P_{\mathcal{K}^\circ}(z).

Hence LaTeX: P_{\mathcal{K}^\circ}(z)=z-x=y, thus LaTeX: y\in\mathcal{K}^\circ. Moreau's theorem also implies that LaTeX: \langle x,y\rangle=0. In conclusion, LaTeX: x\in\mathcal{K}, LaTeX: F(x)=-y\in\mathcal{K}^*, and LaTeX: \langle x,F(x)\rangle=0. Therefore LaTeX: x\, is a solution of LaTeX: NCP(F,\mathcal{K}).

An alternative proof without Moreau's theorem

Variational inequalities

Let LaTeX: \mathcal{C} be a closed convex set in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: F:\,\mathbb{H}\to\mathbb{H} a mapping. The variational inequality defined by LaTeX: \mathcal{C} and LaTeX: F\, is the problem

LaTeX: VI(F,\mathcal{C}):\,\left\{ \begin{array}{l} \textrm{Find }x\in\mathcal{C}\textrm{ such that}\\\\\\ \langle y-x,F(x)\rangle\geq 0,\textrm{ for all }y\in\mathcal{C}. \end{array} \right.

Remark

The next result is not needed for the alternative proof and it can be skipped. However, it is an important property in its own. It was included for the completeness of the ideas.

Every fixed point problem defined on closed convex set is equivalent to a variational inequality

Let LaTeX: \mathcal{C} be a closed convex set in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: T:\,\mathcal{C}\to\mathcal{C} a mapping. Then the fixed point problem LaTeX: Fix(T)\, is equivalent to the variational inequality LaTeX: VI(F,\mathcal{C}), where LaTeX: \,F=I-T.

Proof

Suppose that LaTeX: x\, is a solution of LaTeX: Fix(T)\,. Then, LaTeX: F(x)=0\, and thus LaTeX: x\, is a solution of LaTeX: VI(F,\mathcal{C}).

Conversely, suppose that LaTeX: x\, is a solution of LaTeX: VI(F,\mathcal{C}) and let LaTeX: \,y=T(x). Then, LaTeX: \left\langle y-x,F(x)\right\rangle\geq 0, which is equivalent to LaTeX: -\parallel x-T(x)\parallel^2=0. Hence, LaTeX: \,x=T(x); that is, LaTeX: x\, is a solution of LaTeX: Fix(T)\,.

Every variational inequality is equivalent to a fixed point problem

Let LaTeX: \mathcal{C} be a closed convex set in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: F:\,\mathbb{H}\to\mathbb{H} a mapping. Then the variational inequality LaTeX: VI(F,\mathcal{C}) is equivalent to the fixed point problem LaTeX: Fix(P_{\mathcal{C}}\circ(I-F)).

Proof

LaTeX: x\, is a solution of LaTeX: Fix(P_{\mathcal{C}}\circ(I-F)) if and only if LaTeX: x=P_{\mathcal{C}}(x-F(x)). Via characterization of the projection, the latter equation is equivalent to

LaTeX: \langle x-F(x)-x,y-x\rangle\leq0

for all LaTeX: y\in\mathcal{C}. But this holds if and only if LaTeX: x\, is a solution to LaTeX: VI(F,\mathcal{C}).

Remark

The next section shows that the equivalence of variational inequalities and fixed point problems is much stronger than the equivalence of nonlinear complementarity problems and fixed point problems because each nonlinear complementarity problem is a variational inequality defined on a closed convex cone.

Every variational inequality defined on a closed convex cone is equivalent to a complementarity problem

Let LaTeX: \mathcal{K} be a closed convex cone in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: F:\,\mathbb{H}\to\mathbb{H} a mapping. Then the nonlinear complementarity problem LaTeX: NCP(F,\mathcal{K}) is equivalent to the variational inequality LaTeX: VI(F,\mathcal{K}).

Proof

Suppose that LaTeX: x\, is a solution of LaTeX: NCP(F,\mathcal{K}). Then LaTeX: x\in\mathcal{K}, LaTeX: F(x)\in\mathcal{K}^*, and LaTeX: \langle x,F(x)\rangle=0. Hence

LaTeX: \langle y-x,F(x)\rangle\geq 0

for all LaTeX: y\in\mathcal{K}. Therefore LaTeX: x\, is a solution of LaTeX: VI(F,\mathcal{K}).

Conversely, suppose that LaTeX: x\, is a solution of LaTeX: VI(F,\mathcal{K}). Then LaTeX: x\in\mathcal{K} and

LaTeX: \langle y-x,F(x)\rangle\geq 0

for all LaTeX: y\in\mathcal{K}. Choosing LaTeX: y=0\, and LaTeX: y=2x,\, in particular, we get a system of two inequalities that demands LaTeX: \langle x,F(x)\rangle=0. Thus LaTeX: \langle y,F(x)\rangle\geq 0 for all LaTeX: y\in\mathcal{K}; equivalently, LaTeX: F(x)\in\mathcal{K}^*. In conclusion, LaTeX: x\in\mathcal{K}, LaTeX: F(x)\in\mathcal{K}^*, and LaTeX: \langle x,F(x)\rangle=0. Therefore LaTeX: x\, is a solution to LaTeX: NCP(F,\mathcal{K}).

Concluding the alternative proof

Since LaTeX: \mathcal{K} is a closed convex cone, the nonlinear complementarity problem LaTeX: NCP(F,\mathcal{K}) is equivalent to the variational inequality LaTeX: VI(F,\mathcal{K}) which is equivalent to the fixed point problem LaTeX: Fix(P_{\mathcal{K}}\circ(I-F)).

Another fixed point problem providing solutions for variational inequalities and complementarity problems

Let LaTeX: \mathcal{C} be a closed convex set in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle), LaTeX: I:\,\mathbb{H}\to\mathbb{H} the identity mapping defined by LaTeX: I(x)=x and LaTeX: F:\,\mathbb{H}\to\mathbb{H} a mapping. Then, for any solution LaTeX: x\, of the fixed point problem LaTeX: Fix({(I-F)\circ P_{\mathcal{C}}}), the projection LaTeX: P_{\mathcal{C}}(x) of LaTeX: x\, onto LaTeX: \mathcal{C}is a solution of the variational inequality LaTeX: VI(F,\mathcal{C}).

Proof

Since LaTeX: x\, is a solution of the fixed point problem LaTeX: Fix({(I-F)\circ P_{\mathcal{C}}}), we have LaTeX: P_{\mathcal{C}}(x)-F(P_{\mathcal{C}}(x))=x. Thus, LaTeX: P_{\mathcal{C}}(P_{\mathcal{C}}(x)-F(P_{\mathcal{C}}(x)))=P_{\mathcal{C}}(x). Hence, LaTeX: P_{\mathcal{C}}(x) is a solution of the fixed point problem LaTeX: Fix(P_{\mathcal{C}}\circ(I-F)) and thus a solution of the variational inequality LaTeX: VI(F,\mathcal{C}). In particular, if LaTeX: \mathcal{C} is a closed convex cone, then the variational inequality LaTeX: VI(F,\mathcal{C}) becomes the complementarity problem LaTeX: NCP(F,\mathcal{C}).

Implicit complementarity problems

Let LaTeX: \mathcal K be a closed convex cone in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: F,G:\,\mathbb{H}\to\mathbb{H} two mappings. Recall that the dual cone of LaTeX: \mathcal K is the closed convex cone LaTeX: \mathcal K^*=-\mathcal K^\circ where LaTeX: \mathcal K^\circ is the polar of LaTeX: \mathcal K. The implicit complementarity problem defined by LaTeX: \mathcal K and the ordered pair of mappings LaTeX: (F,G)\, is the problem

LaTeX: ICP(F,G,\mathcal{K}):\,\left\{ \begin{array}{l} \textrm{Find }u\in\mathbb{H}\textrm{ such that}\\\\\\ G(u)\in\mathcal{K},\,\,\,F(u)\in\mathcal{K}^*,\,\,\,\langle G(u),F(u)\rangle=0. \end{array} \right.

Every implicit complementarity problem is equivalent to a fixed point problem

Let LaTeX: \mathcal K be a closed convex cone in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: F,G:\,\mathbb{H}\to\mathbb{H} two mappings. Then the implicit complementarity problem LaTeX: ICP(F,G,\mathcal K) is equivalent to the fixed point problem LaTeX: Fix(I-G+P_{\mathcal K}\circ(G-F)) where LaTeX: I:\,\mathbb{H}\to\mathbb{H} is the identity mapping defined by LaTeX: I(x)=x.\,

Proof

For all LaTeX: u\in\mathbb{H} denote LaTeX: z=G(u)-F(u),\, LaTeX: x=G(u),\, and LaTeX: y=-F(u).\, Then LaTeX: z=x+y.\,

Suppose that LaTeX: u\, is a solution of LaTeX: ICP(F,G,\mathcal K). Then LaTeX: z=x+y\, with LaTeX: x\in\mathcal K, LaTeX: y\in\mathcal K^\circ, and LaTeX: \langle x,y\rangle=0. Via Moreau's theorem, LaTeX: x=P_{\mathcal K}z. Therefore LaTeX: u\, is a solution of LaTeX: Fix(I-G+P_{\mathcal K}\circ(G-F)).

Conversely, suppose that LaTeX: u\, is a solution of LaTeX: Fix(I-G+P_{\mathcal K}\circ(G-F)). Then LaTeX: x\in\mathcal K and, via Moreau's theorem,

LaTeX: z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).

Hence LaTeX: P_{\mathcal K^\circ}(z)=z-x=y, thus LaTeX: y\in\mathcal K^\circ. Moreau's theorem also implies LaTeX: \langle x,y\rangle=0. In conclusion, LaTeX: G(u)=x\in\mathcal K, LaTeX: F(u)=-y\in\mathcal K^*, and LaTeX: \langle G(u),F(u)\rangle=0. Therefore LaTeX: u\, is a solution of LaTeX: ICP(F,G,\mathcal K).

Remark

If LaTeX: \,G=I, in particular, we obtain the result every nonlinear complementarity problem is equivalent to a fixed point problem. But the more general result, every implicit complementarity problem is equivalent to a fixed point problem, has no known connection with variational inequalities. Using Moreau's theorem is therefore essential for proving the latter result.

Nonlinear optimization problems

Let LaTeX: \mathcal C be a closed convex set in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: f:\,\mathbb{H}\to\mathbb{R} a function. The nonlinear optimization problem defined by LaTeX: \mathcal C and LaTeX: f\, is the problem

LaTeX: NOPT(f,\mathcal C):\,\left\{ \begin{array}{l} \textrm{Find }x\in\mathcal C\textrm{ such that}\\\\\\ f(x)\leq f(y)\textrm{ for all }y\in\mathcal C \end{array} \right. </p><p>\,~\equiv~\, </p><p>\begin{array}{ll} \textrm{Minimize} & f(x)\\ \textrm{Subject to} & x\in\mathcal C \end{array} </p><p>

Any solution of a nonlinear optimization problem is a solution of a variational inequality

Let LaTeX: \mathcal C be a closed convex set in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: f:\,\mathbb{H}\to\mathbb{R} a differentiable function. Then any solution of the nonlinear optimization problem LaTeX: NOPT(f,\mathcal C) is a solution of the variational inequality LaTeX: VI(\nabla f,\mathcal C) where LaTeX: \nabla f is the gradient of LaTeX: f.\,

Proof

Let LaTeX: \,x\in\mathcal C be a solution of LaTeX: NOPT(f,\mathcal C) and LaTeX: y\in\mathcal C an arbitrary point. Then by convexity of LaTeX: \mathcal C we have LaTeX: x+t(y-x)\in\mathcal C, hence LaTeX: f(x)\leq f(x+t(y-x)) and

LaTeX: \langle \nabla f(x),y-x\rangle=\displaystyle\lim_{t\searrow 0}\frac{f(x+t(y-x))-f(x)}t\geq0.

Therefore LaTeX: x\, is a solution of LaTeX: VI(\nabla f,\mathcal C).

A convex optimization problem is equivalent to a variational inequality

Let LaTeX: \mathcal C be a closed convex set in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: f:\,\mathbb{H}\to\mathbb{R} a differentiable convex function. Then the nonlinear optimization problem LaTeX: NOPT(f,\mathcal C) is equivalent to the variational inequality LaTeX: VI(\nabla f,\mathcal C) where LaTeX: \nabla f is the gradient of LaTeX: f.\,

Proof

Any solution of LaTeX: NOPT(f,\mathcal C) is a solution of LaTeX: VI(\nabla f,\mathcal C).

Conversely, suppose that LaTeX: x\, is a solution of LaTeX: VI(\nabla f,\mathcal C). By convexity of LaTeX: f\, we have LaTeX: f(y)-f(x)\geq\langle\nabla f(x),y-x\rangle\geq0 for all LaTeX: y\in\mathcal C. Therefore LaTeX: x\, is a solution of LaTeX: NOPT(f,\mathcal C).

Any solution of a nonlinear optimization problem on a closed convex cone is a solution of a nonlinear complementarity problem

Let LaTeX: \mathcal K be a closed convex cone in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: f:\,\mathbb{H}\to\mathbb{R} a differentiable function. Then any solution of the nonlinear optimization problem LaTeX: NOPT(f,\mathcal K) is a solution of the nonlinear complementarity problem LaTeX: NCP(\nabla f,\mathcal K).

Proof

Any solution of LaTeX: NOPT(f,\mathcal K) is a solution of LaTeX: VI(\nabla f,\mathcal K) which is equivalent to LaTeX: NCP(\nabla f,\mathcal K).

A convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem

Theorem NOPT.   Let LaTeX: \mathcal K be a closed convex cone in the Hilbert space LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle) and LaTeX: f:\,\mathbb{H}\to\mathbb{R} a differentiable convex function. Then the nonlinear optimization problem LaTeX: NOPT(f,\mathcal K) is equivalent to the nonlinear complementarity problem LaTeX: NCP(\nabla f,\mathcal K).

Proof

LaTeX: NOPT(f,\mathcal K) is equivalent to LaTeX: VI(\nabla f,\mathcal K) which is equivalent to LaTeX: NCP(\nabla f,\mathcal K).

An implicit complementarity problem can be associated to an optimization problem

It follows from the definition of an implicit complementarity problem that LaTeX: y is a solution of LaTeX: ICP(F,G,\mathcal{K}) if and only the minimal value of LaTeX: \langle G(x),F(x)\rangle subject to LaTeX: G(x)\in\mathcal{K} and LaTeX: F(x)\in\mathcal{K}^* is LaTeX: 0 at LaTeX: y.

Fat nonlinear programming problem

Let LaTeX: f:\,\mathbb{R}^n\to\mathbb{R} be a function, LaTeX: b\in\mathbb{R}^n, and LaTeX: A\in\mathbb{R}^{m\times n} a fat matrix of full rank LaTeX: m\leq n. Then the problem

LaTeX: NP(f,A,b):\,\left\{ \begin{array}{ll} \textrm{Minimize} & f(x)\\ \textrm{Subject to} & Ax\leq b \end{array} \right.

is called fat nonlinear programming problem.

Any solution of a fat nonlinear programming problem is a solution of a nonlinear complementarity problem defined by a polyhedral cone

Let LaTeX: f:\,\mathbb{R}^n\to\mathbb{R} be a differentiable function, LaTeX: b\in\mathbb{R}^m, and LaTeX: A\in\mathbb{R}^{m\times n} a fat matrix of full rank LaTeX: m\leq n. If LaTeX: x\in\mathbb{R}^n is a solution of the fat nonlinear programming problem LaTeX: NP(f,A,b),\, then LaTeX: x-x_0\in\mathbb{R}^n is a solution of the nonlinear complementarity problem LaTeX: NCP(G,\mathcal K) where LaTeX: x_0\!\in\mathbb{R}^n is a particular solution of the linear system of equations LaTeX: Ax=b,\, LaTeX: \mathcal K is the polyhedral cone defined by

LaTeX: \mathcal K=\{x\,\mid\,Ax\leq0\}

and LaTeX: G:\,\mathbb{R}^n\to\mathbb{R}^n is defined by

LaTeX: G(x)=\nabla f(x+x_0)


Proof

Let LaTeX: x\in\mathbb{R}^n be a solution of LaTeX: NP(f,A,b).\, Then it is easy to see that LaTeX: x-x_0\, is a solution of LaTeX: \,NP(g,A,0) where LaTeX: g:\,\mathbb{R}^n\to\mathbb{R} is defined by LaTeX: g(x)=f(x+x_0).\, It follows from Theorem NOPT that LaTeX: x-x_0\, is a solution of LaTeX: NCP(G,\mathcal K) because LaTeX: G(x)=\nabla f(x+x_0)=\nabla g(x).

Remark

If LaTeX: f\, is convex, then the converse of the above results also holds. In other words, LaTeX: NP(f,A,b)\equiv NP(g,A,0)\equiv NOPT(g,\mathcal{K})\equiv NCP(G,\mathcal{K}).

We note that there are also many nonlinear programming problems defined by skinny matrices (i.e., m>n) that can be reduced to complementarity problems.

Since a very large class of nonlinear programming problems can be reduced to nonlinear complementarity problems, the importance of nonlinear complementarity problems on polyhedral cones is obvious both from theoretical and practical point of view.

Mixed complementarity problems

Let LaTeX: \mathbb{I} and LaTeX: \mathbb{J} be two Hilbert spaces and LaTeX: \mathcal{K}=\mathbb{I}\times K, where LaTeX: K is an arbitrary closed convex cone in LaTeX: \mathbb{J}. Let LaTeX: G:\,\mathbb{I}\times\mathbb{J}\rightarrow\mathbb{I}, LaTeX: H:\,\mathbb{I}\times\mathbb{J}\to\mathbb{J} and LaTeX: F=(G,H):\,\mathbb{I}\times\mathbb{J}\to\mathbb{I}\times\mathbb{J}. Then, the nonlinear complementarity problem LaTeX: NCP(F,\mathcal{K}) is equivalent to the mixed complementarity problem LaTeX: MiCP(G,H,K) defined by

LaTeX: G(x,y)=0,\textrm{ }K\ni y\perp H(x,y)\in K^*.

Proof

Indeed, LaTeX: MiCP(G,H,K) is a simple equivalent reformulation of the nonlinear complementarity problem LaTeX: NCP(F,\mathcal{K}), by noting that LaTeX: \mathcal{K}^*=\{0\}\times K^*.

Conic optimization problems

Let LaTeX: f:\mathbb{R}^q\to\mathbb{R} be a function and LaTeX: K\subset\mathbb{R}^q be a closed convex cone, LaTeX: A\in\mathbb{R}^{p\times q} a full row rank matrix and LaTeX: b\in\mathbb{R}^p. Then, the conic optimization problem defined by LaTeX: f, LaTeX: A, LaTeX: b, LaTeX: K is the problem

LaTeX: CO(f,A,b,K):\,\left\{ \begin{array}{ll} \textrm{Minimize} & f(x)\\ \textrm{Subject to} & Ax=b\\ & x\in K \end{array} \right.

Reformulation of a conic optimization problem into a mixed complementarity problem

Let LaTeX: f:\mathbb{R}^q\to\mathbb{R} be a differentiable convex function, LaTeX: K\subset\mathbb{R}^q be a closed convex cone, LaTeX: A\in\mathbb{R}^{p\times q} a full row rank matrix and LaTeX: b\in\mathbb{R}^p. Then, LaTeX: x is a solution of the conic optimization problem LaTeX: CO(f,A,b,K) if and only if LaTeX: (y,x) is a solution of the mixed complementarity problem LaTeX: MiCP(G,H,K), where LaTeX: G(y,x)=b-Ax, LaTeX: H(y,x)=\nabla f(x)-A^\top y, which can be written explicitly as

LaTeX: Ax=b,\textrm{ }K\ni x\perp \nabla f(x)-A^\top y\in K^*.

Proof

Consider the auxiliary problems

LaTeX: (1)\,\,\,\left\{ \begin{array}{ll} \textrm{Minimize}_{\,y,x} & y^\top(b-Ax)+f(x)\\ \textrm{Subject to} & Ax=b,\\ & x\in K \end{array} \right.

and

LaTeX: (2)\,\,\,\left\{ \begin{array}{ll} \textrm{Minimize}_{\,y,x} & y^\top(b-Ax)+f(x)\\ \textrm{Subject to} & x\in K, \end{array} \right.

Then, obviously LaTeX: (y,x) is a solution of (1) if and only if LaTeX: x is a solution of the conic optimization problem LaTeX: CO(f,A,b,K). Let LaTeX: (y,x) be a solution of (1) and LaTeX: (\widetilde{y},\widetilde{x}) be a solution of (2). Denote LaTeX: g(y,x)=y^\top(b-Ax)+f(x) the common objective function of (1) and (2). Then,

LaTeX: (3)\,\,\,g(\widetilde{y},\widetilde{x})\le g(y,x),

because the feasible set of (1) is contained in the feasible set of (2).

Since a convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem, we obtain that LaTeX: (\widetilde{y},\widetilde{x}) is a solution of the nonlinear complementarity problem LaTeX: NCP(\nabla g,\mathbb{R}^p\times K), which is equivalent to the mixed complementarity problem LaTeX: MiCP(G,H,K), where LaTeX: G(y,x)=b-Ax, LaTeX: H(y,x)=\nabla f(x)-A^\top y. Hence, LaTeX: A\widetilde{x}=b and therefore LaTeX: (\widetilde{y},\widetilde{x}) is in the feasible set of (1). Thus,

LaTeX: (4)\,\,\,g(y,x)\le g(\widetilde{y},\widetilde{x}).

From equations (3) and (4) it follows that LaTeX: g(y,x)=g(\widetilde{y},\widetilde{x}). This together with LaTeX: (\widetilde{y},\widetilde{x}) being in the feasible set of (1) implies that (1) and (2) are equivalent. Since a convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem problem (2) is equivalent to the nonlinear complementarity problem LaTeX: NCP(\nabla g,\mathbb{R}^p\times K), which is equivalent to the mixed complementarity problem LaTeX: MiCP(G,H,K). Thus, (1) is equivalent to the mixed complementarity problem LaTeX: MiCP(G,H,K). But we have already remarked that LaTeX: (y,x) is a solution of (1) if and only if LaTeX: x is a solution of the conic optimization problem LaTeX: CO(f,A,b,K). Therefore, LaTeX: x is a solution of the conic optimization problem LaTeX: CO(f,A,b,K) if and only if LaTeX: (y,x) is a solution of the mixed complementarity problem LaTeX: MiCP(G,H,K).

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