Convexification techniques for linear complementarity constraints

Abstract

We develop convexification techniques for mathematical programs with complementarity constraints. Specifically, we adapt the reformulation-linearization technique of Sherali and Adams (SIAM J Discrete Math 3, 411–430, 1990) to problems with linear complementarity constraints and discuss how this procedure reduces to its standard specification for binary mixed-integer programs. Then, we consider specially structured complementarity sets that appear in KKT systems with linear constraints. For sets with a single complementarity constraint, we develop a convexification procedure that generates all nontrivial facet-defining inequalities and has an appealing “cancel-and-relax” interpretation. This procedure is used to describe the convex hull of problems with few side constraints in closed-form. As a consequence, we delineate cases when the factorable relaxation techniques yield the convex hull from those for which they do not. We then discuss how these results extend to sets with multiple complementarity constraints. In particular, we show that a suitable sequential application of the cancel-and-relax procedure produces all nontrivial inequalities of their convex hull. We conclude by illustrating, on a set of randomly generated problems, that the relaxations we propose can be significantly stronger than those available in the literature.

:  Proof of Proposition 17

The fact that all facet-defining inequalities of ${\mathrm{conv}}(C^{1,1})$ are described below is a direct consequence of Theorem 5.

• i. Pick $k \in I^+ \ne \emptyset$. Consider the points $(x^0;y_1^0)=({0};{0}), \; (x^j;y_1^j)=(\epsilon (|\alpha _j| {e_k}+\alpha _k {e_j}); {0})$ for $j \in M.$ These $m+1$ affinely independent points satisfy $y_1\ge 0$ at equality and belong to $C^{1,1} \subseteq {\mathrm{conv}}(C^{1,1})$ for a sufficiently small positive $\epsilon$. It follows that $y_1 \ge 0$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$.

• ii. First we consider the reverse implication. If $(M \setminus \{ i \} ) \cap I^+ \ne \emptyset$, there exists $k \in I^+$ with $k \ne i$. Construct the $m+1$ affinely independent points $(x^0; y_1^0) = ({0}; {0}), \; (x^j; y_1^j) = (\epsilon (|\alpha _j| {e_k} + \alpha _k {e_j}); {0 })$ for $j \in M \setminus \{ i\}$ and $(x^{m+1}; y_1^{m+1}) = ({0}; {1}).$ These points belong to $C^{1,1}$ for a sufficiently small positive $\epsilon$. It follows that $x_i \ge 0$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$. If $M \setminus \{ i \} = I^0$, we construct the affinely independent points $(x^0; y_1^0) = ({0}; {0}), \; (x^j; y_1^j) = ({e_j}; {0})$ for $j \in M \setminus \{ i\} = I^0$ and $(x^{m+1}; y_1^{m+1}) = ({0} ;{1}).$ These $m+1$ points belong to $C^{1,1}$, showing that $x_i \ge 0$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$. Conversely, if $(M \setminus \{ i \} )\cap I^+ = \emptyset$, it must be that $I^+ = \{ i\}$ since $I^+ \ne \emptyset$. Further, since $M \setminus \{ i \} \ne I^0$, it must be that $I^- \ne \emptyset$ . In this case $\sum _{j \in M} \alpha _j x_j \ge 0$ reduces to

  $$\begin{aligned} \alpha _i x_i + \sum _{j \in I^-} \alpha _j x_j \ge 0. \end{aligned}$$

  (A.1)

  Since $x_i \ge 0$ can be obtained as a conic combination of (A.1) and of $x_j \ge 0$ for $j \in I^-$ with nonnegative weights $\frac{1}{\alpha _i}$ and $-\frac{\alpha _j}{\alpha _i}$ respectively, it does not define a facet of ${\mathrm{conv}}(C^{1,1})$ as ${\mathrm{conv}}(C^{1,1})$ is full-dimensional.

• iii. Assume that $I^- \ne \emptyset$. Select $l \in I^- \ne \emptyset$, $k \in I^+ \ne \emptyset$, and consider the points $(x^0;y_1^0)=({0}; {0}), \; (x^j;y_1^j)=(\epsilon (-\alpha _l {e_j}+\alpha _j {e_l}); {0})$ for $j \in I^+ \cup I^0, \; (x^j;y_1^j)=(\epsilon (-\alpha _j {e_k} +\alpha _k {e_j}); {0})$ for $j \in I^- \backslash \{l\}$ and $(x^{m+1}; y_1^{m+1}) = ({0}; {1}).$ These $m+1$ affinely independent points belong to $C^{1,1}$ for a sufficiently small positive $\epsilon$ and satisfy $\sum _{i \in M} \alpha _i x_i \ge 0$ at equality. It follows that $\sum _{i \in M} \alpha _i x_i \ge 0$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$. If $I^-= \emptyset$ and $I^+ = \{ k\}$ then $\sum _{i \in M} \alpha _i x_i \ge 0$ reduces to $x_k \ge 0$. Since $M \setminus \{ k\} = I^0$ in this case, it follows from (ii) that $x_k \ge 0$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$. We now prove the direct implication by contraposition. If $I^-= \emptyset$ and $|I^+| \ge 2$ then $\sum _{i \in M} \alpha _i x_i \ge 0$ can be obtained as a conic combination with weight $\alpha _i$ of $x_i \ge 0$ for $i \in I^+$ and is therefore not facet-defining for ${\mathrm{conv}}(C^{1,1})$ since ${\mathrm{conv}}(C^{1,1})$ is full-dimensional.

• iv.

  • a. Assume first that $i \in I^0$. Pick $k \in I^+ \ne \emptyset$ and consider the points $(x^0; y_1^0) = ({e_i}; {0}), \; (x^j; y_1^j) = ({e_i} + \epsilon (|\alpha _j| {e_k} + \alpha _k {e_j}); {0})$ for $j \in M \setminus \{ i\}$ and $(x^{m+1}; y_1^{m+1}) = ({e_i}; {1}).$ These $m+1$ affinely independent points belong to $C^{1,1}$ for a sufficiently small positive $\epsilon$, satisfy $x_i \le 1$ at equality. It follows that $x_i \le 1$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$.

  • b. Assume now that $i\in I^+$. For the reverse implication, the given condition implies that $I^- \ne \emptyset$. Construct the points $(x^0; y_1^0) = ({e_i}; {0}), \; (x^j; y_1^j) = ({e_i} + \epsilon {e_j} ; {0})$ for $j \in M \setminus \{i\}$ and $(x^{m+1}; y_1^{m+1}) = \left( {e_i} + \sum _{j \in I^-} \left( \frac{- \alpha _i}{ \sum _{k \in I^-} \alpha _k}\right) {e_j}; {1}\right) .$ The points $(x^j;y_1^j)$ for $j \in M \setminus \{ i\}$ belong to $C^{1,1}$ for a sufficiently small positive $\epsilon$. Further, $(x^{m+1}; y_1^{m+1})$ satisfies $\sum _{i \in M} \alpha _i x_i \ge 0$ at equality and also satisfies the upper bound constraints on variables $x_j$ for $j \in I^-$ since $\alpha _i + \sum _{j \in I^-} \alpha _j \le 0$. It therefore belongs to $C^{1,1}$. Since these points are affinely independent and satisfy $x_i \le 1$ at equality, we conclude that $x_i \le 1$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$. Conversely, if $\alpha _i + \sum _{j \in I^-} \alpha _j> 0$ then setting $x_i =1$ implies that $\sum _{j \in M} \alpha _j x_j \ge \alpha _i + \sum _{j \in I^-} \alpha _j x_j \ge \alpha _i + \sum _{j \in I^-} \alpha _j > 0$, which in turn requires that $y_1=0$. We conclude that the face of ${\mathrm{conv}}(C^{1,1})$ induced by $x_i \le 1$ has dimension less than or equal to $m-1$.

  • c. Assume finally that $i \in I^-$. We know from assumption that $I^+ \ne \emptyset$. For the reverse implication, construct the points $(x^0; y_1^0) = \left( {e_i} + \sum _{j \in I^+} \left( \frac{-\alpha _i + \epsilon }{\sum _{k \in I^+} \alpha _k}\right) {e_j}; {0} \right) ,$ $(x^j; y_1^j) = (x^0 + {\tilde{\epsilon }}{e_j}; {0})$ for $j \in M\setminus \{ i\}$ and $(x^{m+1}; y^{m+1}) = \left( {e_i} + \sum _{j \in I^+} \left( \frac{- \alpha _i}{ \sum _{k \in I^+} \alpha _k}\right) {e_j} ; {1}\right) .$ For $\epsilon$ positive but sufficiently small, the point $(x^0, y_1^0)$ is feasible to $C^{1,1}$. It satisfies $\sum _{i \in M} \alpha _i x_i \ge 0$ and the upper bound constraints of $C^{1,1}$ strictly at inequality since $\epsilon >0$ and $\alpha _i + \sum _{j \in I^+} \alpha _j> 0$. It follows that the points $(x^j; y_1^j)$, for $j \in M \setminus \{i\}$ belong to $C^{1,1}$ when ${\tilde{\epsilon }}$ is sufficiently small. Finally, $(x^{m+1}, y_1^{m+1})$ also belongs to $C^{1,1}$ as we assumed that $\alpha _i + \sum _{j \in I^+} \alpha _j > 0$. Since these $m+1$ points are affinely independent and satisfy $x_i \le 1$ at equality, we conclude that $x_i \le 1$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$. Conversely, if $\alpha _i + \sum _{j \in I^+} \alpha _j \le 0$ then setting $x_i =1$ implies that $0 \le \sum _{j \in M} \alpha _j x_j \le \alpha _i + \sum _{j \in I^+} \alpha _j x_j \le \alpha _i + \sum _{j \in I^+} \alpha _j \le 0$, which in turn requires that $\sum _{j \in M} \alpha _j x_j =0$. We conclude that the face of ${\mathrm{conv}}(C^{1,1})$ induced by $x_i \le 1$ has dimension less than or equal to $m-1$.

• v.

  • a. We first observe that if $I^- = \emptyset$ and $|S^+|=1$, say $S^+ = \{ k\}$, (3.13) reduces to

    $$\begin{aligned} \alpha _k y_1+ \alpha _kx_k \le \alpha _k \end{aligned}$$

    (A.2)

    Inequality (A.2) holds at equality for the feasible points $(x^0; y_1^0) = ({0}; {1})$, $(x^j; y_1^j) = ({e_k + e_j};{0})$ for $j \in M \setminus \{ k\}$ and $(x^{m+1}; y_1^{m+1}) = ({e_k}; {0})$ showing that (A.2) is facet-defining for ${\mathrm{conv}}(C^{1,1})$. Conversely, if $|S^+| \ge 2$, (3.13) can be obtained as a conic combination of valid inequalities $\alpha _i y_1+ \alpha _i x_i \le \alpha _i$ for $i \in S^+$ showing that it is not facet-defining for ${\mathrm{conv}}(C^{1,1})$. If $|S^+| = 0$, inequality (3.13) is trivial and is clearly not facet-defining for ${\mathrm{conv}}(C^{1,1})$.

  • b. Assume now that $I^- \ne \emptyset$. We prove the reverse implications first. Assume first that $S^+ = \{ k\}$, $S^- =\emptyset$ and $\alpha _k + \sum _{j \in I^-} \alpha _j =0$. Then, (3.13) reduces to $x_k \le 1$. It follows from (iv)(b) that $x_k \le 1$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$. If $S^+ = \emptyset$, $S^- =\{k\}$ and $\alpha _k + \sum _{j \in I^-\backslash S^-} \alpha _j =0$ then (3.13) reduces to $x_k \ge 0$. It follows from (ii) that $x_k \ge 0$ is facet-defining for ${\mathrm{conv}}(C^{1,1})$. Assume next that $\sum _{i \in S^+} \alpha _i + \sum _{i \in I^- \setminus S^-} \alpha _i> 0$. Clearly $S^+ \ne \emptyset$. Construct the points $({\check{x}}; {\check{y}}_1 ) = (\sum _{j \in S^+} {e_j} ; { 0})$, $({\bar{x}}^k; {\bar{y}}_1^k ) = (\sum _{j \in S^+} {e_j} + {e_k} ; { 0})$ for $k \in I^+ \setminus S^+$, $({\dot{x}}^{l}; {\dot{y}}_1^{l} ) = (\sum _{j \in S^+} {e_j} + \epsilon {e_l}; {0})$ for $l \in I^-\setminus S^-$, $({\tilde{x}}^k; {\tilde{y}}_1^k ) = (\sum _{j \in S^+} {e_j} + {e_k} ; { 0})$ for $k \in I^0$, $({\hat{x}};{\hat{y}}_1) = (\delta \sum _{j \in S^+ } {e_j} + \sum _{j \in I^- \setminus S^-} {e_j} ; {1})$, $(x^{k,l}; y_1^{k,l})= (\delta \sum _{j \in S^+} {e_j} + \sum _{j \in I^- \setminus S^-} {e_j} + \frac{\epsilon }{\alpha _k} {e_k} - \frac{\epsilon }{\alpha _l} {e_l} ;{1})$ for $k,l \in S^+$, $({\ddot{x}}^{k,l}; {\ddot{y}}_1^{k,l}) = (\delta \sum _{j \in S^+} {e_j} + \sum _{j \in I^- \setminus S^-} {e_j} + \frac{\epsilon }{\alpha _k} {e_k} - \frac{\epsilon }{\alpha _l} {e_l} ; {1})$ for $k \in S^+, \; l \in S^-$ where $\delta = - \frac{\sum _{j \in I^- \setminus S^-} \alpha _j}{\sum _{i \in S^+} \alpha _i} \in [0,1)$. These points belong to $C^{1,1}$ if $\epsilon$ is a sufficiently small positive number. Further, these points belong to the face of ${\mathrm{conv}}(C^{1,1})$ induced by (3.13). Consider now any valid inequality $\sum _{i \in M} \beta _i x_i + \beta _0 y_1 \le \gamma$ for ${\mathrm{conv}}(C^{1,1})$ whose associated face contains the face of ${\mathrm{conv}}(C^{1,1})$ that (3.13) defines. It must therefore be satisfied at equality by the aforementioned points. The coefficients $\beta _i$ and $\gamma$ satisfy

    $$\begin{aligned}&\sum _{i \in S^+} \beta _i = \gamma \end{aligned}$$

    (A.3)
    $$\begin{aligned}&\sum _{i \in S^+} \beta _i + \beta _k = \gamma , \quad \forall k \in I^+ \setminus S^+ \end{aligned}$$

    (A.4)
    $$\begin{aligned}&\sum _{i \in S^+} \beta _i + \epsilon \beta _l = \gamma , \quad \forall l \in I^- \setminus S^- \end{aligned}$$

    (A.5)
    $$\begin{aligned}&\sum _{i \in S^+} \beta _i + \beta _k = \gamma , \quad \forall k \in I^0 \end{aligned}$$

    (A.6)
    $$\begin{aligned}&\delta \sum _{i \in S^+} \beta _i + \sum _{j \in I^- \setminus S^-} \beta _j + \beta _0 = \gamma \end{aligned}$$

    (A.7)
    $$\begin{aligned}&\delta \sum _{i \in S^+} \beta _i + \sum _{j \in I^- \setminus S^-}\beta _j +\frac{\epsilon }{\alpha _k} \beta _k - \frac{\epsilon }{\alpha _l} \beta _l + \beta _0 = \gamma , \forall k,l \in S^+ \end{aligned}$$

    (A.8)
    $$\begin{aligned}&\delta \sum _{i \in S^+} \beta _i + \sum _{j \in I^-\setminus S^-} \beta _j + \frac{\epsilon }{\alpha _k} \beta _k - \frac{\epsilon }{\alpha _l} \beta _l + \beta _0 =\gamma , \forall k \in S^+, \forall l \in S^- . \end{aligned}$$

    (A.9)

    We next show that $\sum _{i \in M} \beta _i x_i + \beta _0 y_1 \le \gamma$ is a scalar multiple of (3.13), proving that (3.13) is facet-defining for ${\mathrm{conv}}(C^{1,1})$. From (A.3) and (A.4), we obtain that $\beta _k = 0, \, \forall k \in I^+ \setminus S^+.$ Similarly, we obtain that $\beta _l =0, \, \forall l \in I^- \setminus S^-$ from (A.3) and (A.5), and we obtain that $\beta _k =0, \, \forall k \in I^0$ from (A.3) and (A.6). It then follows from (A.7) that $\delta \gamma +\beta _0=\gamma$, i.e., $\beta _0= \gamma (1- \delta ) = \gamma \Big ( \frac{ \sum _{i \in S^+} \alpha _i + \sum _{j \in I^- \setminus S^-} \alpha _j}{\sum _{i \in S^+} \alpha _i} \Big )$. Note that this conclusion holds both when $I^- \backslash S^-$ is empty and when it is nonempty.

    From (A.7) and (A.8), we conclude there exists $\theta \in {\mathbb {R}}$ such that $\frac{\beta _k}{\alpha _k} =\theta$ for $k \in S^+$. We then see that $\frac{\gamma }{\sum _{i \in S^+} \alpha _i} = \frac{ \sum _{i \in S^+}\beta _i}{\sum _{i \in S^+} \alpha _i} = \frac{ \theta \sum _{i \in S^+} \alpha _i}{ \sum _{i \in S^+} \alpha _i} = \theta$ where the first equality holds because of (A.3). Therefore, $\beta _k = \frac{\gamma \alpha _k}{ \sum _{i \in S^+} \alpha _i}$ for $k \in S^+$. Further, from (A.7) and (A.9), we establish that $\frac{\beta _k}{\alpha _k} = \frac{\beta _l}{\alpha _l} = \theta$ for $k \in S^+$ and $l \in S^-$. This implies that $\beta _l = \frac{\gamma \alpha _l}{\sum _{i \in S^+} \alpha _i }$ for $l \in S^-$. It follows that (3.13) is facet-defining for ${\mathrm{conv}}(C^{1,1})$.

    Conversely, assume that ($\sum _{i \in S^+} \alpha _i + \sum _{i \in I^- \setminus S^-} \alpha _i \le 0$) and ($|S^+|+|S^-| \ne 1$ or $\sum _{i \in S^+} \alpha _i + \sum _{i \in I^-\setminus S^-} \alpha _i \ne 0.)$ If ($\sum _{i \in S^+} \alpha _i + \sum _{i \in I^- \setminus S^-} \alpha _i < 0$), (3.13) is satisfied at equality only when $y_1=0$, showing that the dimension of the face that it induces is less than or equal to $m-1$. If $\sum _{i \in S^+} \alpha _i + \sum _{i \in I^-\setminus S^-} \alpha _i =0$, then (3.13) reduces to

    $$\begin{aligned} \sum _{i \in S^+} \alpha _i x_i+ \sum _{i \in S^-} \alpha _i x_i \le \sum _{i \in S^+} \alpha _i. \end{aligned}$$

    (A.10)

    If $|S^-| \ne 0$, then (A.10) can only be tight if $x_i =0$ for $i \in S^-$, showing that (A.10) is facet-defining for ${\mathrm{conv}}(C^{1,1})$ if and only if $|S^+|=0$ and $|S^-|=1$. We now assume that $|S^-| =0$. If $|S^+| \ge 2$ then (A.10) is not facet-defining since it can be obtained as a conic combination of inequalities $x_i \le 1$ with positive weights $\alpha _i$ for $i \in S^+$. If $|S^+| = 0$ as well, then (A.10) is trivial and clearly not facet-defining. $\square$