# A global QP-free algorithm for mathematical programs with complementarity constraints

## Abstract

In this paper, a primal–dual interior point QP-free algorithm for mathematical programs with complementarity constraints is presented. Firstly, based on Fischer–Burmeister function and smoothing techniques, the investigated problem is approximated by a smooth nonlinear constrained optimization problem. Secondly, combining with an effective penalty function technique and working set, a QP-free algorithm is proposed to solve the smooth constrained optimization problem. At each iteration, only two reduced linear equations with the same coefficient matrix are solved to obtain the search direction. Under some mild conditions, the proposed algorithm possesses global convergence. Finally, some numerical results are reported.

## Introduction

In this paper, we discuss the following mathematical programming problem with complementarity constraints (MPCC for short):

$$\begin{gathered}\min\quad f(x,y) \\ \quad\text{s.t.}\quad g(x,y)\leq0, \\ \quad\phantom{\text{s.t.}}\quad0\leq F(x,y)\perp y\geq0,\end{gathered}$$
(1)

where $$f:\mathbb{R}^{n+m}\rightarrow\mathbb{R}$$, $$g=(g_{1},\ldots ,g_{m_{g}})^{\mathrm{T}}:\mathbb{R}^{n+m}\rightarrow\mathbb {R}^{{m_{g}}}$$, $$F=(F_{1},\ldots,F_{m})^{\mathrm{T}}:\mathbb {R}^{n+m}\rightarrow\mathbb{R}^{m}$$ are continuously differentiable functions. “$$F(x,y)\perp y$$” means that the vectors $$F(x,y)$$ and y are perpendicular to each other.

MPCC (1) has a broad of applications in real world, such as engineering design, traffic transportation, game theory and so on. A detailed overview of MPCC applications can be found in  and the monographs .

Since MPCC (1) is a nonconvex optimization problem and the standard Mangasarian–Fromovitz constraint qualification (MFCQ) is violated at any feasible point, the well-developed algorithms for the standard nonlinear programs (for example, ) typically have severe difficulties if they are directly used to solve the MPCC (1). Hence, MPCC-tailed algorithms are desired.

It is well known that the QP-free method is one of the efficient methods for nonlinear programming (see [14, 15, 1925] ). The nice properties of the QP-free method are as follows: (a) The search directions are determined only by solving systems of linear equations rather than solving QP-subproblems. As a consequence, the computational cost is decreased greatly. (b) Compared with the SQP method, the size of the investigated problem solved by QP-free method is larger. It is worth mentioning that the primal–dual interior point QP-free algorithm in  has an improvement, that is, the strict restriction that the Lagrangian Hessian estimate is uniformly positive is relaxed. It is noticed that the active set identifying technique is incorporated into the QP-free algorithms in [22, 23]. Consequently, the computational cost is further decreased and the numerical performance in  is encouraging.

Although there are many OP-free algorithms for nonlinear programming, to the best of our knowledge, there are few MPCC-tailed QP-free algorithms for the MPCC (1). Motivated by the ideas of the algorithms in [21, 2427] and combining with smoothing techniques, we propose a primal–dual interior point QP-free algorithm for the MPCC (1). The proposed algorithm possesses the following nice properties:

1. (a)

The technique of active set is incorporated into the algorithm. As a consequence, the size of SLEs becomes smaller and the computational cost is decreased.

2. (b)

At each iteration, the search direction is obtained by solving two SLEs with the same coefficient matrix, which further decreases the computational cost.

3. (c)

The uniformly positive definiteness on the Lagrangian Hessian estimate $${H_{k}}$$ is relaxed.

4. (d)

The algorithm possesses global convergence without assuming that the stationary points are isolated.

## Preliminaries and reformulation

In this section, for completeness, we first restate some definitions and results about the MPCC (1), then deduce an approximation problem of the MPCC (1) by Fischer–Burmeister function and smoothing techniques.

Denoted by $$X_{0}$$ the feasible set of the MPCC (1), and denoted by $${I^{\mathcal{C}}}=\{1,2,\ldots,m\}$$ and $$I^{g}=\{ 1,2,\ldots,{m_{g}}\}$$ are the index sets of the complementarity constraints and the inequality constraints, respectively.

### Definition 1

()

Given $$(x^{*},y^{*})\in X_{0}$$, if

$$\bigl(y^{*}_{i}, F_{i} \bigl(x^{*},y^{*} \bigr) \bigr)\neq(0,0),\quad \forall i\in I^{\mathcal{C}},$$
(2)

then we call that the lower-level strict complementarity (LLSC) is satisfied at $$(x^{*},y^{*})$$.

### Definition 2

()

A point $$(x^{*},y^{*})\in X_{0}$$ is said to be a KKT stationary point of the MPCC (1) if there exists a KKT multiplier vector $$(\lambda^{*},\omega^{*},\mu^{*})\in\mathbb{R}^{m+m+m_{g}}$$ such that

\begin{aligned} & \nabla f\bigl(x^{*},y^{*}\bigr)+\nabla g\bigl(x^{*},y^{*}\bigr)\lambda^{*}- \nabla F\bigl(x^{*},y^{*}\bigr)\omega^{*}- \left ( \textstyle\begin{array}{c} 0_{n\times m} \\ E_{m} \end{array}\displaystyle \right ) \mu^{*} =0, \\ &0\leq-g \bigl(x^{*}, y^{*} \bigr)\perp\lambda^{*}\geq0;\qquad F \bigl(x^{*}, y^{*} \bigr) \geq0,\quad y^{*}\geq0,\qquad F_{i} \bigl(x^{*}, y^{*} \bigr)y_{i}^{*}=0; \\ &\omega^{*}_{i}=0,\quad \mbox{if } F_{i} \bigl(x^{*},y^{*} \bigr)>0; \qquad\mu^{*}_{i}=0, \quad\mbox{if } y^{*}_{i}>0; \\ &\omega^{*}_{i}\geq0, \qquad\mu^{*}_{i}\geq0, \quad\mbox{if } F_{i} \bigl(x^{*},y^{*} \bigr)=y^{*}_{i}=0, \end{aligned}

where $$E_{m}\in\mathbb{R}^{m\times m}$$ is an mth-order identity matrix. The vector $$((x^{*},y^{*}),\lambda^{*},\omega^{*},\mu^{*})$$ is said to be a KKT stationary pair of the MPCC (1).

### Proposition 1

()

Suppose that $$(x^{*}, y^{*})\in X_{0}$$satisfies LLSC, then $$((x^{*},y^{*}),\lambda^{*},\omega^{*},\mu^{*})$$is a KKT stationary pair of the MPCC (1) if and only if there exists a vector $$\iota ^{*}=(\iota^{*}_{i},i\in I^{\mathcal{C}})\in\mathbb{R}^{m}$$, such that

\begin{aligned} & \left ( \textstyle\begin{array}{c} \nabla_{x} f(x^{*}, y^{*}) \\ \nabla_{y} f(x^{*}, y^{*}) \\ 0_{m\times1} \end{array}\displaystyle \right ) + \left ( \textstyle\begin{array}{c} \nabla_{x} g(x^{*}, y^{*}) \\ \nabla_{y} g(x^{*}, y^{*}) \\ 0_{m\times l} \end{array}\displaystyle \right ) \lambda^{*}- \left ( \textstyle\begin{array}{c} \nabla_{x} F(x^{*}, y^{*}) \\ \nabla_{y} F(x^{*}, y^{*}) \\ -E_{m} \end{array}\displaystyle \right ) \omega^{*} - \left ( \textstyle\begin{array}{c} 0_{n\times m} \\ V^{*} \\ Y^{*} \end{array}\displaystyle \right ) \iota^{*} =0, \\ & 0\leq-g \bigl(x^{*}, y^{*} \bigr)\perp\lambda^{*}\geq0,\qquad 0\leq F \bigl(x^{*}, y^{*} \bigr) \perp y^{*}\geq0, \\ & \iota^{*}_{i}=\left \{ \textstyle\begin{array}{l@{\quad}l} \mu^{*}_{i}/F_{i}(x^{*},y^{*}), &\textit{if } F_{i}(x^{*},y^{*})>0,\\ \omega^{*}_{i}/y^{*}_{i}, &\textit{if } y^{*}_{i}>0, \end{array}\displaystyle \right . \end{aligned}

where $$V^{*}=\operatorname{diag}(F_{i}(x^{*},y^{*}), i\in I^{\mathcal{C}})$$, $$Y^{*}=\operatorname{diag}(y^{*}_{i},i\in I^{\mathcal{C}})$$.

It is well known that the Fischer–Burmeister function is a complementarity function, which is defined by

\begin{aligned} \psi(a,b)=a+b-\sqrt{a^{2}+b^{2}}. \end{aligned}

Obviously, ψ satisfies the following basic property:

$$\psi(a,b)=0\quad \Longleftrightarrow\quad a \geq0, b \geq0, ab=0.$$
(3)

And ψ is continuous differentiable in $$\{(a,b)\in\mathbb{R}^{2}| (a,b)\neq(0,0)\}$$, namely

\begin{aligned}& \psi^{\prime}_{a}(a,b) =1-\frac{a}{\sqrt{a^{2}+b^{2}}}; \qquad \psi^{\prime}_{b}(a,b) =1-\frac{b}{\sqrt{a^{2}+b^{2}}},\quad (a,b)\neq(0,0). \end{aligned}

Let $$v=F(x,y)$$, according to the property (3), the MPCC (1) is equivalently transformed into the following nonsmooth nonlinear optimization problem:

$$\begin{gathered}\min\quad f(x,y) \\ \quad \mbox{s.t.} \quad v-F(x,y)=0, \\ \phantom{\quad \mbox{s.t.} \quad} \varPsi(y,v)=0, \\ \phantom{\quad \mbox{s.t.} \quad} g(x,y)\leq0,\end{gathered}$$
(4)

where $$\varPsi(y,v)=(\psi(y_{i},v_{i}),i\in{I^{\mathcal{C}}})$$.

Obviously, the function ψ is not differentiable at the point $$(0,0)$$. Borrowing ideas from , we define the functions as follows:

\begin{aligned} &\psi_{\varepsilon}(y_{i},v_{i})=\left \{ \textstyle\begin{array}{l@{\quad}l} \psi(y_{i},v_{i}), &i \in{I^{\mathcal{C}}} \setminus{I^{\mathcal {C}}}(y,v,\varepsilon),\\ \frac{2\varepsilon- y_{i}}{2\varepsilon}y_{i} + \frac{2\varepsilon- v_{i}}{2\varepsilon}v_{i} - \frac{\varepsilon}{2}, &i \in{I^{\mathcal {C}}}(y,v,\varepsilon), \end{array}\displaystyle \right . \\ &\overline{\psi}_{\varepsilon}(y_{i},v_{i}) = \left \{ \textstyle\begin{array}{l@{\quad}l} 0, &i \in{I^{\mathcal{C}}} \setminus{I^{\mathcal {C}}}(y,v,\varepsilon),\\ \frac{ ( \sqrt{y^{2}_{i}+v^{2}_{i}} - \varepsilon )^{2}}{2\varepsilon}, &i \in{I^{\mathcal{C}}}(y,v,\varepsilon), \end{array}\displaystyle \right . \end{aligned}

where the index set $${I^{\mathcal{C}}}(y,v,\varepsilon)= \{i\in {I^{\mathcal{C}}}\mid\sqrt{y^{2}_{i}+v^{2}_{i}} <\varepsilon, \varepsilon >0 \}$$. Moreover, define

\begin{aligned} \varPsi_{\varepsilon}(y,v)= \bigl(\psi_{\varepsilon}(y_{i},v_{i}), i \in {I^{\mathcal{C}}} \bigr),\qquad \overline{\varPsi}_{\varepsilon}(y,v)= \bigl( \overline{\psi}_{\varepsilon}(y_{i},v_{i}), i \in{I^{\mathcal{C}}} \bigr), \end{aligned}

so we have

$$\varPsi(y,v)=\varPsi_{\varepsilon}(y,v)+\overline{ \varPsi}_{\varepsilon}(y,v).$$
(5)

For any $$\varepsilon>0$$, the function $$\psi_{\varepsilon}(y_{i},v_{i})$$ is differentiable, and it follows that

\begin{aligned} &\psi^{\prime}_{\varepsilon,y_{i}}(y_{i},v_{i})= \frac {\partial\psi_{\varepsilon}(y_{i},v_{i})}{\partial y_{i}} =\left \{ \textstyle\begin{array}{l@{\quad}l} 1-\frac{y_{i}}{\sqrt{y_{i}^{2}+v_{i}^{2}}}, &i \in I^{\mathcal{C}} \setminus I^{\mathcal{C}}(y,v,\varepsilon),\\ 1-\frac{y_{i}}{\varepsilon}, &i \in I^{\mathcal{C}}(y,v,\varepsilon), \end{array}\displaystyle \right . \\ &\psi^{\prime}_{\varepsilon,v_{i}}(y_{i},v_{i})= \frac{\partial\psi _{\varepsilon}(y_{i},v_{i})}{\partial v_{i}} =\left \{ \textstyle\begin{array}{l@{\quad}l} 1-\frac{v_{i}}{\sqrt{y_{i}^{2}+v_{i}^{2}}}, &i\in I^{\mathcal{C}} \setminus I^{\mathcal{C}}(y,v,\varepsilon),\\ 1-\frac{v_{i}}{\varepsilon}, &i \in I^{\mathcal{C}}(y,v,\varepsilon). \end{array}\displaystyle \right . \end{aligned}

So the smooth problem

$$\begin{gathered}\min\quad f(x,y) \\ \quad\mbox{s.t.} \quad v-F(x,y)=0, \\ \phantom{\quad\mbox{s.t.} \quad}\varPsi_{\varepsilon}(y,v)=0, \\ \phantom{\quad\mbox{s.t.} \quad}g(x,y)\leq0\end{gathered}$$
(6)

will be used as an approximation of the MPCC (1). Obviously, if $$I^{\mathcal{C}}(y^{*},v^{*},\varepsilon)=\emptyset$$, then the problem (6) is equivalent with the problem (4). Under some mild conditions, $$I^{\mathcal {C}}(y^{*},v^{*},\varepsilon)=\emptyset$$ is guaranteed, where $$(x^{*},y^{*},v^{*})$$ is an accumulation point of the iterative sequence $$\{ (x^{k},y^{k},v^{k})\}$$ (see the proof of Theorem 1 in the sequel).

Define a penalty function $$f_{r}(x,y)$$ for the problem (6) by

\begin{aligned} f_{r}(x,y)=f(x,y)-r\sum_{i\in I^{\mathcal{C}}} \bigl(v_{i}-F_{i}(x,y) \bigr) -r\sum _{i\in I^{\mathcal{C}}}\psi_{\varepsilon}(y_{i},v_{i}), \end{aligned}

where $$r>0$$ is a penalty parameter. Similar to , we can convert the problem (6) to the following smoothing optimization only with inequality constraints:

$$\begin{gathered}\min\quad f_{r}(x,y) \\ \quad\mbox{s.t.} \quad v_{i}-F_{i}(x,y)\leq0, \quad i\in I^{\mathcal{C}}, \\ \phantom{\quad\mbox{s.t.} \quad} \psi_{\varepsilon}(y_{i},v_{i})\leq0,\quad i\in I^{\mathcal{C}}, \\ \phantom{\quad\mbox{s.t.} \quad} g_{i}(x,y)\leq0, \quad i\in I^{g}.\end{gathered}$$
(7)

For simplicity, we use the following notations throughout this paper:

\begin{aligned} &z=(x,y,v),\qquad u=(x,y),\qquad w=(y,v),\\ &dz=(dx,dy,dv),\qquad du=(dx,dy),\qquad dw=(dy,dv), \\ &z^{k}= \bigl(x^{k},y^{k},v^{k} \bigr),\qquad u^{k}= \bigl(x^{k},y^{k} \bigr),\qquad w^{k}= \bigl(y^{k},v^{k} \bigr),\\ &dz^{k}= \bigl(dx^{k},dy^{k},dv^{k} \bigr),\qquad du^{k}= \bigl(dx^{k},dy^{k} \bigr), \qquad dw^{k}= \bigl(dy^{k},dv^{k} \bigr),\\ &e_{I}=(1,1,\ldots,1)^{T}\in\mathbb {R}^{2m+m_{g}},\qquad e_{I^{\mathcal{C}}}=(1,1, \ldots,1)^{T}\in\mathbb{R}^{m}, \\ &X:= \bigl\{ z\in\mathbb{R}^{n+2m} \mid v_{i}-F_{i}(x,y)=0,i \in I^{\mathcal {C}}; \psi_{\varepsilon}(y_{i},v_{i})=0,i \in I^{\mathcal{C}}; g_{i}(x,y)\leq0,i\in I^{g} \bigr\} , \\ &\widetilde{X}:= \bigl\{ z\in\mathbb{R}^{n+2m}\mid v_{i}-F_{i}(x,y) \leq 0,i\in I^{\mathcal{C}}; \psi_{\varepsilon}(y_{i},v_{i}) \leq0,i\in I^{\mathcal{C}}; g_{i}(x,y)\leq0,i\in I^{g} \bigr\} , \\ &\widetilde{X}_{0}:= \bigl\{ z\in\mathbb{R}^{n+2m}\mid v_{i}-F_{i}(x,y)< 0,i\in I^{\mathcal{C}}; \psi_{\varepsilon}(y_{i},v_{i})< 0,i\in I^{\mathcal{C}}; g_{i}(x,y)< 0,i\in I^{g} \bigr\} , \\ &\varphi(x,y)=\max \bigl\{ 0;g_{i}(x,y),i\in I^{g} \bigr\} ,\qquad I(x,y)= \bigl\{ i\in I^{g} \mid g_{i}(x,y)= \varphi(x,y) \bigr\} , \\ &I^{F}_{0}(z)= \bigl\{ i\in I^{\mathcal{C}}\mid v_{i}-F_{i}(x,y)=0 \bigr\} ,\qquad I^{\varPsi_{\varepsilon}}_{0}(z)= \bigl\{ i\in I^{\mathcal{C}}\mid \psi_{\varepsilon}(y_{i},v_{i})=0 \bigr\} , \\ &I^{g}_{0}(z)= \bigl\{ i\in I^{g}\mid g_{i}(x,y)=0 \bigr\} . \end{aligned}

The following proposition shows the equivalence between the problem (6) and the problem (7).

### Proposition 2

If $$(z,\lambda^{r})$$is a KKT pair of the problem (7), and $$v-F(x,y)=0$$, $$\varPsi_{\varepsilon}(y,v)=0$$, then $$(z,\lambda)$$with multiplier $$\lambda=\lambda^{r}-r\tilde{e}$$is a KKT pair of the problem (6), where $$\lambda ^{r}:=(\lambda^{r}_{F,I^{\mathcal{C}}},\lambda^{r}_{{{\varPsi_{\varepsilon}}},I^{\mathcal{C}}},\lambda^{r}_{g,I^{g}})$$, $$\lambda:=(\lambda_{F,I^{\mathcal{C}}},\lambda_{{{\varPsi_{\varepsilon}}},I^{\mathcal{C}}},\lambda_{g,I^{g}})$$, $$\tilde{e}=(1,\ldots,1_{{ m\textit{th}}},1,\ldots,1_{(2m)\textit{th}},0,\ldots ,0_{(2m+m_{g})\textit{th}})^{\mathrm{T}}$$.

### Proof

Since $$(z,\lambda^{r})$$ is a KKT pair of the problem (7), the vector pair $$(z,\lambda ^{r})$$ satisfies the following relations:

\begin{aligned} &\nabla_{z} f_{r}(x,y)+ \nabla_{z} \bigl(v-F(x,y) \bigr)\lambda^{r}_{F,I^{\mathcal{C}}}+ \nabla_{z} \varPsi _{\varepsilon}(y,v) \lambda^{r}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}} + \nabla_{z} g(x,y)\lambda^{r}_{g,I^{g}}=0, \\ &\lambda^{r}_{F,i}\geq0, v_{i}-F_{i}(x,y) \leq0, \quad\lambda^{r}_{F,i} \bigl(v_{i}-F_{i}(x,y) \bigr)=0, \quad i\in I^{\mathcal{C}}, \\ &\lambda^{r}_{{\varPsi_{\varepsilon}},i}\geq0, \psi_{\varepsilon}(y_{i},v_{i})\leq0, \quad\lambda^{r}_{{\varPsi_{\varepsilon}},i} \psi_{\varepsilon}(y_{i},v_{i})=0, \quad i\in I^{\mathcal{C}}, \\ &\lambda^{r}_{g,i}\geq0, g_{i}(x,y)\leq0,\quad \lambda ^{r}_{g,i}g_{i}(x,y)=0,\quad i\in I^{g}. \end{aligned}

Note that

\begin{aligned} \nabla_{z} f_{r}(x,y)=\nabla_{z} f(x,y)-r \sum_{i\in I^{\mathcal {C}}}\nabla_{z} \bigl(v_{i}-F_{i}(x,y) \bigr) -r\sum _{i\in I^{\mathcal{C}}}\nabla_{z}\psi_{\varepsilon}(y_{i},v_{i}), \end{aligned}

we get

\begin{aligned} \nabla_{z}f(x,y)+\nabla_{z} \bigl(v-F(x,y) \bigr) \bigl( \lambda^{r}_{F,I^{\mathcal{C}}}-re_{I^{\mathcal{C}}} \bigr) + \nabla_{z}\varPsi_{\varepsilon}(y,v) \bigl(\lambda^{r}_{{\varPsi_{\varepsilon }},I^{\mathcal{C}}} -re_{I^{\mathcal{C}}} \bigr) +\nabla_{z} g(x,y)\lambda^{r}_{g,I^{g}}=0. \end{aligned}

Let $$\lambda_{F,I^{\mathcal{C}}}=\lambda^{r}_{F,I^{\mathcal{C}}}-re_{I^{\mathcal{C}}}$$, $$\lambda_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}=\lambda^{r}_{{\varPsi _{\varepsilon}},I^{\mathcal{C}}}-re_{I^{\mathcal{C}}}$$, $$\lambda _{g,I^{g}}=\lambda^{r}_{g,I^{g}}$$, we obtain

\begin{aligned} \nabla_{z}f(x,y)+\nabla_{z} \bigl(v-F(x,y) \bigr) \lambda_{F,I^{\mathcal{C}}}+\nabla _{z}\varPsi_{\varepsilon}(y,v) \lambda_{\varPsi_{\varepsilon},I^{\mathcal{C}}} +\nabla_{z}g(x,y)\lambda_{g,I^{g}}=0. \end{aligned}

In view of $$v-F(x,y)=0$$, $$\varPsi_{\varepsilon}(y,v)=0$$, $$g(x,y)\leq0$$, and

\begin{aligned} \lambda_{g,i}\geq0,\quad \lambda_{g,i}g_{i}(x,y)=0,\quad i\in I^{g}. \end{aligned}

Let $$\lambda:=(\lambda_{F,I^{C}},\lambda_{\varPsi_{\varepsilon },I^{C}},\lambda_{g,I^{g}})=\lambda^{r}-r\tilde{e}$$, so $$(z,\lambda)$$ is a KKT pair of the problem (6). □

## Description of the algorithm

For the sake of theoretical analysis, we make a basic assumption throughout this paper.

### Assumption 1

1. (1)

For any $$(x,y)\in\mathbb{R}^{n+m}$$, the matrix $$\nabla_{y}F(x,y)$$ is a $$P_{0}$$ matrix, i.e., all the principal minors of $$\nabla_{y}F(x,y)$$ are nonnegative.

2. (2)

For any $$(x,y)\in\mathbb{R}^{n+m}$$, the submatrix $$(\nabla _{y}F(x,y))_{J^{*}J^{*}}$$ of the matrix $$\nabla_{y}F(x,y)$$ is nonsingular, where the index set $$J^{*}=\{i\in I^{C} \mid\psi'_{\varepsilon ,y_{i}}(y_{i},v_{i})=0\}$$.

In order to construct the coefficient matrix of linear equations conveniently, denote

\begin{aligned} \varOmega_{J}:=\varOmega_{J}(z, \varepsilon)= \left ( \textstyle\begin{array}{c} v-F(x,y)\\ \varPsi_{\varepsilon}(y,v)\\ g_{J}(x,y) \end{array}\displaystyle \right ), \end{aligned}

and denote by $$A_{J}$$ the gradient matrix of $$\varOmega_{J}$$, that is,

\begin{aligned} A_{J}:=A_{J}(z,\varepsilon) =\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} \textstyle\begin{array}{c} -\nabla_{x}F(x,y)\\ -\nabla_{y}F(x,y)\\ E_{m} \end{array}\displaystyle & \textstyle\begin{array}{c} 0 \\ \nabla_{y}\varPsi_{\varepsilon}(y,v)\\ \nabla_{v}\varPsi _{\varepsilon}(y,v) \end{array}\displaystyle &\textstyle\begin{array}{c} \nabla_{x} g_{J}(x,y)\\ \nabla_{y}g_{J}(x,y)\\ 0 \end{array}\displaystyle \end{array}\displaystyle \right ), \end{aligned}

where $$J\subseteq I^{g}$$, the diagonal matrix

\begin{aligned} \nabla_{y}\varPsi_{\varepsilon}(y,v)=\operatorname{diag} \bigl( \psi '_{\varepsilon,y_{i}}(y_{i},v_{i}),i \in I^{C} \bigr),\qquad \nabla_{v}\varPsi _{\varepsilon}(y,v)= \operatorname{diag} \bigl(\psi'_{\varepsilon ,v_{i}}(y_{i},v_{i}),i \in I^{C} \bigr). \end{aligned}

Define the matrix

$$U:=U(z,\varepsilon)=\left ( \textstyle\begin{array}{c@{\quad}c} \textstyle\begin{array}{c} -\nabla_{y}F(x,y)\\ E_{m} \end{array}\displaystyle &\textstyle\begin{array}{c} \nabla_{y}\varPsi_{\varepsilon}(y,v)\\ \nabla_{v}\varPsi _{\varepsilon}(y,v) \end{array}\displaystyle \end{array}\displaystyle \right ).$$
(8)

Similar to the proof in [30, 31], we can prove that the following proposition is true.

### Proposition 3

1. (1)

Suppose that the matrix U is nonsingular. Then the matrix $$A_{I(x,y)}$$is full of column rank, if and only if the matrix

\begin{aligned} \nabla_{x}g_{I(x,y)}(x,y)-\nabla_{x}F(x,y) \bigl(U^{-1} \bigr)_{m}\nabla_{y}g_{I(x,y)}(x,y) \end{aligned}

is full of column rank, where $$(U^{-1})_{m}$$is the mth-order principal submatrix of $$U^{-1}$$, which consists of the first m rows and m columns of $$U^{-1}$$.

2. (2)

Suppose that Assumption 1holds, then the matrix U is nonsingular.

### Assumption 2

For any $$z\in\mathbb{R}^{n+2m}$$, the matrix

\begin{aligned} \nabla_{x}g_{I(x,y)}(x,y)-\nabla_{x}F(x,y) \bigl(U^{-1} \bigr)_{m}\nabla_{y}g_{I(x,y)}(x,y) \end{aligned}

is full of column rank.

### Remark 1

According to Proposition 3, if Assumption 1 is true, then Assumption 2 is equivalent to the LICQ of the problem (7).

Based on Proposition 2, we know that if one can construct an efficient algorithm for the problem (7) and adjust penalty parameter r to force the iterate to asymptotically satisfy $$v-F(x,y)=0$$, $$\varPsi _{\varepsilon}(y,v)=0$$, then the solution to the problem (6) can be yielded.

Define the following optimal identification function:

$$\rho(z,\lambda)= \bigl\Vert \varPhi(z,\lambda) \bigr\Vert ^{\vartheta},$$
(9)

where

\begin{aligned} \varPhi(z,\lambda) =\left ( \textstyle\begin{array}{c} \nabla_{z} L(z,\lambda)\\ v-F(x,y)\\ \varPsi_{\varepsilon}(y,v)\\ \min\{-g(x,y),\lambda_{g,I^{g}}\} \end{array}\displaystyle \right ), \end{aligned}

and $$\lambda=(\lambda_{F,I^{\mathcal{C}}},\lambda_{{{\varPsi _{\varepsilon}}},I^{\mathcal{C}}},\lambda_{g,I^{g}})$$, the parameter $$\vartheta\in(0,1)$$$$\|\cdot\|$$ is the Euclidean norm, and we have the Lagrangian function

\begin{aligned} L(z,\lambda)=f(x,y)+\sum_{i\in I^{\mathcal{C}}} \lambda_{F,i} \bigl(v_{i}-F_{i}(x,y) \bigr) + \sum_{i\in I^{\mathcal{C}}}\lambda_{{\varPsi_{\varepsilon}},i} \psi _{\varepsilon}(y_{i},v_{i}) +\sum _{i\in I^{g}}\lambda_{g,i} g_{i}(x,y). \end{aligned}

Obviously, ρ is nonnegative and continuous. It follows from  that $$\rho(z^{*},\lambda^{*})=0$$ if and only if $$(z^{*},\lambda^{*})$$ is a KKT pair of the problem (6).

For the current iterate $$z^{k}\in\widetilde{X}_{0}$$, we define the corresponding multiplier vector $$\lambda^{k}=(\lambda^{k}_{F,I^{\mathcal{C}}},\lambda^{k}_{{{\varPsi_{\varepsilon}}},I^{\mathcal{C}}},\lambda^{k}_{g,I^{g}})$$ in (9) by

$$\lambda^{0}=s^{0},\qquad \lambda^{k}=\widehat{\lambda}^{k-1}-r_{k-1} \tilde {e},\quad k\geq1,$$
(10)

where $$s^{0}>0$$, and $$(\widehat{\lambda}^{k-1},r_{k-1})$$ is computed in the previous $$(k-1)$$th iteration, we construct the working set $$I_{k}$$ as follows:

$$I_{k}= \bigl\{ i\in I^{g}\mid g_{i} \bigl(x^{k},y^{k} \bigr)+\rho \bigl(z^{k},\lambda^{k} \bigr)\geq0 \bigr\} .$$
(11)

When $$(z^{k},\lambda^{k})$$ is sufficiently close to a KKT pair $$(z^{*},\lambda^{*})$$ of the problem (6), and the second-order sufficient conditions and the MFCQ hold at $$(z^{*},\lambda ^{*})$$, the set $$I_{k}$$ equals the precise identification set $$I^{g}_{0}(z^{*})$$. Therefore, only inequality constraints associated with the working set $$I_{k}$$ need to be considered.

For the current iterate $$z^{k}\in\mathbb{R}^{n+2m}$$, the first SLE in our algorithm is constructed as follows:

$$\mathrm{SLE1}{:}\quad M_{I_{k}} \left ( \textstyle\begin{array}{c} dz \\ \lambda_{I_{k}} \end{array}\displaystyle \right ) = \left ( \textstyle\begin{array}{c} -\nabla_{z} f_{r_{k}}(x^{k},y^{k}) \\ 0 \end{array}\displaystyle \right ),$$
(12)

where the coefficient matrix $$M_{I_{k}}$$ is defined by

$$M_{I_{k}}:=M_{I_{k}} \bigl(z^{k},S_{I_{k}} \bigr)= \left ( \textstyle\begin{array}{c@{\quad}c} \textstyle\begin{array}{c} H_{k}\\ S_{I_{k}}(A_{I_{k}})^{\mathrm{T}} \end{array}\displaystyle &\textstyle\begin{array}{c} A_{I_{k}} \\ G_{I_{k}} \end{array}\displaystyle \end{array}\displaystyle \right )$$
(13)

with

\begin{aligned} G_{I_{k}}=\operatorname{diag} \bigl( v^{k}-F\bigl(x^{k},y^{k}\bigr), \varPsi_{\varepsilon}\bigl(y^{k},v^{k}\bigr), g_{I_{k}}\bigl(x^{k},y^{k}\bigr) \bigr), \end{aligned}

the matrix $$H_{k}\in\mathbb{R}^{n+2m}$$ is an approximation of the Lagrangian Hessian, the matrix $$S_{I_{k}}:=\operatorname{diag}(s^{k}_{I_{k}})$$, where the vector $$s^{k}_{I_{k}}:=(s^{k}_{F,I^{\mathcal{C}}}, s^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}, s^{k}_{g,I_{k}})\in\mathbb{R}^{2m+|I_{k}|}$$ is an approximation of $$\lambda^{k}_{I_{k}}:=(\lambda^{k}_{F,I^{\mathcal {C}}},\lambda^{k}_{\varPsi_{\varepsilon},I^{\mathcal{C}}},\lambda^{k}_{g,I_{k}})$$.

It is not difficult to prove that the following result is true.

### Lemma 1

Suppose that Assumptions 12hold, and the symmetric matrix $$H_{k}$$satisfies the following relation:

\begin{aligned} H_{k} \succ&\sum_{i\in I^{\mathcal{C}}} \frac{s^{k}_{F,i}}{v^{k}_{i}-F_{i}(x^{k},y^{k})} \nabla_{z} \bigl(v^{k}_{i}-F_{i} \bigl(x^{k},y^{k} \bigr) \bigr)\nabla_{z} \bigl(v^{k}_{i}-F_{i} \bigl(x^{k},y^{k} \bigr) \bigr)^{\mathrm {T}} \\ &{}+\sum_{i\in I^{\mathcal{C}}}\frac{s^{k}_{{\varPsi_{\varepsilon }},i}}{{\psi_{\varepsilon}}(y^{k}_{i},v^{k}_{i})} \nabla_{z}{\psi_{\varepsilon}} \bigl(y^{k}_{i},v^{k}_{i} \bigr) \nabla_{z}{\psi_{\varepsilon}} \bigl(y^{k}_{i},v^{k}_{i} \bigr)^{\mathrm{T}} \\ &{}+\sum_{i\in I_{k}}\frac{s^{k}_{g,i}}{g_{i}(x^{k},y^{k})} \nabla_{z} g_{i} \bigl(x^{k},y^{k} \bigr)\nabla_{z} g_{i} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}, \end{aligned}
(14)

then the coefficient matrix $$M_{I_{k}}$$defined by (13) is invertible, where the matrix order $$A\succ B$$means that $$A-B$$is positive definite.

We now describe our algorithm as follows.

### Algorithm A

Step 0. (Initialization)

Choose an initial point $$z^{0}\in\widetilde{X}_{0}$$; $$\alpha\in (0,\frac{1}{2})$$; $$\vartheta, \beta, \theta\in(0,1)$$; $$r_{0}, \varepsilon_{0}, C, \nu, \gamma_{1}, \gamma_{2}, \gamma_{3}, \gamma _{4}>0$$; $${\tau>1}$$; $$q>2$$; termination accuracy $$\epsilon>0$$; the vector $$s^{0}:=(s^{0}_{F,I^{\mathcal{C}}},s^{0}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}},s^{0}_{g,I^{g}})\in\mathbb{R}^{2m+m_{g}}$$ with $$s^{0}_{i}\in[s_{\min },s_{\max}]$$, where $$s_{\max}>s_{\min}>0$$. Set $$k:=0$$.

Step 1. (Compute the working set)

Yield $$\lambda^{k}$$ by (10), then compute $$\rho (z^{k},\lambda^{k})$$ by (9). If $$\rho(z^{k},\lambda^{k})\leq\epsilon$$, and $$I^{\mathcal{C}}(y^{k},v^{k},\varepsilon_{k})=\emptyset$$, then $$z^{k}$$ is a KKT point of the MPCC (1), stop; otherwise, compute the working set $$I_{k}$$ by (11).

Step 2. (Yield a matrix $$H_{k}$$)

Generate a symmetric matrix $$H_{k}$$ such that it is an approximation of the Lagrangian Hessian of the problem (7) and satisfies the condition (14).

Step 3. (Generate a search direction)

1. (1)

Solve the first SLE1 (12). Let its solution be $$(\widehat{dz}^{k},\widehat{\lambda}^{k}_{I_{k}})$$, with $$\widehat{\lambda}^{k}_{I_{k}}:=(\widehat{\lambda}^{k}_{F,I^{\mathcal{C}}},\widehat{\lambda}^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}, \widehat{\lambda}^{k}_{g,I_{k}})\in\mathbb{R}^{2m+|I_{k}|}$$. Set $$\widehat{\lambda}^{k} :=(\widehat{\lambda}^{k}_{F,I^{\mathcal{C}}}, \widehat{\lambda}^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}, \widehat{\lambda}^{k}_{g,I^{g}})$$, where $$\widehat{\lambda}^{k}_{g,I^{g}} :=(\widehat{\lambda}^{k}_{g,I_{k}},0_{I^{g}\setminus I_{k}})$$.

2. (2)

If the following conditions hold:

1. (i)

$$\|\widehat{dz}^{k}\|\leq\gamma_{1}$$,

2. (ii)

$$\widehat{\lambda}^{k}\geq-\gamma_{2}e_{I}$$,

3. (iii)

$$\widehat{\lambda}^{k}_{F,I^{\mathcal{C}}}\ngtr\gamma _{3}e_{I^{\mathcal{C}}}$$,

4. (iv)

$$\widehat{\lambda}^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}\ngtr\gamma_{4}e_{I^{\mathcal{C}}}$$,

5. (v)

$$I^{\mathcal{C}}(y^{k},v^{k},\varepsilon_{k})\neq\emptyset$$, then update the parameter $$r_{k}$$ by $$r_{k+1}=\tau r_{k}$$, set $$z^{k+1}=z^{k}$$, $$s^{k+1}=s^{k}$$, $$H_{k+1}=H_{k}$$, $$I_{k+1}=I_{k}$$, $$\varepsilon _{k+1}=\frac{1}{2}\varepsilon_{k}$$, and go back to Step 3(1); otherwise, set $$r_{k+1}=r_{k}$$, go to Step 3(3).

3. (3)

Find $$\phi^{k}_{{F},I^{\mathcal{C}}}=(\phi^{k}_{{F},i})\in\mathbb{R}^{m}$$, $$D^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}=(D^{k}_{{\varPsi _{\varepsilon}},i})\in\mathbb{R}^{m}$$, $$Q^{k}_{g,I_{k}}=(Q^{k}_{g,i})\in\mathbb{R}^{|I_{k}|}$$, respectively, as follows:

\begin{aligned} &\begin{aligned}[b]\phi^{k}_{F,i}& :=\phi_{i} \bigl(z^{k},\widehat{\lambda}^{k}_{F,i} \bigr) \\&=\min \bigl\{ 0,- \bigl(\max \bigl\{ -\widehat{\lambda}^{k}_{F,i},0 \bigr\} \bigr)^{\nu} -C \bigl(v^{k}_{i}-F_{i} \bigl(x^{k},y^{k} \bigr) \bigr) \bigr\} ,\quad i\in I^{\mathcal{C}},\end{aligned} \end{aligned}
(15)
\begin{aligned} &D^{k}_{{\varPsi_{\varepsilon}},i} :=D_{i} \bigl(z^{k}, \widehat{\lambda}^{k}_{{\varPsi_{\varepsilon}},i} \bigr) =\min \bigl\{ 0,- \bigl( \max \bigl\{ -\widehat{\lambda}^{k}_{{\varPsi_{\varepsilon }},i},0 \bigr\} \bigr)^{\nu} -C\psi_{\varepsilon} \bigl(y^{k}_{i},v^{k}_{i} \bigr) \bigr\} ,\quad i\in I^{\mathcal {C}}, \end{aligned}
(16)
\begin{aligned} &Q^{k}_{g,i} :=Q_{i} \bigl(z^{k}, \widehat{\lambda}^{k}_{g,i} \bigr) =\min \bigl\{ 0,- \bigl( \max \bigl\{ -\widehat{\lambda}^{k}_{g,i},0 \bigr\} \bigr)^{\nu }-Cg_{i} \bigl(x^{k},y^{k} \bigr) \bigr\} ,\quad i\in I_{k}. \end{aligned}
(17)

Denote $$\varpi^{k}=(\phi^{k}_{F, I^{C}}, D^{k}_{\varPsi_{\varepsilon}, I^{C}}, Q^{k}_{g, I_{k} })$$, then compute

\begin{aligned} &\zeta^{k}=\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\, \widehat{dz}^{k} -\sum_{i\in I^{\mathcal{C}}} \frac{\widehat{\lambda }^{k}_{F,i}}{s^{k}_{F,i}}\phi^{k}_{F,i} -\sum _{i\in I^{\mathcal{C}}}\frac{\widehat{\lambda}^{k}_{\varPsi _{\varepsilon},i}}{s^{k}_{\varPsi_{\varepsilon},i}}D^{k}_{\varPsi _{\varepsilon},i} -\sum_{i\in I_{k}}\frac{\widehat{\lambda }^{k}_{g,i}}{s^{k}_{g,i}}Q^{k}_{g,i}, \end{aligned}
(18)
\begin{aligned} &b_{k}= \bigl( \bigl\Vert \widehat{dz}^{k} \bigr\Vert ^{q}+ \bigl\Vert \varpi^{k} \bigr\Vert \bigr) \biggl(\sum_{i\in I^{\mathcal{C}}}\widehat{\lambda}^{k}_{F,i} +\sum_{i\in I^{\mathcal{C}}}\widehat{\lambda}^{k}_{\varPsi _{\varepsilon}i} +\sum_{i\in I_{k}}\widehat{\lambda}^{k}_{g,i} \biggr) \\ &\phantom{b_{k}=}{} +\sum_{i\in I^{\mathcal{C}}}\frac{\widehat{\lambda }^{k}_{F,i}}{s^{k}_{F,i}} \phi^{k}_{F,i} +\sum_{i\in I^{\mathcal{C}}} \frac{\widehat{\lambda}^{k}_{{\varPsi _{\varepsilon}},i}}{s^{k}_{{\varPsi_{\varepsilon}},i}}D^{k}_{{\varPsi _{\varepsilon}},i} +\sum _{i\in I_{k}}\frac{\widehat{\lambda }^{k}_{g,i}}{s^{k}_{g,i}}Q^{k}_{g,i}, \end{aligned}
(19)
\begin{aligned} &\varrho_{k}= \left \{ \textstyle\begin{array}{l@{\quad}l} 1, &\mbox{if } b_{k}\leq0,\\ \min\{\frac{(1-\theta) \vert \zeta^{k} \vert }{b_{k}},1\}, &\mbox{if } b_{k}>0. \end{array}\displaystyle \right . \end{aligned}
(20)
4. (4)

Solve the second SLE as follows:

$$\mathrm{SLE2}{:}\quad M_{I_{k}} \left ( \textstyle\begin{array}{c} dz \\ \lambda_{I_{k}} \end{array}\displaystyle \right ) = \left ( \textstyle\begin{array}{c} -\nabla_{z} f_{r_{k}}(x^{k},y^{k}) \\ \eta^{k}_{I_{k}} \end{array}\displaystyle \right ),$$
(21)

where the perturbation vector $$\eta^{k}_{I_{k}}=(\eta^{k}_{F,I^{\mathcal{C}}},\eta^{k}_{\varPsi_{\varepsilon},I^{\mathcal{C}}},\eta^{k}_{g,I_{k}})$$ is determined by the following convex combination:

\begin{aligned} \eta^{k}_{I_{k}}=(1- \varrho_{k})\varpi^{k}+\varrho_{k} \bigl(- \bigl\Vert \widehat{dz}^{k} \bigr\Vert - \bigl\Vert \varpi^{k} \bigr\Vert \bigr)s^{k}_{I_{k}} \end{aligned}
(22)

with $$s^{k}_{I_{k}}:=(s^{k}_{F,I^{\mathcal{C}}},s^{k}_{{\varPsi_{\varepsilon }},I^{\mathcal{C}}},s^{k}_{g,I_{k}})\in\mathbb{R}^{2m+|I_{k}|}$$.

Let its solution be $$(dz^{k},\lambda^{k}_{I_{k}})$$ with $$\lambda^{k}_{I_{k}}:=(\lambda^{k}_{F,I^{\mathcal{C}}},\lambda^{k}_{{\varPsi _{\varepsilon}},I^{\mathcal{C}}}, \lambda^{k}_{g,I_{k}})\in\mathbb{R}^{2m+|I_{k}|}$$. Set $$\lambda^{k}:= (\lambda^{k}_{F,I^{\mathcal{C}}},\lambda^{k}_{{\varPsi_{\varepsilon }},I^{\mathcal{C}}},\lambda^{k}_{g,I^{g}})$$ with $${\lambda}^{k}_{g,I^{g}} =({\lambda}^{k}_{g,I_{k}},0_{I^{g}\setminus I_{k}})$$.

Step 4. (Perform line search) Compute the step size $$t_{k}$$, which is the first number t of the sequence $$\{1,\beta,\beta^{2},\ldots \}$$ satisfying

\begin{aligned} &f_{r_{k}} \bigl(z^{k}+t\,dz^{k} \bigr)\leq f_{r_{k}} \bigl(z^{k} \bigr)+\alpha t\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\,dz^{k}, \end{aligned}
(23)
\begin{aligned} & \bigl(v^{k}_{i}+t\,dv^{k} \bigr)-F_{i} \bigl(u^{k}+t\,du^{k} \bigr)< 0,\quad \forall i\in I^{\mathcal{C}}, \end{aligned}
(24)
\begin{aligned} &\psi_{\varepsilon}\bigl(w^{k}_{i}+t\,dw^{k} \bigr)< 0,\quad \forall i\in I^{\mathcal{C}}, \end{aligned}
(25)
\begin{aligned} &g_{i} \bigl(u^{k}+t\,du^{k} \bigr)< 0,\quad \forall i\in I^{g}. \end{aligned}
(26)

Step 5. (Update) Let $$z^{k+1}=z^{k}+t_{k} \,dz^{k}$$, and compute

\begin{aligned} &s^{k+1}_{i}=\min \bigl\{ \max \bigl\{ \bigl\Vert dz^{k} \bigr\Vert ^{2}+s_{\min}, \lambda^{k}_{i} \bigr\} ,s_{\max } \bigr\} , \\ &\xi_{k+1}=\min \bigl\{ \sqrt{ \bigl(y^{k+1}_{i} \bigr)^{2}+ \bigl(v^{k+1}_{i} \bigr)^{2}},i\in I^{\mathcal{C}} \bigr\} , \\ &\varepsilon_{k+1}= \left \{ \textstyle\begin{array}{l@{\quad}l} \varepsilon_{k}, &\mbox{if } \varepsilon_{k}\leq\xi_{k+1},\\ \frac{1}{2}\varepsilon_{k}, &\mbox{otherwise}. \end{array}\displaystyle \right . \end{aligned}
(27)

Step 6. Let $$k:=k+1$$, and go back to Step 1.

### Remark 2

The correction technique of penalty parameter $$r_{k}$$ in Step 3(2) is from , when the conditions (i)–(ii) in Step 3(2) are satisfied, the current iteration point $$z^{k}$$ is close to a KKT point of the problem (7). However, the conditions (iii)–(iv) in Step 3(2) indicate that $$z^{k}$$ is far away from the feasible area of the problem (6), and the penalty parameter needs to be increased.

For convenience, denote

\begin{aligned} \varLambda_{k} =&H_{k}-\sum _{i\in I^{\mathcal{C}}} \frac {s^{k}_{F,i}}{v^{k}_{i}-F_{i}(x^{k},y^{k})}\nabla_{z} \bigl(v^{k}_{i}-F_{i} \bigl(x^{k},y^{k} \bigr) \bigr)\nabla _{z} \bigl(v^{k}_{i}-F_{i} \bigl(x^{k},y^{k} \bigr) \bigr)^{\mathrm{T}} \\ &{}-\sum_{i\in I^{\mathcal{C}}}\frac{s^{k}_{{\varPsi_{\varepsilon }},i}}{{\psi_{\varepsilon}}(y^{k}_{i},v^{k}_{i})} \nabla_{z}{\psi_{\varepsilon}} \bigl(y^{k}_{i},v^{k}_{i} \bigr)\nabla_{z}{\psi _{\varepsilon}} \bigl(y^{k}_{i},v^{k}_{i} \bigr)^{\mathrm{T}} \\ &{}-\sum_{i\in I_{k}}\frac{s^{k}_{g,i}}{g_{i}(x^{k},y^{k})} \nabla_{z} g_{i} \bigl(x^{k},y^{k} \bigr)\nabla_{z} g_{i} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}. \end{aligned}

### Lemma 2

For the directions $$\widehat{dz}^{k}$$and $$dz^{k}$$found in Step 3(1)(4), the following conclusions hold:

\begin{aligned} &\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\,\widehat{dz}^{k} =- \bigl( \widehat{dz}^{k} \bigr)^{\mathrm{T}}\varLambda_{k}\, \widehat{dz}^{k}\leq0,\quad \forall k\geq0, \end{aligned}
(28)
\begin{aligned} &\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\, dz^{k}\leq\theta\zeta^{k} \leq0, \quad\forall k\geq0. \end{aligned}
(29)

Furthermore, when iteration goes into Step 3(3)(4), we have $$\widehat {dz}^{k}\neq0$$and $$\zeta^{k}<0$$, so $$dz^{k}$$is a feasible descent direction of the problem (7) at $$z^{k}$$, hence, Algorithm Ais well defined.

### Proof

It follows from the $$\mathrm{SLE1}$$ (12) that

\begin{aligned} &H_{k}\,\widehat{dz}^{k}+A_{I_{k}}\widehat{ \lambda}^{k}_{I_{k}}=-\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr), \end{aligned}
(30)
\begin{aligned} &S_{I_{k}}(A_{I_{k}})^{\mathrm{T}}\, \widehat{dz}^{k}+G_{I_{k}}\widehat{\lambda }_{I_{k}}=0. \end{aligned}
(31)

By (30)–(31), we have

\begin{aligned} \nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr) =&\sum_{i\in I^{\mathcal{C}}}\frac{s^{k}_{F,i}}{v^{k}_{i}-F_{i}(x^{k},y^{k})} \nabla_{z} \bigl(v^{k}_{i}-F_{i} \bigl(x^{k},y^{k} \bigr) \bigr)\nabla_{z} \bigl(v^{k}_{i}-F_{i} \bigl(x^{k},y^{k} \bigr) \bigr)^{\mathrm {T}}\,\widehat{dz}^{k} \\ &{} +\sum_{i\in I^{\mathcal{C}}}\frac{s^{k}_{{\varPsi_{\varepsilon }},i}}{{\psi_{\varepsilon}}(y^{k}_{i},v^{k}_{i})} \nabla_{z}{\psi_{\varepsilon}} \bigl(y^{k}_{i},v^{k}_{i} \bigr) \nabla_{z}{\psi_{\varepsilon}} \bigl(y^{k}_{i},v^{k}_{i} \bigr)^{\mathrm{T}}\,\widehat {dz}^{k} \\ &{} +\sum_{i\in I_{k}}\frac{s^{k}_{g,i}}{g_{i}(x^{k},y^{k})} \nabla_{z} g_{i} \bigl(x^{k},y^{k} \bigr)\nabla_{z} g_{i} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\,\widehat {dz}^{k}-H_{k}\, \widehat{dz}^{k}, \end{aligned}

which indicates $$\nabla_{z} f_{r_{k}}(x^{k},y^{k})^{\mathrm{T}}\,\widehat{dz}^{k}= -(\widehat{dz}^{k})^{\mathrm{T}}\varLambda_{k}\,\widehat{dz}^{k}\leq0$$.

By (15)–(18), we obtain

$$\phi_{F,i}\widehat{\lambda}^{k}_{F,i} \geq0,\quad i\in I^{\mathcal{C}};\qquad D_{\varPsi_{\varepsilon},i}\widehat{\lambda}^{k}_{\varPsi_{\varepsilon },i} \geq0,\quad i\in I^{\mathcal{C}};\qquad Q_{g,i}\widehat{\lambda}^{k}_{g,i} \geq0,\quad i\in I_{k}.$$
(32)

It follows from (18) and (32) that

\begin{aligned} \zeta_{k} \leq \nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\,\widehat{dz}^{k}\leq0. \end{aligned}
(33)

Taking into account SLE1 (12) and SLE2 (21), we have

\begin{aligned} &H_{k}\,\widehat{dz}^{k}+A_{I_{k}}\widehat{ \lambda}^{k}_{I_{k}}=-\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr), \end{aligned}
(34)
\begin{aligned} &S_{I_{k}}(A_{I_{k}})^{\mathrm{T}}\,\widehat{dz}^{k}+G_{I_{k}} \widehat {\lambda}^{k}_{I_{k}}=0, \end{aligned}
(35)
\begin{aligned} &H_{k}\,dz^{k}+A_{I_{k}}\lambda^{k}_{I_{k}}=- \nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr), \end{aligned}
(36)
\begin{aligned} &S_{I_{k}}(A_{I_{k}})^{\mathrm{T}}\,dz^{k}+G_{I_{k}} \lambda ^{k}_{I_{k}}=\eta^{k}_{I_{k}}. \end{aligned}
(37)

It follows from (34) and (36) as well as the symmetry of the matrix $$H_{k}$$ that

\begin{aligned} &\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\, dz^{k}-\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}} \,\widehat{dz}^{k} \\ &\quad= \bigl(\widehat{dz}^{k} \bigr)^{\mathrm{T}} A_{I_{k}} \lambda^{k}_{I_{k}} - \bigl(dz^{k} \bigr)^{\mathrm{T}} A_{I_{k}}\widehat{ \lambda}^{k}_{I_{k}} \\ &\quad= \bigl(\lambda^{k}_{I_{k}} \bigr)^{\mathrm{T}}(A_{I_{k}})^{\mathrm{T}}\, \widehat{dz}^{k} - \bigl(\widehat{\lambda}^{k}_{I_{k}} \bigr)^{\mathrm{T}}(A_{I_{k}})^{\mathrm{T}} \,dz^{k}. \end{aligned}

From (35) and (37), we have

\begin{aligned} & \bigl(\lambda^{k}_{I_{k}} \bigr)^{\mathrm{T}}(A_{I_{k}})^{\mathrm{T}}\, \widehat{dz}^{k} - \bigl(\widehat{\lambda}^{k}_{I_{k}} \bigr)^{\mathrm{T}}(A_{I_{k}})^{\mathrm{T}} \,dz^{k} \\ &\quad=-\sum_{i\in{I^{\mathcal{C}}}} \frac{\widehat{\lambda}^{k}_{F,i}\eta^{k}_{F,i}}{s^{k}_{F,i}} -\sum _{i\in{I^{\mathcal{C}}}} \frac{\widehat{\lambda}^{k}_{\varPsi_{\varepsilon},i}\eta^{k}_{\varPsi _{\varepsilon},i}}{s^{k}_{\varPsi_{\varepsilon},i}} -\sum _{i\in{I_{k}}} \frac{\widehat{\lambda}^{k}_{g,i}\eta^{k}_{g,i}}{s^{k}_{g,i}}. \end{aligned}

By (18) and (19), one gets

\begin{aligned} &\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}} \,dz^{k} \\ &\quad=\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\,\widehat{dz}^{k} -\sum _{i\in{I^{\mathcal{C}}}}\frac{\widehat{\lambda}^{k}_{F,i}\eta ^{k}_{F,i}}{s^{k}_{F,i}} -\sum _{i\in{I^{\mathcal{C}}}}\frac{\widehat{\lambda}^{k}_{\varPsi _{\varepsilon},i}\eta^{k}_{\varPsi_{\varepsilon},i}}{s^{k}_{\varPsi _{\varepsilon},i}} -\sum _{i\in{I_{k}}}\frac{\widehat{\lambda}^{k}_{g,i}\eta ^{k}_{g,i}}{s^{k}_{g,i}} \\ &\quad=\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\,\widehat{dz}^{k} -\sum _{i\in I^{\mathcal{C}}}\frac{\widehat{\lambda}^{k}_{F,i}}{s^{k}_{F,i}} \bigl[(1-\varrho_{k}) \phi^{k}_{F,i}+\varrho_{k} \bigl(- \bigl\Vert \widehat{dz}^{k} \bigr\Vert ^{q}- \bigl\Vert \varpi^{k} \bigr\Vert \bigr)s^{k}_{F,i} \bigr] \\ & \qquad{}-\sum_{i\in I^{\mathcal{C}}}\frac{\widehat{\lambda}^{k}_{{\varPsi _{\varepsilon}},i}}{s^{k}_{{\varPsi_{\varepsilon}},i}} \bigl[(1- \varrho_{k})D^{k}_{{\varPsi_{\varepsilon}},i}+\varrho_{k} \bigl(- \bigl\Vert \widehat{dz}^{k} \bigr\Vert ^{q}- \bigl\Vert \varpi^{k} \bigr\Vert \bigr)s^{k}_{{\varPsi_{\varepsilon }},i} \bigr] \\ &\quad\quad {}-\sum_{i\in I_{k}}\frac{\widehat{\lambda}^{k}_{g,i}}{s^{k}_{g,i}} \bigl[(1- \varrho_{k})Q^{k}_{g,i}+\varrho_{k} \bigl(- \bigl\Vert \widehat{dz}^{k} \bigr\Vert ^{q}- \bigl\Vert \varpi^{k} \bigr\Vert \bigr)s^{k}_{g,i} \bigr] \\ &\quad= \biggl(\nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\,\widehat{dz}^{k} -\sum _{i\in I^{\mathcal{C}}}\frac{\widehat{\lambda }^{k}_{F,i}}{s^{k}_{F,i}}\phi^{k}_{F,i} -\sum_{i\in I^{\mathcal{C}}}\frac{\widehat{\lambda}^{k}_{{\varPsi _{\varepsilon}},i}}{s^{k}_{{\varPsi_{\varepsilon}},i}}D^{k}_{{\varPsi _{\varepsilon}},i} -\sum_{i\in I_{k}}\frac{\widehat{\lambda }^{k}_{g,i}}{s^{k}_{g,i}}Q^{k}_{g,i} \biggr) \\ &{}\qquad +\varrho_{k} \biggl[\sum_{i\in I^{\mathcal{C}}} \frac{\widehat{\lambda }^{k}_{F,i}}{s^{k}_{F,i}}\phi^{k}_{F,i} +\sum _{i\in I^{\mathcal{C}}}\widehat{\lambda}^{k}_{F,i} \bigl( \bigl\Vert \, \widehat{dz}^{k} \bigr\Vert ^{q}+ \bigl\Vert \varpi^{k} \bigr\Vert \bigr) \biggr] \\ & {}\quad\quad+\varrho_{k} \biggl[\sum_{i\in I^{\mathcal{C}}} \frac{\widehat{\lambda }^{k}_{{\varPsi_{\varepsilon}},i}}{s^{k}_{{\varPsi_{\varepsilon }},i}}D^{k}_{{\varPsi_{\varepsilon}},i} +\sum _{i\in I^{\mathcal{C}}}\widehat{\lambda}^{k}_{\varPsi_{\varepsilon},i} \bigl( \bigl\Vert \widehat{dz}^{k} \bigr\Vert ^{q}+ \bigl\Vert \varpi^{k} \bigr\Vert \bigr) \biggr] \\ &\qquad{} +\varrho_{k} \biggl[\sum_{i\in I_{k}} \frac{\widehat{\lambda }^{k}_{g,i}}{s^{k}_{g,i}}Q^{k}_{g,i} +\sum _{i\in I_{k}}\widehat{\lambda}^{k}_{g,i} \bigl( \bigl\Vert \,\widehat{dz}^{k} \bigr\Vert ^{q}+ \bigl\Vert \varpi^{k} \bigr\Vert \bigr) \biggr] \\ &\quad=\zeta^{k}+\varrho_{k}b_{k}. \end{aligned}
(38)

In view of (20), if $$b_{k}\leq0$$, then $$\varrho _{k}b_{k}=b_{k}\leq0$$. Furthermore, we have $$\zeta^{k}+\varrho _{k}b_{k}\leq\zeta^{k}\leq\theta\zeta^{k}$$. If $$b_{k}>0$$, then we have $$\zeta^{k}+\varrho_{k}b_{k}=\zeta^{k}+\min\{(1-\theta)|\zeta ^{k}|,b_{k}\}\leq\zeta^{k}+(\theta-1)\zeta^{k}=\theta\zeta^{k}$$. In conclusion, we obtain $$\zeta^{k}+\varrho_{k}b_{k}\leq\theta\zeta ^{k}$$. From (33) and (38), it follows that

\begin{aligned} \nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\, dz^{k}=\zeta^{k}+\varrho _{k}b_{k}\leq\theta\zeta^{k}\leq0, \end{aligned}

which indicates that the inequality (29) is true.

If $$\widehat{dz}^{k}=0$$, taking into account SLE1 (12), from $$v^{k}-F(x^{k},y^{k})<0$$, $$\varPsi_{\varepsilon}(y^{k},v^{k})<0$$, $$g(x^{k},y^{k})<0$$, we have $$\widehat{\lambda}^{k}_{I_{k}}=0$$, so the iterate k does not go into Step 3(3)(4). Thus, $$\widehat{dz}^{k}\neq0$$ when the iterative process goes into Step 3(3)(4).

By $$\nabla_{z} f_{r_{k}}(x^{k},y^{k})^{\mathrm{T}}\,dz^{k}\leq0$$, we know that $$dz^{k}$$ is a feasible descent direction of the problem (7) at point $$z^{k}$$. Together with $$v^{k}-F(x^{k},y^{k})<0$$, $$\varPsi_{\varepsilon}(y^{k},v^{k})<0$$, $$g(x^{k},y^{k})<0$$ and $$\zeta^{k}<0$$, Step 4 in Algorithm A can be finished by finite calculations, hence Algorithm A is well defined. □

## Global convergence

In this paper, we assume that Algorithm A generates an infinite iteration sequence $$\{z^{k}\}$$. For the sake of convenience, let $$z^{*}$$ be an accumulation point of the iteration sequence $$\{z^{k}\}$$. First, we show that the penalty parameter $$r_{k}$$ can be augmented only in a finite number of steps. Then we prove that $$z^{*}$$ is a KKT point of the problem (7). Finally, we show that $$z^{*}$$ is a KKT point of the MPCC (1). More precisely, we show that Algorithm A is globally convergent. For this purpose, the following assumptions are necessary.

### Assumption 3

Suppose that the sequences $$\{H_{k}\}$$ and $$\{z^{k}\}$$ are all bounded, each accumulation point of $$\{z^{k}\}$$ satisfies LLSC (2). Assume that there exists a positive constant σ such that

$$d^{\mathrm{T}} \varLambda_{k}d\geq\sigma \Vert d \Vert ^{2},\quad \forall k, \forall d\in \mathbb{R}^{n+2m}.$$
(39)

### Assumption 4

For any $$z\in\widetilde{X}$$, if $$z\notin X$$, then no scalars $$\lambda _{F,i}\geq0$$, $$i\in I^{F}_{0}(z)$$; $$\lambda_{\varPsi_{\varepsilon},i}\geq 0$$, $$i\in I^{\varPsi_{\varepsilon}}_{0}(z)$$ and $$\lambda_{g,i}\geq0$$, $$i\in I^{g}_{0}(z)$$, exist such that

\begin{aligned} &\sum_{i\in I^{\mathcal{C}}}\nabla_{z} \bigl(v_{i}-F_{i}(x,y) \bigr)+\sum _{i\in I^{\mathcal{C}}}\nabla_{z}\psi_{\varepsilon}(y_{i},v_{i}) \\ &\quad=\sum_{i\in I^{F}_{0}(z)}\lambda_{F,i} \nabla_{z} \bigl(v_{i}-F_{i}(x,y) \bigr)+ \sum_{i\in I^{\varPsi_{\varepsilon}}_{0}(z)}\lambda_{{\varPsi_{\varepsilon }},i} \nabla_{z}\psi_{\varepsilon}(y_{i},v_{i})+ \sum_{i\in I^{g}_{0}(z)}\lambda_{g,i} \nabla_{z} g_{i}(x,y). \end{aligned}

### Lemma 3

Suppose that Assumptions 14hold, then the penalty parameter $$r_{k}$$can be augmented only in a finite number of steps.

### Proof

By contradiction, suppose that $$r_{k}$$ is increased infinitely many times, i.e., there exists an infinite index set K such that $$r_{k+1}>r_{k}$$ for any $$k\in K$$. The increase of $$r_{k}$$ must satisfy the following conditions for any $$k\in K$$:

\begin{aligned} &\text{(i)}\quad \bigl\Vert \,\widehat{dz}^{k} \bigr\Vert \leq\gamma_{1};\qquad \text{(ii)} \quad\widehat {\lambda}^{k}\geq- \gamma_{2}e_{I}; \qquad\text{(iii)} \quad\widehat{\lambda }^{k}_{{F},I^{\mathcal{C}}}\ngtr\gamma_{3}e_{I^{\mathcal{C}}}; \\ &\text{(iv)} \quad\widehat{\lambda}^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}\ngtr \gamma_{4}e_{I^{\mathcal{C}}}; \qquad\text{(v)}\quad I^{C} \bigl(y^{k},v^{k},\varepsilon_{k} \bigr)\neq{ \emptyset}. \end{aligned}

Denote

\begin{aligned} &A^{k}_{I^{g}}=\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} \textstyle\begin{array}{c} -\nabla_{x}F(x^{k},y^{k})\\ -\nabla_{y}F(x^{k},y^{k})\\ E_{m} \end{array}\displaystyle & \textstyle\begin{array}{c} 0 \\ \nabla_{y}\varPsi_{\varepsilon}(y^{k},v^{k})\\ \nabla _{v}\varPsi_{\varepsilon}(y^{k},v^{k}) \end{array}\displaystyle & \textstyle\begin{array}{c} \nabla_{x} g(x^{k},y^{k})\\ \nabla_{y}g(x^{k},y^{k})\\ 0 \end{array}\displaystyle \end{array}\displaystyle \right ), \\ &G^{k}_{I^{g}}=\operatorname{diag} \bigl( v^{k}-F\bigl(x^{k},y^{k}\bigr), \varPsi_{\varepsilon}\bigl(y^{k},v^{k}\bigr), g\bigl(x^{k},y^{k}\bigr) \bigr), \end{aligned}

and $$S^{k}_{I^{g}}=\operatorname{diag}(s^{k}_{I^{g}})$$, where $$s^{k}_{I^{g}}=(s^{k}_{F,I^{\mathcal{C}}},s^{k}_{\varPsi_{\varepsilon},I^{\mathcal{C}}},s^{k}_{g,I^{g}})$$. According to the SLE

\begin{aligned} \left ( \textstyle\begin{array}{c@{\quad}c} \textstyle\begin{array}{c} H_{k}\\ S^{k}_{I^{g}}(A^{k}_{I^{g}})^{\mathrm{T}} \end{array}\displaystyle & \textstyle\begin{array}{c} A^{k}_{I^{g}}\\ G^{k}_{I^{g}} \end{array}\displaystyle \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{c} \widehat{dz}^{k} \\ \widehat{\lambda}^{k} \end{array}\displaystyle \right ) = \left ( \textstyle\begin{array}{c} -\nabla_{z} f_{r_{k}}(x^{k},y^{k}) \\ 0 \end{array}\displaystyle \right ) , \end{aligned}

we have

\begin{aligned} &H_{k}\,\widehat{dz}^{k}+A^{k}_{I^{g}} \widehat{\lambda}^{k}+ \nabla_{z} f_{r_{k}} \bigl(x^{k},y^{k} \bigr)=0, \end{aligned}
(40)
\begin{aligned} &S^{k}_{I^{g}} \bigl(A^{k}_{I^{g}} \bigr)^{\mathrm{T}}\,\widehat {dz}^{k}+G^{k}_{I^{g}} \widehat{\lambda}^{k}=0. \end{aligned}
(41)

Since $$\{r_{k}\}$$ tends to infinity, and $$\lambda^{k}=\widehat {\lambda}^{k-1}-r_{k-1}\tilde{e}$$, $$k\geq1$$, such that $$\{\|\widehat {\lambda}^{k}\|_{\infty}\}$$ tends to infinity as well, we have

\begin{aligned} \beta_{k}=\max \bigl\{ \bigl\Vert \widehat{\lambda}^{k}_{F,I^{\mathcal{C}}} \bigr\Vert _{\infty}, \bigl\Vert \widehat{\lambda}^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}} \bigr\Vert _{\infty}, \bigl\Vert \widehat{\lambda}^{k}_{g,I^{g}} \bigr\Vert _{\infty},1 \bigr\} \stackrel{K}{\longrightarrow}\infty. \end{aligned}

Define

$$\bar{\lambda}^{k}_{F,I^{\mathcal{C}}} = \beta^{-1}_{k}\widehat{\lambda}^{k}_{F,I^{\mathcal{C}}},\qquad \bar{\lambda}^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}} =\beta^{-1}_{k} \widehat{\lambda}^{k}_{{\varPsi_{\varepsilon }},I^{\mathcal{C}}},\qquad \bar{\lambda}^{k}_{g, I^{g}}= \beta^{-1}_{k}\widehat{\lambda}^{k}_{g,I^{g}},$$
(42)

for k (K) large enough, it follows that

$$\max \bigl\{ \bigl\Vert \bar{\lambda}^{k}_{F,I^{\mathcal{C}}} \bigr\Vert _{\infty}, \bigl\Vert \bar{\lambda}^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}} \bigr\Vert _{\infty}, \bigl\Vert \bar{\lambda}^{k}_{g,I^{g}} \bigr\Vert _{\infty} \bigr\} =1.$$

Since the sequence $$\{z^{k}\}_{k\in K}$$ is bounded by Assumption 3, there exists an infinite index set $$K'\subseteq K$$, $$z^{*}\in\mathbb{R}^{n+2m}$$, and $$(\bar{\lambda}^{*}_{F,I^{\mathcal{C}}}, \bar{\lambda}^{*}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}, \bar{\lambda}^{*}_{g,I^{g}})\neq0$$, for all $$k\in K'$$, such that

$$\lim_{k\rightarrow\infty}z^{k}=z^{*},\qquad \lim _{k\rightarrow\infty}\bar{\lambda}^{k}_{F,I^{\mathcal{C}}} =\bar{ \lambda}^{*}_{F,I^{\mathcal{C}}}, \qquad\lim_{k\rightarrow\infty}\bar{ \lambda}^{k}_{{\varPsi_{\varepsilon }},I^{\mathcal{C}}} =\bar{\lambda}^{*}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}},\qquad \lim_{k\rightarrow\infty}\bar{\lambda}^{k}_{g,I^{g}} = \bar{\lambda}^{*}_{g,I^{g}}.$$

In view of the boundedness of $$\{z^{k}\}$$ and the continuity of the functions, it follows that

$$\bigl\{ \nabla_{z} \bigl(v^{k}-F \bigl(x^{k},y^{k} \bigr) \bigr) \bigr\} ,\qquad \bigl\{ \nabla_{z}\varPsi_{\varepsilon} \bigl(y^{k},v^{k} \bigr) \bigr\} ,\qquad \bigl\{ \nabla_{z} g \bigl(x^{k},y^{k} \bigr) \bigr\} ,$$

are bounded. Further, $$\{S^{k}_{I^{g}}\}$$ are bounded by construction. Multiplying both sides of (41) by $$\beta^{-1}_{k}$$, we have

$$\beta^{-1}_{k}S^{k}_{I^{g}} \bigl(A^{k}_{I^{g}} \bigr)^{\mathrm{T}}\, \widehat{dz}^{k} +\beta^{-1}_{k}G^{k}_{I^{g}} \widehat{\lambda}^{k}=0.$$

Letting $$k\stackrel{K'}{\longrightarrow}\infty$$, the condition (i) shows that $$\beta^{-1}_{k}G^{k}_{I^{g}}\widehat{\lambda}^{k}\stackrel{K'}{\longrightarrow}0$$, implying that $$\beta^{-1}_{k}\widehat{\lambda }^{k}_{F,i}\stackrel{K'}{\longrightarrow}0$$ for $$i\in I^{\mathcal{C}}\setminus I^{F}_{0}(z^{*})$$. Similarly, $$\beta^{-1}_{k}\widehat {\lambda}^{k}_{\varPsi_{\varepsilon},i}\stackrel{K'}{\longrightarrow}0$$ for $$i\in I^{\mathcal{C}}\setminus I^{\varPsi_{\varepsilon}}_{0}(z^{*})$$.

From (10) and (42), we know that $$\{ \frac{r_{k}}{\beta_{k}}\}$$ converges to $$a\geq0$$ as $$k\stackrel{K'}{\longrightarrow}\infty$$, with

$$\bar{\lambda}^{*}_{F,i}=a,\quad \forall i\in I^{\mathcal{C}}\setminus I^{F}_{0} \bigl(z^{*} \bigr);\qquad \bar{\lambda}^{*}_{\varPsi_{\varepsilon},i}=a, \quad\forall i\in I^{\mathcal{C}} \setminus I^{\varPsi_{\varepsilon}}_{0} \bigl(z^{*} \bigr).$$
(43)

Then it follows from the condition (ii) and (42) that

$$\bar{\lambda}^{*}_{F,i}\leq a, \quad\forall i\in I^{F}_{0} \bigl(z^{*} \bigr);\qquad \bar{\lambda}^{*}_{\varPsi_{\varepsilon},i} \leq a,\quad \forall i\in I^{\varPsi _{\varepsilon}}_{0} \bigl(z^{*} \bigr); \qquad\bar{ \lambda}^{*}_{g,i}\leq0,\quad \forall i\in I^{g}_{0} \bigl(z^{*} \bigr).$$
(44)

And from (41), we obtain

$$s^{k}_{g,i}\nabla_{z} g_{i} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}} \,\widehat {dz}^{k}+g_{i} \bigl(x^{k},y^{k} \bigr)\widehat{\lambda}^{k}_{g,i}=0,\quad i\in I^{g},$$

furthermore, multiplying both sides of this equation by $$\beta ^{-1}_{k}$$, together with the condition (i) and taking the limit as $$k\stackrel{K'}{\longrightarrow}\infty$$, we obtain

\begin{aligned} g_{i} \bigl(x^{*},y^{*} \bigr)\widehat{\lambda}^{*}_{g,i}=0,\quad i \in I^{g}, \end{aligned}

so $$\widehat{\lambda}^{*}_{g,i}=0$$ for all $$i\in I^{g}\setminus I^{g}_{0}(z^{*})$$.

In view of the condition (i) and the continuity of the functions, multiplying (40) by $$\beta^{-1}_{k}$$ and taking the limit as $$k\stackrel{K'}{\rightarrow}\infty$$, we get $$\beta ^{-1}_{k}A^{k}_{I^{g}}\widehat{\lambda}^{k}\stackrel{K'}{\longrightarrow}0$$, i.e.,

$$\sum_{i\in I^{\mathcal{C}}}\bar{ \lambda}^{*}_{F,i}\nabla _{z} \bigl(v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr) \bigr) +\sum_{i\in I^{\mathcal{C}}}\bar{ \lambda}^{*}_{\varPsi_{\varepsilon }i}\nabla_{z}\psi_{\varepsilon} \bigl(y^{*}_{i},v^{*}_{i} \bigr) +\sum _{i\in I^{g}_{0}(z^{*})}\bar{\lambda}^{*}_{g,i}\nabla_{z} g_{i} \bigl(x^{*},y^{*} \bigr)=0.$$
(45)

Since $$\bar{\lambda}^{*}_{F,I^{\mathcal{C}}}$$, $$\bar{\lambda}^{*}_{\varPsi _{\varepsilon},I^{\mathcal{C}}}$$, and $$\bar{\lambda}^{*}_{g,I^{g}}$$ are not zero, combining with (45) and Assumption 2, it follows that

\begin{aligned} I^{F}_{0} \bigl(z^{*} \bigr)\neq\{1,\ldots,m\},\qquad I^{\varPsi_{\varepsilon}}_{0} \bigl(z^{*} \bigr)\neq\{1,\ldots,m\}, \end{aligned}

i.e., $$z^{*}\notin X$$ and $$a>0$$.

Based on (43), dividing both sides of (45) by a, we obtain

\begin{aligned} &\sum_{i\in I^{\mathcal{C}}\setminus I^{F}_{0}(z^{*})}\nabla _{z} \bigl(v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr) \bigr) +\sum _{i\in I^{\mathcal{C}}\setminus I^{\varPsi_{\varepsilon}}_{0}(z^{*})} \nabla_{z}\psi_{\varepsilon} \bigl(y^{*}_{i},v^{*}_{i} \bigr) \\ &\quad=-\sum_{i\in I^{F}_{0}(z^{*})}\frac{\bar{\lambda}^{*}_{F,i}}{a} \nabla_{z} \bigl(v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr) \bigr) -\sum_{i\in I^{\varPsi_{\varepsilon}}_{0}(z^{*})} \frac{\bar{\lambda }^{*}_{{\varPsi_{\varepsilon}},i}}{a} \nabla_{z}\psi_{\varepsilon} \bigl(y^{*}_{i},v^{*}_{i} \bigr) \\ &{}\qquad -\sum_{i\in I^{g}_{0}(z^{*})}\frac{\bar{\lambda}^{*}_{g,i}}{a}\nabla _{z} g_{i} \bigl(x^{*},y^{*} \bigr). \end{aligned}
(46)

Adding both sides of the above equality to

\begin{aligned} \sum_{i\in I^{F}_{0}(z^{*})}\nabla_{z} \bigl(v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr) \bigr)+\sum _{i\in I^{\varPsi_{\varepsilon}}_{0}(z^{*})}\nabla_{z}\psi_{\varepsilon} \bigl(y^{*}_{i},v^{*}_{i} \bigr), \end{aligned}

we obtain

\begin{aligned} &\sum_{i\in I^{\mathcal{C}}}\nabla_{z} \bigl(v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr) \bigr) + \sum _{i\in I^{\mathcal{C}}}\nabla_{z}\psi_{\varepsilon } \bigl(y^{*}_{i},v^{*}_{i} \bigr) \\ &\quad=\sum_{i\in I^{F}_{0}(z^{*})}\lambda_{F,i} \nabla_{z} \bigl(v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr) \bigr) +\sum_{i\in I^{\varPsi_{\varepsilon}}_{0}(z^{*})} \lambda_{{\varPsi _{\varepsilon}},i} \nabla_{z}\psi_{\varepsilon} \bigl(y^{*}_{i},v^{*}_{i} \bigr) \\ &{}\qquad +\sum_{i\in I^{g}_{0}(z^{*})}\lambda_{g,i} \nabla_{z} g_{i} \bigl(x^{*},y^{*} \bigr), \end{aligned}

where $$\lambda_{F,i}=1-\frac{\bar{\lambda}^{*}_{F,i}}{a}$$, $$i\in I^{F}_{0}(z^{*})$$; $$\lambda_{{\varPsi_{\varepsilon}},i}=1-\frac{\bar {\lambda}^{*}_{{\varPsi_{\varepsilon}},i}}{a}$$, $$i\in I^{\varPsi_{\varepsilon }}_{0}(z^{*})$$; $$\lambda_{g,i}=-\frac{\bar{\lambda}^{*}_{g,i}}{a}$$, $$i\in I^{g}_{0}(z^{*})$$. In view of (44) and $$z^{*}\notin X$$. This contradicts Assumption 4, so the penalty parameter $$r_{k}$$ is updated at most a finite number of times. □

### Lemma 4

Suppose that Assumptions 14hold, then there exists an integer $$k_{0}$$, such that $$\varepsilon_{k}\equiv\varepsilon_{k_{0}}$$for $$k\geq k_{0}$$.

### Proof

By Step 3(2) and Step 5 of Algorithm A, the sequence $$\{\varepsilon_{k}\}$$ is monotonically decreasing and bounded from below, so the sequence $$\{\varepsilon_{k}\}$$ is convergent. Let

\begin{aligned} &K_{1}=\{k\in K\mid\varepsilon_{k}> \xi_{k+1}\}, \\ &K_{2}= \bigl\{ k\in K\mid\widehat{dz}^{k}=0, \widehat{ \lambda}^{k}_{g,I_{k}}\geq0, I^{\mathcal{C}} \bigl(y^{k},v^{k},\varepsilon_{k} \bigr)\neq \emptyset \bigr\} , \end{aligned}

where $$K=\{0,1,2,\ldots\}$$. Now we turn to prove that $$K_{1}$$ and $$K_{2}$$ are finite sets.

By contradiction, if $$K_{1}$$ is infinite set, then $$\lim_{k\rightarrow\infty}\varepsilon_{k}=0$$. There exist $$i\in I^{\mathcal{C}}$$ and $$\overline{K}_{1}\subseteq K_{1}$$, such that

$$\xi_{k+1}=\sqrt{ \bigl(y^{k+1}_{i} \bigr)^{2}+ \bigl(v^{k+1}_{i} \bigr)^{2}}< \varepsilon_{k},\quad k\in \overline{K}_{1} \subseteq K_{1}.$$
(47)

Together with the boundedness of $$\{z^{k+1}\}_{\overline{K}_{1}}$$, we suppose that $$z^{k+1}\stackrel{\overline{K}_{1}}{\longrightarrow}z^{*}$$. From (47), we have $$(y^{*}_{i})^{2}+(v^{*}_{i})^{2}=0$$, which contradicts Assumption 3. So $$K_{1}$$ is a finite set. Similarly, we show that $$K_{2}$$ is finite. Based on the finiteness of $$K_{1}$$ and $$K_{2}$$, from the update rule (27) of $$\varepsilon_{k}$$ and Step 3(2), we can conclude that Lemma 4 holds. □

Based on Lemmas 34, in the following analysis, we suppose, without loss of generality, that $$r_{k}\equiv\widehat{r}$$, $$\varepsilon_{k}\equiv \varepsilon$$.

### Lemma 5

Suppose that Assumptions 14hold. Then

1. (1)

The coefficient matrix sequence $$\{M_{I_{k}}\}$$is unified invertible, and there exists a positive constant L such that $$\| (M_{I_{k}})^{-1}\|\leq L$$, $$\forall k\geq0$$.

2. (2)

The sequences $$\{(\widehat{dz}^{k},\widehat{\lambda}^{k})\}$$and $$\{ ({dz}^{k},{\lambda}^{k})\}$$are bounded.

### Proof

(1) By contradiction, suppose that there exists an infinite subset K such that $$\|(M_{I_{k}})^{-1}\|\stackrel{K}{\longrightarrow}\infty$$. Since $$\{z^{k}\}$$ and $$\{H_{k}\}$$ are bounded, and the finite choice of $$I_{k}$$, without loss of generality, for all $$k\in K$$, let

$$I_{k}\equiv I', \qquad z^{k}\stackrel{K}{\longrightarrow}z^{*}, \qquad H_{k}\stackrel{K}{\longrightarrow}H_{*},\qquad s^{k}\stackrel{K}{\longrightarrow}s^{*}\geq s_{\min }e_{I}>0.$$

Denote

\begin{aligned} &A^{*}_{I'}=\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} \textstyle\begin{array}{c} -\nabla_{x}F(x^{*},y^{*})\\ -\nabla_{y}F(x^{*},y^{*})\\ E_{m} \end{array}\displaystyle & \textstyle\begin{array}{c} 0_{n\times m} \\ \nabla_{y}\varPsi_{\varepsilon}(y^{*},v^{*})\\ \nabla_{v}\varPsi_{\varepsilon}(y^{*},v^{*}) \end{array}\displaystyle &\textstyle\begin{array}{c} \nabla_{x}g_{I'}(x^{*},y^{*})\\ \nabla_{y}g_{I'}(x^{*},y^{*})\\ 0_{m\times m_{g}} \end{array}\displaystyle \end{array}\displaystyle \right ), \\ &G^{*}_{I'}=\operatorname{diag} \bigl(v^{*}-F \bigl(x^{*},y^{*} \bigr), \varPsi_{\varepsilon}\bigl(y^{*},v^{*} \bigr),g_{I'} \bigl(x^{*},y^{*} \bigr) \bigr), \end{aligned}

then

$$M_{I_{k}}\stackrel{K}{\longrightarrow}M^{*}_{I'}=\left ( \textstyle\begin{array}{c@{\quad}c} \textstyle\begin{array}{c} H_{*}\\ S^{*}_{I'}(A^{*}_{I'})^{\mathrm{T}} \end{array}\displaystyle &\textstyle\begin{array}{c} A^{*}_{I'} \\ G^{*}_{I'} \end{array}\displaystyle \end{array}\displaystyle \right ),$$
(48)

where the matrix $$S^{*}_{I'}=\operatorname{diag}(s^{*}_{I'})$$ with $$s^{*}_{I'}=(s^{*}_{F,I^{\mathcal{C}}}, s^{*}_{\varPsi_{\varepsilon},I^{\mathcal{C}}},s^{*}_{g,I'})$$.

Under the Assumptions, it can be shown easily that $$M^{*}_{I'}$$ is nonsingular. So we obtain

\begin{aligned} \bigl\Vert (M_{I_{k}})^{-1} \bigr\Vert \stackrel{K}{\longrightarrow}\bigl\Vert \bigl(M^{*}_{I'} \bigr)^{-1} \bigr\Vert < \infty, \end{aligned}

which contradicts $$\|(M_{I_{k}})^{-1}\|\stackrel{K}{\longrightarrow}\infty$$.

(2) In view of SLE1 (12), and Lemma 3 as well as $$r_{k}\equiv\widehat{r}$$, we see that $$\{(\widehat{dz}^{k},\widehat{\lambda}^{k})\}$$ is bounded. By (15)–(22) and the boundedness of $$\{(\widehat {dz}^{k},\widehat{\lambda}^{k})\}$$ as well as the boundedness of $$\{s^{k}\}$$, we know that $$\{\eta^{k}_{I_{k}}\}$$ is bounded. Hence, the boundedness of $$\{(dz^{k},\lambda^{k})\}$$ follows by SLE2 (21). □

### Lemma 6

Suppose that Assumptions 14hold, $$z^{*}$$is an accumulation point of the infinite sequence $$\{z^{k}\}$$generated by Algorithm A, and $$\{z^{k}\} _{K}$$converges to $$z^{*}$$. If $$\{\zeta^{k}\}_{K}\rightarrow0$$, then $$z^{*}$$is a KKT point of the problem (7), and both $$\{\widehat{\lambda}^{k}\}_{K}$$and $$\{\lambda ^{k}\}_{K}$$converge to the unique multiplier vector $$\lambda^{*}$$corresponding to $$z^{*}$$.

### Proof

Let $$(\widehat{\lambda}^{*};\lambda^{*})$$ be any accumulation point of $$\{(\widehat{\lambda}^{k};\lambda^{k})\}_{K}$$. First, we verify that $$(z^{*},\widehat{\lambda}^{*})$$ is a KKT pair of the problem (7). Taking into account Assumption 3, Lemma 5, and the finite choice of $$I_{k}$$, there exists an infinite index $$K'\subseteq K$$ such that

\begin{aligned} \begin{gathered}I_{k}\equiv I',\qquad \bigl(\widehat{ \lambda}^{k};\lambda^{k} \bigr)\stackrel{K'}{\rightarrow}\bigl(\widehat{\lambda}^{*};\lambda^{*} \bigr),\qquad H_{k} \stackrel{K'}{\rightarrow}H_{*},\\ \widehat{dz}^{k} \stackrel{K'}{\rightarrow}\widehat{dz}^{*},\qquad s^{k} \stackrel{K'}{\rightarrow}s^{*}\geq s_{\min}e_{I}.\end{gathered} \end{aligned}
(49)

It follows from (33), (28), Assumption 3 and $$\{\zeta^{k}\} _{K}\rightarrow0$$ that $$\widehat{dz}^{*}=0$$. Further, taking the limit in SLE1 (12) for $$k\in K'$$, one obtains

\begin{aligned} \nabla_{z} f_{\widehat{r}} \bigl(x^{*},y^{*} \bigr)+A^{*}_{I'}\widehat{\lambda}^{*}_{I'}=0,\qquad G^{*}_{I'}\widehat{\lambda}^{*}_{I'}=0. \end{aligned}
(50)

Next we will prove $$\widehat{\lambda}^{*}_{I'}=(\widehat{\lambda}^{*}_{F,I^{\mathcal{C}}}, \widehat{\lambda}^{*}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}, \widehat{\lambda}^{*}_{g,I'})\geq0$$, and $$\widehat{\lambda}^{*} =(\widehat{\lambda}^{*}_{I'}, 0_{I^{g}\setminus{I'}})\geq0$$. From $$\widehat{\lambda}^{*}_{F,i}(v^{*}_{i}-F_{i}(x^{*},y^{*}))=0$$, for $$i\in I^{\mathcal{C}}\setminus I^{F}_{0}(z^{*})$$, we have $$\widehat{\lambda }^{*}_{F,I^{\mathcal{C}}\setminus I^{F}_{0}(z^{*})}=0$$. Similarly,

\begin{aligned} \widehat{\lambda}^{*}_{\varPsi_{\varepsilon},I^{\mathcal{C}}\setminus I^{\varPsi_{\varepsilon}}_{0}(z^{*})}=0, \qquad\widehat{\lambda}^{*}_{g,I^{'}\setminus I^{g}_{0}(z^{*})}=0. \end{aligned}

By (18) and $$(\zeta^{k},\widehat{dz}^{k})\stackrel{K'}{\longrightarrow}(0,0)$$, one obtains

\begin{aligned} \sum_{i\in I^{\mathcal{C}}}\frac{\widehat{\lambda }^{k}_{F,i}}{s^{k}_{F,i}} \phi^{k}_{F,i} +\sum_{i\in I^{\mathcal{C}}} \frac{\widehat{\lambda}^{k}_{\varPsi _{\varepsilon},i}}{ s^{k}_{\varPsi_{\varepsilon},i}}D^{k}_{\varPsi_{\varepsilon},i} +\sum _{i\in I'}\frac{\widehat{\lambda}^{k}_{g,i}}{s^{k}_{g,i}}Q^{k}_{g,i} \stackrel{K'}{\longrightarrow}0. \end{aligned}

Furthermore, by (32), we have $$\frac{\widehat{\lambda}^{k}_{F,i}\phi^{k}_{F,i}}{s^{k}_{F,i}}\geq 0$$ for every $$i\in I^{\mathcal{C}}$$. Combining with (49), we have $$\widehat{\lambda}^{k}_{F,i}\phi^{k}_{F,i}\stackrel{K'}{\longrightarrow}0$$, $$i\in I^{\mathcal{C}}$$. So, in view of (15), one has

\begin{aligned} \widehat{\lambda}^{*}_{F,i} \min \bigl\{ 0,- \bigl(\max \bigl\{ - \widehat{ \lambda}^{*}_{F,i},0 \bigr\} \bigr)^{\nu} -C \bigl(v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr) \bigr) \bigr\} =0,\quad i\in I^{\mathcal{C}}, \end{aligned}

which implies that $$\widehat{\lambda}^{*}_{F,i}\geq0$$ for $$i\in I^{\mathcal{C}}\cap I^{F}_{0}(z^{*})$$. Similarly, one obtains

\begin{aligned} \widehat{\lambda}^{*}_{\varPsi_{\varepsilon},i}\geq0 \quad\mbox{for } i\in I^{\mathcal{C}} \cap I^{\varPsi_{\varepsilon}}_{0} \bigl(z^{*} \bigr);\qquad \widehat{ \lambda}^{*}_{g,i}\geq0 \quad\mbox{for } i\in I'\cap I^{g}_{0} \bigl(z^{*} \bigr). \end{aligned}

So we can claim that $$\widehat{\lambda}^{*}_{I'}\geq0$$. Obviously, $$\widehat{\lambda}^{*}=(\widehat{\lambda}^{*}_{I'},0_{I^{g}\setminus {I'}})\geq0$$ is immediate.

Therefore, in view of $$z^{*}\in\widetilde{X}$$ and (50), we see that $$(z^{*},\widehat{\lambda}^{*})$$ is a KKT pair of the problem (7). Moreover, the above analysis indicates that the sequence $$\{\widehat{\lambda}^{k}\}_{K}$$ possesses a unique accumulation point $$\widehat{\lambda}^{*}$$, so $$\lim_{k\in K}\widehat{\lambda}^{k}=\widehat{\lambda}^{*}$$.

Last, noticing that $$(\widehat{dz}^{k},\widehat{\lambda}^{k})\stackrel{K'}{\longrightarrow}(0,\widehat{\lambda}^{*})\geq0$$, (15)–(17) and (22), we have

\begin{aligned} \phi^{k}_{F,I^{\mathcal{C}}}\stackrel{K'}{\longrightarrow}0,\qquad D^{k}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}}\stackrel{K'}{\longrightarrow}0,\qquad Q^{k}_{g,I'}\stackrel{K'}{\longrightarrow}0,\qquad \eta^{k}_{I'}= \bigl( \eta^{k}_{F,I^{\mathcal{C}}},\eta^{k}_{{\varPsi_{\varepsilon }},I^{\mathcal{C}}}, \eta^{k}_{g,I'} \bigr)\stackrel{K'}{\longrightarrow}0. \end{aligned}

Denote

\begin{aligned} M_{I_{k}} \left ( \textstyle\begin{array}{c} dz^{k}\\ \lambda^{k}_{I'} \end{array}\displaystyle \right ) = \left ( \textstyle\begin{array}{c} -\nabla_{z} f_{\widehat{r}}(x^{k},y^{k}) \\ \eta^{k}_{I'} \end{array}\displaystyle \right ), \end{aligned}
(51)

so it follows from the above SLE and SLE1 (12) that

$$M_{I_{k}} \left ( \textstyle\begin{array}{c} dz^{k}-\widehat{dz}^{k} \\ \lambda^{k}_{I'}-\widehat {\lambda}^{k}_{I'} \end{array}\displaystyle \right ) = \left ( \textstyle\begin{array}{c} 0 \\ \eta^{k}_{I'} \end{array}\displaystyle \right ) \stackrel{K'}{\longrightarrow} \left ( \textstyle\begin{array}{c} 0 \\ 0 \end{array}\displaystyle \right ),$$
(52)

which together with Lemma 5(1) gives

\begin{aligned} \lambda^{*}=\lim_{k\in K'}{\lambda}^{k}=\lim _{k\in K'}\widehat{\lambda}^{k} =\widehat{\lambda}^{*}. \end{aligned}

□

Based on the lemmas above, we now present the global convergence of Algorithm A.

### Theorem 1

Suppose that Assumptions 14hold. $$z^{*}$$is an accumulation point of $$\{z^{k}\}$$generated by Algorithm A, then $$z^{*}$$is a KKT point of the MPCC (1).

### Proof

A. We first show that $$z^{*}$$ is a KKT point of the problem (6) (called Conclusion A).

Let $$K'$$ be an infinite index set such that

\begin{aligned} \begin{gathered} I_{k}\equiv I',\qquad \bigl(\widehat{ \lambda}^{k};\lambda^{k} \bigr)\stackrel{K'}{\longrightarrow}\bigl(\widehat{\lambda}^{*};\lambda^{*} \bigr),\qquad H_{k} \stackrel{K'}{\longrightarrow}H_{*},\\ \widehat{dz}^{k} \stackrel{K'}{\longrightarrow}\widehat{dz}^{*},\qquad s^{k} \stackrel{K'}{\longrightarrow}s^{*}\geq s_{\min}e_{I}.\end{gathered} \end{aligned}

Denote

\begin{aligned} &A^{*}_{I'}=\left ( \textstyle\begin{array}{c@{\quad}c@{\quad}c} \textstyle\begin{array}{c} -\nabla_{x}F(x^{*},y^{*})\\ -\nabla_{y}F(x^{*},y^{*})\\ E_{m} \end{array}\displaystyle & \textstyle\begin{array}{c} 0 \\ \nabla_{y}\varPsi_{\varepsilon}(y^{*},v^{*})\\ \nabla _{v}\varPsi_{\varepsilon}(y^{*},v^{*}) \end{array}\displaystyle &\textstyle\begin{array}{c} \nabla_{x}g_{I'}(x^{*},y^{*})\\ \nabla_{y}g_{I'}(x^{*},y^{*})\\ 0 \end{array}\displaystyle \end{array}\displaystyle \right ), \\ &G^{*}_{I'}=\operatorname{diag} \bigl(v^{*}-F \bigl(x^{*},y^{*} \bigr), \varPsi_{\varepsilon}\bigl(y^{*},v^{*} \bigr),g_{I'} \bigl(x^{*},y^{*} \bigr) \bigr), \\ &M^{*}_{I'}=\left ( \textstyle\begin{array}{c@{\quad}c} \textstyle\begin{array}{c} H_{*}\\ S^{*}_{I'}(A^{*}_{I'})^{\mathrm{T}} \end{array}\displaystyle &\textstyle\begin{array}{c} A^{*}_{I'} \\ G^{*}_{I'} \end{array}\displaystyle \end{array}\displaystyle \right ), \end{aligned}

where the matrix $$S^{*}_{I'}=\operatorname{diag}(s^{*}_{I'})$$ with $$s^{*}_{I'}=(s^{*}_{F,I^{\mathcal{C}}},s^{*}_{\varPsi_{\varepsilon},I^{\mathcal {C}}},s^{*}_{g,I'})$$.

By contradiction, suppose that $$z^{*}$$ is not a KKT point of the problem (6). From Lemma 5, without loss of generality, suppose that $$\lambda ^{k}=\widehat{\lambda}^{k-1}-\widehat{r}\tilde{e}\stackrel{K'}{\longrightarrow}\widehat{\lambda}'$$. Therefore, it follows that $$(z^{*},\widehat{\lambda}')$$ is not a KKT pair of the problem (6), which implies $$\rho(z^{*},\widehat{\lambda}')>0$$ and $$I^{g}_{0}(z^{*})\subseteq I_{k}$$, $$k\in K'$$ large enough. The following two cases are discussed.

Case I. $$z^{*}$$ is a KKT point of the problem (7). Then there exists a multiplier $$\widehat {\lambda}''=(\widehat{\lambda}''_{I_{*}},0_{I^{g}\setminus I'})\geq 0$$, such that the KKT condition of the problem (7) is satisfied at $$(z^{*},\widehat{\lambda }''_{I_{*}})$$. For $$k\in K'$$ large enough, we have $$I^{g}_{0}(z^{*})\subseteq I_{k}\equiv I'$$. It is easy to see that $$(0,\widehat{\lambda }''_{I_{*}})$$ is a solution to the following SLE:

$$M^{*}_{I'} \left ( \textstyle\begin{array}{c} dz \\ \lambda_{I_{k}} \end{array}\displaystyle \right ) = - \left ( \textstyle\begin{array}{c} \nabla_{z} f_{\widehat{r}}(x^{*},y^{*}) \\ 0 \end{array}\displaystyle \right ).$$
(53)

Passing to the limit in SLE1 (12), we know that $$(\widehat {dz}^{*},\widehat{\lambda}^{*}_{I_{*}})$$ is the solution to (53) above. Let $$\widehat{\lambda}^{*}=(\widehat{\lambda}^{*}_{I_{*}},0_{I^{g}\setminus I'})$$. In view of Lemma 5 (1), taking into account the nonsingularity of the matrix $$M^{*}_{I'}$$, we see that the solution of (53) is unique. Hence, we obtain $$\widehat{dz}^{*}=0$$, $$\widehat{\lambda}^{*}_{I_{*}}=\widehat{\lambda }''_{I_{*}}\geq0$$, which implies $$\widehat{\lambda}^{*}=\widehat{\lambda}''\geq0$$. The conditions (i) and (ii) in Step 3 (2) are satisfied for $$k\in K'$$ large enough. Moreover, in view of $$r_{k}\equiv\widehat{r}<\infty$$ for k large enough, Step 3 (2) implies $$\widehat{\lambda}^{k}_{F,I^{\mathcal{C}}} >\gamma_{3}e_{I^{\mathcal{C}}}$$, $$\widehat{\lambda}^{k}_{{\varPsi _{\varepsilon}},I^{\mathcal{C}}} >\gamma_{4}e_{I^{\mathcal{C}}}$$ for $$k\in K'$$ large enough, which further implies that

\begin{aligned} \widehat{\lambda}^{*}_{F,I^{\mathcal{C}}} \geq\gamma_{3}e_{I^{\mathcal{C}}}>0,\qquad \widehat{\lambda}^{*}_{{\varPsi_{\varepsilon}},I^{\mathcal{C}}} \geq\gamma_{4}e_{I^{\mathcal{C}}}>0. \end{aligned}

Since the pair $$(z^{*},\widehat{\lambda}'')$$ satisfies the complementary slackness, one gets

\begin{aligned} &\widehat{\lambda}^{''}_{F,i} \bigl(v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr) \bigr) =\widehat{\lambda}^{*}_{F,i} \bigl(v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr) \bigr)=0,\quad i\in I^{\mathcal{C}}, \\ &\widehat{\lambda}^{''}_{\varPsi_{\varepsilon},i}\psi_{\varepsilon } \bigl(y^{*}_{i},v^{*}_{i} \bigr) =\widehat{ \lambda}^{*}_{\varPsi_{\varepsilon},i}\psi_{\varepsilon } \bigl(y^{*}_{i},v^{*}_{i} \bigr)=0,\quad i\in I^{\mathcal{C}}. \end{aligned}

So $$v^{*}-F(x^{*},y^{*})=0$$, $$\varPsi_{\varepsilon}(y^{*},v^{*})=0$$, which together with Proposition 2 shows that $$z^{*}$$ is a KKT point of the problem (6), which contradicts the assumption that $$z^{*}$$ is not a KKT point of the problem (6).

Case II. $$z^{*}$$ is not a KKT point of the problem (7). It follows from Lemma 6 and $$\zeta^{k}\leq0$$ that $$\zeta^{k}\stackrel{K'}{\rightarrow}\bar{\zeta}<0$$. By (18), (28) and Assumption 3, we have

\begin{aligned} \lim_{k\in K'} \bigl( \bigl\Vert \widehat{dz}^{k} \bigr\Vert ^{q}+ \bigl\Vert \varpi^{k} \bigr\Vert \bigr)>0, \end{aligned}

so there exist a subset $$K^{''}\subseteq K'$$ and a positive constant such that

\begin{aligned} \zeta^{k}\leq\frac{\bar{\zeta}}{2}< 0; \qquad\bigl( \bigl\Vert \widehat{dz}^{k} \bigr\Vert ^{q}+ \bigl\Vert \varpi^{k} \bigr\Vert \bigr)\geq\bar{c}>0, \quad k\in K^{''}. \end{aligned}

In the following we first show that there exists a positive constant , such that $$t_{k}\geq\bar{t}$$ for all $$k\in K''$$.

(i) By (24)–(26), we have

\begin{aligned} \begin{gathered} v^{*}_{i}-F_{i} \bigl(x^{*},y^{*} \bigr)< 0,\quad i\notin I^{F}_{0} \bigl(z^{*} \bigr);\qquad \psi_{\varepsilon} \bigl(y^{*}_{i},v^{*}_{i} \bigr)< 0,\quad i\notin I^{\varPsi_{\varepsilon}}_{0} \bigl(z^{*} \bigr); \\ g_{i} \bigl(x^{*},y^{*} \bigr)< 0,\quad i\notin I^{g}_{0} \bigl(z^{*} \bigr).\end{gathered} \end{aligned}

Based on the boundedness of $$\{dz^{k}\}_{K''}$$ and the continuity of functions, for all $$k\in K''$$ large enough and $$t>0$$ sufficiently small, one gets

\begin{aligned} \bigl(v^{k}_{i}+t\,dv^{k} \bigr)-F_{i} \bigl(u^{k}+t\,du^{k} \bigr)< 0,\qquad \psi_{\varepsilon} \bigl(w^{k}_{i}+t\,dw^{k} \bigr)< 0,\qquad g_{i} \bigl(u^{k}+t\,du^{k} \bigr)< 0. \end{aligned}

In view of $$i\in I^{F}_{0}(z^{*})$$, we have $$v^{*}_{i}-F_{i}(u^{*})=0$$. Similarly, $$\psi_{\varepsilon}(w^{*}_{i})=0$$ for $$i\in I^{\varPsi_{\varepsilon }}_{0}(z^{*})$$, and $$g_{i}(u^{*})=0$$ for $$i\in I^{g}_{0}(z^{*})$$. From (22) and SLE2 (21), we have

\begin{aligned} S_{I_{k}}(A_{I_{k}})^{\mathrm{T}}\, dz^{k}+G_{I_{k}} \lambda^{k}_{I_{k}}=\eta^{k}_{I_{k}}, \end{aligned}

that is,

\begin{aligned} &\nabla_{z} \bigl(v^{k}_{i}-F_{i} \bigl(u^{k} \bigr) \bigr)^{\mathrm{T}}\, dz^{k}= \frac{\eta^{k}_{F,i} -\lambda^{k}_{F,i}(v^{k}_{i} -F_{i}(u^{k}))}{s^{k}_{F,i}},\quad i\in I^{\mathcal{C}}, \\ &\nabla_{z}\psi_{\varepsilon} \bigl(w^{k}_{i} \bigr)^{\mathrm{T}}\, dz^{k} =\frac{\eta^{k}_{\varPsi_{\varepsilon},i} -\lambda^{k}_{\varPsi_{\varepsilon},i}\psi_{\varepsilon }(w^{k}_{i})}{s^{k}_{\varPsi_{\varepsilon},i}},\quad i\in I^{\mathcal{C}}, \\ &\nabla_{z} g_{i} \bigl(u^{k} \bigr)^{\mathrm{T}} \,dz^{k}=\frac{\eta^{k}_{g,i}-\lambda ^{k}_{g,i}g(w^{k}_{i})}{s^{k}_{g,i}},\quad i\in I_{k}. \end{aligned}

By Taylor’s expansion and $$\|\widehat{dz}^{k}\|\geq0$$, for $$t>0$$ sufficient small, we get

\begin{aligned} & \bigl(v^{k}_{i}+t\,dv^{k} \bigr)-F_{i} \bigl(u^{k}+t\,du^{k} \bigr) \\ &\quad= \bigl(v^{k}_{i}-F_{i} \bigl(u^{k} \bigr) \bigr)+t\nabla_{z} \bigl(v^{k}_{i}-F_{i} \bigl(u^{k} \bigr) \bigr)^{\mathrm{T}}\,dz^{k}+o(t) \\ &\quad= \bigl(v^{k}_{i}-F_{i} \bigl(u^{k} \bigr) \bigr)+t\frac{\eta^{k}_{F,i}-\lambda ^{k}_{F,i}(v^{k}_{i}-F_{i}(u^{k}))}{s^{k}_{F,i}}+o(t) \\ &\quad= \biggl(1-t\frac{\lambda ^{k}_{F,i}}{s^{k}_{F,i}} \biggr) \bigl(v^{k}_{i}-F_{i} \bigl(u^{k} \bigr) \bigr)+t\frac{1-\varrho _{k}}{s^{k}_{F,i}}\phi^{k}_{F,i}-t \varrho_{k} \bigl( \bigl\Vert \widehat{dz}^{k} \bigr\Vert ^{q}+ \bigl\Vert \varpi^{k} \bigr\Vert \bigr) \\ &\quad\leq-t\varrho_{k} \bigl( \bigl\Vert \widehat{dz}^{k} \bigr\Vert ^{q}+ \bigl\Vert \varpi^{k} \bigr\Vert \bigr)+o(t). \end{aligned}
(54)

Furthermore, we consider $$\zeta^{k}\leq\frac{\bar{\zeta}}{2}<0$$, by (19), (20), and the boundedness of $$b_{k}$$ by Lemma 5, we see that there exists a constant $$\bar{a}>0$$ such that $$\varrho_{k}\geq\bar{a}>0$$, $$k\in K''$$. For $$k\in K''$$ large enough and $$t>0$$ small enough, we have

\begin{aligned} \bigl(v^{k}_{i}+t\,dv^{k} \bigr)-F_{i} \bigl(u^{k}+t\,du^{k} \bigr)\leq- \bar{a}\bar{c} t+o(t)< 0,\quad i\in I^{F}_{0} \bigl(z^{*} \bigr). \end{aligned}

Similarly, for $$k\in K''$$ large enough and $$t>0$$ small enough, we have

\begin{aligned} &\psi_{\varepsilon} \bigl(w^{k}_{i}+t\,dw^{k} \bigr)\leq-\bar{a}\bar{c} t+o(t)< 0,\quad i\in I^{\varPsi_{\varepsilon}}_{0} \bigl(z^{*} \bigr), \\ &g \bigl(u^{k}_{i}+t\,du^{k} \bigr)\leq- \bar{a}\bar{c} t+o(t)< 0,\quad i\in I^{g}_{0} \bigl(z^{*} \bigr). \end{aligned}

(ii) Consider the inequality (23). By (29) and Taylor expansion, for $$k\in K''$$ large enough and $$t>0$$ small enough, it follows that

\begin{aligned} \begin{gathered} f_{\widehat{r}} \bigl(z^{k}+t\,dz^{k} \bigr)-f_{\widehat{r}} \bigl(z^{k} \bigr)-\alpha t \nabla_{z} f_{\widehat{r}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\,dz^{k} \\ \quad=(1-\alpha)t\nabla_{z} f_{\widehat{r}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}} \,dz^{k}+o(t) \\ \quad\leq(1-\alpha)t\theta\zeta^{k}+o(t)\leq(1-\alpha)t\theta \frac {\bar{\zeta}}{2}+o(t)\leq0.\end{gathered} \end{aligned}

By summarizing the above discussion, we can conclude that the inequalities (23)–(26) hold for $$k\in K''$$ large enough and $$t>0$$ small enough, that is, $$\bar{t}=\inf\{t_{k}:k\in K''\}>0$$.

In what follows, based on $$t_{k}\geq\bar{t}>0$$ ($$k\in K''$$) we will deduce a contradiction. In view of the monotone decreasing property of $$\{f_{\widehat{r}}(z^{k})\}$$ and $$\lim_{k\in K''}f_{\widehat {r}}(z^{k})=f_{\widehat{r}}(z^{*})$$, we have $$\lim_{k\rightarrow\infty }f_{\widehat{r}}(z^{k})=f_{\widehat{r}}(z^{*})$$. Moreover, from (23) and (29), for $$k\in K''$$ large enough, we have

\begin{aligned} f_{\widehat{r}} \bigl(z^{k+1} \bigr)-f_{\widehat{r}} \bigl(z^{k} \bigr)\leq\alpha t_{k}\nabla _{z} f_{\widehat{r}} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}\, dz^{k}\leq\alpha t_{k}\theta \zeta^{k}\leq\alpha\bar{t}\theta\frac{\bar{\zeta}}{2}. \end{aligned}

Taking the limit $$k\stackrel{K''}{\rightarrow}\infty$$ in the inequality above, we obtain a contradiction.

From the above analysis, we can conclude that $$z^{*}$$ is a KKT point of the problem (6).

B. We now show that $$I^{\mathcal{C}}(y^{*},v^{*},\varepsilon )=\emptyset$$ (called Conclusion B).

From Lemma 4 and the updating rule of $$\varepsilon_{k}$$ in Step 5, we see that

\begin{aligned} \sqrt{ \bigl(y^{*}_{i} \bigr)^{2}+ \bigl(v^{*}_{i} \bigr)^{2}}\geq\varepsilon,\quad \forall i\in I^{\mathcal{C}}, \end{aligned}

which implies that $$I^{\mathcal{C}}(y^{*},v^{*},\varepsilon)=\emptyset$$.

Based on Conclusions A and B, and combining with Proposition 1, we can conclude that $$z^{*}$$ is a KKT point of the MPCC (1). □

## Numerical experiments

In this section, a preliminary implementation of Algorithm A is given on a Intel(R) Core(TM) i5-7200U CPU (@ 2.50 GHz, 2.71 GHz) and RAM (4 GB). A Matlab code (Version R2014a) is written corresponding to this implementation.

In our tests, referring to , we initialize $$H_{0}=E_{n}$$, and for $$k=1,2,\dots$$, the approximate Hessian matrix $$H_{k}$$ is determined by the following equation:

\begin{aligned} H_{k}:=\nabla^{2}_{zz}L_{r_{k}} \bigl(z^{k},s^{k} \bigr)+h_{k}E_{n+2m}, \end{aligned}

where $$E_{n+2m}$$ is an $$(n+2m)$$th-order identity matrix, $$h_{k}$$ is chosen as follows:

\begin{aligned} h_{k}:= \left \{ \textstyle\begin{array}{l@{\quad}l} 0, &\mbox{if } \lambda_{\min}>s_{\max};\\ -\lambda_{\min}+s_{\min}, &\mbox{if } \vert \lambda_{\min} \vert \leq s_{\max };\\ 2 \vert \lambda_{\min} \vert , &\mbox{otherwise}, \end{array}\displaystyle \right . \end{aligned}

where $$s_{\min}>0$$ and $$s_{\max}>0$$ are sufficiently small and sufficiently large, respectively, $$\lambda_{\min}$$ is the smallest eigenvalue of the following matrix:

\begin{aligned} \begin{aligned} \mathcal{B}_{k}&=\nabla^{2}_{zz}L_{r_{k}} \bigl(z^{k},s^{k} \bigr)-\sum _{i\in I^{\mathcal{C}}} \frac{s^{k}_{F,i}}{v_{i}-F_{i}(x^{k},y^{k})}\nabla_{z} \bigl(v_{i}-F_{i} \bigl(x^{k},y^{k} \bigr) \bigr)\nabla _{z} \bigl(v_{i}-F_{i} \bigl(x^{k},y^{k} \bigr) \bigr)^{\mathrm{T}} \\ &\quad -\sum_{i\in I^{\mathcal{C}}}\frac{s^{k}_{\Psi_{\varepsilon},i}}{\psi_{\varepsilon}(y^{k}_{i},v^{k}_{i})}\nabla _{z} \bigl(\psi_{\varepsilon} \bigl(y^{k}_{i},v^{k}_{i} \bigr) \bigr)\nabla_{z} \bigl(\psi_{\varepsilon} \bigl(y^{k}_{i},v^{k}_{i} \bigr) \bigr)^{\mathrm{T}} \\ &\quad -\sum_{i\in I_{g}}\frac{s^{k}_{g,i}}{g_{i}(x^{k},y^{k})} \nabla_{z} g_{i} \bigl(x^{k},y^{k} \bigr)\nabla_{z} g_{i} \bigl(x^{k},y^{k} \bigr)^{\mathrm{T}}.\end{aligned} \end{aligned}

In the numerical experiments, the parameters are chosen as follows:

\begin{aligned} &\alpha=0.49,\quad\quad \varepsilon=10^{-5},\qquad \beta=0.5,\qquad \theta=0.99,\qquad q=3,\qquad \tau=2, \\ &r=1,\qquad C=100, \qquad\vartheta=0.5, \qquad\nu=1, \qquad\gamma_{1}=\gamma_{2}= \gamma_{3}=\gamma _{4}=1,\\ & s_{\max}=10^{20},\qquad s_{\min}=10^{-5}. \end{aligned}

Algorithm A stops if one of the following termination criteria is satisfied:

(1) $$\rho(z^{k},\lambda^{k})\leq10^{-5}$$ and $$I^{\mathcal{C}}(y^{k},v^{k},\varepsilon_{k})=\emptyset$$.

(2) $$\|\widehat{dz}^{k}\|\leq10^{-5}$$, $$\max\{-\widehat{\lambda }^{k}_{g,i},i\in I^{g}\}<10^{-5}$$ and $$I^{\mathcal{C}}(y^{k},v^{k},\varepsilon _{k})=\emptyset$$.

The following test problems are selected from . Algorithm A can find their solutions within a small number of iterations. We compared Algorithm A with four algorithms, i.e., filtermpec, snopt, loqo, and knitro given in . The computational results are given in Table 1. The meaning of some notations in Table 1 are as follows:

Problem: The problem in .

dim::

the dimensions of the variables $$(x,y)$$;

itr::

the number of iterations;

$$f_{\mathrm{opt}}$$::

the optimal value given in ;

$$f^{*}$$::

the optimal value obtained by Algorithm A.

## Conclusions

In this paper, based on Fischer–Burmeister function and working set techniques, a primal–dual interior point QP-free algorithm for mathematical programs with complementarity constraints is presented. At each iteration, only two reduced linear equations with the same coefficient matrix are solved to yield the search direction. The use of the working set decreases the computational cost. Moreover, the uniformly positive definiteness on the Lagrangian Hessian estimate $${H_{k}}$$ is relaxed. Under some mild conditions, the proposed algorithm is globally convergent. The preliminary numerical results show that the proposed algorithm is effective.

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Not applicable.

## Funding

This research is supported by the National Natural Science Foundation (11561005) of China.

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Correspondence to Jian Ling Li.

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