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Abstract and Applied Analysis
Volume 2012 (2012), Article ID 591612, 12 pages
http://dx.doi.org/10.1155/2012/591612
Research Article

Homotopy Method for a General Multiobjective Programming Problem under Generalized Quasinormal Cone Condition

1Department of Mathematics, Beihua University, Jilin 132013, China
2Department of Mathematics, Jilin University, Changchun 130001, China
3Institute of Applied Mathematics, Changchun University of Technology, Changchun 130012, China

Received 14 February 2012; Accepted 24 July 2012

Academic Editor: Irena Rachůnková

Copyright © 2012 X. Zhao et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

A combined interior point homotopy continuation method is proposed for solving general multiobjective programming problem. We prove the existence and convergence of a smooth homotopy path from almost any interior initial interior point to a solution of the KKT system under some basic assumptions.

1. Introduction

In this paper, for any two vectors 𝑦=(𝑦1,𝑦2,,𝑦𝑛)𝑇 and 𝑧=(𝑧1,𝑧2,,𝑧𝑛)𝑇 in 𝑅𝑛, we use the following conventions: 𝑦=𝑧,i𝑦𝑖=𝑧𝑖,𝑖=1,2,,𝑛;𝑦<𝑧,i𝑦𝑖<𝑧𝑖,𝑖=1,2,,𝑛;𝑦𝑧,i𝑦𝑖𝑧𝑖,𝑖=1,2,,𝑛;𝑦𝑧,i𝑦𝑖𝑧𝑖,𝑦𝑧,𝑖=1,2,,𝑛.(1.1)

We consider the following multiobjective programming problem:

(MOP) min𝑓(𝑥),s.t.𝑔(𝑥)0,(𝑥)=0,(1.2) where 𝑓=(𝑓1,𝑓2,,𝑓𝑝)𝑇𝑅𝑛𝑅𝑝, 𝑔=(𝑔1,𝑔2,,𝑔𝑚)𝑇𝑅𝑛𝑅𝑚, and =(1,2,,𝑠)𝑇𝑅𝑛𝑅𝑠.

For 𝜆=(𝜆1,𝜆2,,𝜆𝑝)𝑇𝑅𝑝, let 𝑅𝑛+ and 𝑅𝑛++ denote the nonnegative and positive orthant of 𝑅𝑛. Respectively, let Ω={𝑥𝑅𝑛𝑔(𝑥)0,(𝑥)=0},Ω0={𝑥𝑅𝑛Ω𝑔(𝑥)<0,(𝑥)=0},𝜕Ω=Ω0,Λ+=𝜆𝑅𝑝+𝑝𝑖=1𝜆𝑖=1,Λ++=𝜆𝑅𝑝++𝑝𝑖=1𝜆𝑖,=1𝐼={1,2,,𝑚},𝐽={1,2,,𝑠},(1.3) and let 𝐵(𝑥)=𝑖{1,2,,𝑚}𝑔𝑖(𝑥)=0(1.4) denote the active index set at a given point.

MOP has important application in many practical fields like production planning, structural designing, portfolio selection, and so forth. Research on it can be traced back to Pareto [1], Von Neumann and Morgenstern [2], and Koopmans [3] or even earlier. Especially, more and more attention has been paid to the homotopy method since Kellogg et al. [4], Smale [5], and Chow et al. [6] published the remarkable papers. The homotopy method now becomes an important tool for numerically solving complementary, variational inequalities, convex multiobjective programming, and nonlinear mathematical programming et al. [712] as a globally convergent method.

Among many methods, the weighed sum method is popular and efficient. It transforms the MOP to a single-objective programming [13]: min𝜆𝑇𝑓(𝑥),s.t.𝑔(𝑥)0,𝑝𝑖=1𝜆𝑖=1,(1.5) where 𝜆 is the weight vector.

Recently, Song and Yao [14] generalize the combined homotopy interior point method to the general multi-objective programming problem under the so-called normal cone condition instead of the convexity condition about the feasible set. In that paper, they proved the existence of the homotopy path under the following assumptions:(A1)Ω0 is nonempty and bounded;(A2) for all 𝑥Ω, the vectors {𝑔𝑖(𝑥),𝑖𝐵(𝑥),𝑗(𝑥),𝑗𝐽} are linearly independent;(A3) for all 𝑥Ω, {𝑥+𝑖𝐵(𝑥)𝑢𝑖𝑔𝑖(𝑥)+𝑗𝐽𝑧𝑗𝑗(𝑥)𝑧𝑗𝑅,𝑢𝑖0,𝑗𝐽,𝑖𝐵(𝑥)}Ω={𝑥}.

In [14], the combined homotopy method was given as follows: 𝐻𝜔,𝜔(0)=,𝑡(1𝑡)(𝑓(𝑥)𝜆+𝑔(𝑥)𝑢)+(𝑥)𝑧+𝑡𝑥𝑥(0)(𝑥)𝑈×𝑔(𝑥)𝑡𝑈(0)𝑥×𝑔(0)(1𝑡)1𝑝𝑖=1𝜆𝑖𝑒𝑡𝜆𝜆(0)=0,(1.6) where 𝑥(0)Ω0, 𝑢(0)>0, 𝜆(0)>0, and 𝑝𝑖=1𝜆𝑖(0)=1. However, the solution simply yields 𝜆=𝜆(0) for all 𝑡(0,1]. That is, 𝜆 is fixed. In fact, from the last equation, we have 𝑝(1𝑡)+(𝑝𝑡𝑝𝑡)𝑝𝑖=1𝜆𝑖+𝑡=0. According to this, we know that 𝜆𝜆0 for all of 𝑡[0,1].

That is, these methods are all solving the single-objective programming problem.

In [15], they present the concept of “positive linear independent” and weaken the assumptions than the ones in [14]. But in order to extend their results to a broader class of nonconvex multi-objective programming problems, we construct a new homotopy equation under generalized quasinorm cone condition in this paper and 𝜆 is not fixed in the calculation process.

The paper is organized as follows. In Section 2, we recall some notations and preliminaries results. In Section 3, we construct a new combined homotopy mapping and prove the existence and convergence of a smooth homotopy path from almost any interior initial point to the KKT points of MOP under some assumptions.  In Section 4, numerical results are given,which show that the method is feasible and effective.

2. Some Definitions and Properties

Definition 2.1. Let 𝑈𝑅𝑛 be an open set, and let 𝜑𝑈𝑅𝑃 be a smooth mapping. If Range[𝜕𝜑(𝑥)/𝜕𝑥]=𝑅𝑝 for all 𝑥𝜑1(𝑦), then 𝑦𝑅𝑝 is a regular value and 𝑥𝑅𝑛 is a regular point.

Definition 2.2. Let 𝜂𝑖𝑅𝑛𝑅𝑛(𝑖=1,2,,𝑚) and 𝛽𝑗𝑅𝑛𝑅𝑛(𝑗=1,2,,𝑠). For any 𝑥Ω,{𝑔𝑖(𝑥),𝜂𝑖(𝑥)𝑖𝐵(𝑥)} is said to be positive linear independent with respect to 𝛽(𝑥), if 𝛽(𝑥)𝑧+𝑖𝐵(𝑥)𝑦𝑖𝑔𝑖(𝑥)+𝑢𝑖𝜂𝑖(𝑥)=0,𝑧𝑅𝑠,𝑦𝑖0,𝑢𝑖0(2.1) implies that 𝑧=0,𝑦𝑖=0,𝑢𝑖=0(𝑖𝐵(𝑥)),(2.2) where 𝛽(𝑥)=(𝛽1(𝑥),,𝛽𝑠(𝑥)).

Lemma 2.3 (parametric form of the Sard theorem on a smooth manifold; see [16]). Let 𝑄,𝑁,𝑃 be smooth manifolds of dimensions 𝑞,𝑚,𝑝. Respectively, let 𝜑𝑄×𝑁𝑃 be a 𝐶𝑟 map, where 𝑟>max{0,𝑚𝑝}. If 0𝑃 is a regular value of 𝜑, then for almost all 𝛼𝑄, 0 is a regular value of 𝜑(𝛼,).

Lemma 2.4 (inverse image theorem; see [17]). If 0 is a regular value of the mapping 𝜑𝛼()𝜑(𝛼,), then 𝜑𝛼1(0) consists of some smooth manifolds.

Lemma 2.5 (classification theorem of one-dimensional manifold; see [17]). A one-dimensional smooth manifold is diffeomorphic to a unit circle or a unit interval.

The following four basic assumptions are commonly used in this paper:(C1)Ω0 is nonempty and bounded;(C2) for any 𝑥Ω and 𝑡[0,1], there exists map 𝜂(𝑥) and 𝛽(𝑥), such that {𝑔𝑖(𝑥),𝜂𝑖(𝑥)𝑖𝐵(𝑥)} is positive linear independent with respect to (𝑥)+𝑡(𝛽(𝑥)(𝑥)); (C3) for any 𝑥𝜕Ω, 𝑥+𝑖𝐵(𝑥)𝑢𝑖𝜂𝑖(𝑥)+𝛽(𝑥)𝑧𝑧𝑅𝑠,𝑢𝑖0,𝑖𝐵(𝑥)Ω={𝑥}(2.3) (generalized quasinormal cone condition);(C4) for any 𝑥Ω, (𝑥)𝑇𝛽(𝑥) is nonsingular.

Remark 2.6. If Ω satisfies the assumptions (A1)–(A3), then it necessarily satisfies the assumptions (C1)–(C4).
In fact, if we choose 𝜂(𝑥)=𝑔(𝑥) and 𝛽(𝑥)=(𝑥), then it is easy to get the result. Clearly, if Ω satisfies the assumptions (C1)–(C4), then it does not necessarily satisfies the assumptions (A1)–(A3).

3. Main Results

Let 𝑥Ω be a KKT point of MOP; our aim is to find (𝜆,𝑢,𝑧)𝑅+𝑝+𝑚×𝑅𝑠, such that 𝑓(𝑥)𝜆+𝑔(𝑥)𝑢+(𝑥)𝑧=0,(3.1a)𝑈𝑔(𝑥)=0,(3.1b)1𝑝𝑖=1𝜆𝑖=0,(3.1c)where 𝑓(𝑥)=(𝑓1(𝑥),,𝑓𝑝(𝑥))𝑅𝑛×𝑝, 𝑔(𝑥)=(𝑔1(𝑥),,𝑔𝑚(𝑥))𝑅𝑛×𝑚, (𝑥)=(1(𝑥),,𝑚(𝑥))𝑅𝑛×𝑠.

Meanwhile, the KKT system of MOP is (3.1a)–(3.1c).

For a convex multi-objective programming problem, the solution of the MOP can be obtained from the KKT system. And for a nonconvex multi-objective programming problem, it is significant that we can obtain a solution of the KKT system.

To solve the KKT system (3.1a)–(3.1c), we construct a homotopy equation as follows: 𝐻𝜔,𝜔(0)=,𝑡(1𝑡)𝑓(𝑥)𝜆+𝑔(𝑥)𝑢+𝑡𝜂(𝑥)𝑢2+[](𝑥)+𝑡(𝛽(𝑥)(𝑥))𝑧+𝑡𝑥𝑥(0)(𝑥)𝑈×𝑔(𝑥)𝑡𝑈(0)𝑥×𝑔(0)(1𝑡)1𝑝𝑖=1𝜆𝑖𝜆𝑒𝑡5/2𝜆(0)5/2=0,(3.2) where 𝜔(0)=(𝑥(0),𝜆(0),𝑢(0),𝑧(0))Ω0×Λ++×𝑅𝑚++×{0}, 𝜔=(𝑥,𝜆,𝑢,𝑧)Ω×𝑅𝑝+𝑚+𝑠, 𝑢2=(𝑢21,𝑢22,𝑢2𝑚)𝑇𝑅𝑚, 𝜆5/2=(𝜆15/2,𝜆25/2,,𝜆𝑝5/2)𝑇𝑅𝑝, 𝑈=diag(𝑢), 𝑒=(1,1,,1)𝑇𝑅𝑝, and 𝑡[0,1].

As 𝑡=1, the homotopy equation (3.2) becomes 𝛽(𝑥)𝑧+𝑥𝑥(0)=0,(3.3a)(𝑥)=0,(3.3b)𝑈𝑔(𝑥)𝑈(0)𝑔𝑥(0)𝜆=0,(3.3c)5/2=𝜆(0)5/2.(3.3d)

By the assumption (C3), we get 𝑧=0, 𝑥=𝑥(0). Since 𝑔(𝑥(0))<0 and 𝑥=𝑥(0), (3.3c) implies that 𝑢=𝑢(0). Equation (3.3d) shows that 𝜆=𝜆(0). That is, 𝐻(𝜔,𝜔(0),1)=0 with respect to 𝜔 has only one solution 𝜔=𝜔(0)=(𝑥(0),𝜆(0),𝑢(0),0).

As 𝑡=0, 𝐻(𝜔,𝜔(0),𝑡)=0 turns to the KKT system (3.1a)–(3.1c).

For a given 𝜔(0), rewrite 𝐻(𝜔,𝜔(0),𝑡) as 𝐻𝜔(0)(𝜔,𝑡). The zero-point set of 𝐻𝜔(0) is 𝐻𝜔1(0)=(𝜔,𝑡)Ω×𝑅𝑝+𝑚++×𝑅𝑠]×(0,1𝐻𝜔,𝜔(0).,𝑡=0(3.4)

Theorem 3.1. Suppose 𝑓,𝑔, and are three times continuous differentiable functions. In addition, let the assumptions (C1)-(C2) hold and 𝜂𝑖,𝛽𝑗 twice times continuously differentiable functions. Then for almost all initial points 𝜔(0)Ω0×Λ++×𝑅𝑚++×{0}, 0 is a regular value of 𝐻𝜔(0) and 𝐻𝜔1(0) consists of some smooth curves. Among them, a smooth curve, say Γ𝜔(0), is starting from (𝜔(0),1).

Proof. Denote the Jacobi matrix of 𝐻(𝜔,𝜔(0),𝑡) by 𝐷𝐻(𝜔,𝜔(0),𝑡). For any 𝜔(0)Ω0×Λ++×𝑅𝑚++×{0} and 𝑡[0,1], we have 𝐷𝐻(𝜔,𝜔(0),𝑡)=(𝜕𝐻/𝜕𝜔,𝜕𝐻/𝜕𝜔(0),𝜕𝐻/𝜕𝑡). Now, we consider the submatrix of 𝐷𝐻(𝜔,𝜔(0),𝑡).
For any (𝑥,𝑥(0),𝜆(0),𝑢(0))𝑅𝑛×Ω0×Λ++×𝑅𝑚++, 𝜕𝐻𝜕𝑥,𝑥(0),𝜆(0),𝑢(0)=𝑄𝑡𝐼𝑛00(𝑥)𝑇000𝑈𝑔(𝑥)𝑇𝑡𝑈(0)𝑥𝑔(0)𝑇𝑔𝑥0𝑡diag(0)5002𝑡𝜆(0)3/2𝐼𝑝0,(3.5) where 𝑄=(1𝑡)(𝑝𝑖=1𝜆𝑖2𝑓𝑖(𝑥)+𝑚𝑗=1𝑢𝑗2𝑔𝑗(𝑥)+𝑡𝑚𝑗=1𝑢2𝑗𝜂𝑗(𝑥))+[2(𝑥)+𝑡(𝛽(𝑥)2(𝑥))]𝑧+𝑡𝐼𝑛.
We obtain that rank𝜕𝐻𝜕𝑥,𝑥(0),𝜆(0),𝑢(0)=𝑛+𝑝+𝑚+𝑠.(3.6)
That is, 0 is a regular value of 𝐻. By parametric form of the Sard theorem, for almost all 𝜔(0)Ω0×Λ++×𝑅𝑚++×{0}, 0 is a regular value of 𝐻𝜔(0). By inverse image theorem, 𝐻𝜔1(0)(0) consists of some smooth curves. Since 𝐻(𝜔(0),𝜔(0),1)=0, there must be a smooth curve, denoted by Γ𝜔(0), that starts from (𝜔(0),1).

Theorem 3.2. Let assumptions (C1)-(C2) hold. For a given 𝜔(0)=(𝑥(0),𝜆(0),𝑢(0),𝑧(0))Ω0×Λ++×𝑅𝑚++×{0}, if 0 is a regular value of 𝐻𝜔(0), then the projection of the smooth curve Γ𝜔(0) on the component 𝜆 is bounded.

Proof. Suppose that the conclusion does not hold. Since (0,1] is bounded, there exists a sequence {(𝜔(𝑘),𝑡𝑘)}Γ𝜔(0), such that 𝑡𝑘𝑡,𝜆(𝑘).(3.7) From the last equality of (3.2), we have 1𝑡𝑘1𝑡𝑘1𝑡𝑘1𝑡𝑘𝜆1(𝑘)+1𝑡𝑘𝑖1𝜆𝑖(𝑘)+𝑡𝑘𝜆1(𝑘)5/21𝑡𝑘𝜆2(𝑘)+1𝑡𝑘𝑖2𝜆𝑖(𝑘)+𝑡𝑘𝜆2(𝑘)5/21𝑡𝑘𝜆𝑝(𝑘)+1𝑡𝑘𝑖𝑝𝜆𝑖(𝑘)+𝑡𝑘𝜆𝑝(𝑘)5/2𝑡𝑘𝜆1(0)5/2𝜆2(0)5/2𝜆𝑝(0)5/2=0.(3.8) If we assume 𝜆(𝑘)+(𝑘), this hypothesis implies 𝑖{1,2,,𝑝}lim𝑘𝜆𝑖(𝑘)=Φ.(3.9)
Since 𝑡𝑘𝑡, 𝜆(𝑘)>0, it follows that the second part in the left-hand side of some equations in (3.8) tends to infinity as 𝑘. But the other two parts are bounded. This is impossible. Thus, the component 𝜆 is bounded.

Theorem 3.3. Let 𝑓,𝑔, and be three times continuous differentiable functions. In addition, let the assumptions (C1)–(C4) hold and 𝜂𝑖,𝛽𝑗 twice times continuously differentiable functions. Then, for almost all of 𝜔(0)Ω0×Λ++×𝑅𝑚++×{0}, 𝐻𝜔1(0)(0) contains a smooth curve Γ𝜔(0)Ω×𝑅𝑝+×𝑅𝑚+×𝑅𝑠×(0,1], which starts from (𝜔(0),1). As 𝑡0, the limit set 𝑇×{0}Ω×Λ+×𝑅𝑚+×𝑅𝑠×{0} of Γ𝜔(0) is nonempty and every point in 𝑇 is a solution of the KKT system (3.1a)–(3.1c).

Proof. From the homotopy equation (3.2), it is easy to see that Γ𝜔(0)Ω×𝑅𝑝+×𝑅𝑚+×𝑅𝑠×(0,1]. By Theorem 3.1, for almost all 𝜔(0)Ω0×Λ++×𝑅𝑚++×{0}, 0 is a regular value of 𝐻𝜔(0) and 𝐻𝜔1(0) contains a smooth curve Γ𝜔(0) starting from (𝜔(0),1). By the classification theorem of one-dimensional smooth manifolds, Γ𝜔(0) is diffeomorphic to a unit circle or the unit interval (0,1].
Noticing that ||||𝜕𝐻𝜔(0)𝜔(0),1||||=|||||||||||||||𝐼𝜕𝜔𝑛𝑥00𝛽(0)𝑥(0)𝑇𝑈000(0)𝑥𝑔(0)𝑇𝑔𝑥0diag(0)0502𝜆(0)3/2𝐼𝑝|||||||||||||||=00(1)𝑠||𝑔𝑥diag(0)|||||52𝜆(0)3/2𝐼𝑝||||||𝑥(0)𝑇𝛽𝑥(0)|||.(3.10)
Because 𝑔(𝑥(0))<0, 𝜆(0)Λ++ and by the assumption (C4), we know that [𝜕𝐻𝜔(0)(𝜔(0),1)/𝜕𝜔] is nonsingular. Therefore, the smooth curve Γ𝜔(0) starts from (𝜔(0),1) diffeomorphic to (0,1].
Let (𝜔,𝑡) be a limit point of Γ𝜔(0); only three cases are possible:(a)(𝜔,𝑡)Ω×Λ+×𝑅𝑚+×𝑅𝑠×{0}; (b)(𝜔,𝑡)𝜕(Ω0×𝑅+𝑝+𝑚)×𝑅𝑠×(0,1];(c)(𝜔,𝑡)Ω×𝑅+𝑝+𝑚×𝑅𝑠×{1}.
Because 𝐻(𝜔(0),𝜔(0),1)=0 has a unique solution (𝜔(0),1), the case (c) will not happen.
In case (b), because Ω and (0,1] are bounded sets and by the assumption (C2), for any 𝑥Ω and 𝑡[0,1], there exists map 𝜂(𝑥) and 𝛽(𝑥) such that, {𝑔𝑖(𝑥),𝜂𝑖(𝑥)𝑖𝐵(𝑥)} is positive linear independent with respect to (𝑥)+𝑡(𝛽(𝑥)(𝑥)). From the first equation of (3.2), we get that the component 𝑧 of Γ𝜔(0) is bounded.
If case (b) holds, then there exists a sequence {(𝜔(𝑘),𝑡𝑘)}Γ𝜔(0), such that 𝜔(𝑘),𝑡𝑘.(3.11)
Because Ω and (0,1] are bounded, there exists a subsequence (denoted also by {(𝜔(𝑘),𝑡𝑘)}Γ𝜔(0)) such that 𝑥(𝑘)𝑥,𝜆(𝑘)𝜆,𝑢(𝑘),𝑧(𝑘)𝑧,𝑡𝑘𝑡,as𝑘.(3.12)
By the third equation of (3.2), we have 𝑔𝑥(𝑘)=𝑡𝑘𝑈(𝑘)1𝑈(0)𝑔𝑥(0).(3.13) Hence, the active index set 𝐵(𝑥) is nonempty.
From the first equation of (3.2), it follows that 1𝑡𝑘𝑥𝑓(𝑘)𝜆(𝑘)𝑥+𝑔(𝑘)𝑢(𝑘)+𝑡𝑘𝜂𝑥(𝑘)𝑢(𝑘)2+𝑥(𝑘)+𝑡𝑘𝛽𝑥(𝑘)𝑥(𝑘)𝑧(𝑘)+𝑡𝑘𝑥(𝑘)𝑥(0)=0.(3.14)
(i) When 𝑡=1, rewrite (3.14) as 𝑗𝐵(𝑥)1𝑡𝑘𝑔𝑗𝑥(𝑘)𝑢𝑗(𝑘)+𝑡𝑘𝜂𝑗𝑥(𝑘)𝑢𝑗(𝑘)2+𝑡𝑘𝛽𝑥(𝑘)𝑧(𝑘)+1𝑡𝑘𝑥(𝑘)𝑧(𝑘)+𝑥(𝑘)𝑥(0)=1𝑡𝑘𝑗𝐵(𝑥)𝑔𝑗𝑥(𝑘)𝑢𝑗(𝑘)+𝑡𝑘𝜂𝑗𝑥(𝑘)𝑢𝑗(𝑘)2𝑥+𝑓(𝑘)𝜆(𝑘)𝑥(𝑘)𝑥(0).(3.15)
From the fact that 𝑢𝑗(𝑘) is bounded for 𝑗𝐵(𝑥) and by the assumptions (C1) and (C2), when 𝑘, we observe that 𝑗𝐵(𝑥)𝑔𝑗𝑥lim𝑘1𝑡𝑘𝑢𝑗(𝑘)+𝜂𝑗𝑥lim𝑘1𝑡𝑘𝑡𝑘𝑢𝑗(𝑘)2𝛽𝑥=𝑧+𝑥𝑥(0).(3.16)
It is easy to see that the right-hand side of the equation is bounded. By the assumption (C2), we have lim𝑘1𝑡𝑘𝑢𝑗(𝑘)=0,lim𝑘1𝑡𝑘𝑡𝑘𝑢𝑗(𝑘)2=𝛼𝑗𝑥,𝑗𝐵,(3.17) where 𝛼𝑗0.
Then, we have 𝑥(0)=𝑥𝑥+𝛽𝑧+𝑗𝐵(𝑥)𝛼𝑗𝜂𝑗𝑥,(3.18) which contradicts the assumption (C3).
(ii) When 𝑡[0,1), rewrite (3.14) as 𝑗𝐵(𝑥)1𝑡𝑘𝑔𝑗𝑥(𝑘)𝑢𝑗(𝑘)+𝑡𝑘𝜂𝑗𝑥(𝑘)𝑢𝑗(𝑘)2=1𝑡𝑘𝑗𝐵(𝑥)𝑔𝑗𝑥(𝑘)𝑢𝑗(𝑘)+𝑡𝑘𝜂𝑗𝑥(𝑘)𝑢𝑗(𝑘)2𝑥+𝑓(𝑘)𝜆(𝑘)𝑡𝑘𝑥(𝑘)𝑥(0)𝑥(𝑘)+𝑡𝑘𝛽𝑥(𝑘)𝑥(𝑘)𝑧(𝑘).(3.19)
We know that, since Ω and 𝑢𝑗(𝑘), 𝑗𝐵(𝑥) are bounded as 𝑘, the right-hand side of (3.19) is bounded. But by the assumption (C2), if 𝑢𝑗(𝑘)(𝑗𝐵(𝑥)), then the left-hand side of (3.19) is infinite; this is a contradiction.
As a conclusion, (a) is the only possible case, and 𝜔 is a solution of the KKT system.
Let 𝑠 be the arc-length of Γ𝜔(0). We can parameterize Γ𝜔(0) with respect to 𝑠.

Theorem 3.4. The homotopy path Γ𝜔(0) is determined by the following initial-value problem for the ordinary differential equation 𝐷𝐻𝜔(0)𝜔𝜇(𝜔(𝑠),𝑡(𝑠))=0,𝜔(0)=𝜔(0),𝑡(0)=1.(3.20) The component 𝜔 of the solution point (𝜔(𝑠),𝑡(𝑠)),   for 𝑡(𝑠)=0, is the solution of the KKT system.

4. Algorithm and Numerical Example

Algorithm 4.1. MOP’s Euler-Newton method.

Step 1. Give an initial point (𝜔(0),1)Ω0×𝑅𝑝+𝑚++×{0}×{1}, an initial step length 0>0, and three small positive numbers 𝜀1,𝜀2,𝜀3. Let 𝑘=0.

Step 2. Compute the direction 𝛾(𝑘) of the predictor step.(a)Compute a unit tangent vector 𝜉(𝑘)𝑅𝑛+𝑝+𝑚+𝑠+1 of Γ𝜔0 at (𝜔(0),𝑡𝑘).(b)Determine the direction 𝛾(𝑘) of the predictor step.If the sign of the determinant |||𝐷𝐻𝜔(0)(𝜔(𝑘),𝑡𝑘)𝜉𝑇(𝑘)||| is (1)𝑝+𝑚+𝑠+𝑝𝑚+𝑝𝑠+𝑚𝑠+1, take 𝛾(𝑘)=𝜉(𝑘).If the sign of the determinant |||𝐷𝐻𝜔(0)(𝜔(𝑘),𝑡𝑘)𝜉𝑇(𝑘)||| is (1)𝑝+𝑚+𝑠+𝑝𝑚+𝑝𝑠+𝑚𝑠, take 𝛾(𝑘)=𝜉(𝑘).

Step 3. Compute a corrector point (𝜔(𝑘+1),𝑡𝑘+1): 𝜔(𝑘),𝑡𝑘=𝜔(𝑘),𝑡𝑘+𝑘𝛾(𝑘),𝜔(𝑘+1),𝑡𝑘+1=𝜔(𝑘),𝑡𝑘𝐷𝐻𝜔(0)𝜔(𝑘),𝑡𝑘+𝐻𝜔(0)𝜔(𝑘),𝑡𝑘,(4.1) where 𝐷𝐻𝜔(0)(𝜔,𝑡)+=𝐷𝐻𝜔(0)(𝜔,𝑡)𝑇𝐷𝐻𝜔(0)(𝜔,𝑡)𝐷𝐻𝜔(0)(𝜔,𝑡)𝑇1(4.2) is the Moore-Penrose inverse of 𝐷𝐻𝜔(0)(𝜔,𝑡).If 𝐻𝜔(0)(𝜔(𝑘+1),𝑡𝑘+1)𝜀1, let 𝑘+1=min{0,2𝑘}, and go to Step 4.If 𝐻𝜔(0)(𝜔(𝑘+1),𝑡𝑘+1)(𝜀1,𝜀2), let 𝑘+1=𝑘, and go to Step 4.If 𝐻𝜔(0)(𝜔(𝑘+1),𝑡𝑘+1)𝜀2, let 𝑘+1=max{(1/2)0,(1/2)𝑘}, and go to Step 3.

Step 4. If 𝜔(𝑘+1)Ω×𝑅+𝑝+𝑚×𝑅𝑠 and 𝑡𝑘+1>𝜀3, let 𝑘=𝑘+1 and go to Step 2.
If 𝜔(𝑘+1)Ω×𝑅+𝑝+𝑚×𝑅𝑠 and 𝑡𝑘+1<𝜀3, let 𝑘=𝑘(𝑡𝑘/(𝑡𝑘𝑡𝑘+1)), go to Step 3, and recompute (𝜔(𝑘+1),𝑡𝑘+1) for the initial point (𝜔(𝑘),𝑡𝑘).
If 𝜔(𝑘+1)Ω×𝑅+𝑝+𝑚×𝑅𝑠, let 𝑘=(𝑘/2)(𝑡𝑘/(𝑡𝑘𝑡𝑘+1)), go to Step 3, and recompute (𝜔(𝑘+1),𝑡𝑘+1) for the initial point (𝜔(𝑘),𝑡𝑘).
If 𝜔(𝑘+1)Ω×𝑅+𝑝+𝑚×𝑅𝑠, and |𝑡𝑘+1|𝜀3, then stop.

Example 4.2 (see [9]). Consider𝑥min𝑓=min21+𝑥22+𝑥23+𝑥24+𝑥25,3𝑥1+2𝑥213𝑥3𝑥+0.014𝑥53;s.t.𝑔1(𝑥)=𝑥21+𝑥22+𝑥23+𝑥24100;1(𝑥)=4𝑥12𝑥2+0.8𝑥3+0.6𝑥4+0.5𝑥25=0;2(𝑥)=𝑥1+2𝑥2𝑥30.5𝑥4+𝑥52=0.(4.3)
The results are shown in Table 1.

tab1
Table 1: Results for Example 4.2.

Example 4.3. Consider𝑥min𝑓=min21+𝑥22,𝑥1;s.t.𝑔1(𝑥)=𝑥2𝑔60;2(𝑥)=𝑥260;1(𝑥)=𝑥1𝑥223=0.(4.4)
(1) Since (𝑥)=(1,2𝑥2)𝑇, it is easy to see that the assumption (A3) in [14] and the assumption (C3) in [15] are not satisfied at most points in feasible set. Hence, we introduce the functions 𝜂𝑖(𝑥)(𝑖=1,2) and 𝛽(𝑥).
(2) Let 𝜂𝑖(𝑥)=𝑔𝑖(𝑥)(𝑖=1,2) and 𝛽(𝑥)=(10,0)𝑇. It is easily verified that the feasible set satisfies the assumptions (C1)–(C4). The results are shown in Table 2.

tab2
Table 2: Results for Example 4.3.

Acknowledgments

This work was supported by National Natural Science Foundation of China (Grant no. 11171003), Key Project of Chinese Ministry of Education (Grant no. 211039), and the Jilin Province Natural Science Foundation (Grant no. 20101597).

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