Homotopy Interior-Point Method for a General Multiobjective Programming Problem

Abstract

We present a combined homotopy interior-point method for a general multiobjective programming problem. For solving the KKT points of the multiobjective programming problem, the homotopy equation is constructed. We prove the existence and convergence of a smooth homotopy path from almost any initial interior point to a solution of the KKT system under some basic assumptions.

1. Introduction

We consider the following multiobjective programming problem:s.mtℎ.i(𝑔n𝑥(𝑓)𝑥=)(𝑥≤0),,0,(MOP) where 𝑓=(𝑓₁,𝑓₂,…,𝑓_𝑝)^𝑇∶𝑅^𝑛→𝑅^𝑝, 𝑔=(𝑔₁,𝑔₂,…,𝑔_𝑚)^𝑇∶𝑅^𝑛→𝑅^𝑚 and ℎ=(ℎ₁,ℎ₂,…,ℎ_𝑠)^𝑇∶𝑅^𝑛→𝑅^𝑠.

Since Kellogg et al. [1] and Smale [2] have published their remarkable papers, more and more attention has been paid to the homotopy method. As a globally convergent method, the homotopy method now becomes an important tool for numerically solving complementary problem and nonlinear mathematical programming problem [3–5].

Among most interior-point methods for solving mathematical programming, one of the main ideas is numerically tracing the center path generated by the optimal solution set of the logarithmic barrier function. Usually, the strict convexity of the logarithmic barrier function or boundedness of the solution set is needed [6, 7]. Lin et al. [8] presented a new interior-point method, combined homotopy interior-point method (CHIP method), for convex nonlinear programming. Subsequently, Lin et al. [9] generalized the CHIP method to convex multi-objective programming with only inequality constraints. Recently, Song and Yao [10] generalized the combined homotopy interior-point method to the general multiobjective programming problem under the so-called normal cone condition. In that paper, they proved the existence of the homotopy path under the following assumptions:(A₁)Ω₀ is nonempty and bounded;(A₂)for all 𝑥∈Ω, the vectors {∇𝑔_𝑖(𝑥),𝑖∈𝐵(𝑥),∇ℎ_𝑗(𝑥),𝑗∈𝐽} are linearly independent;(A₃)for all 𝑥∈Ω, {𝑥+∑_{𝑖∈𝐵(𝑥)}𝑢_𝑖∇𝑔_𝑖(𝑥)+∑_𝑗∈𝐽𝑧_𝑗∇ℎ_𝑗(𝑥)∶𝑧=(𝑧_𝑗)∈𝑅^𝑠,𝑢_𝑖≥0,𝑖∈𝐵(𝑥)}⋂Ω={𝑥},

where  Ω𝑔={𝑥∈𝑅^𝑛∣𝑔(𝑥)≦0,ℎ(𝑥)=0},Ω₀={𝑥∈𝑅^𝑛∣𝑔(𝑥)<0,ℎ(𝑥)=0},and𝐵(𝑥)={𝑖∈{1,2,…,𝑚}∣_𝑖(𝑥)=0}.

In [10], the combined homotopy method was given as follows:𝐻𝜔,𝜔⁽⁰⁾,𝑡=⎡⎢⎢⎢⎢⎢⎣(1−𝑡)(∇𝑓(𝑥)𝜆+∇𝑔(𝑥)𝑢)+∇ℎ(𝑥)𝑧+𝑡𝑥−𝑥⁽⁰⁾𝑈×𝑔(𝑥)−ℎ𝑡𝑈(𝑥)₍₀₎(1−𝑡)1−_𝑝×𝑔𝑥⁽⁰⁾_𝑖=1𝜆_𝑖𝑒−𝑡𝜆−𝜆⁽⁰⁾⎤⎥⎥⎥⎥⎥⎦=0,(1.1) where 𝑥⁽⁰⁾∈Ω₀,𝑢⁽⁰⁾>0,𝜆⁽⁰⁾>0, and ∑^𝑝𝑖=1𝜆_𝑖(0)=1. However, the solution simply yields 𝜆=𝜆⁽⁰⁾ for all 𝑡∈(0,1]. In fact, from the last equation, we have 𝑝(1−𝑡)+(𝑝𝑡−𝑝−𝑡)∑^𝑝𝑖=1𝜆_𝑖+𝑡=0. According to this, we know that 𝜆≡𝜆⁰ for all of 𝑡∈[0,1]. That is, the method given in [10] just solves the single-objective programming problem.

The purpose of this paper is to generalize the combined homotopy interior-point method for a general multiobjective programming problem (MOP) under quasinorm cone condition that weakens the assumptions more than the ones in [10] and constructs a new homotopy equation which is much different and simpler the that one given in [9].

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 a KKT point of MOP under some assumptions. In Section 4, numerical results are given.

2. Some Definitions and Properties

Let 𝑅^𝑛+ and 𝑅^𝑛++ denote the nonnegative and positive orthants of 𝑅^𝑛, respectively. For any two vectors 𝑦=(𝑦₁,𝑦₂,…,𝑦_𝑛)^𝑇 and 𝑧=(𝑧₁,𝑧₂,…,𝑧_n)^𝑇 in 𝑅^𝑛, we use the following conventions:𝑦𝑦<=𝑧𝑧,,iiffff𝑦𝑦_𝑖𝑖=𝑧_𝑖,𝑖=1,2,…,𝑛,𝑦≦𝑧,iff𝑦_𝑖<𝑧_𝑖,𝑖=1,2,…,𝑛,𝑦≤𝑧,iff𝑦_𝑖≤𝑧_𝑖,𝑖=1,2,…,𝑛,≤𝑧_𝑖,𝑦≠𝑧,𝑖=1,2,…,𝑛.(2.1) Suppose that 𝑓=(𝑓₁,𝑓₂,…,𝑓_𝑝)^𝑇∶𝑅^𝑛→𝑅^𝑝, 𝑔=(𝑔₁,𝑔₂,…,𝑔_𝑚)^𝑇∶𝑅^𝑛→𝑅^𝑚, and ℎ=(ℎ₁,ℎ₂,…,ℎ_𝑠)^𝑇∶𝑅^𝑛→𝑅^𝑠 are three times continuously differentiable functions. LetΩ={𝑥∈𝑅^𝑛∣𝑔(𝑥)≦0,ℎ(𝑥)=0},Ω₀={𝑥∈𝑅^𝑛∣𝑔(𝑥)<0,ℎ(𝑥)=0},𝜕Ω=ΩΩ₀Λ₊,=𝜆∈𝑅^𝑝+∣^𝑝_𝑖=1𝜆_𝑖=1,Λ⁺⁺=𝜆∈𝑅^𝑝++∣^𝑝_𝑖=1𝐼={1,2,…,𝑚},𝐽={1,2,…,𝑠},𝜆_𝑖=1,(2.2) and let𝐵(𝑥)=𝑖∈{1,2,…,𝑚}𝑔_𝑖(𝑥)=0(2.3) denote the active index set at a given point.

Definition 2.1. A point 𝑥∈Ω is said to be an efficient solution to multiobjective programming problem (MOP), if there is no 𝑦∈Ω such that 𝑓(𝑦)≤𝑓(𝑥) holds.

Definition 2.2. Let 𝑈⊂𝑅^𝑛 be an open set, and let 𝜑∶𝑈→𝑅^𝑃 be a smooth mapping. If Range [𝜕𝜑(𝑥)/𝜕𝑥]=𝑅^𝑝 for all 𝑥∈𝜑⁻¹(𝑦), then 𝑦∈𝑅^𝑝 is a regular value and 𝑥∈𝑅^𝑛 is a regular point.

Definition 2.3. Let 𝜂_𝑖∶𝑅^𝑛→𝑅^𝑛(𝑖=1,2,…,𝑚). For any 𝑥∈Ω,{𝜂_𝑖(𝑥)∶𝑖∈𝐵(𝑥)} is said to be positive linear independent with respect to ∇𝑔(𝑥)𝑛 and ∇ℎ(𝑥) if ∇ℎ(𝑥)𝑧+_{𝑖∈𝐵(𝑥)}𝑦_𝑖∇𝑔_𝑖(𝑥)+𝑢_𝑖𝜂_𝑖(𝑥)=0,𝑧∈𝑅^𝑠,𝑦_𝑖≥0,𝑢_𝑖≥0(𝑖∈𝐵(𝑥))(2.4) implies that 𝑧=0,𝑦_𝑖=0,𝑢_𝑖=0(𝑖∈𝐵(𝑥)).(2.5)

Lemma 2.4 (parametric form of the Sard Theorem on a smooth manifold; see [11]). Let 𝑄,𝑁and 𝑃 be smooth manifolds of dimensions 𝑞,𝑚 and 𝑝. Respectively, let 𝜑∶𝑄×𝑁→𝑃 be a 𝐶^r map, where 𝑟>max{0,𝑚−𝑝}. If 0∈𝑃 is a regular value of 𝜑, then for almost all 𝛼∈𝑄, 0 is a regular value of 𝜑(𝛼,⋅).

Lemma 2.5 (inverse image theorem; see [12]). If 0 is a regular value of the mapping 𝜑_𝛼(⋅)≜𝜑(𝛼,⋅), then 𝜑^𝛼−1(0) consists of some smooth manifolds.

Lemma 2.6 (classification theorem of one-dimensional manifold; see [12]). A one-dimensional smooth manifold is diffeomorphic to a unit circle or a unit interval.
The following three basic assumptions are commonly used in this paper: (C₁)Ω₀ is nonempty and bounded;(C₂)for any 𝑥∈Ω, there exists a positive linear independent map 𝜂(𝑥) with respect to ∇𝑔(𝑥) and ∇ℎ(𝑥) such that, {𝜂_i(𝑥)∶𝑖∈𝐵(𝑥)} is positive linear independent with respect to ∇𝑔(𝑥) and ∇ℎ(𝑥);(C₃)for any 𝑥∈𝜕Ω, there exists a positive linear independent map 𝜂(𝑥) with respect to ∇𝑔(𝑥) and ∇ℎ(𝑥), such that𝑥+_{𝑖∈𝐵(𝑥)}𝑢_i𝜂_i(𝑥)+_𝑗∈𝐽𝑧_j∇ℎ_j(𝑥)∶𝑧_j∈𝑅,𝑗∈{𝐽},𝑢_i≥0,𝑖∈𝐵(𝑥)Ω={𝑥}(2.6) (quasinormal cone condition).

Remark 2.7. If Ω satisfies assumptions (A₁)–(A₃), then it necessarily satisfies assumptions (C₁)–(C₃).
In fact, if we choose 𝜂(𝑥)=∇𝑔(𝑥), then it is easy to get the result. Clearly, if Ω satisfies assumptions (C₁)–(C₃), then it does not necessarily satisfies assumptions (A₁)–(A₃).

3. Main Results

Let 𝑥∈Ω. We say that 𝑥 is a KKT point of (MOP) if there exists (𝜆,𝑢,𝑧)∈𝑅^+𝑝+𝑚×𝑅^𝑠, such that∇𝑓(𝑥)𝜆+1∇𝑈−𝑔𝑔((𝑥_𝑝𝑥))𝑢=+0∇,ℎ(𝑥)𝑧=0,((33..11ba))_𝑖=1𝜆_𝑖=0,(3.1c)where ∇𝑓(𝑥)=(∇𝑓₁(𝑥),…,∇𝑓_𝑝(𝑥))∈𝑅^𝑛×𝑝, ∇𝑔(𝑥)=(∇𝑔₁(𝑥),…,∇𝑔_𝑚(𝑥))∈𝑅^𝑛×𝑚, ∇ℎ(𝑥)=(∇ℎ₁(𝑥),…,∇ℎ_𝑚(𝑥))∈𝑅^𝑛×𝑠.

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

For a convex multiobjective programming problem, the solution of the MOP can be obtained from the KKT system, and for a non-convex multi-objective programming problem, it is significant when we get a solution of the KKT system.

To solve the KKT system (3.1a)–(3.1c),  we construct a homotopy equation as follows:𝐻𝜔,𝜔⁽⁰⁾,𝑡=⎡⎢⎢⎢⎢⎢⎣(1−𝑡)∇𝑓(𝑥)𝜆+∇𝑔(𝑥)𝑢+𝑡𝜂(𝑥)𝑢²𝑈×𝑔(𝑥)−ℎ𝑡𝑈(𝑥)₍₀₎+∇ℎ(𝑥)𝑧+𝑡𝑥−𝑥⁽⁰⁾(1−𝑡)1−_𝑝×𝑔𝑥⁽⁰⁾_𝑖=1𝜆_𝑖𝑒−𝑡𝜆^5/2−𝜆⁽⁰⁾^5/2⎤⎥⎥⎥⎥⎥⎦=0,(3.2) where 𝜔⁽⁰⁾=(𝑥⁽⁰⁾,𝜆⁽⁰⁾,𝑢⁽⁰⁾,𝑧⁽⁰⁾)∈Ω₀×Λ⁺⁺×𝑅^𝑚++×{0}, 𝜔=(𝑥,𝜆,𝑢,𝑧)∈𝑅^𝑛×𝑅^𝑝×𝑅^𝑚×𝑅^𝑠, 𝑢²=(𝑢²¹,𝑢²²…,𝑢^2𝑚)^𝑇∈𝑅^𝑚, 𝜆^5/2=(𝜆^15/2,𝜆^25/2,…,𝜆^𝑝5/2)^𝑇∈𝑅^𝑝, 𝑈=diag(𝑢), 𝑒=(1,1,…,1)^𝑇∈𝑅^𝑝 and 𝑡∈[0,1].

When 𝑡=1, the homotopy equation (3.2) becomes∇𝑈ℎ𝑔((𝑥𝑥))𝑧−ℎ+𝑈(𝑥𝑥)₍₀₎=−0𝑥,⁽⁰⁾=0,((33..33ba))𝜆𝑔𝑥⁽⁰⁾=0,(3.3c)_5/2=𝜆⁽⁰⁾^5/2.(3.3d)

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

When 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 𝐻_𝜔(0)(𝜔,𝑡). The zero-point set of 𝐻_𝜔(0) is𝐻^𝜔−1₍₀₎=(𝜔,𝑡)∈Ω×𝑅^𝑝+𝑚++×𝑅^𝑠×(0,1]∶𝐻𝜔,𝜔⁽⁰⁾,𝑡=0.(3.4)

Theorem 3.1. Suppose that 𝐻 is defined as in (3.2) and let 𝑓,𝑔, and ℎ be three times continuously differentiable functions. In addition, let assumptions (C₁)–(C₃)  hold, and let 𝜂_i be two times continuously differentiable function. Then, for almost all initial points 𝜔⁽⁰⁾∈Ω₀×Λ⁺⁺×𝑅^𝑚++×{0}, 0 is a regular value of 𝐻_𝜔(0) and 𝐻^𝜔−1₍₀₎ consists of some smooth curves. Among them, a smooth curve, say Γ_𝜔(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⁾∈Ω 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⁾)∈𝑅^𝑛×Ω₀×Λ⁺⁺×𝑅^𝑚++, 𝜕𝐻𝜕𝑥,𝑥⁽⁰⁾,𝜆⁽⁰⁾,𝑢⁽⁰⁾=⎡⎢⎢⎢⎢⎣𝑄−𝑡𝐼^𝑛∇ℎ(𝑥)_𝑇00𝑈∇𝑔(𝑥)_𝑇0000−𝑡𝑈⁽⁰⁾∇0𝑔𝑥⁽⁰⁾^𝑇50−𝑡diag𝑔𝑥⁽⁰⁾2𝑡𝜆⁽⁰⁾^3/2𝐼_𝑝0⎤⎥⎥⎥⎥⎦,(3.5) where 𝑄=(1−𝑡)(∑^𝑝𝑖=1𝜆_𝑖∇²𝑓_𝑖(𝑥)+∑^𝑚𝑗=1𝑢_𝑗∇²𝑔_𝑗(𝑥)+𝑡∑^𝑚𝑗=1𝑢^2𝑗∇𝜂_𝑗(𝑥))+∑^𝑠𝑘=1𝑧_𝑘∇²ℎ_𝑘(𝑥)+𝑡𝐼_𝑛.
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 the parametric form of the Sard theorem, for almost all 𝜔⁽⁰⁾∈Ω₀×Λ⁺⁺×𝑅^𝑚++×{0}, 0 is a regular value of 𝐻_𝜔(0). By the inverse image theorem, 𝐻^𝜔−1₍₀₎(0) consists of some smooth curves. Since 𝐻(𝜔⁽⁰⁾,𝜔⁽⁰⁾,1)=0, there must be a smooth curve, denoted by Γ_𝜔(0), starting 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⁺⁺×𝑅^m₊₊×{0}, if 0 is a regular value of 𝐻_𝜔(0), then the projection of the smooth curve Γ_𝜔(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),𝑡_𝑘)}⊂Γ_𝜔(0) such that 𝑡_𝑘⟶𝑡_∗,‖‖𝜆^(𝑘)‖‖⟶∞.(3.7)
From the last equality of (3.2), we have ⎛⎜⎜⎝11−−𝑡𝑡^𝑘_𝑘1⋯−𝑡_𝑘⎞⎟⎟⎠−⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝1−𝑡^𝑘𝜆^1(𝑘)+1−𝑡_𝑘_𝑖≠1𝜆^𝑖(𝑘)+𝑡_𝑘𝜆^1(𝑘)^5/21−𝑡_𝑘𝜆^2(𝑘)+1−𝑡_𝑘_𝑖≠2𝜆^𝑖(𝑘)+𝑡_𝑘𝜆^2(𝑘)^5/21−𝑡_𝑘⋯𝜆^𝑝(𝑘)+1−𝑡_𝑘_𝑖≠𝑝𝜆^𝑖(𝑘)+𝑡_𝑘𝜆^𝑝(𝑘)^5/2⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠−𝑡_𝑘⎛⎜⎜⎜⎜⎜⎜⎝𝜆¹⁽⁰⁾^5/2𝜆²⁽⁰⁾^5/2𝜆⋯_𝑝(0)^5/2⎞⎟⎟⎟⎟⎟⎟⎠=0.(3.8)
If we assume ‖𝜆^(𝑘)‖→+∞(𝑘→∞), this hypothesis implies that 𝑖∈{1,2,…,𝑝}∶lim_𝑘→∞𝜆^𝑖(𝑘)=∞≠Φ.(3.9)
Since 𝑡_𝑘→𝑡_∗, 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 ℎ be three times continuously differentiable functions. In addition, let assumptions (C₁)–(C₃) hold, and let 𝜂_𝑖 be two times continuously differentiable function. Then, for almost all of 𝜔⁽⁰⁾∈Ω₀×Λ⁺⁺×𝑅^𝑚++×{0}, 𝐻^𝜔−1₍₀₎(0) contains a smooth curve Γ_𝜔(0)⊂Ω×𝑅^𝑝+×𝑅^𝑚+×𝑅^𝑠×(0,1], which starts from (𝜔⁽⁰⁾,1). As 𝑡→0, the limit set 𝑇×{0}⊂Ω×Λ⁺×𝑅^𝑚+×𝑅^𝑠×{0}  of Γ_𝜔(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 Γ_𝜔(0)⊂Ω×𝑅^𝑝+×𝑅^𝑚+×𝑅^𝑠×(0,1].  By Theorem 3.1, for almost all 𝜔⁽⁰⁾∈Ω₀×Λ⁺⁺×𝑅^𝑚++×{0},  0 is a regular value of 𝐻_𝜔(0) and 𝐻^𝜔−1₍₀₎ contains of a smooth curve Γ_𝜔(0) starting from (𝜔⁽⁰⁾,1). By the classification theorem of one-dimensional smooth manifolds, Γ_𝜔(0) is diffeomorphic to a unit circle or the unit interval (0,1].
Noticing that ||||𝜕𝐻_𝜔(0)𝜔⁽⁰⁾,1𝜕𝜔||||=|||||||||𝐼^𝑛∇ℎ𝑥₍₀₎00∇ℎ𝑥⁽⁰⁾𝑈^𝑇000₍₀₎∇𝑔0𝑥⁽⁰⁾^𝑇−50diag𝑔𝑥⁽⁰⁾02𝜆⁽⁰⁾^3/2=(−1)_𝑠𝐼_𝑝00||||||||||||diag𝑔𝑥₍₀₎‖‖‖−52𝜆⁽⁰⁾^3/2𝐼_𝑝‖‖‖∇ℎ𝑥₍₀₎^𝑇∇ℎ𝑥⁽⁰⁾|||,(3.10)
By assumption (C₂), we know that |∇ℎ(𝑥⁽⁰⁾)^𝑇∇ℎ(𝑥⁽⁰⁾)|≠0. And because 𝑔(𝑥⁽⁰⁾)<0,𝜆⁽⁰⁾∈Λ⁺⁺, we obtain that [𝜕𝐻_𝜔(0)(𝜔⁽⁰⁾,1)/𝜕𝜔]  is nonsingular. Therefore, the smooth curve Γ_𝜔(0), which starts from (𝜔⁽⁰⁾,1), is diffeomorphic to (0,1].
Let (𝜔^∗,𝑡_∗) be a limit point of Γ_𝜔(0). Only three cases are possible:(a)(𝜔^∗,𝑡_∗)∈Ω×Λ⁺×𝑅^𝑚+×𝑅^𝑠×{0}, (b)(𝜔^∗,𝑡_∗)∈𝜕(Ω₀×𝑅^+𝑝+𝑚)×𝑅^𝑠×(0,1], (c)(𝜔^∗,𝑡_∗)∈Ω×𝑅^+𝑝+𝑚×𝑅^𝑠×{1}.
Because 𝐻(𝜔⁽⁰⁾,𝜔⁽⁰⁾,1)=0 has a unique solution (𝜔⁽⁰⁾,1), case (c) will not happen.
In case (b), because Ω and (0,1] are bounded sets and by assumption (C₂), for any 𝑥∈Ω, there exists a positive linear independent map 𝜂(𝑥) with respect to ∇𝑔(𝑥) and ∇ℎ(𝑥) such that {𝜂_𝑖(𝑥)∶𝑖∈𝐵(𝑥)} is positive linear independent with respect to ∇𝑔(𝑥) and ∇ℎ(𝑥). From the first equality of (3.2), we get that the component 𝑧 of Γ_𝜔(0) is bounded.
If case (b) holds, then there exists a sequence {(𝜔^(𝑘),𝑡_𝑘)}⊂Γ_𝜔(0) such that ‖‖𝜔^(𝑘),𝑡_𝑘‖‖⟶∞.(3.11)
Because Ω and (0,1] are bounded and by Theorem 3.2, there exists a sequence (denoted also by {(𝜔^(𝑘),𝑡_𝑘)}⊂Γ_𝜔(0)) such that 𝑥^(𝑘)⟶𝑥^∗,𝜆^(𝑘)⟶𝜆^∗,‖‖𝑢^(𝑘)‖‖⟶∞,𝑧^(𝑘)⟶𝑧^∗,𝑡_𝑘⟶𝑡_∗,as𝑘⟶∞.(3.12)
From the third equality of (3.2), we have 𝑔𝑥^(𝑘)=𝑡_𝑘𝑈^(𝑘)⁻¹𝑈⁽⁰⁾𝑔𝑥⁽⁰⁾.(3.13) Hence, the active index set 𝐵(𝑥^∗) is nonempty.
From the homotopy equation (3.2), it follows that 1−+𝑡_𝑘∇ℎ∇𝑥𝑓_(𝑘)𝑥^(𝑘)𝜆^(𝑘)+∇𝑔𝑥^(𝑘)𝑢^(𝑘)+𝑡_𝑘𝜂𝑥^(𝑘)𝑢^(𝑘)²𝑧^(𝑘)+𝑡_𝑘𝑥^(𝑘)−𝑥⁽⁰⁾=0.(3.14) (i)When 𝑡_∗=1, rewrite (3.14) as_𝑗∈𝐵∑_(𝑥∗)1−𝑡_𝑘∇𝑔_j𝑥^(𝑘)𝑢^𝑗(𝑘)+𝑡_𝑘𝜂_𝑗𝑥^(𝑘)𝑢^𝑗(𝑘)²+∇ℎ𝑥^(𝑘)𝑧^(𝑘)=−1−𝑡_𝑘+𝑥^(𝑘)−𝑥⁽⁰⁾∑_{𝑗∉𝐵(𝑥∗)}∇𝑔_𝑗𝑥^(𝑘)𝑢^𝑗(𝑘)+𝑡_𝑘𝜂_𝑗𝑥^(𝑘)𝑢^𝑗(𝑘)²+∇𝑓𝑥^(𝑘)𝜆^(𝑘)−𝑥^(𝑘)−𝑥⁽⁰⁾.(3.15)
Because {𝑢^𝑗(𝑘)},𝑗∉𝐵(𝑥^∗), are bounded and by assumption (𝐶₁), when 𝑘→∞, we observe that lim_𝑘→∞_{𝑗∈𝐵(𝑥∗)}1−𝑡_𝑘∇𝑔_𝑗𝑥^(𝑘)𝑢^𝑗(𝑘)+𝑡_𝑘𝜂_𝑗𝑥^(𝑘)𝑢^𝑗(𝑘)²+∇ℎ𝑥^(𝑘)𝑧^(𝑘)+𝑥^(𝑘)−𝑥⁽⁰⁾=0.(3.16)
Using 𝑥^(𝑘)→𝑥^∗ and 𝑧^(𝑘)→𝑧^∗(𝑘→∞), we have from (3.16) that _𝑗∈_{𝐵(𝑥∗)}∇𝑔^𝑗𝑥^∗lim_𝑘→∞1−𝑡_𝑘𝑢^𝑗(𝑘)+𝜂_𝑗𝑥^∗lim_𝑘→∞1−𝑡_𝑘𝑡_𝑘𝑢^𝑗(𝑘)²=−∇ℎ𝑥^∗𝑧^∗+𝑥^∗−𝑥⁽⁰⁾.(3.17)
It is easy to see that the right-hand side of (3.17) is bounded. By assumption (C₂) and (3.17), we get lim_𝑘→∞1−𝑡_𝑘𝑢^𝑗(𝑘)=0,lim_𝑘→∞1−𝑡_𝑘𝑡_𝑘𝑢^𝑗(𝑘)²=𝛼_𝑗,𝑗∈𝐵𝑥^∗,(3.18) where 𝛼_𝑗≥0.
Then, we have 𝑥⁽⁰⁾=𝑥^∗+∇ℎ𝑥^∗𝑧^∗+_{𝑗∈𝐵(𝑥∗)}𝛼_𝑗𝜂_𝑗𝑥^∗,(3.19) which contradicts assumption (𝐶₃).
(ii)When 𝑡_∗∈[0,1), rewrite (3.14) as_𝑗∈_{𝐵(𝑥∗)}1−𝑡_𝑘∇𝑔_𝑗𝑥^(𝑘)𝑢^𝑗(𝑘)+𝑡_𝑘𝜂_𝑗𝑥^(𝑘)𝑢^𝑗(𝑘)²=−1−𝑡_𝑘_{𝑗∉𝐵(𝑥∗)}∇𝑔^𝑗𝑥^(𝑘)𝑢^𝑗(𝑘)+𝑡_𝑘𝜂_𝑗𝑥^(𝑘)𝑢^𝑗(𝑘)²+∇𝑓𝑥^(𝑘)𝜆^(𝑘)−𝑡_k𝑥^(𝑘)−𝑥⁽⁰⁾−∇ℎ𝑥^(𝑘)𝑧^(𝑘).(3.20)
We know that, since Ω and {𝑢^𝑗(𝑘)},𝑗∉𝐵(𝑥^∗) are bounded as 𝑘→∞, the right-hand side of (3.20) is bounded. But by assumptions (C₂) and (𝐶₃), if 𝑢^𝑗(𝑘)→∞(𝑗∈𝐵(𝑥^∗)), then the left-hand side of (3.20) is infinite, this is a contradiction.
As a conclusion, (a) is the only possible case, and 𝜔^∗ is a solution of the KKT system (3.1a)–(3.1c).
Let 𝑠 be the arc-length of Γ_𝜔(0). We can parameterize Γ_𝜔(0) with respect to 𝑠.

Theorem 3.4. The homotopy path Γ_𝜔(0) is determined by the following initial-value problem for the ordinary differential equation: 𝐷𝜔𝐻(0)_𝜔(0)=(𝜔𝜔(𝑠)₍₀₎,𝑡(𝑠))⎡⎢⎢⎣𝜔𝜇⋅⋅⎤⎥⎥⎦=0,,𝑡(0)=1.(3.21) The component 𝜔^∗ of the solution point (𝜔(𝑠^∗),𝑡(𝑠^∗)), for 𝑡(𝑠^∗)=0, is a solution of the KKT system.

4. Algorithm and Example

The following algorithm describes the numerical computation for Pareto optimal solutions.

Algorithm 4.1 ((MOP) Euler-Newton method). Step 1. Give an initial point (𝜔⁽⁰⁾,1)∈Ω₀×𝑅^𝑝+𝑚++×{0}×{1}, an initial step length ℎ₀>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)∈𝑅^{𝑛+𝑝+𝑚+𝑠+1}  of Γ_𝜔0 at (𝜔⁽⁰⁾,𝑡_𝑘);(b)determine the direction 𝛾^(𝑘) of the predictor step.
If the sign of the determinant |||^𝐷𝐻_𝜔(0)^{(𝜔(𝑘),𝑡}_𝑘^)𝜉_𝑇(𝑘)|||   is (−1)^{𝑝+𝑚+𝑠+𝑝𝑚+𝑝𝑠+𝑚𝑠+1}, take 𝛾^(𝑘)=𝜉^(𝑘).
If the sign of the determinant |||^𝐷𝐻_𝜔(0)^{(𝜔(𝑘),𝑡}_𝑘^)𝜉_𝑇(𝑘)||| is (−1)^{𝑝+𝑚+𝑠+𝑝𝑚+𝑝𝑠+𝑚𝑠}, take 𝛾^(𝑘)=−𝜉^(𝑘).Step 3. Compute a corrector point (𝜔^(𝑘+1),𝑡_𝑘+1)∶𝜔_(𝑘+1)𝜔^∗^(𝑘),𝑡_∗_𝑘=𝜔^(𝑘),𝑡_𝑘+ℎ_𝑘𝛾^(𝑘),,𝑡_𝑘+1=𝜔^∗^(𝑘),𝑡_∗_𝑘−𝐷𝐻^𝜔(0)𝜔^∗^(𝑘),𝑡_∗_𝑘⁺𝐻_𝜔(0)𝜔^∗^(𝑘),𝑡_∗_𝑘,(4.1) where 𝐷𝐻_𝜔(0)(𝜔,𝑡)⁺=𝐷𝐻_𝜔(0)(𝜔,𝑡)^𝑇𝐷𝐻_𝜔⁽⁰⁾(𝜔,𝑡)𝐷𝐻_𝜔(0)(𝜔,𝑡)^𝑇⁻¹(4.2) is the Moore-Penrose inverse of 𝐷𝐻_𝜔(0)(𝜔,𝑡).
If ‖𝐻_𝜔(0)(𝜔^(𝑘+1),𝑡_𝑘+1)‖≤𝜀₁, let ℎ_𝑘+1=min{ℎ₀,2ℎ_𝑘} and go to Step 4.
If ‖𝐻_𝜔(0)(𝜔^(𝑘+1),𝑡_𝑘+1)‖∈(𝜀₁,𝜀₂), let ℎ_𝑘+1=ℎ_𝑘 and go to Step 4.
If ‖𝐻_𝜔(0)(𝜔^(𝑘+1),𝑡_𝑘+1)‖≥𝜀₂, let ℎ_𝑘+1=max{(1/2)ℎ₀,(1/2)ℎ_𝑘} and go to Step 3.Step 4. If 𝜔^(𝑘+1)∈Ω×𝑅^+𝑝+𝑚×𝑅^𝑠 and 𝑡_𝑘+1>𝜀₃, let 𝑘=𝑘+1 and go to Step 2.
If 𝜔^(𝑘+1)∈Ω×𝑅^+𝑝+𝑚×𝑅^𝑠 and 𝑡_𝑘+1<−𝜀₃, let ℎ_𝑘∶=ℎ_𝑘(𝑡_𝑘/(𝑡_𝑘−𝑡_𝑘+1)),  go to Step 3, and recompute (𝜔^(𝑘+1),𝑡_𝑘+1)  for the initial point (𝑤^(𝑘),𝑡_𝑘).
If 𝜔^(𝑘+1)∉Ω×𝑅^+𝑝+𝑚×𝑅^𝑠, let ℎ_𝑘∶=(ℎ_𝑘/2)(𝑡_𝑘/(𝑡_𝑘−𝑡_𝑘+1)),  go to Step 3, and recompute (𝜔^(𝑘+1),𝑡_𝑘+1) for the initial point (𝜔^(𝑘),𝑡_𝑘).
If 𝜔^(𝑘+1)∈Ω×𝑅^+𝑝+𝑚×𝑅^𝑠 and |t_𝑘+1|≤𝜀₃, then stop.

Remark 4.2. In Algorithm 4.1, the arc length parameter is not computed explicitly. The tangent vector at a point on Γ_𝜔(0) has two opposite directions, one (the positive direction) makes 𝑠 increase and the other (the negative direction) makes 𝑠 decrease. The negative direction will lead us back to the initial point, so we must go along the positive direction. The criterion in Step 2 (b) of Algorithm 4.1 that determines the positive direction is based on a basic theory of homotopy method, that is, the positive direction 𝛾 at any point (𝜔,𝑡) on Γ_𝜔(0) keeps the sign of the determinant |||^𝐷𝐻_𝜔(0)^{𝛾(𝜔,𝑡)}_𝑇||| invariant.

Example 4.3. We have msi.nt.𝑓𝑔=₁min𝑥²¹+𝑥²²,𝑥₁+3²+𝑥²²,𝑔(𝑥)=𝑥₁−3²+𝑥²²−64≤0,₂ℎ((𝑥𝑥))==𝑥₁𝑥₁−5²+𝑥²²−9≤0,−𝑥²²−3=0.(4.3) The results are shown in Table 1.

Table 1 

Results of Example 4.3.

Example 4.4. We have min𝑓=min𝑥²¹+𝑥²²+𝑥²³+𝑥²⁴+𝑥²⁵,3𝑥₁+2𝑥₂−13𝑥³s.t.𝑔₁+0.01𝑥₄−𝑥₅³,ℎ(𝑥)=𝑥²¹+𝑥²²+𝑥²³+𝑥²⁴−10≤0,₁ℎ(𝑥)=4𝑥₁−2𝑥₂+0.8𝑥₃+0.6𝑥₄+0.5𝑥²⁵=0,₂(𝑥)=𝑥₁+2𝑥₂−𝑥₃−0.5𝑥₄+𝑥₅−2=0.(4.4) The results are shown in Table 2.

Table 2 

Results of Example 4.4.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (Grant no. 10771020) and the Jilin Province Natural Science Foundation (Grant no. 20101597).