An interior-point method for nonlinear optimization problems with locatable and separable nonsmoothness

Abstract

Many real-world optimization models comprise nonconvex and nonsmooth functions leading to very hard classes of optimization models. In this article, a new interior-point method for the special, but practically relevant class of optimization problems with locatable and separable nonsmooth aspects is presented. After motivating and formalizing the problems under consideration, modifications and extensions to a standard interior-point method for nonlinear programming are investigated to solve the introduced problem class. First theoretical results are given and a numerical study is presented that shows the applicability of the new method for real-world instances from gas network optimization.

Appendix A: Low-dimensional test problems

1.1 A.1 Problem 1a (Fig. 6)

$$\begin{aligned} \min \quad&f(x) = -x_1 - x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 - |x_1| \le 0,\\&c_2 (x_1,x_2) \mathrel {{\mathop :}{=}}x_2 + x_1^2 - 1 \le 0,\\&x_2 \ge -1 \end{aligned}$$

Fig. 6

Illustration of test problem 1a

Localization functions: $\ell _1(x_1,x_2) = x_1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (0.61803,0.61803)$

Objective value: $f(x^*) = -1.23606$.

1.2 A.2 Problem 1b (Fig. 7)

$$\begin{aligned} \min \quad&f(x) = -x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 - |x_1| \ge 0,\\&c_2(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 + x_1^2 -1 \le 0 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (0, 1)$

Objective value: $f(x^*) = -1$.

Fig. 7

Illustration of test problem 1b

1.3 A.3 Problem 2a (Fig. 8)

$$\begin{aligned} \min \quad&f(x) = x_1\\ \text {s.t.}\quad&c_1(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 - x_1^2 \ge 0,\\&c_2(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 + |x_1| -1 \le 0 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (-0.61804, 0.38197)$

Objective value: $f(x^*) = -0.61804$.

Fig. 8

Illustration of test problem 2a

1.4 A.4 Problem 2b (Fig. 9)

$$\begin{aligned} \min \quad&f(x) = x_1\\ \text {s.t.}\quad&c_1(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 - x_1^2 \le 0,\\&c_2(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 + |x_1| -1 \le 0,\\&x_2 \ge -1 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (-2, -1)$

Objective value: $f(x^*) = -2$.

Fig. 9

Illustration of test problem 2b

1.5 A.5 Problem 3a (Fig. 10)

$$\begin{aligned} \min \quad&f(x) = -x_1 - x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 - \max \{0,x_1\} \ge 0,\\&c_2(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 + x_1^2 -1 \le 0 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (0.5, 0.75)$

Objective value: $f(x^*) = -1.25$.

Fig. 10

Illustration of test problem 3a

1.6 A.6 Problem 3b (Fig. 11)

$$\begin{aligned} \min \quad&f(x) = -x_1 - x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 - \max \{0, x_1\} \le 0,\\&c_2(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 + x_1^2 -1 \le 0,\\&x_2 \ge -1 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (0.61803, 0.61803)$

Objective value: $f(x^*) = -1.23606$.

Fig. 11

Illustration of test problem 3b

1.7 A.7 Problem 4a (Fig. 12)

$$\begin{aligned} \min \quad&f(x) = -x_1\\ \text {s.t.}\quad&c_1(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 - \min \{0, -x_1\} \ge 0,\\&c_2(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 + x_1^2 -1 \le 0 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (1.6180, -1.6180)$

Objective value: $f(x^*) = 1.6180$.

Fig. 12

Illustration of test problem 4a

1.8 A.8 Problem 4b (Fig. 13)

$$\begin{aligned} \min \quad&f(x) = -x_1\\ \text {s.t.}\quad&c_1(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 - \min \{0, -x_1\} \le 0,\\&c_2(x_1,x_2) \mathrel {{\mathop :}{=}}x_2 + x_1^2 -1 \le 0,\\&x_2 \ge -1 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (1, -1)$

Objective value: $f(x^*) = -1$.

Fig. 13

Illustration of test problem 4b

1.9 A.9 Problem 5a (Fig. 14)

$$\begin{aligned} \min \quad&f(x) = x_1\\ \text {s.t.}\quad&c_1(x_1,x_2) \ge 0, \quad c_1(x_1,x_2) \mathrel {{\mathop :}{=}}{\left\{ \begin{array}{ll} x_2 - (x_1-1)^2 + 1 , &{} \quad x_1 < 1, \\ x_2 + x_1 , &{} \quad x_1 \ge 1, \end{array}\right. }\\&x_1 \le 2, \\&x_2 \le 2 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1 - 1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (-0.73205, 2)$

Objective value: $f(x^*) = -0.73205$.

Fig. 14

Illustration of test problem 5a

1.10 A.10 Problem 5b (Fig. 15)

$$\begin{aligned} \min \quad&f(x) = x_1\\ \text {s.t.}\quad&c_1(x_1,x_2) \le 0, \quad c_1(x_1,x_2) \mathrel {{\mathop :}{=}}{\left\{ \begin{array}{ll} x_2 - (x_1-1)^2 + 1 , &{} x_1 < 1, \\ x_2 + x_1 , &{} x_1 \ge 1, \end{array}\right. },\\&x_1 \ge -2, \\&-2 \le x_2 \le 2 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1-1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (-2, x_2^*)$ with $-2 \le x_2^2 \le 2$

Objective value: $f(x^*) = -2$.

Fig. 15

Illustration of test problem 5b

1.11 A.11 Problem 6a (Fig. 16)

$$\begin{aligned} \min \quad&f(x) = -x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \le 0, \quad c_1(x_1,x_2) \mathrel {{\mathop :}{=}}{\left\{ \begin{array}{ll} x_2 + (x_1-1)^2 + 1 , &{} x_1 \le 1, \\ x_2 + x_1 , &{} x_1 > 1, \end{array}\right. }\\&x_2 \ge -2 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1-1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (1, -1)$

Objective value: $f(x^*) = 1$.

Fig. 16

Illustration of test problem 6a

1.12 A.12 Problem 6b (Fig. 17)

$$\begin{aligned} \min \quad&f(x) = x_1 + x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \ge 0, \quad c_1(x_1,x_2) \mathrel {{\mathop :}{=}}{\left\{ \begin{array}{ll} x_2 + (x_1-1)^2 + 1 , &{} x_1 \le 1, \\ x_2 + x_1 , &{} x_1 > 1, \end{array}\right. }\\&0 \le x_1 \le 2, \\&x_2 \le 0 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1-1$

Global optimal solution: $x^* = (x_1^*, x_2^*) = (0, -2)$

Global optimal objective value: $f(x^*) = -2$

Local optimal solution: $x^* = (x_1^*, x_2^*)$ with $x_1^* \in (1,2]$ and $x_2^* = -x_1^*$.

Local optimal objective value: $f(x^*) = 0$.

Fig. 17

Illustration of test problem 6b

1.13 A.13 Problem 7a (Fig. 18)

$$\begin{aligned} \min \quad&f(x) = x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \ge 0, \quad c_1(x_1,x_2) \mathrel {{\mathop :}{=}}{\left\{ \begin{array}{ll} x_2 - (x_1-1)^2 - 1 , &{} x_1 \le 1, \\ x_2 - x_1 , &{} x_1 > 1, \end{array}\right. }\\&x_2 \le 2 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1-1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (1, 1)$

Objective value: $f(x^*) = 1$.

Fig. 18

Illustration of test problem 7a

1.14 A.14 Problem 7b (Fig. 19)

$$\begin{aligned} \min \quad&f(x) = -x_1 - x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \le 0, \quad c_1(x_1,x_2) \mathrel {{\mathop :}{=}}{\left\{ \begin{array}{ll} x_2 - (x_1-1)^2 - 1 , &{} x_1 \le 1, \\ x_2 - x_1 , &{} x_1 > 1, \end{array}\right. }\\&0 \le x_1 \le 2, \\&x_2 \ge 0 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1-1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (2, 2)$

Objective value: $f(x^*) = -4$.

Fig. 19

Illustration of test problem 7b

1.15 A.15 Problem 8a (Fig. 20)

$$\begin{aligned} \min \quad&f(x) = -x_1 - x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \ge 0, \quad c_1(x_1,x_2) \mathrel {{\mathop :}{=}}{\left\{ \begin{array}{ll} x_2 + (x_1+1)^2 + 1 , &{} x_1 \le -1, \\ x_2 - x_1 , &{} x_1 > -1, \end{array}\right. }\\&x_1 \ge -2, \\&x_2 \le 2 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1+1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (2, 2)$

Objective value: $f(x^*) = -4$.

Fig. 20

Illustration of test problem 8a

Fig. 21

Illustration of test problem 8b

1.16 A.16 Problem 8b (Fig. 21)

$$\begin{aligned} \min \quad&f(x) = x_1 + x_2\\ \text {s.t.}\quad&c_1(x_1,x_2) \le 0, \quad c_1(x_1,x_2) \mathrel {{\mathop :}{=}}{\left\{ \begin{array}{ll} x_2 + (x_1+1)^2 + 1 , &{} x_1 \le -1, \\ x_2 - x_1 , &{} x_1 > -1, \end{array}\right. }\\&x_1 \le 2, \\&x_2 \ge -2 \end{aligned}$$

Localization functions: $\ell _1(x_1,x_2) = x_1+1$

Optimal solution: $x^* = (x_1^*, x_2^*) = (-2, -2)$

Objective value: $f(x^*) = -4$.