An improved differential evolution algorithm for optimization including linear equality constraints

Abstract

A differential evolution algorithm (DE) is proposed to exactly satisfy the linear equality constraints present in a continuous optimization problem that may also include additional non-linear equality and inequality constraints. The proposed DE technique, denoted by DELEqC-II, is an extension of a previous method developed by the authors. In contrast to the previous approach, it uses both mutation and crossover strategies that maintain feasibility with respect to the linear equality constraints. Also, a procedure to correct numerical errors detected in the previous approach was incorporated in DELEqC-II. In the numerical experiments, scalable test-problems with linear equality constraints are used to analyze the performance of the new proposal.

Appendices

Appendix A: Test-problems

Test-problems 1–5 were taken from [10]. Also note that those problems are subject to the same set of linear constraints, described in Fig. 13. Following, the description and the optimal solution value of each problem.

Problem 1 (Sphere):

:

$$\begin{aligned} \min _{x \in \mathcal{E}} \quad \sum ^{n}_{i=1} x_i^2. \end{aligned}$$

The feasible set $\mathcal{E}$ is given by the linear equality constraints in Fig. 13 and $f(x^*) = 32.137$.

Problem 2 (Quadratic):

:

$$\begin{aligned}&\min _{x \in \mathcal{E}} \quad \sum ^{n}_{i=1}\sum ^{n}_{j=1} e^{-(x_i - x_j)^2}x_ix_j + \sum ^{n}_ {i=1}x_i\\&f(x^*) = 35.377 \end{aligned}$$

Problem 3 (Rastrigin):

:

$$\begin{aligned}&\min _{x \in \mathcal{E}} \quad \sum ^{n}_{i=1} x_i^2 + 10 - 10\cos (2\pi x_i)\\&f(x^*) = 36.975. \end{aligned}$$

Problem 4 (Rosenbrock):

:

$$\begin{aligned}&\min _{x \in \mathcal{E}} \quad \sum ^{n-1}_{i=1} 100(x_{i+1} - x_i^{2})^2 + (x_i - 1)^2\\&f(x^*) = 21485.3. \end{aligned}$$

Problem 5 (Griewank):

:

$$\begin{aligned}&\min _{x \in \mathcal{E}} \quad \dfrac{1}{4000}\sum ^{n}_{i=1} x_i^2 - \prod ^{n}_{i=10}\cos (\dfrac{x_i}{\sqrt{i}}) + 1\\&f(x^*) = 0.151. \end{aligned}$$

Appendix B: Scaling the test-problems

Problems 1–5 can be scaled by varying the number of variables (n) and the number of constraints (m). Let $G \in \mathbb {R}^{5 \times 10}$ and $p \in \mathbb {R}^{5}$ be the coefficient matrix and vector, respectively, associated to the set of linear equality constraints in Fig. 13. Hence

$$\begin{aligned} G = \begin{bmatrix} 0&-3&-1&0&0&2&-6&0&-4&-2 \\ -1&-3&-1&0&0&0&-5&-1&-7&-2 \\ 0&0&1&0&0&1&3&0&-2&2 \\ 2&6&2&2&0&0&4&6&16&4 \\ -1&-6&-1&-2&-2&3&-6&-5&-13&-4 \end{bmatrix} \end{aligned}$$

and $p^{T} = \begin{bmatrix} 3&0&9&-16&30 \end{bmatrix}$.

Now, let $0 \in \mathbb {R}^{5 \times 10}$ be a matrix with all values equal to zero. By varying n and m we can construct new sets of linear equality constraints for problems 1–5. Those sets are described as follow.

Constraint-set 1 In (7), setting $m = 5$ and $n = 20$ we obtain $x \in \mathbb {R}^{20}$ and $E = \begin{bmatrix} G&0 \end{bmatrix} \quad \text{ and } \quad c = \begin{matrix} p \end{matrix}$. The matrix $E \in \mathbb {R}^{5 \times 20}$ and the vector $c \in \mathbb {R}^{5}$ now describes the set of linear equality constraints, where E is of the special form $[G \quad 0]$, that is, its columns can be partitioned into two sets; the first 10 columns make up the original G matrix and the last 10 columns make up the 0 matrix. The next constraint-sets, follow the same construction rule.

Constraint-set 2 Setting $m = 10$ and $n = 20$ we obtain $E = \begin{bmatrix} G&0 \\ 0&G \end{bmatrix} \quad \text{ and } \quad c = \begin{bmatrix} p \\ p \end{bmatrix}$.

Constraint-set 3 Setting $m = 5$ and $n = 30$, we obtain $E = \begin{bmatrix} G&0&0 \\ \end{bmatrix} \quad \text{ and } \quad c = \begin{matrix} p \end{matrix}$.

Constraint-set 4 Setting $m = 15$ and $n = 30$, we obtain $E = \begin{bmatrix} G&0&0 \\ 0&G&0 \\ 0&0&G \\ \end{bmatrix} \quad \text{ and } \quad c = \begin{bmatrix} p \\ p \\ p \end{bmatrix}$.

Constraint-set 5 Setting $m = 5$ and $n = 40$, we obtain $E = \begin{bmatrix} G&0&0&0 \\ \end{bmatrix} \quad \text{ and } \quad c = \begin{matrix} p \end{matrix}$.

Constraint-set 6 Setting $m = 20$ and $n = 40$, we obtain $E = \begin{bmatrix} G&0&0&0 \\ 0&G&0&0 \\ 0&0&G&0 \\ 0&0&0&G \\ \end{bmatrix}$ and $c = \begin{bmatrix} p \\ p \\ p \\ p \end{bmatrix}$.