A genetic algorithm with a self-reproduction operator to solve systems of nonlinear equations

Abstract

A genetic algorithm for solving systems of nonlinear equations that uses a self-reproduction operator bases on residual approaches is presented and analyzed. To ensure convergence the elitist model is used. A convergence analysis is given. With the aim of showing the advantages of the proposed genetic algorithm an extensive set of numerical experiments with standard test problems and some specific applications are reported.

Appendix A Test problems

This appendix provides the test functions $F({\mathbf {x}})=(f_{1}({\mathbf {x}}),\ldots ,f_{n}({\mathbf {x}}))^{\top }$, and the vectors ${\mathbf {l}}=(l_{1},\ldots ,l_{n})^{\top }$ and ${\mathbf {u}}=(u_{1},\ldots ,u_{n})^{\top }$ used to generate the initial population $P_{0}$ of the genetic algorithms.

• 1. Function 1 [15]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= x_1 - x_2 + 1 + \frac{1}{9}\mid x_1 - 1\mid , \\ f_{2}({\mathbf {x}})&= x_2^2 + x_1 - 7 + \frac{1}{9}\mid x_2\mid ,\\ {\mathbf {l}}&= (-2,-2)^{\top },\quad {\mathbf {u}}=(6,6)^{\top } \end{aligned}$$

• 2. Function 2 [43]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= \cos (2x_1) - \cos (2x_2) - 0.4, \\ f_{2}({\mathbf {x}})&= 2(x_2-x_1) + \sin (2x_2) - \sin (2x_1) - 1.2,\\ {\mathbf {l}}&= (-1,-1)^{\top },\quad {\mathbf {u}}=(1,1)^{\top } \end{aligned}$$

• 3. Function 3 [43]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= e^{x_1} + x_1\,x_2 - 1, \\ f_{2}({\mathbf {x}})&= \sin (x_1\,x_2) + x_1 + x_2 - 1,\\ {\mathbf {l}}&= (-2,-2)^{\top },\quad {\mathbf {u}}=(2,2)^{\top } \end{aligned}$$

• 4. Function 4 [15]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= 3x_1^2 + \sin (x_1\,x_2) - x_3^2 + 2, \\ f_{2}({\mathbf {x}})&= 2x_1^2 - x_2^2 - x_3 + 3,\\ f_{3}({\mathbf {x}})&= \sin (2x_1) + \cos (x_1\,x_2) + x_2 - 1,\\ {\mathbf {l}}&= (-5,-1,-5)^{\top },\quad {\mathbf {u}}=(5,3,5)^{\top } \end{aligned}$$

• 5. Interval Arithmetic Benchmark [44]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= x_1 - 0.25428722 - 0.18324757x_4x_3x_9, \\ f_{2}({\mathbf {x}})&= x_2 - 0.37842197 - 0.16275449x_1x_{10}x_6,\\ f_{3}({\mathbf {x}})&= x_3 - 0.27162577 - 0.16955071x_1x_2x_{10},\\ f_{4}({\mathbf {x}})&= x_4 - 0.19807914 - 0.15585316x_7x_1x_6,\\ f_{5}({\mathbf {x}})&= x_5 - 0.44166728 - 0.19950920x_7x_6x_3,\\ f_{6}({\mathbf {x}})&= x_6 - 0.14654113 - 0.18922793x_8x_5x_{10},\\ f_{7}({\mathbf {x}})&= x_7 - 0.42937161 - 0.21180486x_2x_5x_8,\\ f_{8}({\mathbf {x}})&= x_8 - 0.07056438 - 0.17081208x_1x_7x_6,\\ f_{9}({\mathbf {x}})&= x_9 - 0.34504906 - 0.19612740x_{10}x_6x_8,\\ f_{10}({\mathbf {x}})&= x_{10} - 0.42651102 - 0.21466544x_4x_8x_1,\\ {\mathbf {l}}&=(-5,-5,\ldots ,-5)^{\top },\quad {\mathbf {u}}=(5,5,\ldots ,5)^{\top } \end{aligned}$$

• 6. Combustion problem [45]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= x_2 + 2x_6 + x_9 + 2x_{10} - 10^{-5}, \\ f_{2}({\mathbf {x}})&= x_3 + x_8 - 3\cdot 10^{-5},\\ f_{3}({\mathbf {x}})&= x_1 + x_3 + 2x_5 + 2x_8 + x_9 + x_{10} - 5\cdot 10^{-5},\\ f_{4}({\mathbf {x}})&= x_4 + 2x_7 - 10^{-5},\\ f_{5}({\mathbf {x}})&= 0.5140437\cdot 10^{-7} - x_1^2,\\ f_{6}({\mathbf {x}})&= 0.1006932\cdot 10^{-6}x_6 - 2x_2^2,\\ f_{7}({\mathbf {x}})&= 0.7816278\cdot 10^{-15}x_7 - x_4^2,\\ f_{8}({\mathbf {x}})&= 0.1496236\cdot 10^{-6}x_8 - x_1x_3,\\ f_{9}({\mathbf {x}})&= 0.6194411\cdot 10^{-7}x_9 - x_1x_2,\\ f_{10}({\mathbf {x}})&= 0.2089296\cdot 10^{-14}x_{10} - x_1x_2^2,\\ {\mathbf {l}}&=\left( -\frac{1}{4},-\frac{1}{4},\ldots ,-\frac{1}{4}\right) ^{\top },\quad {\mathbf {u}}=\left( \frac{1}{4},\frac{1}{4},\ldots ,\frac{1}{4}\right) ^{\top } \end{aligned}$$

• 7. Neurophysiology problem [45]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= x_1^2 + x_3^2 - 1, \\ f_{2}({\mathbf {x}})&= x_2^2 + x_4^2 - 1,\\ f_{3}({\mathbf {x}})&= x_5x_3^3 + x_6x_4^3,\\ f_{4}({\mathbf {x}})&= x_5x_1^3 + x_6x_2^3,\\ f_{5}({\mathbf {x}})&= x_5x_1x_3^2 + x_6x_4^2x_2,\\ f_{6}({\mathbf {x}})&= x_5x_1^2x_3 + x_6x_2^2x_4,\\ {\mathbf {l}}&=\left( -1,-1,\ldots ,-1\right) ^{\top },\quad {\mathbf {u}}=\left( 1,1,\ldots ,1\right) ^{\top } \end{aligned}$$

• 8. Chemical equilibrium problem [45]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= x_1x_2 + x_1 - 3x_5, \\ f_{2}({\mathbf {x}})&= 2x_1x_2 + x_1 + x_2x_3^2 + R_8x_2 - Rx_5 + 2R_{10}x_2^2\\&\qquad + R_7x_2x_3 + R_9x_2x_4,\\ f_{3}({\mathbf {x}})&= 2x_2x_3^2 + 2R_5x_3^2 - 8x_5 + R_6x_3 + R_7x_2x_3,\\ f_{4}({\mathbf {x}})&= R_9x_2x_4 + 2x_4^2 - 4Rx_5,\\ f_{5}({\mathbf {x}})&= x_1(x_2 + 1) + R_{10}x_2^2 + x_2x_3^2 + R_8x_2\\&\qquad +R_5x_3^2 + x_4^2 - 1 + R_6x_3 + R_7x_2x_3 + R_9x_2x_4,\\ {\mathbf {l}}&=\left( -1,-1,\ldots ,-1\right) ^{\top },\quad {\mathbf {u}}=\left( 1,1,\ldots ,1\right) ^{\top }, \end{aligned}$$

  where

  $$\begin{aligned} \begin{array}{llll} R = 10, &{} R_5 = 0.193, &{} R_6 = \frac{0.002597}{\sqrt{40}}, &{} R_7 = \frac{0.003448}{\sqrt{40}}, \\ R_8 = \frac{0.00001799}{40}, &{} R_9 = \frac{0.0002155}{\sqrt{40}}, &{} R_{10} = \frac{0.00003846}{40} \end{array} \end{aligned}$$

• 9. Kinematic application [45]

  $$\begin{aligned} f_{i}({\mathbf {x}})&= x_i^2 + x_{i+1}^2 - 1,\quad \text{ for } =1,2,3,4,\\ f_{4+i}({\mathbf {x}})&= a_{1i}x_1x_3 + a_{2i}x_1x_4 + a_{3i}x_2x_3 + a_{4i}x_2x_4\\&\qquad + a_{5i}x_2x_7 + a_{6i}x_5x_8 + a_{7i}x_6x_7 \\&\quad + a_{8i}x_6x_8 + a_{9i}x_1 + a_{10i}x_2 + a_{11i}x_3 + a_{12i}x_4 \\&\qquad + a_{13i}x_5 + a_{14i}x_6 + a_{15i}x_7 \\&\quad +a_{16i}x_8 + a_{17i},\quad \text{ for } =1,2,3,4,\\ {\mathbf {l}}&= \left( -1,-1,\ldots ,-1\right) ^{\top },\quad {\mathbf {u}}=\left( 1,1,\ldots ,1\right) ^{\top }, \end{aligned}$$

  where the matrix $A=(a_{ki})$ is given by:

  $$\begin{aligned} A = \begin{pmatrix} -0.249150680 &{} 0.125016350 &{} -0.635550077 &{} 1.48947730 \\ 1.609135400 &{} -0.686607360 &{} -0.115719920 &{} 0.23062341 \\ 0.279423430 &{} -0.119228110 &{} -0.666404480 &{} 1.32810730 \\ 1.434801600 &{} -0.719940470 &{} 0.110362110 &{} -0.25864503 \\ 0.000000000 &{} -0.432419270 &{} 0.290702030 &{} 1.16517200 \\ 0.400263840 &{} 0.000000000 &{} 1.258776700 &{} -0.26908494 \\ -0.800527680 &{} 0.000000000 &{} -0.629388360 &{} 0.53816987 \\ 0.000000000 &{} -0.864838550 &{} 0.581404060 &{} 0.58258598 \\ 0.074052388 &{} -0.037157270 &{} 0.195946620 &{} -0.20816985 \\ -0.083050031 &{} 0.035436896 &{} -1.228034200 &{} 2.68683200 \\ -0.386159610 &{} -0.085383482 &{} -0.000000000 &{} -0.69910317 \\ -0.755266030 &{} 0.000000000 &{} -0.079034221 &{} 0.35744413 \\ 0.504201680 &{} -0.039251961 &{} 0.026187877 &{} 1.24991170 \\ -1.091628700 &{} 0.000000000 &{} -0.057131430 &{} 1.46773600 \\ 0.000000000 &{} -0.432419270 &{} -1.162808100 &{} 1.16517200 \\ 0.049207290 &{} 0.000000000 &{} 1.258776100 &{} 1.07633970 \\ 0.049207290 &{} 0.013873010 &{} 2.162575000 &{} -0.69686809 \end{pmatrix} \end{aligned}$$

• 10. Economics Modeling Application [46]

  $$\begin{aligned} f_{i}({\mathbf {x}})&= \left( x_{i} + \sum _{k=1}^{n-k-1}x_{k}x_{k+i}\right) x_{n} - c_{i},\quad \text{ for } =1,\ldots ,n-1,\\ f_{n}({\mathbf {x}})&= \sum _{k=1}^{n-1}x_{k} + 1,\\ {\mathbf {l}}&= \left( -1,-1,\ldots ,-1\right) ^{\top },\quad {\mathbf {u}}=\left( 1,1,\ldots ,1\right) ^{\top }, \end{aligned}$$

  where the constants $c_i$ can be randomly chosen. In the numerical experiment, set $c_{i}=0$.

• 11. Discrete boundary value problem [47]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= 2x_1 + 0.5h^2(x_1+h)^3 - x_2, \\ f_{i}({\mathbf {x}})&= 2x_i + 0.5h^2(x_i+hi)^3 - x_{i-1} + x_{i+1},\quad \text{ for } i=2,\ldots ,n-1 \\ f_{n}({\mathbf {x}})&= 2x_n + 0.5h^2(x_n+hn)^3 - x_{n-1}, \\ h&= 1/(n+1),\\ {\mathbf {l}}&= (-5,-5,\ldots ,-5)^{\top },\quad {\mathbf {u}}=(5,5,\ldots ,5)^{\top } \end{aligned}$$

• 12. Brown’s almost linear system

  $$\begin{aligned} f_i({\mathbf {x}})&= x_i + \sum _{j=1}^n x_j - (n+1),\quad \text{ for } i=1,2,\dots ,n-1, \\ f_n(x)&= \prod _{j=1}^n x_j - 1,\\ {\mathbf {l}}&=(-2,-2,\ldots ,-2)^{\top },\quad {\mathbf {u}}=(2,2,\ldots ,2)^{\top } \end{aligned}$$

• 13. Function 13

  $$\begin{aligned} f_{i}({\mathbf {x}})&= \frac{1}{2n}\left( i + \sum _{k=1}^{n} x_{i}^{3}\right) - x_{i}, \quad i = 1,2,\ldots ,n,\\ {\mathbf {l}}&= (-2,-2,\ldots ,-5)^{\top },\quad {\mathbf {u}}=(2,2,\ldots ,5)^{\top } \end{aligned}$$

• 14. Exponential function 1 [18]

  $$\begin{aligned} f_1({\mathbf {x}})&= e^{x_1-1}-1, \\ f_i({\mathbf {x}})&= i\left( e^{x_i-1}-x_i\right) ,\quad \text {for }i=2,3,\dots ,n, \\ {\mathbf {l}}&=(-2,-2,\ldots ,-2)^{\top },\quad {\mathbf {u}}=(2,2,\ldots ,2)^{\top } \end{aligned}$$

• 15. Exponential function 2 [18]

  $$\begin{aligned} f_1({\mathbf {x}})&=e^{x_1}-1, \\ f_i({\mathbf {x}})&=\frac{i}{10}\left( e^{x_i}+x_{i-1}-1\right) ,\quad \text {for }i=2,3,\dots ,n, \\ {\mathbf {l}}&=(-2,-2,\ldots ,-2)^{\top },\quad {\mathbf {u}}=(2,2,\ldots ,2)^{\top } \end{aligned}$$

• 16. Exponential function 3 [18]

  $$\begin{aligned} f_i({\mathbf {x}})&=\frac{i}{10}\left( 1-x_i^2-e^{-x_i^2}\right) ,\quad \text {for }i=2,3,\dots ,n-1, \\ f_n({\mathbf {x}})&=\frac{n}{10}\left( 1-e^{-x_n^2}\right) ,\\ {\mathbf {l}}&=(-2,-2,\ldots ,-2)^{\top },\quad {\mathbf {u}}=(2,2,\ldots ,2)^{\top } \end{aligned}$$

• 17. Extended Rosenbrock function (n is even) [48, p. 89]

  $$\begin{aligned} f_{2i-1}({\mathbf {x}})&= 10\left( x_{2i}-x_{2i-1}^2\right) ,\quad \text {for }i=1,2,\ldots ,n/2, \\ f_{2i}({\mathbf {x}})&= 1-x_{2i-1},\quad \text {for }i=1,2,\ldots ,n/2, \\ {\mathbf {l}}&=(-2,-2,\ldots ,-2)^{\top },\quad {\mathbf {u}}=(2,2,\ldots ,2)^{\top } \end{aligned}$$

• 18. Chandrasekhar’s H-equation [2, p. 198]

  $$\begin{aligned} f_{i}({\mathbf {x}})&= x_i-\left( 1-\frac{c}{2n}\sum _{j=1}^n\frac{\mu _i x_j}{\mu _i+\mu _j}\right) ^{-1}, \quad \text {for }i=1,2,\dots ,n,\\ {\mathbf {l}}&=(-5,-5,\ldots ,-5)^{\top },\quad {\mathbf {u}}=(5,5,\ldots ,5)^{\top }, \end{aligned}$$

  where $c\in [0,1)$ and $\mu _i=(i-1/2)/n$, for $1\le i\le n$. (In the numerical experiment, set $c=0.9$).

• 19. Badly scaled augmented Powell’s function (n is a multiple of 3) [48, p. 89]

  $$\begin{aligned} f_{3i-2}({\mathbf {x}})&= 10^4x_{3i-2}x_{3i-1}-1,\quad \text {for }i=1,2,\ldots ,n/3, \\ f_{3i-1}({\mathbf {x}})&= \exp \left( -x_{3i-2}\right) +\exp \left( -x_{3i-1}\right) -1.0001, \quad \text {for }i=1,2,\ldots ,n/3,\\ f_{3i}({\mathbf {x}})&= \phi (x_{3i}),\quad \text {for }i=1,2,\ldots ,n/3,\\ {\mathbf {l}}&=(-1,-1,\ldots ,-1)^{\top },\quad {\mathbf {u}}=(1,1,\ldots ,1)^{\top }, \end{aligned}$$

  where

  $$\begin{aligned} \phi (t)={\left\{ \begin{array}{ll} 0.5t-2 &{} t\le -11\\ \left( -592t^3+888t^2+4551t-1924\right) /1998, &{} -1<t<2\\ 0.5t+2 &{} t\ge 2. \end{array}\right. } \end{aligned}$$

• 20. Trigonometric function

  $$\begin{aligned} f_{i}({\mathbf {x}})&= 2\left( n+i(1-\cos x_i)-\sin x_i-\sum _{j=1}^n\cos x_j\right) \left( 2\sin x_i-\cos x_i\right) ,\\&\text{ for } i=1,2,\dots ,n,\\ {\mathbf {l}}&=(-1,-1,\ldots ,-1)^{\top },\quad {\mathbf {u}}=(1,1,\ldots ,1)^{\top } \end{aligned}$$

• 21. Singular function [18]

  $$\begin{aligned} f_1({\mathbf {x}})&= \frac{1}{3} x_1^3+\frac{1}{2} x_2^2 \\ f_{i}({\mathbf {x}})&= -\frac{1}{2} x_i^2+\frac{i}{3}x_i^3+\frac{1}{2} x_{i+1}^2, \quad \text {for }i=2,3,\dots ,n-1,\\ f_n(x)&= -\frac{1}{2} x_n^2+\frac{n}{3} x_n^3, \\ {\mathbf {l}}&=(-10,-10,\ldots ,-10)^{\top },\quad {\mathbf {u}}=(10,10,\ldots ,10)^{\top } \end{aligned}$$

• 22. Broyden Tridiagonal function [49, pp. 471-472]

  $$\begin{aligned} f_1({\mathbf {x}})&=(3-0.5x_1)x_1-2x_2+1,\\ f_{i}({\mathbf {x}})&= (3-0.5x_i)x_i-x_{i-1}-2x_{i+1}+1, \quad \text {for }i=2,3,\dots ,n-1,\\ f_n({\mathbf {x}})&= (3-0.5x_n)x_n-x_{n-1}+1,\\ {\mathbf {l}}&=(-1,-1,\ldots ,-1)^{\top },\quad {\mathbf {u}}=(1,1,\ldots ,1)^{\top } \end{aligned}$$

• 23. Trigexp function [49, p. 473]

  $$\begin{aligned} f_1({\mathbf {x}})&= x_1^3+2x_2-5+\sin (x_1-x_2)\sin (x_1+x_2),\\ f_{i}({\mathbf {x}})&= -x_{i-1}e^{(x_{i-1}-x_i)}+x_i(4+3x_i^2)+2x_{i+1}+ \sin (x_i-x_{i+1})\sin (x_i+x_{i+1})-8,\\&\text{ for } i=2,3,\dots ,n-1,\\ f_n({\mathbf {x}})&= -x_{n-1}e^{(x_{n-1}-x_n)}+4x_n-3,\\ {\mathbf {l}}&= (-1,-1,\ldots ,-1)^{\top },\quad {\mathbf {u}}=(1,1,\ldots ,1)^{\top } \end{aligned}$$

• 24. Strictly convex function 1 [50, p. 29]. F is is the gradient of $h({\mathbf {x}})=\sum _{i=1}^n\left( e^{x_1}-x_i\right)$.

  $$\begin{aligned} f_{i}({\mathbf {x}})&= e^{x_i}-1,\quad \text {for }i=1,2,\dots ,n,\\ {\mathbf {l}}&= (-2,-2,\ldots ,-2)^{\top },\quad {\mathbf {u}}=(2,2,\ldots ,2)^{\top } \end{aligned}$$

• 25. Strictly convex function 2 [50, p. 30]. F is is the gradient of $h({\mathbf {x}})=\sum _{i=1}^n\frac{i}{10}\left( e^{x_1}-x_i\right)$.

  $$\begin{aligned} f_{i}({\mathbf {x}})&= \frac{i}{10}\left( e^{x_i}-1\right) ,\quad \text {for }i=1,2,\dots ,n,\\ {\mathbf {l}}&= (-2,-2,\ldots ,-2)^{\top },\quad {\mathbf {u}}=(2,2,\ldots ,2)^{\top } \end{aligned}$$

• 26. Zero Jacobian function [18]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= \sum _{j=1}^n x_j^2, \\ f_{i}({\mathbf {x}})&= -2x_1x_i,\quad \text{ for } i=2,\dots ,n,\\ {\mathbf {l}}&= (-1,-1,\ldots ,-1)^{\top },\quad {\mathbf {u}}=(1,1,\ldots ,1)^{\top } \end{aligned}$$

• 27. Freudenstein and Roth extended function (n par) [51]

  $$\begin{aligned} f_{2i-1}({\mathbf {x}})&= x_{2i-1} + ((5-x_{2i})x_{2i}-2) x_{2i} - 13,\quad \text{ for } i=2,\dots ,n/2,\\ f_{2i}({\mathbf {x}})&= x_{2i-1} + ((x_{2i}+1))x_{2i} - 14) x_{2i} - 29,\quad \text{ for } i=2,\dots ,n/2,\\ {\mathbf {l}}&= (-1,-1,\ldots ,-1)^{\top },\quad {\mathbf {u}}=(1,1,\ldots ,1)^{\top } \end{aligned}$$

• 28. Cragg and Levy’s extended problem (n múltiplo de 4) [47]

  $$\begin{aligned} f_{4i-3}({\mathbf {x}})&= (\exp (x_{4i-3})-x_{4i-2})^2,\quad \text{ for } i=2,\dots ,n/4,\\ f_{4i-2}({\mathbf {x}})&= 10(x_{4i-2}-x_{4i-1})^3,\quad \text{ for } i=2,\dots ,n/4,\\ f_{4i-1}({\mathbf {x}})&= \tan ^2(x_{4i-1}-x_{4i}),\quad \text{ for } i=2,\dots ,n/4,\\ f_{4i}({\mathbf {x}})&= x_{4i} - 1,\quad \text{ for } i=2,\dots ,n/4,\\ {\mathbf {l}}&= (-5,-5,\ldots ,-5)^{\top },\quad {\mathbf {u}}=(5,5,\ldots ,5)^{\top } \end{aligned}$$

• 29. Wood’s extended problem (n múltiplo de 4) [52]

  $$\begin{aligned} f_{4i-3}({\mathbf {x}})&= -200x_{4i-3}\left( x_{4i-2}-x_{4i-3}^2\right) -(1-x_{4i-3}),\quad \text{ for } i=2,\dots ,n/4,\\ f_{4i-2}({\mathbf {x}})&= 200\left( x_{4i-2}-x_{4i-3}^2\right) +20(x_{4i-2}-1)+19.8(x_{4i}-1),\quad \text{ for } i=2,\dots ,n/4,\\ f_{4i-1}({\mathbf {x}})&= -180x_{4i-1}\left( x_{4i}-x_{4i-1}^2\right) -(1-x_{4i-1}),\quad \text{ for } i=2,\dots ,n/4,\\ f_{4i}({\mathbf {x}})&= 180 \left( x_{4i}-x_{4i-1}^2\right) + 20.2(x_{4i}-1) + 19.8(x_{4i-2}-1),\quad \text{ for } i=2,\dots ,n/4,\\ {\mathbf {l}}&= (-2,-2,\ldots ,-2)^{\top },\quad {\mathbf {u}}=(2,2,\ldots ,2)^{\top } \end{aligned}$$

• 30. Exponential tridiagonal problem [51]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= x_1 - \exp (\cos (h(x_1+x_2))), \\ f_{i}({\mathbf {x}})&= x_i - \exp (\cos (h(x_{i-1}+x_i+x_{i+1}))),\quad \text{ for } i=2,\ldots ,n-1 \\ f_{n}({\mathbf {x}})&= x_n - \exp (\cos (h(x_{n-1}+x_n))), \\ h&= 1/(n+1),\\ {\mathbf {l}}&= (-5,-5,\ldots ,-5)^{\top },\quad {\mathbf {u}}=(5,5,\ldots ,5)^{\top } \end{aligned}$$

• 31. Brent problem [53]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= 3x_1 (x_2-2x_1) + x_2^2/4, \\ f_{i}({\mathbf {x}})&= 3x_i(x_{i+1}-2x_i+x_{i-1})+(x_{i+1}-x_{i-1})^2/4,\quad \text{ for } i=2,\ldots ,n-1,\\ f_{n}({\mathbf {x}})&= 3x_n(20-2x_n+x_{n-1}) + (20-x_{n-1})^2/4, \\ h&= 1/(n+1),\\ {\mathbf {l}}&= (-5,-5,\ldots ,-5)^{\top },\quad {\mathbf {u}}=(5,5,\ldots ,5)^{\top } \end{aligned}$$

• 32. Troesch problem [54]

  $$\begin{aligned} f_{1}({\mathbf {x}})&= 2x_1 + \rho h^2\sinh (\rho x_1) - x_2, \\ f_{i}({\mathbf {x}})&= 2x_i + \rho h^2\sinh (\rho x_i) - x_{i-1}-x_{i+1},\quad \text{ for } i=2,\ldots ,n-1,\\ f_{n}({\mathbf {x}})&= 2x_n + \rho h^2\sinh (\rho x_n) - x_{n-1}, \\ \rho&= 10,\;\;h = 1/(n+1),\\ {\mathbf {l}}&= (-1,-1,\ldots ,-1)^{\top },\quad {\mathbf {u}}=(1,1,\ldots ,1)^{\top } \end{aligned}$$

• 33. Function 33 [55]

  $$\begin{aligned} f_{i}({\mathbf {x}})&= 2x_i - \sin \mid x_i\mid ,\quad \text{ for } i=1,2,\ldots ,n,\\ {\mathbf {l}}&= (-1,-1,\ldots ,-1)^{\top },\quad {\mathbf {u}}=(1,1,\ldots ,1)^{\top } \end{aligned}$$

• 34. Function 34

  $$\begin{aligned} f_{i}({\mathbf {x}})&= \min \left\{ \left( x_{i}-\frac{i}{n}\right) ^2,\, \mid e^{x_{i}-\frac{i}{n}}-1\mid \right\} , \quad i = 1,2,\ldots ,n,\\ {\mathbf {l}}&= (-5,-5,\ldots ,-5)^{\top },\quad {\mathbf {u}}=(5,5,\ldots ,5)^{\top } \end{aligned}$$