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Kaj Madsen - DTU Informatics•Head of Department

Kaj Madsen
Head of Department
DTU Informatics
Professor, dr. techn.
km@imm.dtu.dk
Tel.: +45 4525 3370
Fax.: +45 4593 0360

Last modified Monday, June 16, 2008

 
Forskning•IMM

Test problems for global optimization - described here

In the following descriptions of test problems the notation will be used:

n - dimension,
X - feasible region,
f(x) - objective function,
G(x) - Gradient function,
f* - global minimum,
X* - set of global minimizers,
nloc - number of local minimizers.

The considered test problems are:

Rosenbrock function

n=2, X=[-2,2]n

f(x) = 100(x2-x12)2+(1-x1)2

f* = 0, X* = (1,1) , nloc = 1

 

McCormic function

n=2, X=([-1.5,4],[-3,4])

\begin{displaymath} f(x) = \sin(x_1+x_2)+(x_1-x_2)^2-1.5x_1+2.5x_2+1 \end{displaymath}

f* = -1.9133, X* = (-0.54719,-1.54719) , nloc = 1

 

Box and Betts exponential quadratic sum

n=3, X=([0.9,1.2],[9,11.2],[0.9,1.2])

\begin{displaymath} f(x) = \sum^{10}_{i=1}g(x)^2 \end{displaymath}

where:

\begin{displaymath} g(x) = (\exp(-0.1ix_1)-\exp(-0.1ix_2)-(\exp(-0.1i)-\exp(-i))x_3) \end{displaymath}

f* = 0, X* = (1,10,1) , nloc = 1

 

Paviani function

n=10, X=[2.001,9.999]n

\begin{displaymath} f(x) = \sum^{n}_{i=1}\CENTER(\ln^2(x_i-2)+\ln^2(10-x_i)\right)- \CENTER(\prod_{i=1}^nx_i\right)^{0.2} \end{displaymath}

\begin{displaymath} G_j(x) = \frac{2\ln(x_j-2)}{x_j-2}-\frac{2\ln(10-x_j)}{10-x_... ...c{0.2\prod_{i=1}^nx_i}{x_j\CENTER(\prod_{i=1}^nx_i\right)^{0.8}} \end{displaymath}
f* = -45.778470, $X^* = { (9.350266,9.350266,\dots,9.350266 ) }$,nloc = 1

 

Generalized Rosenbrock function

n=30, X=[-n,n]n

\begin{displaymath} f(x) = \sum^{n-1}_{i=1}\CENTER[100(x_{i+1}-x_i^2)^2+(x_i-1)^2 \right] \end{displaymath}

f* = 0, $X^* = { (1,\dots,1) } $, nloc = 1

 

Goldstein and Price function

n=2, X=[-2,2]n

f* = 3, X* = (0,-1) , nloc = 4

 

Shekel function

n=4, X=[0,10]n

\begin{displaymath} f(x) = -\sum_{i=1}^{m}\frac{1}{(x-A_i)(x-A_i)^T+c_i} \end{displaymath}

where :

\begin{displaymath} \begin{array} {cc} A = \CENTER[ \begin{array} {rrrr} 4 & 4 & 4... ...\ 0.3 \\ 0.7 \\ 0.5 \\ 0.5 \end{array} \right] \end{array} \end{displaymath}

\begin{displaymath} G_j(x) = \sum_{i=1}^m\frac{2(x_j-a_{i,j})} {\CENTER((x-A_i)(x-A_i)^T+c_i\right)^2}\end{displaymath}

for m=5: f* = -10.1532, X* = (4.00004,4.00013,4.00004,4.00013 ) , nloc = 5

for m=7: f* = -10.4029, X* = (4.00057,4.00069,3.99949,3.99961 ) , nloc = 7

for m=10: f* = -10.5364, X* = (4.00075,4.00059,3.99966,3.99951 ) , nloc = 10

 

Levy function

n=4, 5, 6, 7
for n=4: X=[-10,10]n
for n=5,6,7: X=[-5,5]n

\begin{displaymath} f(x) = \sin^2(3\pi x_1)+\sum_{i=1}^{n-1}(x_i-1)^2(1+\sin^2(3\pi x_{i+1})) + (x_n-1)(1+\sin^2(2\pi x_n)) \end{displaymath}

for n=4: f* = -21.502356, X* = (1,1,1,-9.752356 ) , nloc = 71000
for n=5,6,7: f* = -11.504403, $X^* = { (1,\dots,1,-4.754402 ) }$,nloc = 105, 106, 108

 

Grienwank function

n=10, X=[-500,700]n

\begin{displaymath} f(x) = \sum^{n}_{i=1}\frac{x_i^2}{4000} - \prod_{i=1}^n\cos\CENTER(\frac{x_i}{\sqrt{i}}\right)+1 \end{displaymath}

\begin{displaymath} G_j(x) = \frac{x_j}{2000} + \frac{\prod_{i=1}^{n}\cos\CENTER(\... ...rt{j}}\right)} {\sqrt{j}\cos\CENTER(\frac{x_j}{\sqrt{j}}\right)} \end{displaymath}

f* = 0, $X^* = { (0,0,\dots,0 ) }$, nloc = 103

 

Six Hump Camel Back function

n=2, X=[-5,5]n

\begin{displaymath} f(x) = 4x_1^2-2.1x_1^4+\frac{1}{3}x_1^6+x_1x_2-4x_2^2+4x_2^4 \end{displaymath}

f* = -1.03163, $ X^* = \CENTER\{ ( 0.08984, -0.71266 ), ( -0.08984, 0.71266 ) \right\} $,nloc = 6

 

Branin function

n=2, X=[-10,10]n

\begin{displaymath} f(x) = \CENTER(1-2x_2+\frac{1}{20}\sin 4\pi x_2-x_1\right)^2+ \CENTER(x_2-\frac{1}{2}\sin 2\pi x_1\right)^2 \end{displaymath}

f* = 0

\begin{displaymath} X^* = \CENTER\{ \begin{array} {rrrr} ( & 1, & 0 & ), \\ ( & 0.... ...08 & ), \\ ( & 1.85130, & -0.402086 & ) \\ \end{array}\right\} \end{displaymath}

 

Shubert function

n=2, X=[-10,10]n

\begin{displaymath} f(x) = -\sum_{i=1}^n\sum_{j=1}^5j\sin((j+1)x_i+j) \end{displaymath}

\begin{displaymath} G_j(x) = -\sum_{i=1}^5i(i+1)\cos((i+1)x_j+i) \end{displaymath}

f* = -24.062499

\begin{displaymath} X^* = \CENTER\{ \begin{array} {rrrr} ( & -6.774576, & -6.77457... ...91 & ), \\ ( & 5.791794, & 5.791794 & ) \\ \end{array}\right\} \end{displaymath}

nloc = 400

 

Hansen function

n=2, X=[-10,10]n

\begin{displaymath} f(x) = \sum_{i=1}^5i\cos((i-1)x_1+i)\sum_{j=1}^5j\cos((j+1)x_2+j) \end{displaymath}

f* = -176.541793

\begin{displaymath} X^* = \CENTER\{ \begin{array} {rrrr} ( & -7.589893, & -7.70831... ...28 & ), \\ ( & 4.976478, & 4.858057 & ) \\ \end{array}\right\} \end{displaymath}

nloc = 760

Cola function

n=17, U=([0,4],[-4,4]n-1)

x1 = y1 = y2 = 0, x2 = u1, xi = u2(i-2), yi = u2(i-2)+1

\begin{displaymath} f(x,y) = \sum_{j<i}(r_{i,j}-d_{i,j})^2 \end{displaymath}

\begin{displaymath} d = \CENTER\{ \begin{array} {rrrrrrrrrr} \dots \\ 1.27 & \dots... ... & 2.85 & 2.81 & 2.56 & 2.91 & 2.97 & \dots\end{array}\right\} \end{displaymath}

\begin{displaymath} r_{i,j} = [(x_i-x_j)^2+(y_i-y_j)^2]^{\frac{1}{2}} \end{displaymath}

f* = 11.7464

 

Practical fitting problem

This problem was supplied by PhD student Per R. Andersen at IMM, DTU. It is a parameter estimation problem within a growth model of the human mandible (the lower jar). The data of the problem are the coordinates of 271 points of equivalent morphology from 3 mandibles of the same patient in the age of 9 months, 21 month and 7 years. The goal of the problem is to position three mandibles in the space so that the sum of distances of points from the middle mandible to the lines connecting points from the first and third mandibles would be minimal. The middle mandible is fixed and every of both others has 6 degrees of freedom: 3 angles of rotation and 3 directions of translation.

n=12, $X = [-\pi, \pi]^6 \times [-120,120]^6$. The best known function value found previously using a multistart strategy is f* = 205.104. The numbers of local and global minimizers are not known.

The data are available here.