Example 2: Bounded Least Squares

The nonlinear least-squares problem

$\min \frac{1}{2}\sum\limits_{i = 0}^1 {f_i \left( x \right)^2 }$

$- 2 \le x_0 \le 0.5$

$-1 \le x_1 \le 2$

where

$f_0 (x) = 10(x_1 - x_0^2 ) \,\, {\rm{and}} \,\, f_1 (x) = (1 - x_0 )$

is solved. An initial guess (-1.2, 1.0) is supplied, as well as the analytic Jacobian. The residual at the approximate solution is returned.

import com.imsl.math.*;

public class BoundedLeastSquaresEx2 {
    public static void main(String args[]) throws Exception {
        int m = 2;
        int n = 2;
        int ibtype = 0;
        double[] xlb = {-2.0, -1.0};
        double[] xub = {0.5, 2.0};
        double[] xguess = {-1.2, 1.0};
        
        BoundedLeastSquares.Function rosbck =
        new BoundedLeastSquares.Function() {
            public void compute(double[] x, double[] f) {
                f[0] = 10.0*(x[1] - x[0]*x[0]);
                f[1] = 1.0 - x[0];
            }
        };
        
        BoundedLeastSquares.Jacobian jacob =
        new BoundedLeastSquares.Jacobian() {
            public void compute(double[] x, double[] fjac) {
                fjac[0] = -20.0*x[0];
                fjac[1] = 10.0;
                fjac[2] = -1.0;
                fjac[3] = 0.0;
            }
        };
        
        BoundedLeastSquares zf =
        new BoundedLeastSquares(rosbck, m, n, ibtype, xlb, xub);
        
        zf.setJacobian(jacob);
        zf.setGuess(xguess);
        zf.solve();
        new PrintMatrix("Solution").print(zf.getSolution());
        new PrintMatrix("Residuals").print(zf.getResiduals());
    }
}

Output

Solution
    0    
0  0.5   
1  0.25  

Residuals
    0   
0  0    
1  0.5

Link to Java source.