Nonlinear Data-Fitting

This example shows how to fit a nonlinear function to data using several Optimization Toolbox™ algorithms.

Problem Setup

The problem is to fit a function to the following data.

xdata = ...
  [0.0000    5.8955
   0.1000    3.5639
   0.2000    2.5173
   0.3000    1.9790
   0.4000    1.8990
   0.5000    1.3938
   0.6000    1.1359
   0.7000    1.0096
   0.8000    1.0343
   0.9000    0.8435
   1.0000    0.6856
   1.1000    0.6100
   1.2000    0.5392
   1.3000    0.3946
   1.4000    0.3903
   1.5000    0.5474
   1.6000    0.3459
   1.7000    0.1370
   1.8000    0.2211
   1.9000    0.1704
   2.0000    0.2636];

Plot the data points.

t = xdata(:,1);
y = xdata(:,2);
plot(t,y,"o")
title("Data points")

Try to fit the function

$$y(t)=c_1\exp (-{\lambda }_{1}t)+c_2\exp (-{\lambda }_{2}t)$$

to the data.

Solution Approach Using lsqcurvefit

To fit a function to data using lsqcurvefit, map the parameters in the $$y(t)$$ function to a variable x as follows:

x(1) = $$c_1$$

x(2) = $${\lambda }_1$$

x(3) = $$c_2$$

x(4) = $${\lambda }_2$$

Define the parameters to fit, x, in a function handle along with the data xdata.

F = @(x,xdata)x(1)*exp(-x(2)*xdata) + x(3)*exp(-x(4)*xdata);

Arbitrarily set the initial point x0 as follows: $$c_1$$ = 1, $${\lambda }_1$$ = 1, $$c_2$$ = 1, $${\lambda }_2$$ = 0.

x0 = [1 1 1 0];

Run the lsqcurvefit solver and plot the resulting fit.

[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,t,y)

Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

<stopping criteria details>

x = 1×4

    3.0068   10.5869    2.8891    1.4003

resnorm = 
0.1477

exitflag = 
3

output = struct with fields:
      firstorderopt: 7.8830e-06
         iterations: 6
          funcCount: 35
       cgiterations: 0
          algorithm: 'trust-region-reflective'
           stepsize: 0.0096
            message: 'Local minimum possible.↵↵lsqcurvefit stopped because the final change in the sum of squares relative to ↵its initial value is less than the value of the function tolerance.↵↵<stopping criteria details>↵↵Optimization stopped because the relative sum of squares (r) is changing↵by less than options.FunctionTolerance = 1.000000e-06.'
       bestfeasible: []
    constrviolation: []

hold on
plot(t,F(x,t))
hold off

Solution Approach Using fminunc

To solve the problem using fminunc, define the objective function as the sum of squares of the residuals.

Fsumsquares = @(x)sum((F(x,t) - y).^2);
[xunc,ressquared,eflag,outputu] = ...
    fminunc(Fsumsquares,x0)

Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.

<stopping criteria details>

xunc = 1×4

    2.8890    1.4003    3.0069   10.5862

ressquared = 
0.1477

eflag = 
1

outputu = struct with fields:
       iterations: 30
        funcCount: 185
         stepsize: 0.0017
     lssteplength: 1
    firstorderopt: 2.9662e-05
        algorithm: 'quasi-newton'
          message: 'Local minimum found.↵↵Optimization completed because the size of the gradient is less than↵the value of the optimality tolerance.↵↵<stopping criteria details>↵↵Optimization completed: The first-order optimality measure, 9.409720e-07, is less ↵than options.OptimalityTolerance = 1.000000e-06.'

Notice that fminunc finds the same solution as lsqcurvefit, but takes many more function evaluations to do so. The parameters for fminunc are in the opposite order as those for lsqcurvefit; the larger $$\lambda$$ is $${\lambda }_2$$, not $${\lambda }_1$$. This should not be surprising, because the order of variables is arbitrary.

fprintf(['There were %d iterations using fminunc,' ...
    ' and %d using lsqcurvefit.\n'], ...
    outputu.iterations,output.iterations)

There were 30 iterations using fminunc, and 6 using lsqcurvefit.

fprintf(['There were %d function evaluations using fminunc,' ...
    ' and %d using lsqcurvefit.'], ...
    outputu.funcCount,output.funcCount)

There were 185 function evaluations using fminunc, and 35 using lsqcurvefit.

Splitting the Linear and Nonlinear Problems

Notice that the fitting problem is linear in the parameters $$c_1$$ and $$c_2$$. This means for any values of $${\lambda }_1$$ and $${\lambda }_2$$, you can use the backslash operator to find the values of $$c_1$$ and $$c_2$$ that solve the least-squares problem.

Rework the problem as a two-dimensional problem, searching for the best values of $${\lambda }_1$$ and $${\lambda }_2$$. The values of $$c_1$$ and $$c_2$$ are calculated at each step using the backslash operator as described above.

function yEst = fitvector(lam,xdata,ydata)
%FITVECTOR  Used by DATDEMO to return value of fitting function.
%   yEst = FITVECTOR(lam,xdata) returns the value of the fitting function, y
%   (defined below), at the data points xdata with parameters set to lam.
%   yEst is returned as a N-by-1 column vector, where N is the number of
%   data points.
%
%   FITVECTOR assumes the fitting function, y, takes the form
%
%     y =  c(1)*exp(-lam(1)*t) + ... + c(n)*exp(-lam(n)*t)
%
%   with n linear parameters c, and n nonlinear parameters lam.
%
%   To solve for the linear parameters c, we build a matrix A
%   where the j-th column of A is exp(-lam(j)*xdata) (xdata is a vector).
%   Then we solve A*c = ydata for the linear least-squares solution c,
%   where ydata is the observed values of y.

A = zeros(length(xdata),length(lam));  % Build the A matrix.
for j = 1:length(lam)
   A(:,j) = exp(-lam(j)*xdata);
end
c = A\ydata; % Solve A*c = y for linear parameters c.
yEst = A*c; % Return the estimated response based on c.
end

Solve the problem using lsqcurvefit, starting from a two-dimensional initial point [$${\lambda }_1{\lambda }_2$$]:

x02 = [1 0];
F2 = @(x,t) fitvector(x,t,y);

[x2,resnorm2,~,exitflag2,output2] = lsqcurvefit(F2,x02,t,y)

Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

<stopping criteria details>

x2 = 1×2

   10.5861    1.4003

resnorm2 = 
0.1477

exitflag2 = 
3

output2 = struct with fields:
      firstorderopt: 4.4087e-06
         iterations: 10
          funcCount: 33
       cgiterations: 0
          algorithm: 'trust-region-reflective'
           stepsize: 0.0080
            message: 'Local minimum possible.↵↵lsqcurvefit stopped because the final change in the sum of squares relative to ↵its initial value is less than the value of the function tolerance.↵↵<stopping criteria details>↵↵Optimization stopped because the relative sum of squares (r) is changing↵by less than options.FunctionTolerance = 1.000000e-06.'
       bestfeasible: []
    constrviolation: []

The efficiency of the two-dimensional solution is similar to that of the four-dimensional solution:

fprintf(['There were %d function evaluations using the 2-d ' ...
    'formulation, and %d using the 4-d formulation.'], ...
    output2.funcCount,output.funcCount)

There were 33 function evaluations using the 2-d formulation, and 35 using the 4-d formulation.

Split Problem is More Robust to Initial Guess

Choosing a bad starting point for the original four-parameter problem leads to a local solution that is not global. However, choosing a starting point with the same bad $${\lambda }_1$$ and $${\lambda }_2$$ values for the split two-parameter problem leads to the global solution. To demonstrate this result, re-run the original problem with a start point that leads to a relatively bad local solution, and compare the resulting fit with the global solution.

x0bad = [5 1 1 0];
[xbad,resnormbad,~,exitflagbad,outputbad] = ...
    lsqcurvefit(F,x0bad,t,y)

Local minimum possible.

lsqcurvefit stopped because the final change in the sum of squares relative to 
its initial value is less than the value of the function tolerance.

<stopping criteria details>

xbad = 1×4

  -22.9036    2.4792   28.0273    2.4791

resnormbad = 
2.2173

exitflagbad = 
3

outputbad = struct with fields:
      firstorderopt: 0.0056
         iterations: 32
          funcCount: 165
       cgiterations: 0
          algorithm: 'trust-region-reflective'
           stepsize: 0.0021
            message: 'Local minimum possible.↵↵lsqcurvefit stopped because the final change in the sum of squares relative to ↵its initial value is less than the value of the function tolerance.↵↵<stopping criteria details>↵↵Optimization stopped because the relative sum of squares (r) is changing↵by less than options.FunctionTolerance = 1.000000e-06.'
       bestfeasible: []
    constrviolation: []

hold on
plot(t,F(xbad,t),'g')
legend("Data","Global fit","Bad local fit","Location","NE")
hold off

fprintf(['The residual norm at the good ending point is %f,\r' ...
   'and the residual norm at the bad ending point is %f.'], ...
   resnorm,resnormbad)

The residual norm at the good ending point is 0.147723,
and the residual norm at the bad ending point is 2.217300.