%% Nonlinear Data-Fitting
% This example shows how to fit a nonlinear function to data using several Optimization Toolbox™ algorithms.
%% Problem Setup
% Consider the following data:

Data = ...
  [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];

%
% Let's plot these data points.

t = Data(:,1);
y = Data(:,2);
% axis([0 2 -0.5 6])
% hold on
plot(t,y,'ro')
title('Data points')
% hold off

% We would like to fit the function
%    y =  c(1)*exp(-lam(1)*t) + c(2)*exp(-lam(2)*t)
% to the data.
%
%% Solution Approach Using lsqcurvefit
% The lsqcurvefit function solves this type of problem easily.
% To begin, define the parameters in terms of one variable x:
% x(1) = c(1)
% x(2) = lam(1)
% x(3) = c(2)
% x(4) = lam(2)
% Then define the curve as a function of the parameters x and the data t:

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

%
% We arbitrarily set our initial point x0 as follows: c(1) = 1, lam(1) = 1, c(2) = 1, lam(2) = 0:

x0 = [1 1 1 0];

%
% We run the solver and plot the resulting fit.

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

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

%
%% Solution Approach Using fminunc
% To solve the problem using fminunc, we set the objective function as the sum of squares of the residuals.

Fsumsquares = @(x)sum((F(x,t) - y).^2);
opts = optimoptions('fminunc','Algorithm','quasi-newton');
[xunc,ressquared,eflag,outputu] = ...
    fminunc(Fsumsquares,x0,opts)

%
% Notice that fminunc found the same solution as lsqcurvefit, but took many more function evaluations to do so. The parameters for fminunc are in the opposite order as those for lsqcurvefit; the larger lam is lam(2), not lam(1). This is not surprising, the order of variables is arbitrary.

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

%
%% 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 lam(1) and lam(2), we can use the backslash operator to find the values of c(1) and c(2) that solve the least-squares problem.
% We now rework the problem as a two-dimensional problem, searching for the best values of lam(1) and lam(2). The values of c(1) and c(2) are calculated at each step using the backslash operator as described above.

type fitvector

%
% Solve the problem using lsqcurvefit, starting from a two-dimensional initial point lam(1), lam(2):

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

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

%
% 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)

%
%% 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. Choosing a starting point with the same bad lam(1) and lam(2) values for the split two-parameter problem leads to the global solution. To show this we 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)

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,' ...
   ' and the residual norm at the bad ending point is %f.'], ...
   resnorm,resnormbad)

% Copyright 1990-2012 The MathWorks, Inc.