Optimization Toolbox 

This example shows how to fit a nonlinear function to data using several Optimization Toolbox™ algorithms.
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Solution Approach Using lsqcurvefit Solution Approach Using fminunc 
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
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the default value of the function tolerance. x = 3.0068 10.5869 2.8891 1.4003 resnorm = 0.1477 exitflag = 3 output = firstorderopt: 7.8802e06 iterations: 6 funcCount: 35 cgiterations: 0 algorithm: 'trustregionreflective' message: 'Local minimum possible. lsqcurvefit stopped because t...'
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','quasinewton'); [xunc,ressquared,eflag,outputu] = ... fminunc(Fsumsquares,x0,opts)
Local minimum found. Optimization completed because the size of the gradient is less than the default value of the function tolerance. xunc = 2.8890 1.4003 3.0069 10.5862 ressquared = 0.1477 eflag = 1 outputu = iterations: 30 funcCount: 185 stepsize: 1 firstorderopt: 2.9476e05 algorithm: 'mediumscale: QuasiNewton line search' message: 'Local minimum found. Optimization completed because t...'
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)
There were 30 iterations using fminunc, and 6 using lsqcurvefit. 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 lam(1) and lam(2), we can use the backslash operator to find the values of c(1) and c(2) that solve the leastsquares problem.
We now rework the problem as a twodimensional 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
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 Nby1 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 jth column of A is exp(lam(j)*xdata) (xdata is a vector). % Then we solve A*c = ydata for the linear leastsquares solution c, % where ydata is the observed values of y. A = zeros(length(xdata),length(lam)); % build 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
Solve the problem using lsqcurvefit, starting from a twodimensional 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)
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the default value of the function tolerance. x2 = 10.5861 1.4003 resnorm2 = 0.1477 exitflag2 = 3 output2 = firstorderopt: 4.4085e06 iterations: 10 funcCount: 33 cgiterations: 0 algorithm: 'trustregionreflective' message: 'Local minimum possible. lsqcurvefit stopped because t...'
The efficiency of the twodimensional solution is similar to that of the fourdimensional solution:
fprintf(['There were %d function evaluations using the 2d ' ... 'formulation, and %d using the 4d formulation.'], ... output2.funcCount,output.funcCount)
There were 33 function evaluations using the 2d formulation, and 35 using the 4d formulation.
Split Problem is More Robust to Initial Guess
Choosing a bad starting point for the original fourparameter 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 twoparameter problem leads to the global solution. To show this we rerun 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)
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the default value of the function tolerance. xbad = 22.9036 2.4792 28.0273 2.4791 resnormbad = 2.2173 exitflagbad = 3 outputbad = firstorderopt: 0.0058 iterations: 32 funcCount: 165 cgiterations: 0 algorithm: 'trustregionreflective' message: 'Local minimum possible. lsqcurvefit stopped because t...' The residual norm at the good ending point is 0.147723, and the residual norm at the bad ending point is 2.217300.