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Problem definition and solution technique

This example shows how to solve a nonlinear least squares problem in two ways. It first shows the solution without using a Jacobian function. Then it shows how to include a Jacobian, and it shows the efficiency improvement that the Jacobian gives.

The problem has 10 terms with 2 unknowns: find x, a two-dimensional vector, that minimizes

starting at the point x0 = [0.3,0.4].

Because lsqnonlin assumes that the sum of squares is not explicitly formed in the user function, the function passed to lsqnonlin should compute the vector valued function

for k = 1 to 10 (that is, F should have 10 components).

Step 1: Write a file myfun.m that computes the objective function values.

function F = myfun(x)
k = 1:10;
F = 2 + 2*k-exp(k*x(1))-exp(k*x(2));

Step 2: Call the nonlinear least-squares routine.

x0 = [0.3,0.4]; % Starting guess
[x,resnorm,res,eflag,output1] = lsqnonlin(@myfun,x0); % Invoke optimizer

Because the Jacobian is not computed in myfun.m, lsqnonlin calls the trust-region reflective algorithm with full finite differencing. Note that the SpecifyObjectiveGradient option in options is set to false by default.

After 72 function evaluations, this example gives the solution

x,resnorm

x = 
     0.2578   0.2578

resnorm = 
     124.3622

Most computer systems can handle much larger full problems, say into the hundreds of equations and variables. But if there is some sparsity structure in the Jacobian (or Hessian) that can be taken advantage of, the large-scale methods always runs faster if this information is provided.

Step 3: Include a Jacobian.

The objective function is simple enough to calculate its Jacobian. Following the definition in Jacobians of Vector Functions, a Jacobian function represents the matrix

Here, F_k(x is the kth component of the objective function. This example has

so

Modify the objective function file.

function [F,J] = myfun(x)
k = 1:10;
F = 2 + 2*k-exp(k*x(1))-exp(k*x(2));
if nargout > 1
    J = zeros(10,2);
    J(k,1) = -k.*exp(k*x(1));
    J(k,2) = -k.*exp(k*x(2));
end

Set options so the solver uses the Jacobian.

opts = optimoptions(@lsqnonlin,'SpecifyObjectiveGradient',true);

Run the solver.

x0 = [0.3 0.4]; % Starting guess
[x,resnorm,res,eflag,output2] = lsqnonlin(@myfun,x0,[],[],opts);

The solution is the same as before.

x,resnorm

x = 
     0.2578   0.2578

resnorm = 
     124.3622

The advantage to using a Jacobian is that the solver takes fewer function evaluations, 24 instead of 72.

[output1.funcCount,output2.funcCount]

ans =
    72    24