Documentation |
On this page… |
---|
Problem definition and solution technique Step 1: Write a file myfun.m that computes the objective function values. |
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
$$\sum _{k=1}^{10}{\left(2+2k-{e}^{k{x}_{1}}-{e}^{k{x}_{2}}\right)}^{2}},$$
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
$${F}_{k}(x)=2+2k-{e}^{k{x}_{1}}-{e}^{k{x}_{2}},$$
for k = 1 to 10 (that is, F should have 10 components).
function F = myfun(x) k = 1:10; F = 2 + 2*k-exp(k*x(1))-exp(k*x(2));
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, and no Jacobian sparsity pattern is provided by the JacobPattern option in options, lsqnonlin calls the trust-region reflective algorithm with JacobPattern set to Jstr = sparse(ones(10,2)). This is the default for lsqnonlin. Note that the Jacobian option in options is set to 'off' by default.
When the finite-differencing routine is called initially, it detects that Jstr is actually a dense matrix, i.e., no speed benefit is derived from storing it as a sparse matrix. From then on, the finite-differencing routine uses Jstr = ones(10,2) (a full matrix) for the optimization computations.
After about 24 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.
The objective function is simple enough to calculate its Jacobian. Following the definition in Jacobians of Vector Functions, a Jacobian function represents the matrix
$${J}_{kj}(x)=\frac{\partial {F}_{k}(x)}{\partial {x}_{j}}.$$
Here, F_{k}(x is the kth component of the objective function. This example has
$${F}_{k}(x)=2+2k-{e}^{k{x}_{1}}-{e}^{k{x}_{2}},$$
so
$$\begin{array}{l}{J}_{k1}(x)=-k{e}^{k{x}_{1}}\\ {J}_{k2}(x)=-k{e}^{k{x}_{2}}.\end{array}$$
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,'Jacobian','on');
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