This example show how to solve a nonlinear problem with bounds using the `fmincon`

`trust-region-reflective`

algorithm. This algorithm has provisions for added efficiency when the problem is sparse, has an analytic gradient, and has known structure such as its Hessian pattern.

For a given $$n$$ that is a positive multiple of 4, the objective function is

$$f(x)=1+\sum _{i=1}^{n}{\left|\left(3-2{x}_{i}\right){x}_{i}-{x}_{i-1}-{x}_{i+1}+1\right|}^{p}+\sum _{i=1}^{n/2}{\left|{x}_{i}+{x}_{i+n/2}\right|}^{p},$$

where $$p=7/3$$, $${x}_{0}=0$$, and $${x}_{n+1}=0$$. The `tbroy`

function at the end of this example implements the objective function, including its gradient.

The problem has bounds: $$-10\le {x}_{i}\le 10$$ for all $$i$$. Use $$n$$ = 800.

n = 800; lb = -10*ones(n,1); ub = -lb;

The sparsity pattern of the Hessian matrix has been predetermined and stored in the file `tbroyhstr.mat`

. The sparsity structure for the Hessian of this problem is banded, as you can see in the following spy plot.

```
load tbroyhstr
spy(Hstr)
```

In this plot, the center stripe is itself a five-banded matrix. The following plot shows the matrix more clearly.

spy(Hstr(1:20,1:20))

Set options to use the `trust-region-reflective`

algorithm. This algorithm requires you to set the `SpecifyObjectiveGradient`

option to `true`

.

Also, use `optimoptions`

to set the `HessPattern`

option to `Hstr`

. When a problem as large as this has obvious sparsity structure, not setting the `HessPattern`

option requires a huge amount of unnecessary memory and computation. This is because `fmincon`

attempts to use finite differencing on a full Hessian matrix of 640,000 nonzero entries.

options = optimoptions('fmincon','SpecifyObjectiveGradient',true,'HessPattern',Hstr,... 'Algorithm','trust-region-reflective');

Set the initial point to –1 for odd indices and +1 for even indices.

x0 = -ones(n,1); x0(2:2:n) = 1;

There are no linear or nonlinear constraints, so set those parameters to `[]`

.

A = []; b = []; Aeq = []; beq = []; nonlcon = [];

Call `fmincon`

to solve the problem.

[x,fval,exitflag,output] = ... fmincon(@tbroyfg,x0,A,b,Aeq,beq,lb,ub,nonlcon,options);

Local minimum possible. fmincon stopped because the final change in function value relative to its initial value is less than the value of the function tolerance.

Examine the exit flag, objective function value, first-order optimality measure, and number of solver iterations.

disp(exitflag);

3

disp(fval)

270.4790

disp(output.firstorderopt)

0.0163

disp(output.iterations)

7

`fmincon`

did not take very many iterations to reach a solution. However, the solution has a relatively high first-order optimality measure, which is the reason that the exit flag is not the more preferable value of 1.

Try using a five-banded preconditioner instead of the default diagonal preconditioner. Using `optimoptions`

, set the `PrecondBandWidth`

option to 2 and solve the problem again. (The bandwidth is the number of upper (or lower) diagonals, not counting the main diagonal.)

options.PrecondBandWidth = 2; [x2,fval2,exitflag2,output2] = ... fmincon(@tbroyfg,x0,A,b,Aeq,beq,lb,ub,nonlcon,options);

Local minimum possible. fmincon stopped because the final change in function value relative to its initial value is less than the value of the function tolerance.

disp(exitflag2);

3

disp(fval2)

270.4790

disp(output2.firstorderopt)

7.5340e-05

disp(output2.iterations)

9

The exit flag and objective function value do not appear to change. However, the number of iterations increased, and the first-orer optimality measure decreased considerably. Compute the difference in objective function value.

disp(fval2 - fval)

-2.9005e-07

The objective function value decreased by a tiny amount. The improvement in the solution is only an improvement in the first-order optimality measure, not in the objective function.

This code creates the `tbroy`

function.

function [f,grad] = tbroyfg(x,dummy) %TBROYFG Objective function and its gradients for nonlinear minimization % with bound constraints and banded preconditioner. % Documentation example. % Copyright 1990-2008 The MathWorks, Inc. n = length(x); % n should be a multiple of 4 p = 7/3; y=zeros(n,1); i = 2:(n-1); y(i) = abs((3-2*x(i)).*x(i) - x(i-1) - x(i+1)+1).^p; y(n) = abs((3-2*x(n)).*x(n) - x(n-1)+1).^p; y(1) = abs((3-2*x(1)).*x(1) - x(2)+1).^p; j = 1:(n/2); z = zeros(length(j),1); z(j) = abs(x(j) + x(j+n/2)).^p; f = 1 + sum(y) + sum(z); % % Evaluate the gradient. if nargout > 1 g = zeros(n,1); t = zeros(n,1); i = 2:(n-1); t(i) = (3-2*x(i)).*x(i) - x(i-1) - x(i+1) + 1; g(i) = p*abs(t(i)).^(p-1).*sign(t(i)).*(3-4*x(i)); g(i-1) = g(i-1) - p*abs(t(i)).^(p-1).*sign(t(i)); g(i+1) = g(i+1) - p*abs(t(i)).^(p-1).*sign(t(i)); tt = (3-2*x(n)).*x(n) - x(n-1) + 1; g(n) = g(n) + p*abs(tt).^(p-1).*sign(tt).*(3-4*x(n)); g(n-1) = g(n-1) - p*abs(tt).^(p-1).*sign(tt); tt = (3-2*x(1)).*x(1)-x(2)+1; g(1) = g(1) + p*abs(tt).^(p-1).*sign(tt).*(3-4*x(1)); g(2) = g(2) - p*abs(tt).^(p-1).*sign(tt); j = 1:(n/2); t(j) = x(j) + x(j+n/2); g(j) = g(j) + p*abs(t(j)).^(p-1).*sign(t(j)); jj = j + (n/2); g(jj) = g(jj) + p*abs(t(j)).^(p-1).*sign(t(j)); grad = g; end end