Example 2

The second example⁵⁹ solves the following 3-dimensional constrained optimization, printing the details of the solver’s progress:

First, create a Matlab function to evaluate the objective function and its gradients,⁶⁰

function [f, df, d2f] = f2(x)
f = -x(1)*x(2) - x(2)*x(3);
if nargout > 1           %% gradient is required
    df = -[x(2); x(1)+x(3); x(2)];
    if nargout > 2       %% Hessian is required
        d2f = -[0 1 0; 1 0 1; 0 1 0];   %% actually not used since
    end                                 %% 'hess_fcn' is provided
end

one to evaluate the constraints, in this case inequalities only, and their gradients,

function [h, g, dh, dg] = gh2(x)
h = [ 1 -1 1; 1 1 1] * x.^2 + [-2; -10];
dh = 2 * [x(1) x(1); -x(2) x(2); x(3) x(3)];
g = []; dg = [];

function Lxx = hess2(x, lam, cost_mult)
if nargin < 3, cost_mult = 1; end %% allows to be used with 'fmincon'
mu = lam.ineqnonlin;
Lxx = cost_mult * [0 -1 0; -1 0 -1; 0 -1 0] + ...
[2*[1 1]*mu 0 0; 0 2*[-1 1]*mu 0; 0 0 2*[1 1]*mu];

Then create a problem struct with handles to these functions, a starting value for $x$ and an option to print the solver’s progress. Finally, pass this struct to mips to solve the problem and print some of the return values to get the output below.

function example2
problem = struct( ...
    'f_fcn',    @(x)f2(x), ...
    'gh_fcn',   @(x)gh2(x), ...
    'hess_fcn', @(x, lam, cost_mult)hess2(x, lam, cost_mult), ...
    'x0',       [1; 1; 0], ...
    'opt',      struct('verbose', 2) ...
);
[x, f, exitflag, output, lambda] = mips(problem);
fprintf('\nf = %g   exitflag = %d\n', f, exitflag);
fprintf('\nx = \n');
fprintf('   %g\n', x);
fprintf('\nlambda.ineqnonlin =\n');
fprintf('   %g\n', lambda.ineqnonlin);

>> example2
MATPOWER Interior Point Solver -- MIPS, Version 1.3.1, 20-Jun-2019
(using built-in linear solver)
it    objective   step size   feascond     gradcond     compcond     costcond
----  ------------ --------- ------------ ------------ ------------ ------------
  0            -1                       0          1.5            5            0
  1    -5.3250167     1.6875            0     0.894235     0.850653      2.16251
  2    -7.4708991    0.97413     0.129183   0.00936418     0.117278     0.339269
  3    -7.0553031    0.10406            0   0.00174933    0.0196518    0.0490616
  4    -7.0686267   0.034574            0   0.00041301    0.0030084   0.00165402
  5    -7.0706104  0.0065191            0  1.53531e-05  0.000337971  0.000245844
  6    -7.0710134 0.00062152            0  1.22094e-07  3.41308e-05  4.99387e-05
  7    -7.0710623 5.7217e-05            0  9.84879e-10  3.41587e-06  6.05875e-06
  8    -7.0710673 5.6761e-06            0  9.73527e-12  3.41615e-07  6.15483e-07
Converged!

f = -7.07107   exitflag = 1

x =
   1.58114
   2.23607
   1.58114

lambda.ineqnonlin =
   0
   0.707107

A.2 Example 2