Note: This page has been translated by MathWorks. Click here to see

To view all translated materials including this page, select Country from the country navigator on the bottom of this page.

To view all translated materials including this page, select Country from the country navigator on the bottom of this page.

This example shows how to use the Optimization app with the `fmincon`

solver
to minimize a quadratic subject to linear and nonlinear constraints
and bounds.

The Optimization app warns that it will be removed in a future release.

Consider the problem of finding [*x*_{1}, *x*_{2}]
that solves

$$\underset{x}{\mathrm{min}}f(x)={x}_{1}^{2}+{x}_{2}^{2}$$

subject to the constraints

$$\begin{array}{rr}\hfill 0.5\le {x}_{1}& \hfill \text{(bound)}\\ \hfill -{x}_{1}-{x}_{2}+1\le 0& \hfill \text{(linearinequality)}\\ \hfill \begin{array}{r}\hfill -{x}_{1}^{2}-{x}_{2}^{2}+1\le 0\\ \hfill -9{x}_{1}^{2}-{x}_{2}^{2}+9\le 0\\ \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-{x}_{1}^{2}+{x}_{2}\le 0\\ \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-{x}_{2}^{2}+{x}_{1}\le 0\end{array}\}& \hfill \text{(nonlinearinequality)}\end{array}$$

The starting guess for this problem is *x*_{1} =
3 and *x*_{2} = 1.

function f = objecfun(x) f = x(1)^2 + x(2)^2;

function [c,ceq] = nonlconstr(x) c = [-x(1)^2 - x(2)^2 + 1; -9*x(1)^2 - x(2)^2 + 9; -x(1)^2 + x(2); -x(2)^2 + x(1)]; ceq = [];

Enter

`optimtool`

in the Command Window to open the Optimization app.Select

`fmincon`

from the selection of solvers and change the**Algorithm**field to`Active set`

.Enter

`@objecfun`

in the**Objective function**field to call the`objecfun.m`

file.Enter

`[3;1]`

in the**Start point**field.Define the constraints.

Set the bound

`0.5`

≤*x*_{1}by entering`[0.5,-Inf]`

in the**Lower**field. The`-Inf`

entry means there is no lower bound on*x*_{2}.Set the linear inequality constraint by entering

`[-1 -1]`

in the**A**field and enter`-1`

in the**b**field.Set the nonlinear constraints by entering

`@nonlconstr`

in the**Nonlinear constraint function**field.

In the

**Options**pane, expand the**Display to command window**option if necessary, and select`Iterative`

to show algorithm information at the Command Window for each iteration.Click the

**Start**button as shown in the following figure.When the algorithm terminates, under

**Run solver and view results**the following information is displayed:The

**Current iteration**value when the algorithm terminated, which for this example is`7`

.The final value of the objective function when the algorithm terminated:

Objective function value: 2.0000000268595803

The algorithm termination message:

Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.

The final point, which for this example is

1 1

In the Command Window, the algorithm information is displayed for each iteration:

Max Line search Directional First-order Iter F-count f(x) constraint steplength derivative optimality Procedure 0 3 10 2 Infeasible start point 1 6 4.84298 -0.1322 1 -5.22 1.74 2 9 4.0251 -0.01168 1 -4.39 4.08 Hessian modified twice 3 12 2.42704 -0.03214 1 -3.85 1.09 4 15 2.03615 -0.004728 1 -3.04 0.995 Hessian modified twice 5 18 2.00033 -5.596e-05 1 -2.82 0.0664 Hessian modified twice 6 21 2 -5.326e-09 1 -2.81 0.000522 Hessian modified twice Active inequalities (to within options.ConstraintTolerance = 1e-06): lower upper ineqlin ineqnonlin 3 4 Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.