Nonlinear Inequality Constraints

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This example shows how to solve a scalar minimization problem with nonlinear inequality constraints. The problem is to find $x$ that solves

$\min_{x} f (x) = e^{x_{1}} (4 x_{1}^{2} + 2 x_{2}^{2} + 4 x_{1} x_{2} + 2 x_{2} + 1),$

subject to the constraints

$\begin{array}{l} x_{1} x_{2} - x_{1} - x_{2} \leq - 1.5 \\ x_{1} x_{2} \geq - 10 . \end{array}$

Because neither of the constraints is linear, create a function,confun.m, that returns the value of both constraints in a vectorc. Because thefminconsolver expects the constraints to be written in the form $c (x) \leq 0$ , write your constraint function to return the following value:

$c (x) = [\begin{array}{c} x_{1} x_{2} - x_{1} - x_{2} + 1.5 \\ - 10 - x_{1} x_{2} \end{array}]$ .

Create Objective Function

The helper functionobjfunis the objective function; it appears at theend of this example. Set thefunargument as a function handle to theobjfunfunction.

fun = @objfun;

Create Nonlinear Constraint Function

Nonlinear constraint functions must return two arguments:c, the inequality constraint, andceq, the equality constraint. Because this problem has no equality constraint, the helper functionconfunat theend of this examplereturns[]as the equality constraint.

Solve Problem

Set the initial point to[1].

x0 = [1];

The problem has no bounds or linear constraints. Set those arguments to[].

A = []; b = []; Aeq = []; beq = []; lb = []; ub = [];

Solve the problem usingfmincon.

[x,fval] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub,@confun)

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.

x =1×2-9.5473 1.0474

fval = 0.0236

Examine Solution

The exit message indicates that the solution is feasible with respect to the constraints. To double-check, evaluate the nonlinear constraint function at the solution. Negative values indicate satisfied constraints.

[c,ceq] = confun(x)

c =2×110^-4× -0.3179 -0.3063

ceq = []

Both nonlinear constraints are negative and close to zero, indicating that the solution is feasible and that both constraints are active at the solution.

Helper Functions

这段代码创建了objfunhelper function.

functionf = objfun(x) f = exp(x(1))*(4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);end

这段代码创建了confunhelper function.

function[c,ceq] = confun(x)% Nonlinear inequality constraintsc = [1.5 + x(1)*x(2) - x(1) - x(2); -x(1)*x(2) - 10];% Nonlinear equality constraintsceq = [];end