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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
$$\underset{x}{\mathrm{min}}f(x)={e}^{{x}_{1}}\left(4{x}_{1}^{2}+2{x}_{2}^{2}+4{x}_{1}{x}_{2}+2{x}_{2}+1\right).$$ | (6-56) |
subject to the constraints
x_{1}x_{2} – x_{1} – x_{2} ≤
–1.5,
x_{1}x_{2} ≥
–10.
Because neither of the constraints is linear, you cannot pass the constraints to fmincon at the command line. Instead you can create a second file, confun.m, that returns the value at both constraints at the current x in a vector c. The constrained optimizer, fmincon, is then invoked. Because fmincon expects the constraints to be written in the form c(x) ≤ 0, you must rewrite your constraints in the form
function f = objfun(x) f = exp(x(1))*(4*x(1)^2 + 2*x(2)^2 + 4*x(1)*x(2) + 2*x(2) + 1);
function [c, ceq] = confun(x) % Nonlinear inequality constraints c = [1.5 + x(1)*x(2) - x(1) - x(2); -x(1)*x(2) - 10]; % Nonlinear equality constraints ceq = [];
x0 = [-1,1]; % Make a starting guess at the solution options = optimoptions(@fmincon,'Algorithm','sqp'); [x,fval] = ... fmincon(@objfun,x0,[],[],[],[],[],[],@confun,options);
fmincon produces the solution x with function value fval:
x,fval x = -9.5474 1.0474 fval = 0.0236
You can evaluate the constraints at the solution by entering
[c,ceq] = confun(x)
This returns numbers close to zero, such as
c = 1.0e-14 * -0.6661 0.7105 ceq = []
Note that both constraint values are, to within a small tolerance, less than or equal to 0; that is, x satisfies c(x) ≤ 0.