I am trying to use the function fmincon for multiple nonlinear inequality constraints, and the output given does not satisfy all the inequality constraint

objective = @(x) - 16000*x(1)^2 - 32000*x(2)^2 - 40000*x(3)^2;
x0 = [0.05, 0.05, 0.07];
A = [];
b = [];
aeq = [1,1,1];
beq =0.6;
lb = [0.05, 0.05, 0.07];
ub = [];
nonlincon = @nlcon;
fmincon(objective, x0, A,b,aeq,beq,lb,ub,nonlincon,options)

output: 0.2833 0.0500 0.2667

c(1) = 16000*x(1)^2 - 1000 = 284.44444 %(which is not less than 0)

function [c,ceq] = nlcon(x)
c(1)= 16000*x(1)^2 - 1000;
c(2)= 32000*x(2)^2 - 1500;
c(3)= 40000*x(3)^2 - 2000;
end

[Torsten]

And what is the exitflag from fmincon ?

[x,fval,exitflag,output] = fmincon(___)

[Matt J]

Works fine for me:

objective = @(x) - 16000*x(1)^2 - 32000*x(2)^2 - 40000*x(3)^2;
x0 = [0.05, 0.05, 0.07];
A = [];
b = [];
aeq = [1,1,1];
beq =0.6;
lb = [0.05, 0.05, 0.07];
ub = [];
nonlincon = @nlcon;
x=fmincon(objective, x0, A,b,aeq,beq,lb,ub,nonlincon)

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 = 1×3

0.1599 0.2165 0.2236

c1 = 16000*x(1)^2 - 1000

c1 = -590.9768

function [c,ceq] = nlcon(x)
c(1)= 16000*x(1)^2 - 1000;
c(2)= 32000*x(2)^2 - 1500;
c(3)= 40000*x(3)^2 - 2000;
ceq=[];
end

Answer 1

[Matt J]

As a general rule, you should avoid nonlinear constraints when they have a linear equivalent. In your case, the nonlinear constraints can be re-expressed as simple bounds:

objective = @(x) - 16000*x(1)^2 - 32000*x(2)^2 - 40000*x(3)^2;
x0 = [0.05, 0.05, 0.07];
A = [];
b = [];
aeq = [1,1,1];
beq =0.6;

lb = [0.05, 0.05, 0.07];
ub = sqrt( [1000, 1500, 2000]./[16000,32000,40000] );

x=fmincon(objective, x0, A,b,aeq,beq,lb,ub)

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 = 1×3

0.1599 0.2165 0.2236