# problems with fmincon, how to solve?

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fernando brito 2020 年 8 月 28 日

can someone help me solve this problem, I tried to modify the code but it didn't solve it. I need to minimize the following equation.
objective function:
function cosmyfun(x)
cost = @(x) dfghjhgfx(1)^2 + x(2)^2 - 5*x(1) - 6*x(2) + 15;
end
function with restrictions of inequality and equality:
function [c,ceq]=mycon(x)
c(1) = x(1)^2 + x(2)^2 - 4*x(1) - 3*x(2) + 5;
c(2) = -x(1)^2 - x(2)^2 + 4*x(1) + 3*x(2) - 10;
ceq = 2*x(1)^2 + 2*x(2)^2 - 3*x(1) - 3*x(2) - 2;
end
script:
clc
clear all
Aeq=[];
beq=[];
A=[];
b=[];
lb=[-3,-3];
ub=[3,3];
x0=[4,4];
options = optimoptions('fmincon','Algorithm','sqp','Display','iter-detailed','MaxFunctionEvaluations',100000, 'MaxIterations',2000, 'OptimalityTolerance' ,1e-10);
[x,cost] = fmincon(@(x) myfun,x0,A,b,Aeq,beq,lb,ub,mycon)
error:
Error in mycon (line 2)
c(1) = x(1)^2 + x(2)^2 - 4*x(1) - 3*x(2) + 5;
Error in Example (line 11)
[x,cost] = fmincon(@(x) myfun,x0,A,b,Aeq,beq,lb,ub,mycon)

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### 採用された回答

Hussein Ammar 2020 年 8 月 29 日
Your problem is convex, so you can use the CVX package if you want. Also, the solution should be the global minimum; not a local minimum. To use fmincon, do the following:
Equality/nonEquality constraint:
function [c,ceq]=mycon(x)
c(1) = x(1)^2 + x(2)^2 - 4*x(1) - 3*x(2) + 5;
c(2) = -x(1)^2 - x(2)^2 + 4*x(1) + 3*x(2) - 6; % notice the fixed typo
ceq = 2*x(1)^2 + 2*x(2)^2 - 3*x(1) - 3*x(2) - 2;
end
Construct the problem and call fmincon:
clc
clear all
Aeq=[];
beq=[];
A=[];
b=[];
% if you are not sure that vector x is within lb and ub do not restrict it
lb = []; %lb=[-3,-3];
ub = []; %ub=[3,3]; %
x0=[4,4];
myObj = @(x) x(1)^2 + x(2)^2 - 5*x(1) - 6*x(2) + 15;
nonlcon = @mycon;
options = optimoptions('fmincon','Algorithm','sqp','Display','iter-detailed','MaxFunctionEvaluations',100000, 'MaxIterations',2000, 'OptimalityTolerance' ,1e-10);
[x,cost] = fmincon(myObj,x0,A,b,Aeq,beq,lb,ub,nonlcon)
Output (global minimum):
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.
<stopping criteria details>
x =
1.6450 1.9007
cost =
1.6896

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