How can solve constrait optimization?

Hi everyone, I have constrait optimizatıon problem. I am trying to maxımize x, y and z. And some of my constrait are like that

0.2<=y
0.2<= x
0.2<=z<=2
x+y<=z

How can ı express the lass constrait ın matlab? Using A, b ,x0

Answer 1

[Walter Roberson]

A = [1 1 -1] b = 0 for A*X <= b, where X = [x; y; z], expresses x+y-z<=0 , which is x+y<=z

Your other constraints that you posted should be expressed as lb = [0.2, 0.2, 0.2], ub = [inf, inf, 2]

[Sinem Senel]

Thank you so much for your help. I understood now. And also when ı solve the problem manually ı find 1 1 and 2 But in matlab it did not me exact solution. İt give me 0.9993, 0.993, 1.993. But I am pretry sure my manual solution. What should ı do?

[Walter Roberson]

what is your code?

[Walter Roberson]

You can do a bit better by adjusting the tolerance

fun = @(x)-((1-x(1)-x(2))+ log(x(1))+log(x(2)));
lb = [0.2, 0.2,0.2];
ub = [inf,inf,2];
A = [1,1,-1];
b = [0];
Aeq = [];
beq = [];
x0 = [1/4,1/4,1.1];
nonlcon = [];

options = optimoptions('fmincon','Algorithm','interior-point',...
    'OptimalityTolerance',1e-20, ...
    'StepTolerance', 1e-20, ...
    'ConstraintTolerance', 1e-20, ...
    'Display', 'iter' );

[x, fval] = fmincon(fun, x0, A, b, Aeq, beq, lb, ub, nonlcon, options);
disp(x)
disp(fval)

The OptimalityTolerance is the only one that I found made a difference.

Unfortunately the function calls into barrier.p so I cannot trace to find out exactly why it is ending the optimization.

Answer 2

[Sinem Senel]

fun = @(x)-((1-x(1)-x(2))+ log(x(1))+log(x(2)));
lb = [0.2, 0.2,0.2];
ub = [inf,inf,2];
A = [1,1,-1];
b = [0];
Aeq = [];
beq = [];
x0 = [1/4,1/4,1.1]

x = fmincon(fun,x0,A,b,Aeq,beq,lb,ub)


%this is my entire code, i use x(2) as y variable and x(3) as z variable.