Linear constraints not satisfied during optimization using fmincon
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Hi, I am relatively new to optimization and I am trying to optimize a nonlinear cost function with linear constraints. Using fmincon the optimization completes after a few iterations but the ans I recieved for my variables doesnt satisfy the constraints.
Optimization function
function [x, fval, iter] = myrun
x0 = [800;350;200;250];
A = [-1 -1 -1 -1;
1 1 1 1;
-0.59 -15.44 -7.12 -0.62;
-79.77 0 0 -16.15;
-6.65 -59.1 -25.94 -4];
b = [1360;
2268;
70;
560;
114.3];
Aeq = [];
beq = [];
lb = [360,250,150,120];
ub = [1080,500,300,360];
options = optimoptions(@fmincon,'Display','iter');
[x,fval,exitflag, output] = fmincon(@myfun,x0,A,b,Aeq,beq,lb,ub,@nlcon,options);
end
and my objective function is
function f = myfun(x)
f = (0.0114*x(1)+0.01032*x(2)+0.0228*x(3)+0.0081*x(4))+15*((log(0.0114*x(1)))+(log(0.01032*x(2)))+(log(0.0228*x(3)))+(log(0.0081*x(4))));
end
3 Comments
John D'Errico
on 9 Apr 2022
When you post a PICTURE of the code you wrote, anyone who might be tempted to help you must then retype your code from the picture.
If you paste in the TEXT of your code, then they can copy it, and then paste it directly into MATLAB. It is easier for you to paste in your code too. In this case, it becomes far easier for someone to offer to help you.
So is there a good reason why you want to make it more difficult to get help on your homework problem?
MOHAMMAD BAWANY
on 9 Apr 2022
John D'Errico
on 9 Apr 2022
And now I can actually look at your question.
Accepted Answer
John D'Errico
on 9 Apr 2022
Edited: John D'Errico
on 9 Apr 2022
First, learn to use function handles.
myfun = @(x) (0.0114*x(1)+0.01032*x(2)+0.0228*x(3)+0.0081*x(4))+15*((log(0.0114*x(1)))+(log(0.01032*x(2)))+(log(0.0228*x(3)))+(log(0.0081*x(4))));
Test to see if your function is well posed. Does it return a scalar result?
x0 = [800;350;200;250];
myfun(x0)
Next, I'll change the objective slightly. Does it change value when I do?
myfun(x0 + randn(size(x0)))
So things seem to be well posed as an optimization problem. ALWAYS do tests like this.
A = [-1 -1 -1 -1;
1 1 1 1;
-0.59 -15.44 -7.12 -0.62;
-79.77 0 0 -16.15;
-6.65 -59.1 -25.94 -4];
b = [1360;
2268;
70;
560;
114.3];
Aeq = [];
beq = [];
lb = [360,250,150,120];
ub = [1080,500,300,360];
Next, does your function satisfy the constraints? Clearly, your start point satisfies the bound constraints.
[A*x0,b]
The constraints for fmincon require that A*xo <= b. In fact, that is again true. So you have made a good start.
How about the nonlinear constraints? Surprisingly, you have shown NO nonloinear constraints? So why pass what you don't have into fmincon, unless you have them, and just forgot to tell us? Sigh. I'll assume there are none, and that you think you need to pass something in as a placeholder.
options = optimoptions(@fmincon,'Display','iter');
[x,fval,exitflag, output] = fmincon(myfun,x0,A,b,Aeq,beq,lb,ub,[],options);
x
fval
exitflag
Fmincon seems happy with the result, but I notice the solution lives at exactly the lower bounds. That won't necessarily be a good thing.
[A*x,b]
The inequality constraints are satisfied. I fail to see the problem. Perhaps you really have nonlinear constraints in some form, and you forgot to tell us.
2 Comments
MOHAMMAD BAWANY
on 9 Apr 2022
John D'Errico
on 9 Apr 2022
Edited: John D'Errico
on 9 Apr 2022
I never like to see an optimization where all of the parameters end up at exactly the bounds. That is often an issue in my eyes, since often, those bounds may have been chosen fairly arbitrarily.
Anyway, my first question is if you get a different result from some other code, but the solution is clearly at the lower bounds, then have you coded the SAME objective function? I would look at that VERY closely. I really cannot know. But you might check to see if you used the same log function? For example, might you have used log to the base 10 in the other tool?
Looking at your objective, the nice thing about Answers is it gives us a pretty printed result.
syms x y z w
(0.0114*x+0.01032*y+0.0228*z+0.0081*w)+15*((log(0.0114*x))+(log(0.01032*y))+(log(0.0228*z))+(log(0.0081*w)))
Do you see that ALL of the coefficients are positive? As such, we can make it as small as possible, merely by making x,y,z,w ALL as small as possible. Think about it. Is the log function one that is STRICTLY monotone increasing? How about each of those linear terms? Can we make than all as small as possible, merely by using the lower bounds, so by making x,y,z,w all as small as possible? Anyway, the solution would have been identically the same if you used some other log base, since all logs are monotone increasing.
So as long as the lower bounds for your problem satisfy the linear inequality constraints, then the answer is trivial.
A = [-1 -1 -1 -1;
1 1 1 1;
-0.59 -15.44 -7.12 -0.62;
-79.77 0 0 -16.15;
-6.65 -59.1 -25.94 -4];
b = [1360;
2268;
70;
560;
114.3];
lb = [360,250,150,120];
A*lb' <= b
Given that, the answer is trivial. Your lower bounds are indeed the solution to your minimization problem.
Worse, since your objective is of a separable form, where x,y,z,w are all independent, we could have solved this by 4 separate 1 variable optimizations.
What you did in any other code, we cannot possibly guess. And why you got something that you consider reasonable, again, I have no clue. Perhaps you have the signs on the inequality constraints wrong.
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