Global Optimizationproblem using Global Search

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Hello, i am trying to use the globalSearch function to solve the following Optimization Problem:
min - x(3)
s.t. -x(1) -x(2) <= 0
-10*x(1)+x(1)^2-4*x(2)+x(2)^2+x(3) <= 0
3x(1) + x(2) <= 12
2x(1) + x(2) <= 9
x(1) + 2x(2) <= 12
x(1), x(2) >= 0
If you try a bit out you see that x(1) = 4, x(2) = 0, x(3) = 24 is the optimal solution.
But my Matlab Code gives a different solution and i do not know why.
Here my Code:
f=@(x)-x(3);
x0 = [0,0,0];
lb = [0,0,-Inf];
gs = GlobalSearch;
A = [-1 -1 0;
3 1 0;
2 1 0;
1 2 0];
b = [0; 12; 9; 12];
nonlincon = @constr;
problem = createOptimProblem('fmincon','x0',x0,'objective',f,'lb',lb,'Aineq',A,'bineq',b,'nonlcon',nonlincon)
x = run(gs,problem)
I would be thankful if someone could tell me if I did a mistake somewhere.
  7 个评论
Daniela Würmseer
I used fmincon already before but the solution was not the right one so i tried GlobalSearch.
@Torsten could you tell me how your Code looks like to get this result? Iam new to matlab so perhaps i did something wrong?
Thank you for all of your answers.
Torsten
Torsten 2022-1-8
function main
f=@(x)-x(3);
x0 = [0,0,0];
lb = [0,0,-Inf];
A = [-1 -1 0;
3 1 0;
2 1 0;
1 2 0];
b = [0; 12; 9; 12];
nonlcon = @constr;
sol = fmincon (f, x0, A, b, [], [], lb, [], nonlcon)
end
function [c,ceq] = constr(x)
c = -10*x(1)+x(1)^2-4*x(2)+x(2)^2+x(3);
ceq = [];
end

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采纳的回答

Matt J
Matt J 2022-1-8
We can verify that Torsten's solution is feasible as below. Since it gives a better objective function value than your experimental solution, your solution cannot be the correct one.
A = [-1 -1 0;
3 1 0;
2 1 0;
1 2 0];
b = [0; 12; 9; 12];
constr = @(x) -10*x(1)+x(1)^2-4*x(2)+x(2)^2+x(3);
x=[3.5000 1.5000 26.5000]';
b-A*x
ans = 4×1
5.0000 0 0.5000 5.5000
constr(x)
ans = 0
  6 个评论
Matt J
Matt J 2022-1-13
编辑:Matt J 2022-1-13
i still dont get the right solution and I dont know why
What do you mean "still"? I thought we established that you were getting the right solution all along.
That seems to be the case here again. The soluton you've shown is very close to x = (0,0,0,0).
Daniela Würmseer
Daniela Würmseer 2022-1-13
Sorry, i think the word "still" was misplaced here. I was just not sure about my Code but you are right i was not seeing the "10-10 x" in the solution and like this the solution is really close to x = (0,0,0,0,0).
Thank you for the help.

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更多回答(1 个)

Matt J
Matt J 2022-1-8
编辑:Matt J 2022-1-8
In fact, the problem can also be solved with quadprog, since it is equivalent to,
min -10*x(1)+x(1)^2-4*x(2)+x(2)^2
s.t.
3x(1) + x(2) <= 12
2x(1) + x(2) <= 9
x(1) + 2x(2) <= 12
x(1), x(2) >= 0
and since the objective function of the reformulated problem is srictly convex, it establishes that the solution is also unique:
f=@(x)-x(3);
x0 = [0,0,0];
lb = [0,0];
A =[3 1;
2 1;
1 2];
b = [12; 9; 12];
H=2*eye(2);
f=[-10;-4];
[x12,x3]=quadprog(H,f,A,b,[],[]);
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=[x12;-x3]'
x = 1×3
3.5000 1.5000 26.5000

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