Global Optimizationproblem using Global Search

2022 1 7
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更新时间：2022 1 13
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1 个投票

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
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Walter Roberson
Walter Roberson 2022-1-7
-x(1) -x(2) <= 0
okay, so x(1) + x(2) >= 0
x(1), x(2) >= 0
But if x(1) and x(2) are both >= 0, then their sum is certain to be at least 0. So it looks like that first constraint is redundant ?
Torsten
Torsten 2022-1-7
Where do you define the function handle for the quadratic constraint ?
Torsten
Torsten 2022-1-7
编辑：Torsten 2022-1-7
function [c,ceq] = constr(x)
c = -10*x(1)+x(1)^2-4*x(2)+x(2)^2+x(3);
ceq = [];
end
?
I get
x(1) = 3.5000 x(2) = 1.5000 x(3) = 26.5000
as optimal solution.
Matt J
Matt J 2022-1-8
编辑：Matt J 2022-1-8
Since the problem convex, and therefore has no non-global minima, it is strange to use GlobalSearch in this case. Just use fmincon.
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
Ran in:

1 个投票

Ran in:
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
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Hello, thank you for your answer. How am I able to get this answer with matlab? Because Global Search and fmincon gives me a different answer. (Iam new to Matlab so perhaps my Code is wrong?)
I tried (in the upper Code)
x = fmincon(f,x0,A,b,[],[],lb,[],nonlincon)
instead of
problem = createOptimProblem('fmincon','x0',x0,'objective',f,'lb',lb,'Aineq',A,'bineq',b,'nonlcon',nonlincon)
x = run(gs,problem)
Matt J
Matt J 2022-1-8
Ran in:
Ran in:
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];
nonlincon = @constr;
[x,fval]=fmincon(f, [0,0,0],A,b,[],[],lb,[],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
3.5000 1.5000 26.5000
fval = -26.5000
function [c,ceq] = constr(x)
c = -10*x(1)+x(1)^2-4*x(2)+x(2)^2+x(3);
ceq = [];
end
Thank you now it works. I had the same code I think but without the "fval".
Daniela Würmseer
Daniela Würmseer 2022-1-13
Iam still trying to learn matlab better, but by some OptimizationProblems i still dont get the right solution and I dont know why. By the following Problem the optimal solution can be only x = (0,0,0,0) because of the Aeq constraint and the lb but my code gives another solution:
f=@(x)-x(3)-x(4);
x0 = [0,0,0,0];
lb = [0,0,0,0];
A = [3 1 0 0;
2 1 0 0;
1 2 0 0];
b = [12; 9; 12];
Aeq = [-1 -1 -2 -3.5];
beq = 0;
nonlincon = @constr;
[x,fval] = fmincon(f,x0,A,b,Aeq,beq,lb,[],nonlincon)
function [c,ceq] = constr(x)
c = [];
ceq = [-10*x(1) + x(1)^2 -4*x(2)+x(2)^2 - 5*x(3) - 1.5*x(4)];
end
Solution :
x = 1×4
10-10 ×
0.2778 0.2503 0.2008 0.6934
fval = -8.9416e-11
Perhaps one of the guys who answered before @Torsten or @Matt J could look at this code and could spot what iam always doing wrong. Thank you.
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
Ran in:

1 个投票

Ran in:
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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