How to define objective function that is not a direct function of decision variable?

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I want to minimize "Cost" while "a, b, c" are the variables.
F1 = (1/a) + b
F2 = 2/F1
F3 = F2*3/c
Cost = F2 + F3
I was wondering how can I solve for minimum Cost, using a problem-based non-linear optimization approach? Any lead will be greatly appreciated.
  2 comentarios
Matt J
Matt J el 21 de Abr. de 2020
The best advice will depend on what constraints you have on a,b,c. Without constraints, your objective function is unbounded.
Thaneer Malai Narayanan
Thaneer Malai Narayanan el 21 de Abr. de 2020
Thanks, lets assume the variables ae bounded and continuous. My question is how do i express the constaints and obj function of this problem?

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Respuestas (1)

Ameer Hamza
Ameer Hamza el 21 de Abr. de 2020
Editada: Ameer Hamza el 21 de Abr. de 2020
See this example to see how define the objective function and bound on the optimization variables.
F1 = @(a,b,c) (1./a) + b;
F2 = @(a,b,c) 2./F1(a,b,c);
F3 = @(a,b,c) F2(a,b,c).*3./c;
Cost = @(a,b,c) F2(a,b,c) + F3(a,b,c);
lb = [0 0 0]; % lower bounds on a, b, and c
ub = [1 1 1]; % lower bounds on a, b, and c
x0 = rand(1,3); % initial guess
x_sol = fmincon(@(x) Cost(x(1), x(2), x(3)), x0, [], [], [], [], lb, ub);
a_sol = x_sol(1);
b_sol = x_sol(2);
c_sol = x_sol(3);
  2 comentarios
Thaneer Malai Narayanan
Thaneer Malai Narayanan el 21 de Abr. de 2020
Thank you Ameer. This worked. Now I want to add a constrant on F2. Could you please show how to do it?
For example,
F2 <= 10
I was wondering how to change include that as Ax = b? Really appreciate your help.
Ameer Hamza
Ameer Hamza el 22 de Abr. de 2020
Editada: Ameer Hamza el 22 de Abr. de 2020
Thaneer, You cannot include as Ax=b, because your constraint is not linear. You can add it like this.
F1 = @(a,b,c) (1./a) + b;
F2 = @(a,b,c) 2./F1(a,b,c);
F3 = @(a,b,c) F2(a,b,c).*3./c;
Cost = @(a,b,c) F2(a,b,c) + F3(a,b,c);
lb = [0 0 0]; % lower bounds on a, b, and c
ub = [1 1 1]; % lower bounds on a, b, and c
x0 = rand(1,3); % initial guess
x_sol = fmincon(@(x) Cost(x(1), x(2), x(3)), x0, [], [], [], [], lb, ub, @(x) cons(x, F2));
a_sol = x_sol(1);
b_sol = x_sol(2);
c_sol = x_sol(3);
function [c, ceq] = cons(x, F2)
ceq = [];
c = F2(x(1), x(2), x(3)) - 10;
end
Solution:
>> x_sol
x_sol =
0.0000 0.7288 0.8549
Verify of constraint is met:
>> F2(x_sol(1), x_sol(2), x_sol(3))
ans =
4.4348e-09

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