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

4 vues (au cours des 30 derniers jours)
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 commentaires
Matt J
Matt J le 21 Avr 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 le 21 Avr 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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Réponses (1)

Ameer Hamza
Ameer Hamza le 21 Avr 2020
Modifié(e) : Ameer Hamza le 21 Avr 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 commentaires
Thaneer Malai Narayanan
Thaneer Malai Narayanan le 21 Avr 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 le 22 Avr 2020
Modifié(e) : Ameer Hamza le 22 Avr 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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