Constraint with decision variable

5 visualizzazioni (ultimi 30 giorni)
Nana
Nana il 27 Mar 2016
Commentato: Walter Roberson il 29 Mar 2016
How can I write a constraint for optimization which includes the decision variable (x)? The constraint is:
max (0, x(i-1) - P) <= x(i) <= min (C, x(i-1) + D)
  1 Commento
John D'Errico
John D'Errico il 27 Mar 2016
Modificato: John D'Errico il 27 Mar 2016
Don't all constraints for an optimization somehow involve the decision variables? Note that what you have written is actually more than one constraint. There are TWO inequalities there.

Accedi per commentare.

Risposta accettata

Walter Roberson
Walter Roberson il 28 Mar 2016
You should use a couple of different kinds of constraints. But you are only constraining variable #i, so we have to initialize the constraints so the other variables are not affected
Nx = length(x);
%no equality constraints
Aeq = [];
beq = [];
%the constraint has to be for some _specific_ i
i = 13;
%now to construct x(i-1) - P <= x(i) and x(i) <= x(i-1) + D
A = zeros(2, Nx);
b = zeros(2, 1);
A(1, i-1) = 1; A(1, i) = -1; b(1) = P; % (1) * x(i-1) + (-1) * x(i) <= (P)
A(2, i-1) = -1; A(2, i) = 1; b(2) = D; % (-1) * x(i-1) + (1) * x(i) <= (D)
%lower bounds are -inf, upper bounds are +inf for all variables except #i
lb = -inf * ones(1, Nx);
ub = inf * ones(1, Nx);
%but for variable #i, the value cannot be less than 0 or more than C
lb(i) = 0;
ub(i) = C;
fmincon(@objective, x0, A, b, Aeq, beq, lb, ub)
Your constraints cannot be generalized to all of the variables because there is no x(0) to constrain x(1) against.
  2 Commenti
Nana
Nana il 29 Mar 2016
Hi Walter, thanks for your answer!
So basically, I need two columns of b?
Walter Roberson
Walter Roberson il 29 Mar 2016
You need one entry in b for each of the separate constraints. You have two inequality constraints for each i, so two entries in b for each i. It does not matter whether you specify b as a row or as a column vector; a row vector will be transformed into a column vector internally.

Accedi per commentare.

Più risposte (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by