fmincon or linprog use in minimization?

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bus14
bus14 2019 年 4 月 17 日
編集済み: Torsten 2019 年 4 月 17 日
Hi community,
As I am busy trying to optimize the function I encouter quite some difficulties. As I am not certain which optimization function to use( fmincon or linprog)? The letters are vectors and A is a m*n matrix. I am trying to optimize the values of Z & Y. In which antother problem arises as well, since z and y have different dimensions. Z is a 1x1 vector ( a number) and Y is a 2x1 vector. Does someone has a tip on which function to use and how to deal with the dimensional problems of the decision variables? Thankyou
min (l-q).'*z-s.'*y
s.t. y=x-A.'*z
0<z<d , y>0
  1 件のコメント
Torsten
Torsten 2019 年 4 月 17 日
編集済み: Torsten 2019 年 4 月 17 日
x, A, l, q and s are given ?

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採用された回答

Torsten
Torsten 2019 年 4 月 17 日
編集済み: Torsten 2019 年 4 月 17 日
l = 0.25;
q = 2;
s = 1;
A = 1;
x = 20;
d = 15;
f= [-s, l-q];
Aeq = [1, A];
beq = x;
lb = [0, 0];
ub = [Inf, d];
sol = linprog(f,[],[],Aeq,beq,lb,ub);
y = sol(1)
z = sol(2)
  2 件のコメント
bus14
bus14 2019 年 4 月 17 日
thank you a lot, one question that remains is how you came to the formulation of Aeq[1, A] and beq=x as for me it is vague how Aeq and beq are constructed from the given constraint.
Torsten
Torsten 2019 年 4 月 17 日
編集済み: Torsten 2019 年 4 月 17 日
1*y + A*z = x <-> [1, A]*[y;z] = x <-> Aeq*[y;z] = beq

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その他の回答 (2 件)

bus14
bus14 2019 年 4 月 17 日
for the first instance my goal is to let the optimization problem work with the following values for x,A,l,q
L=[0.25]
Q=[2 ]
S=[1 ]
A= [1]
X= [20]

bus14
bus14 2019 年 4 月 17 日
thank you a lot, so linprog is indeed the one to use as my objective function is a linear function.

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