How can I add constraints in a solution of a function?

2 次查看（过去 30 天）
gsourop
gsourop 2018-10-29
评论： Bruno Luong 2018-11-5
Hi everyone,
I would like to solve the following function by adding two constraints.
S=randn(6,6); y=randn(6,1); ONE = ones(6,1); rft=randn(1,1); K = randn(1,1);
x = inv(S)*(y- ONE*rft)*0.1/sqrt(K);
I would like to include 2 additional constraints, -1<sum(x)<2 . I am not sure how I should use fmincon.
  10 个评论
显示 8更早的评论
gsourop
gsourop 2018-10-30
I see. But is there a way to impose the additional constraints without solving the relationship arithmetically?

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Bruno Luong
Bruno Luong 2018-10-30
编辑：Bruno Luong 2018-10-30
n = 6;
L = randn(n);
S = L'*L;
y = randn(n,1);
rft = randn(1,1);
su = 2;
sl = -1;
% T = V*V'
T = S/(0.1^2);
T = 0.5*(T+T');
[V,D] = eig(T);
V = V.*sqrt(diag(D)');
A = inv(V');
% xx = V'*x
% x = V'\xx = W'*xx
% |xx| = 1
yy = V \ (-rft+y);
SX = sum(A,1);
SXu = SX/su;
SXl = SX/sl;
% ReTurn = (1-x'*ONE)*rft+x'*y
% return = rtf - x'*(rtf + y)
% = rft + xx'*yy
% with yy = W*(-rft+y)
% maximize (xx'*yy)
% such that
% |xx| = 1
% SXu*xx <= 2
% SXl*xx <= 1
xx = yy/norm(yy);
if SXu*xx > 1
n2 = (SXu*SXu');
cs2 = 1/n2;
sn = sqrt(1-cs2);
Tu = null(SXu);
yyu = Tu*(Tu'*yy);
xx = cs2*SXu' + (sn/norm(yyu))*yyu;
elseif SXl*xx > 1
n2 = (SXl*SXl');
cs2 = 1/n2;
sn = sqrt(1-cs2);
Tl = null(SXl);
yyl = Tl*(Tl'*yy);
xx = cs2*SXl' + (sn/norm(yyl))*yyl;
end
x = V' \ xx
% Check constraints
sum(x)
x'*S*x
  5 个评论
显示 3更早的评论
gsourop
gsourop 2018-11-5
I am getting an error on the dimensions of the matrix. Hence, I replaces this line with
Scaling = sqrt(diag(D)');
for i = 1:6
for j =1 : 6
V(i,j) = V(i,j).* Scaling(j);
end
end
but the validations at the end do not give me the expected result.
Bruno Luong
Bruno Luong 2018-11-5
"I am getting an error on the dimensions of the matrix"
Then apply my code wrongly. I provide the code working with some fake data
S: sym-def-pos matrix (6 x 6)
y: vector (6 x 1)

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Linear Least Squares 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by