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

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gsourop on 29 Oct 2018
Commented: Bruno Luong on 5 Nov 2018
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.
gsourop on 30 Oct 2018
I see. But is there a way to impose the additional constraints without solving the relationship arithmetically?

Bruno Luong on 30 Oct 2018
Edited: Bruno Luong on 30 Oct 2018
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
Bruno Luong on 5 Nov 2018
"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)