Minimizing norm( a_i x - ( y B_i + c_i ) ) with constraint

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Syam MS 2020 年 7 月 29 日

Hi all,
Given the signal vectors B_i and c_i, I have obtained a new signal vector a_i to find out vector y. Then to fine tune, I introduced a scalar x.
Now, I need to minimize the problem which is defined as follows:
norm( a_i x - ( y B_i + c_i ) ) s.t. norm(y_n)<=1.
In matlab, how to solve this?
Variables here are x and y.
Further, 'a_i' is a vector of order (1 \times N),
x is a scalar (tuning parameter) ,
y is a vector of order (1 \times N), y_n is a term in vector y,
B_i is a matrice of (N \times N),
c_i is a vector of (1 \times N).

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回答 (2 件)

Bruno Luong 2020 年 7 月 29 日

So you want more or less
Minimize | A*z - b |^2 with the constraint | z | <= 1.
where |.| designates the l2 norm.
A := B.' % known
z := y.' % unknown
b := (a*x - c).' % known
To do that, first you do
z = pinv(A)*b % if A is full rank it's z = A\b
Check if |z| <= 1, if yes you are done.
If not then solve the following problem this equality constraint
Minimize | A*z - b |^2 with the constraint | z | = 1.
by using for example this function in File-exchange
z = spherelsq(A,b,1);
Then get back
y = z.'
Put all that together
A = B.';
b = (a*x - c).';
z = pinv(A)*b;
if norm(z,2) > 1
z = spherelsq(A,b,1);
end
y = z.';
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Syam MS 2020 年 7 月 31 日
Function spherelsq(A,b,1) have the constraint |x| = c. Here norm of vector X is taken, my constraint instead is for the each element in the vector X.

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Bruno Luong 2020 年 7 月 29 日

If you have optimization toolbox you can do (see my other answer for definition of A, b and z)
z= fmincon(@(z) norm(A*z-b,2).^2, A\b, [], [], [], [], [], [], @norm1con)
% where this is the constraint
function [c, ceq] = norm1con(z)
c = sum(z.^2,1) - 1;
ceq = [];
end
I can check both of my methods give the same results
% test.m script
N = 10;
A = rand(N);
z = randn(N,1);
z = z/norm(z)*2;
b = A*z;
z = pinv(A)*b;
if norm(z,2) > 1
z = spherelsq(A,b,1);
end
zmincon = fmincon(@(z) norm(A*z-b,2).^2, A\b, [], [], [], [], [], [], @norm1con);
% Check
z
zmincon
norm(z-zmincon) / norm(z)
function [c, ceq] = norm1con(z)
c = sum(z.^2,1) - 1;
ceq = [];
end
Result
>> test
Local minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.
<stopping criteria details>
z =
-0.5826
-0.2210
0.2825
-0.4180
0.4400
0.0441
-0.0615
0.2073
-0.1440
-0.3070
zmincon =
-0.5826
-0.2210
0.2825
-0.4180
0.4400
0.0441
-0.0615
0.2073
-0.1440
-0.3070
ans =
1.1762e-07
Look good to me

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