# unable to optimise the following using linprog

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Malloju Dinesh on 9 Feb 2022
I am trying to implement equation 9(in the image attached below) in linprog .. the v vector has (w,r): w is the weights vector which i have to find so that r is min. . i am trying to generate weights vector from this optmisation to be used in KNN(k-nearest neighbors) algorithm. I have evaluated the A and b matrices from the data set ...
1)The size of my data set is 100*4(4 is the number of features which is D)
2)size of f , v vector is (D+1) which is 5
3)size of A(10000*5), b(10000*1).I have evaluated A and b matrices from the data set.So i have A and b.
4)The vector to be optmised is v which is represented as(w,r).w is weights vector(size 4) which i have to find so that r
is minimum.
5)r is the maximum of r1,r2. r1:maximum of intraclass1 distance. r2: maximum of intraclass2 distance.
The each observation of data set belongs to two classes (1 or 2).
6)distance is the mahalanobis weighted distance between two observations.  Let me know step by step approach on how to solve it .Thanks . if any further information is required let me know ..thanks Abolfazl Chaman Motlagh on 9 Feb 2022
HI. here's how you should do it:
First i create a sample database for my self to use as parameters: (i use 100 points in 4D feature space with 3 label, change every part for your problem)
N = 100; % number of points
D = 4; % Dimension of feature space
Class_count = 3; % number of Classes
%% Create random dataset for parametrization
X = rand(N,D); % Position of Points in feature space
Y = randi([1 Class_count],N,1); % Label of Points
Now let us create Matrix of Coefficients A, right-hand-side vector B and Cost weights f.
f = [zeros(1,D) 1];
A = zeros(N^2,D+1);
b = zeros(N^2,1);
for i=1:N
for j=1:N
I = (i-1)*N + j;
if Y(i)==Y(j)
A(I,:) = [((X(i,:)-X(j,:)).^2) -1];
b(I) = 0;
else
A(I,:) = [-((X(i,:)-X(j,:)).^2) 0];
b(I) = -1;
end
end
end
lb = zeros(D+1,1); % lower bound
Now time to use linprog:
Solution = linprog(f,A,b,[],[],lb,[]) % no equality constraint and no upper bound
Optimal solution found.
Solution = 5×1
60.0201 85.0521 34.1626 70.6886 151.5027
W = Solution(1:D)
W = 4×1
60.0201 85.0521 34.1626 70.6886
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Abolfazl Chaman Motlagh on 9 Feb 2022
Happy to help.