To write perfectly this command: A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];

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Please I need to express this command perfectly:
A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];
In order to Run this:
n = 3
n = 3
h(1) = 0.6
h = 0.6000
h(2) = 0.8
h = 1×2
0.6000 0.8000
h(3) = 1
h = 1×3
0.6000 0.8000 1.0000
T = 4
T = 4
% Objective function: minimize total power transmitted by all users
f = ones(n, 1);
% Inequality constraints: signal-to-interference ratio must exceed threshold
A = zeros(n,n);
b = zeros(n,1);
for i = 1:3
A(i,i) = -h(i);
A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];
b(i) = -T*h(i);
end
Arrays have incompatible sizes for this operation.
% Lower bounds: power transmitted by each user must be non-negative
lb = zeros(n,1);
% Solve using linprog
[x, fval] = linprog(f, A, b, [], [], lb);
% Display results
disp('Optimal power transmitted by each user:');
disp(x);
disp(['Minimum total power transmitted: ', num2str(fval)]);

采纳的回答

Walter Roberson
Walter Roberson 2023-5-2
n = 3
A = zeros(n,n);
So A will be initialized as 3 x 3
for i = 1:3
A(i,i) = -h(i);
A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];
when i = 1 then:
  • literal 0 is a 1 x 1 scalar
  • ones(1,i-1) is ones(1,0) which is 1 x 0 empty
  • literal 0 is 1 x 1 scalar
  • ones(1,n-i) is ones(1,3-1) (for this iteration) which is ones(1,2) which is 1 x 2
Those widths are 1 + 0 + 1 + 2 = 4.
So you are trying to add a 1 x 4 vector to a 1 x 3 vector A(i,:)
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