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

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)]);

Answer 1

[Walter Roberson]

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,:)

[Dalia ElNakib]

Thank you Dr. Walter Roberson

Please how to express this correctly:

A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];

In order to add a 1 x 4 vector to a 1 x 3 vector A(i,:)

Many Thanks for your cooperation

[Walter Roberson]

1 + (i-1) + 1 + (n-i) = 1 + n - i + i = n + 1 . So no matter what your n is, your code attempts to create a vector that is one element wider. If that is what you want, then why not initialize A to be n+1 by n+1 ?

Note by the way,

A(i,i) = -h(i);
A(i,:) = A(i,:) + h(i)*[0, ones(1,i-1), 0, ones(1,n-i)];

Consider A(i,i) . In the first statement you assign it -h(i) specifically. Then [0, ones(1,i-1)] would be a width of 1+(i-1) = i, so [0, ones(1,i-1)] as a leading subset is going to have its 0 if i==1 but otherwise is going to be 1. h(i) times that would be 0 if i == 1, and otherwise would be h(i) . Then with the A(i,:) added to that and A(i,i) having been assigned -h(i) then that is going to be -h(i) + 0 for i == 1, but otherwise is going to be -h(i) + h(i)*1 which would total 0 to within round-off error.

Is that what you want? That the A(i,i) should be left as -h(i) for the case i == 1, but that otherwise through that calculation that afterwards it should just happen to come out A(i,i) == 0 (unless there are problems with round-off error or problems with infinities) ??

Why would you want to write such unclear code??

[Dalia ElNakib]

Thank you so much for your help

I'm trying to solve this attached problem as Assignment using linprog Code

Then it will be solved through:

h(1) = 0.6

h(2) =0.8

h(3) =1

g(1) =0.1

g(2) =0.2

g(3) =0.3

C Matrix

C = [1

1

1]

A Matrix (3*3)

A= [h(1) - g(1)h(2) -g(3)h(3)

-g(2)h(1) h(2) -g(2)h(3)

-g(3)h(1) -g(3)h(2) h(3)]

b Matrix

b = [0

0

0]

Assuming Values for h's and Gamma's as shown above

Do you think we'll need Upper limit and Lower Limit for this example?

[Torsten]

@Dalia ElNakib

Matlab needs a multiplication sign (*) between numbers you want to multiply.

[Walter Roberson]

Assuming that g and h are row vector

A = - h * g.';
A(1:n+1:end) = h;