Getting vertcat error, not using vertcat

Error using vertcat Dimensions of matrices being concatenated are not consistent.

I keep getting this error despite not having vertcat in my code. What can I do to fix this? Code is attached

Answer 1

[John D'Errico]

Your code:

ID=5415;
K=ID/1000;

uts=1;
uph=0.5;

a=0.05;
b=1.5;
for n=1:100
    k5(n)=a+(b-a)*rand(size(n));

      A=[-1 1 1 -1 0;-1 1+K 0 0 0;K 0 K*K (K*K)/(1+(K/K)) 1+K-K;0 0 0 K*(K+(K/(1+(K/K)))) -1;0 0 -1 1 k5/K];
      B=[0;K;uts+uph;0;0];
      x=inv(A)*B;

end

So, look at the LAST row of A. It is:

0 0 -1 1 k5/K

But what is k5?

k5(n)=a+(b-a)*rand(size(n));

After the very first iteration of that loop, you keep inserting new elements at the end of to k5, growing it. k5 is a vector.

That last row of A is your problem. What you really intended, how can I possible guess? n is a SCALAR. So rand(size(n) is ALWAYS just a scalar. Regardless, k5 is a vector, growing in length. So after the first iteration of your loop, k5 is a vector of length 2. Therefore, at iteration 2, the row:

0 0 -1 1 k5/K

now has length 6. NOT 5 as you so strongly claimed. On the nth iteration, that row would have length 4+n, that is, IF you could ever have gotten past n=2.

By the way, you ARE using vertcat. You may not know it, but you are. In fact, you use vertcat TWICE in that loop.

When you use square brackets with a semi-colon between elements or between rows, that is a call to vertcat.

Answer 2

[Mark McBroom]

The assignment to A vertical concatentates 4 row vectors. The first 3 rows have 5 elements, the 4th row only has 4 elements.

[Andrew Tyers]

All 5 rows have 5 elements. I just checked.