Originally Posted by

**Darkprince** But they are two different loops, doesn't make a difference? How that is?

When the second loop is taking place, it "brings" the solution to its calculation? Is the solution stored?

No; you store it when it is initially calculated

Code:

c(i)' = c(1)/b(1);div(i)=1 for i = 1 and div(i)=(b(i) - c(i-1)' * a(i); c(i) / div(i) for i=2 to (n-1)
d(i)' = d(1)/b(1) for i = 1 and (d(i) - d(i-1)' * a(i) ) / div(i) for i=2 to n
The back substitution is x(n)=d(n)' and x(i) = d(i)' - c(i)' * x(i+1) for i=(n-1) to 1

CB