%forwardsubs
% forward substitution to solve L y = b 
% input: L, b, n Output: y 
% L is a lower triangular matrix
% b is the right-hand side vector 
% y is the solution vector 
% n is the number of rows (and columns) of L 


y= zeros(n,1);
temp= b; 
for j=1:n, 
    if (abs(L(j,j)) <= eps) sprintf('small diagonal element'); return;
    else 
       y(j) = temp(j)/ L(j,j); 
       i=j+1:n; 
        temp(i) = temp(i) - L(i,j) * y(j); 
    end;
end;