Simplex Algorithm Cycles How can I show that the linear program
\begin{align*}&\min -2x-3y+z+12w\\
&s.t. 
\\&-2x-9y+z+9w+s=0\\&1/3x+2y-1/3z-2w+t=0\\&x,y,z,w,s,t\geq 0\end{align*}
can induce cycling in the Simplex Algorithm if we use the original pivot rule of Dantzig? I know that I have to show that the algorithm repeats bases which all correspond to the same vertex but I am not sure how to do that here.

Edit: the linear program above is a conversion of the program
\begin{align*}&\min -2x-3y+z+12w\\
&s.t. 
\\&-2x-9y+z+9w\leq0\\&1/3x+2y-1/3z-2w\leq0\\&x,y,z,w\geq 0\end{align*}
into standard form to make it accessible to the Simplex Algorithm.

To simplify things, let’s denote each basis as $B_0,\ldots,B_n$ for the number of bases we visit until we notice a cycle. Thus, with the given model in the question, the first basis will be:
\begin{array} {|c|c|}
\hline \text{Current Basis} & BV & k & x & y & z & w & s_1 & s_2 & RHS & RT \\ \hline  & k & 1 & 2 & 3 & -1 & -12 & 0 & 0 & 0 & - \\ B_0 & s_1 & 0 & -2 & -9 & 1 & 9 & 1 & 0 & 0 & 0 \\  & s_2 & 0 & 1/3 & 2 & -1/3 & -2 & 0 & 1 & 0 & 0 \\ \hline  \end{array}
For this to work, we’ll always pivot the first row that is the minimum row in the minimum-ratio test. Thus, we’ll pivot the $y$ column with the $s_1$ row to produce the next basis:
\begin{array} {|c|c|}
\hline \text{Current Basis} 
    & BV & k & x & y & z & w & s_1 & s_2 & RHS & RT \\ \hline  
    & k & 1 & 4/3 & 0 & -2/3 & -9 & 1/3 & 0 & 0 & - \\ 
B_1 & y & 0 & 2/9 & 1 & -1/9 & -1 & -1/9 & 0 & 0 & 0 \\
  & s_2 & 0 & -1/9 & 0 & -1/9 & 0 & 2/9 & 1 & 0 & 0 \\ \hline  \end{array}
Then we’ll pivot the $x$ column with the $y$ row to produce:
\begin{array} {|c|c|}
\hline \text{Current Basis}
& BV & k & x & y & z & w & s_1 & s_2 & RHS & RT \\ \hline
 & k & 1 & 0 & -6 & 2/3 & -3 & 5/3 & 0 & 0 & - \\
B_2&x& 0 & 1 & 9/2 & -1/2 & -9/2 & -1/2 & 0 & 0 & 0 \\ 
&s_2 & 0 & 0 & 1/2 & -1/6 & -1/2 & 1/6 & 1 & 0 & 0 \\ \hline  \end{array}
Then we’ll pivot the $s_1$ column with the $x$ row to produce:
\begin{array} {|c|c|}
\hline \text{Current Basis} & BV & k & x & y & z & w & s_1 & s_2 & RHS & RT \\ \hline  & k & 1 & 2 & 3 & -1 & -12 & 0 & 0 & 0 & - \\ B_3 & s_1 & 0 & -2 & -9 & 1 & 9 & 1 & 0 & 0 & - \\ 
& s_2 & 0 & 1/3 & 2 & -1/3 & -2 & 0 & 1 & 0 & - \\ \hline  \end{array}
From here, notice that basis $B_0=B_3$ in that every element in the the tableau matches with one another. Thus, if we start pivoting again we’ll be going in the exact same path we did to reach the basis $B_3$. Thus, this is how we prove that cycling is occurring via the Simplex Method in tabular form.
