Show that if $n-m=2$, then the simplex method will not cycle, no matter which pivoting rule is used

Question

Here $n$ is the number of variables and $m$ is the number of constraints. This is the Exercise 3.10 (with asterisk) in the classical textbook Introduction to Linear Optimization by Dimitris Bertsimas and John Tsitsiklis.

The solution given by my professor says "Please find the proof in Marshall and Suurballe (1969)." However, in their 1969 paper "A note on cycling in the simplex method", they actually assumed a specific pivoting rule. They only proved that using that version of simplex method, the claim is true. The proof is highly dependent on the way of choosing entering and leaving variable. I tried to imitate the proof in an arbitrary pivoting rule setting, but failed.

I also searched online for references that are related to this topic, but the most related convincing result is "a basic feasible solution with a single degeneracy cannot reappear in the sequence of bases" in the paper by Saul I. Gass, published in Annals of Operations Research $47(1993)335-342$.

I can easily prove no cycling occurs for any pivoting rule when $n-m=1$ because the variable just entered the basis cannot leave in the next iteration. However, this cannot be generalized to $n-m=2$ because the entered variable can leave basis in the next second iteration.

I would appreciate it if anyone could give some hint on this problem.

Answer 1

Here is a solution manual for that question.

https://math.berkeley.edu/~bernd/HW3solution.pdf

I don’t really understand the arguments, but basically it says when $n-m=2$, either the polyhedron is unbounded or the polyhedron is bounded but all BFS are nondegenerate. In both cases, simplex method terminates in finitely many steps.

Answer 2

Suppose towards a contradiction that some pivoting rule generates a cycle. Without loss of generality, let's assume the tableau looks like below at the beginning of the cycle.

\begin{array}{c| c c c c c c c} -{c_b}^T x_b & 0 & \dots & 0 & \dots & 0 & c_{m+1} & c_{m+2}\\ \hline x_1 & 1 & \dots & 0 & \dots & 0 & u_{1,1} & u_{1,2}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ x_l & 0 & \dots & 1 & \dots & 0 & u_{l,1} & u_{l,2}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ x_m & 0 & \dots & 0 & \dots & 1 & u_{m,1} & u_{m,2}\\ \end{array}

Also, suppose $c_{m+1} < 0$ and the pivoting rule select $x_l$ as the entering variable. Then $l \in r_1 = \{i| 1\leq i \leq m, u_{i,1} > 0\}$. After this iteration, the tableau becomes

\begin{array}{c| c c c c c c c} -{c_b}^T x_b & 0 & \dots & -\frac{c_{m+1}}{u_{l,1}} & \dots & 0 & 0 & c_{m+2}\\ \hline x_1 & 1 & \dots & -\frac{u_{1,1}}{u_{l,1}} & \dots & 0 & 0 & u_{1,2}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ x_{m+1} & 0 & \dots & \frac{1}{u_{l,1}} & \dots & 0 & 1 & u_{l,2}\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ x_m & 0 & \dots & -\frac{u_{m,1}}{u_{l,1}} & \dots & 1 & 0 & u_{m,2}\\ \end{array}

For simplicity, we still notify the updated last column with the same variable names. As the reduced cost of the basic direction $x_l$ is $-\frac{c_{m+1}}{u_{l,1}} < 0$, we must have $c_{m+2}<0$ otherwise the simplex algorithm will terminate. Suppose the exiting variable is $x_j$, we will show 1)$j\neq l$ and 2)$j\in r_1$. First notice that, after this iteration, we will have two nonbasic variables $x_j$ and $x_l$. Since $x_j$ just exits the basis, it cannot reenter the basis in the very next iteration. Hence $x_l$ must be the entering variables in the next iteration. It implies the reduced cost $\bar{c_l}$ in the next iteration must be negative. If $j = l$, then the reduced cost $\bar{c_l} = -\frac{c_{m+1}}{u_{l,1}} + (-\frac{c_{m+2}}{u_{l,2}})\cdot \frac{1}{u_{l,1}} > 0$ as $c_{m+1},c_{m+2}<0$ and $u_{l,1}, u_{l,2} > 0$. Therefore $j \neq l$. To prove 2), suppose that $j \notin r_1$. Then $u_{j,1} \leq 0$. The reduced cost $\bar{c_{l}}$ in the next iteration would be $ \bar{c_l} = -\frac{c_{m+1}}{u_{l,1}} + (-\frac{c_{m+2}}{u_{j,2}}) \cdot (-\frac{u_{j,1}}{u_{l,1}}) > 0$. Hence 2) must be true.

From 1) and 2), we conclude that $j$ must belong the the set $r_2 = \{i| u_{i,2}>0, i\in r_1, i \neq l\}$. As $\lvert r_1 \rvert \leq m$, we have $\lvert r_2 \rvert \leq m -1$. If we continue this process, we will have $\lvert r_{m+1} \rvert \leq 0$. Hence the simplex must terminate after m iterations and therefore it cannot cycle, we reach a contradiction here. As a corollary, we conclude that if $n-m=2$, then the simplex algorithm will terminate after at most $m$ iterations.