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Currently, I am studying total unimodularity. I was reading this link: https://ostad.nit.ac.ir/upload/Integer_Programming_1.pdf, from page 38-41 and I came across the statement:

It is clear that when matrix A is totally unimodular, the linear programming relaxation solves the IP $$\max\left\{cx:Ax \leq b, x \in Z_{+}^n\right\}.$$

I've read and understand about TU (totally unimodular) based on the link above and about LP relaxation. However, I still do not know how do I prove the statement above.

What I know: I know that if $A$ is TU then $(A, I)$ is TU for an identity matrix $I$. Also, as stated in the book that

From linear programming theory, we know that basic feasible solutions take the form:. $x = (x_B,x_N) = (B^{-1} b, 0)$ where $B$ is an $m \times m$ nonsingular submatrix of $(A, I)$ and $I$ is an $m \times m$ identity matrix.

Observation 3.1 (Sufficient Condition) If the optimal basis $B$ has $det(B) = ±1$, then the linear programming relaxation solves IP.'

My Hypothesis: Does a TU matrix $A$ always have a basis B such that $det(B) = \pm 1$? But if $A$ is TU, all its square submatrix is either 0,-1, or 1 and not 1 or -1.

Extra Note: I am working on some shortest path inner problem of a variable $X_{ij}$ (the original model is bi-level) where $(i,j)$ represents the arc in the graph. The paper stated that since $X_{ij}$ is unimodular, we can use LP relaxation in order to use KKT to turn the bi-level into single level.

What I don't know: I do not know anything about polyhedron.

Thanks in advance.

Rodrigo de Azevedo
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Michelle Gunawan
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1 Answers1

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Let $B$ be a basis determining the basic solution $x^*$. Then $A_Bx_B^*=b$ and $x_j^* = 0$ for all $j \notin B$. For each $i \in B$, let $B_i$ denote the matrix obtained by replacing column $i$ from $A_B$ with $b$.

By Cramer's rule, for each $i \in B$, we have $x_i^* = \frac{\det(B_i)}{\det(A_B)}$ which is integral since $\det(A_B)$ is either $1$ or $-1$ as $A_B$ is totally unimodular, and $\det(B_i)$ is integral since all the entries in $B_i$ are integers. Hence, $x^* \in \mathbb{Z}^n$.

m6rco
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