Past exam of the integer programming course of the University of Oxford / 2010 / 2

A totally unimodular matrix has the property that every square submatrix has determinant zero, one or minus one. This includes the one-by-one submatrices, so every entry belongs to $\{0,1,-1\}$.

Its importance is integrality of linear-programming vertices: if $A$ is totally unimodular and $b$ is integral, every vertex of $\{x:Ax\le b,x\ge 0\}$ is integral. Adding rows of the identity preserves total unimodularity. A nonsingular active basis therefore has determinant $\pm 1$ and an integer inverse by the cofactor formula, making its basic solution integral. Thus a finite optimum attained at a vertex of the linear programming relaxation gives an integer optimum without branching.

One sufficient condition is: entries belong to $\{0,1,-1\}$, each column has at most two nonzero entries, and any column with two has one $+1$ and one $-1$. This is the signed column condition for total unimodularity, satisfied by directed vertex-edge incidence matrices. It is sufficient, not necessary.

The constraints give the valid upper boundFor example $(x_1,x_2,x_3,x_4)=(0,1,3,4)$ satisfies every constraint: the first left side is four, the next three are three, five and two, and $x_4\ge 1$. Its objective is eight. Hence the upper bound is attained by an integer point, provingAll optimal solutions can also be identified. Equality forces $x_1=0$, $x_3=3$ and $x_2+x_4=5$. The remaining constraints reduce to $x_2\ge 1$ and $x_4\ge 1$. Thus exactly four integer optima occur:This is an optimality certificate by a valid relaxation bound and a feasible integer solution; no assumption of total unimodularity is needed. The source's ${\mathbb{Z}}_{+}$ denotes nonnegative integers, as is also necessary for its assignment equations in the next question.