Is there an example of an $m\times n$ integer matrix $A$ and an integer vector $b\in \mathbb {Z}^{m}$ such that the polyhedron $P := \{ x\in \mathbb {R}^{n} \mid A x \leq b\}$ is integer, while $A$ is not totally unimodular?
Note that a polyhedron is integer if all of its vertices are integral.
The answer is yes, and the proof is trivial. Recall that a TUM matrix has all coefficients in $\lbrace -1, 0, 1\rbrace.$ Start with any $P := \{ x\in \mathbb {R}^{n} \mid A x \leq b\}$ that is integer. If $A$ happens to be TUM, change $A$ to $2A$ and $b$ to $2b$. The polynomial is unchanged, but the new constraint matrix $2A$ cannot be TUM since it will contain components of $2$ and/or $-2$.
You might want to look at MRP(Material Requirements Planning)/Leontief matrices.
In an MRP/Leontief matrix/model:
Each constraint is an equality,
Every column has exactly one positive coefficient and it is a +1, (the output)
Each column has 0 or more negative coefficients, every one of which is integer, (inputs needed)
Each RHS coefficient is a nonnegative integer. (the demands)
Result: An LP whose complete constraint set is an MRP set has an optimal solution that is integer.
Further, if the objective coefficients are all integer, then there is an optimal solution with integral dual prices.
Ref: Jeroslow, R., K. Martin, R. Rardin, J. Wang(1992), "Gainfree Leontief substitution flow problems," Mathematical Programming, 57, pp. 375-414.;
