For the branch-and-cut method, it is essential to know many facets of the polytopes generated by the problem. However, it is currently one of the hardest problems to actually calculate all facets of such polytopes as they rapidly grow in size.
For an arbitrary optimization problem, the polytope used by branch-and-cut or also by cutting-plane-methods is the convex hull of all feasible vertices. A vertex is an assignment of all variables of the model. As a (very simple) example: if one would maximize $2\cdot x+y$ s.t. $x+y \leq 1$ and $0\leq x,y\leq 1.5$ then the vertices $(0,0)$, $(0,1)$ and $(1,0)$ are feasible vertices. $(1,1)$ violates the inequality $x+y\leq 1.5$ and is therefore not feasible. The (combinatorical) optimization problem would be to choose among the feasible vertices. (In this case, obviously $(1,0)$ is the optimum). The convex hull of these vertices is the triangle with exactly these three vertices. The facets of this simple polytope are $x\geq0$, $y\geq 0$ and $x+y\leq 1$. Note that the description through facets is more accurate than the model. In most hard problems - such as the TSP - the number of facets exceeds the number of model inequalities by several orders of magnitude.
Considering the Travelling Salesman Problem, for which number of nodes is the polytope fully known and how much facets are there. if it is not complete, what are lower bounds on the number of facets?
I'm particularly interested in the so-called hamiltonian path formulation of the TSP:
$$min \sum_{i=0}^{n-1}(\sum_{j=0}^{i-1}c_{i,j}\cdot x_{i,j}+\sum_{j=i+1}^{n-1}c_{i,j}\cdot x_{i,j})$$ s.t.
$$\forall i \neq j:\ \ 0 \leq x_{i,j}\leq 1$$ $$\forall i \neq j\ \ \ x_{i,j}+x_{j,i}\leq 1$$ $$\forall j \ \ \sum_{i=0}^{j-1}x_{i,j}+\sum_{i=j+1}^{n-1}x_{i,j}\leq 1$$ $$\forall j \ \ \sum_{i=0}^{j-1}x_{j,i}+\sum_{i=j+1}^{n-1}x_{j,i}\leq 1$$ $$\sum_{i=0}^{n-1}(\sum_{j=0}^{i-1}x_{i,j}+\sum_{j=i+1}^{n-1}x_{i,j})=n-1$$
If you have any information about polytopes of other formulations of the TSP, feel free to share that too.