I know it's a bit of a stretch, but in some sense the answer is yes. I apologize in advance for the self-promotion as the following is based on a paper I have co-authored. Also I apologize for the rather lengthy reply.
The Capacitated Vehicle Routing Problem (CVRP) can be defined as follows . Let $G = (V, E)$ be an undirected graph with node set $V = \{0, . . . , n\}$ and edge set $E$. Node 0 represents a depot, whereas each of the nodes in $V_c = {1, . . . , n}$ represents a customer. The symmetric cost of travel between nodes $i$ and $j$ is denoted by $c_{ij}$. A number $K$ of identical vehicles, each of capacity $Q > 0$, is available. Each customer $i \in V_c$, has an integer demand $q_i$ ($q_0=0$). Each customer must be served by a single vehicle and and the vehicles' capacities must be respected. The task is to find a set of vehicle routes of minimum cost, where each vehicle used leaves from and returns to the depot. In the paper, we propose a formulation of the CVRP based on decomposing each route into two paths leaving the depot and meeting up in the customer having the largest index on the route. To do so, we introduce the following sets of variables:
- For each $(i,j)\in A$, $x_{ij} = 1$ if a vehicle travels along arc $(i,j)$, otherwise $x_{ij} = 0$.
- For each $(i,j)\in \in A$, if a vehicle travels along arc $(i,j)$, then $f_{ij}$ denotes the total quantity delivered on the path when the vehicle leaves vertex $i$, otherwise $f_{ij} = 0$.
- For each $i\in V_c\setminus\{n\}$, the variable $u_i$ denotes the largest customer index visited on the path from the depot to and including customer $i$. $u_n=n$.
- For each $i\in V_c$, the variable $p_i$ denotes whether or not node $i$ is the customer with the largest index on a route. $p_n=1$.
- For each $i\in V_c$, if $p_i = 1$, then $t_i$ represents the total demand on the route which services node $i$, otherwise $t_i = 0$.
In the paper we propose the following formulation of CVRP
\begin{align}
\min & \sum_{(i,j) \in A} c_{ij} x_{ij} \label{eq:our_obj} & & & \\
\text{s.t.:} & x(\delta^+(i)) + p_i = 1 & & \forall i \in V_c & \\
& x(\delta^-(i)) - p_i = 1 & & \forall i \in V_c & \\
& t_i + f(\delta^+(i)) = f(\delta^-(i)) + q_i & & \forall i \in V_c & \\
& q_i x_{ij} \leq f_{ij} \leq (Q-q_j) x_{ij} & & \forall (i,j) \in A_c & \\
& q_i p_i \leq t_i \leq Q p_i & & \forall i \in V_c & \\
& x(\delta^+(0)) = 2K & & & \\
& \sum_{i \in V_c} p_i = K & & & \\
& i \leq u_i \leq i p_i + (n-1) (1-p_i) & & \forall i \in V_c \setminus \{n\} & \\
& u_i-u_j +(n-j-1)x_{ij} + (n-\max\{i,j\}-1) x_{ji} \leq n-j-1 & & \forall i,j \in V_c \setminus \{n\}, \ i \not= j & \\
&\sum_{i \in S_j} p_i \geq \bigl \lceil \sum_{i \in S_j} q_i / Q \bigr \rceil &&\forall j=1,\ldots,n&\\
& x_{ij} \in \{0,1\} & & \forall (i,j) \in A_c & \\
& x_{0j} \in \{0,1,2\} & & \forall j \in V_c & \\
& p_{i} \in \{0,1\} & & \forall i \in V_c &
\end{align}
For a detailed description of the constraint sets, please see the paper.
Now, when we did the research, the goal was to get rid of the symmetry that comes from the fact that for the symmetric CVRP, it does not matter in which direction you traverse a route. As a by-product of this, we noticed, that the LP relaxation of this formulation was very dependent on the ordering of the customers. To illustrate this, consider the following small example with 6 customers:
The number of vehicles is set to $K=2$, the vehicle capacity to $Q=47$, and the $c_{ij}$'s are the rounded Euclidean distances between the coordinates. If one solves the LP-relaxation of the above program, the optimal objective function value is $z^*=54.786$. If, on the other hand, one sorts the customers according to non-increasing distance to the depot, the optimal objective function value of the LP relaxation is $z^*=49.987$. Obviously, the optimal solution value of the IP is independent of the ordering and is in this case equal to 61. We noticed that sorting the customers in non-decreasing distance to the depot worked well in general, however we were not able to come up with a provable optimal ordering in general.
