Can Cplex prioritize some variables so that they're likely to be chosen together? - optimization

Question

So my problem contains a vehicle that moves from one node to the next. I have a bunch of nodes that may or may not be related to each other. I want the nodes that are similar to each other to be visited by the vehicle as much as possbible.
Is there any possible ways that i can prioritize the related nodes so that they're more likely to be grouped together? I thought to create sets or tuples that represent the different groups, and to have a variable X[i][j] = 1 if the vehicle moves from node i to node j, but i'm stuck at the "prioritize i and j if they come from the same set" part. Is it the boolean value that makes it impossible to render that? Should I modify my formulations somehow?
This is my code for the problem for now, i still haven't come out with the priority part
int nNode = 20;
range N = 1..nNode; //set of locations to visit
range V = 0..nNode; //set of locations plus the depot
range Vehicle = 1..6; //there are six vehicles
range boxType = 1..3; //three types of boxes to be transported
int demand[V][boxType] =...; //demand for a location in terms of different boxes
int timeBox[boxType] =...; //time associated with the actions on a type of box
dvar int+ totalLoad[Vehicle];
dvar int+ load[Vehicle][boxType]; //load in terms of box type
dvar boolean X[V][V][Vehicle]; /*1 if the vehicle Vehicle goes from node V to the
next node, 0 if not*/
dvar int+ t[Vehicle]; //total time a vehicle spends
dvar int time[Vehicle]; /*equals |t[vehicle] - target cycle time|, this is to make sure
each vehicle spends as close to target cycle time as possible*/
minimize sum (v in Vehicle)time[v];
subject to
{
forall (i in V)
sum (j in V, k in Vehicle)X[i][j][k] == 1; /* so that each starting node will have
exactly one destination node, i.e it will belong to exactly 1 route only*/
forall (j in V)
sum (i in V, k in Vehicle)X[i][j][k] == 1; // similar but for ending node
forall (k in Vehicle)
totalLoad[k] == sum(i in V, j in V)X[i][j][k]* (sum(b in boxType)demand[j][b]); /*total
load of a vehicle equals the total boxes collected at each stop on its path */
forall (b in boxType, k in Vehicle)
load[k][b] == sum(i in V, j in V) X[i][j][k]*(sum(j in Vehicle)demand[j][b]); /* calculate
separate number of boxes for each route*/
forall (k in Vehicle)
{
time[k] >= t[k] - 1.5;
time[k] >= - t[k] + 1.5;
time[k] <= t[k] + 1.5;
time[k] <= 2 - t[k] - 1.5; // breakdown of time[k] = |t[k]-1.5|, 1.5 is target cycle time
t[k] == sum(b in boxType) load[k][b]*timeBox[b]; // calculate the total time involved in a route
}
}

Answer 1

You could try adding a term into your objective that penalises giving different values to those sets of variables. Easy enough if there are only two of them but more fiddly if there are bigger subsets and/or lots of subsets to coordinate.

Answer 2

I would do something along the lines of what Tim suggested. Here is a little bit more meat on the bones:
x[i,j,k]=1 => L[g] ≤ k ∀i∈g, ∀j,k A lowerbound on the route k for group g
x[i,j,k]=1 => U[g] ≥ k ∀i∈g, ∀j,k An upperbound on the route k for group g
U[g]-L[g] ≥ 1 => δ[g]=1 δ[g]=1 if g is on different routes
min sum(g,δ[g]) objective
δ[g] ∈ {0,1} δ[g] is a binary variable
One way to implement the first 3 equations is:
L[g] ≤ k⋅x[i,j,k] + M(1-x[i,j,k]) ∀i ∈ g, ∀j,k
U[g] ≥ k⋅x[i,j,k] ∀i ∈ g, ∀j,k
M⋅δ[g] ≥ U[g]-L[g]
here g indicates a group. This makes the problem a multi-objective problem, so you can choose from a few possible approaches for that.

Answer 3

you could use priorities if you do not want to change the model from a logical point of view.
See https://github.com/AlexFleischerParis/zooopl/blob/master/zoopriorities.mod
int nbKids=300;
float costBus40=500;
float costBus30=400;
dvar int+ nbBus40;
dvar int+ nbBus30;
execute
{
nbBus40.priority=100;
nbBus30.priority=0;
}
minimize
costBus40*nbBus40 +nbBus30*costBus30;
subject to
{
40*nbBus40+nbBus30*30>=nbKids;
}
in Making optimization Simple
If you want to change the model from a logical point of view you can change the objective or add a second objective
int nbKids=350;
float costBus40=400;
float costBus30=300;
dvar int+ nbBus40;
dvar int+ nbBus30;
dexpr float absdistancebetweennumbers=abs(nbBus40-nbBus30);
minimize
staticLex(costBus40*nbBus40 +nbBus30*costBus30,absdistancebetweennumbers);
subject to
{
40*nbBus40+nbBus30*30>=nbKids;
}