```
# Sets
set MACHINES;
set WELLS;
set DAYS; # Can be TIMES with a smaller measure rather than Days
# Parameters
param well_operation{WELLS}; # Required operation type for each well
param machine_operation{MACHINES}; # Operation type of each machine
param distance{WELLS, WELLS}; # Distance between wells
param operating_cost{MACHINES}; # Operating cost per day for each machine
param priority{WELLS}; # Priority of each well

# Decision variables
var operates{MACHINES, WELLS, DAYS} binary; # operates[m, i, d] = 1 if machine m is at well i on day d
var finished{WELLS, DAYS} binary; # finished[i, d] = 1 if well i operation is over on day d
var travel{MACHINES, WELLS, WELLS, DAYS} binary; # travel[m, i, j, d] = 1 if machine m travels from well i to well j on day d
var works{MACHINES, DAYS} binary; # works[m, d] = 1 if machine m is operating on day d

subject to Travel_Consistency{m in MACHINES, i in WELLS, j in WELLS, d in DAYS}:
    travel[m, i, j, d] == 1 <==> operates[m, i, d] == 1;
subject to Operation_Consistency{m in MACHINES, i in WELLS, d in DAYS}:
    operates[m, i, d] == 1 <==> works[m,i];
# Less priority implies, higher priority wells have already been finished
subject to Priorities{i in WELLS, j in WELLS: priority[i] < priority[j]}:
    finished[m, i, d] == 1 ==> finished[m,j,d] == 1;
# TODO simplified version of problem req. Best solution would be to have a set of feasible combinations between Machines and Wells
subject to Incompatibilities{m in MACHINES, i in WELLS: machine_operation[m] != well_operation[i]}:
    works[m,i] == 0;

# Defined variables
var total_distance = sum{m in MACHINES, i in WELLS, j in WELLS, d in DAYS} distance[i, j] * travel[m, i, j, d];
var total_renting_cost = sum{m in MACHINES, d in DAYS} operating_cost[m] * works[m, d];

minimize Total_Cost:
    total_distance + total_renting_cost;
```