Schedule the field work of several teams from several bases

Solves the assignment of the units to bases (depots) and their routing jointly, as one multi-depot problem: every route (one team-day) leaves from and returns to one base, each base may send at most teams * days routes, and the objective is the total travel plus the per_route cost of the cost model for every route opened. The routes of each base are then distributed over its teams and the days available, longest routes first, so that the teams finish as evenly as possible. The result is a calendar: which team visits which units on which day, and whether the work fits.

Usage

schedule_fieldwork(
  matrix,
  units,
  teams,
  days,
  bases = NULL,
  max_length = Inf,
  max_stops = Inf,
  service_time = 0,
  cost_model = NULL,
  iterations = 200,
  time_limit = NULL,
  alpha = 0.3,
  seed = 1,
  engine = c("fieldopt", "vrpr"),
  region = NULL
)

Arguments

matrix: A travel_matrix() holding the units and the bases.
units: Names of the units to visit, or a select_units() sample.
teams: Named vector: number of teams per base (names are units of the matrix that act as bases). A single unnamed number with one base is accepted when bases is given.
days: Number of working days available.
bases: Optional names of the bases when teams is unnamed.
max_length, max_stops, service_time: Daily limits of a route, as in route_fieldwork().
cost_model: Optional field_cost_model(); its per_route cost is charged once per team-day and enters the objective.
iterations, time_limit, alpha, seed, engine: Solver settings of route_fieldwork(); both engines solve the multi-depot problem jointly.
region: Optional data frame with columns unit and region covering every unit and base: the schedule is then solved one region at a time (its units with its bases) and the calendars are combined. This is how a national survey is scheduled state by state without a single matrix over everything; matrix may then hold only the pairs within each region (other entries are ignored).

Value

An object of class fieldopt_schedule: calendar (a tibble with one row per route: base, team, day, route, stops, travel, duration, and units as a list column), assignment (unit, base), routes (the route_fieldwork() solution of each base), summary (per base: units, teams, routes, days needed, days available, whether it fits), fits and cost.

Details

When the team-days of the bases are not enough, the routes are solved without the limit per base and the calendar reports how many days each base would need (fits is FALSE).

Examples

set.seed(5)
pts <- data.frame(
  unit = c("base1", "base2", paste0("s", 1:30)),
  x = c(2, 8, runif(30, 0, 10)), y = c(2, 8, runif(30, 0, 10))
)
m <- travel_matrix(pts, method = "euclidean")
sch <- schedule_fieldwork(m,
  units = paste0("s", 1:30), teams = c(base1 = 2, base2 = 1),
  days = 4, max_stops = 4, iterations = 30
)
sch
#> 
#> ── Field schedule ──────────────────────────────────────────────────────────────
#> 30 units, 2 bases, 3 teams, 4 days available: the work fits.
#> "base1": 14 units, 2 teams, 4 routes, 2 of 4 days.
#> "base2": 16 units, 1 team, 4 routes, 4 of 4 days.
sch$calendar
#> # A tibble: 8 × 8
#>   base  team      day route stops travel duration units    
#>   <chr> <chr>   <int> <int> <int>  <dbl>    <dbl> <list>   
#> 1 base1 base1-1     1     1     4  16.9     16.9  <chr [4]>
#> 2 base1 base1-1     2     2     3   6.14     6.14 <chr [3]>
#> 3 base1 base1-2     1     4     4  12.8     12.8  <chr [4]>
#> 4 base1 base1-2     2     3     3   7.44     7.44 <chr [3]>
#> 5 base2 base2-1     1     3     4  13.7     13.7  <chr [4]>
#> 6 base2 base2-1     2     4     4  10.7     10.7  <chr [4]>
#> 7 base2 base2-1     3     1     4   9.32     9.32 <chr [4]>
#> 8 base2 base2-1     4     2     4   8.87     8.87 <chr [4]>