Add maximum number of targets met objective

Set the objective of a conservation planning problem to fulfill as many targets as possible, whilst ensuring that the cost of the solution does not exceed a budget. Note that this objective does not value the partial achievement of a given target—only whether or not the target has been met. For this reason, we generally recommend using the the minimum shortfall objective (add_min_shortfall_objective() instead for budget-limited scenarios.

Usage

add_max_n_targets_met_objective(x, budget)

Arguments

x: problem() object.
budget: numeric value specifying the maximum expenditure permitted for the solution. If x has multiple zones, then budget can be (i) a single numeric value to specify an overall budget for the entire solution or (ii) a numeric vector to specify a budget for each zone (separately) in the solution.

Details

The maximum number of targets met objective is an enhanced version of the maximum coverage objective add_max_cover_objective() because targets can be used to ensure that a certain amount of each feature is required in order for them to be adequately represented (similar to the minimum set objective (see add_min_set_objective()). This objective finds the set of planning units that meets representation targets for as many features as possible while staying within a fixed budget (inspired by Cabeza and Moilanen 2001). Additionally, weights can be used to favor the representation of certain features over other features (see add_feature_weights()). If multiple solutions can meet the same number of weighted targets while staying within budget, the cheapest solution is returned.

Mathematical formulation

This objective can be expressed mathematically for a set of planning units ($I$ indexed by $i$) and a set of features ($J$ indexed by $j$) as:

$$\mathit{Maximize} \space \sum_{j = 1}^{J} y_j w_j \\ \mathit{subject \space to} \\ \sum_{i = 1}^{I} x_i r_{ij} \geq y_j t_j \forall j \in J \\ \sum_{i = 1}^{I} x_i c_i \leq B$$

Here, $x_i$ is the decisions variable (e.g., specifying whether planning unit $i$ has been selected (1) or not (0)), $r_{ij}$ is the amount of feature $j$ in planning unit $i$, $t_j$ is the representation target for feature $j$, $y_j$ indicates if the solution has meet the target $t_j$ for feature $j$, and $w_j$ is the weight for feature $j$ (defaults to 1 for all features; see add_feature_weights() to specify weights). Additionally, $B$ is the budget allocated for the solution, and $c_i$ is the cost of planning unit $i$.

Notes

In previous versions (< 9.0.0), this function was called the add_max_features_objective() and has since been renamed to provide greater clarity. Additionally, it previously had extra terms to help minimize the solution cost. Although these terms have since been removed to reduce solve time, this behavior can still be achieved by building a multi-objective optimization problem and specifying the first problem based on this objective function and the second problem based on minimizing penalties (via add_min_penalties_objective()) with penalties set according to cost values (via add_linear_penalties()).

References

Cabeza M and Moilanen A (2001) Design of reserve networks and the persistence of biodiversity. Trends in Ecology & Evolution, 16: 242–248.

Examples

# load data
sim_pu_raster <- get_sim_pu_raster()
sim_features <- get_sim_features()
sim_zones_pu_raster <- get_sim_zones_pu_raster()
sim_zones_features <- get_sim_zones_features()

# create problem with maximum number of targets met objective
p1 <-
  problem(sim_pu_raster, sim_features) %>%
  add_max_n_targets_met_objective(1800) %>%
  add_relative_targets(0.1) %>%
  add_binary_decisions() %>%
  add_default_solver(verbose = FALSE)

# solve problem
s1 <- solve(p1)

# plot solution
plot(s1, main = "solution", axes = FALSE)


# create multi-zone problem with maximum number of targets met objective,
# 10% representation targets for each feature, and set
# a budget such that the total maximum expenditure in all zones
# cannot exceed 3000
p2 <-
  problem(sim_zones_pu_raster, sim_zones_features) %>%
  add_max_n_targets_met_objective(3000) %>%
  add_relative_targets(matrix(0.1, ncol = 3, nrow = 5)) %>%
  add_binary_decisions() %>%
  add_default_solver(verbose = FALSE)

# solve problem
s2 <- solve(p2)

# plot solution
plot(category_layer(s2), main = "solution", axes = FALSE)


# create multi-zone problem with maximum number of targets met objective,
# 10% representation targets for each feature, and set
# separate budgets for each management zone
p3 <-
  problem(sim_zones_pu_raster, sim_zones_features) %>%
  add_max_n_targets_met_objective(c(3000, 3000, 3000)) %>%
  add_relative_targets(matrix(0.1, ncol = 3, nrow = 5)) %>%
  add_binary_decisions() %>%
  add_default_solver(verbose = FALSE)

# solve problem
s3 <- solve(p3)

# plot solution
plot(category_layer(s3), main = "solution", axes = FALSE)