add_max_utility_objective: Add maximum utility objective

Description

Set the objective of a conservation planning problem to secure as much of the features as possible without exceeding a budget. This type of objective does not use targets, and feature weights should be used instead to increase the representation of different features in solutions. Note that this objective does not aim to maximize as much of each feature as possible and so often results in solutions that are heavily biased towards specific features.

Usage

add_max_utility_objective(x, budget)

Arguments

ConservationProblem-class object.

budget

numeric value specifying the maximum expenditure of the prioritization. For problems with multiple zones, the argument to budget can be a single numeric value to specify a budget for the entire solution or a numeric vector to specify a budget for each each management zone.

Value

ConservationProblem-class object with the objective added to it.

Details

A problem objective is used to specify the overall goal of the conservation planning problem. Please note that all conservation planning problems formulated in the prioritizr package require the addition of objectives---failing to do so will return an error message when attempting to solve problem.

The maximum utility objective seeks to find the set of planning units that maximizes the overall level of representation across a suite of conservation features, while keeping cost within a fixed budget. Additionally, weights can be used to favor the representation of certain features over other features (see add_feature_weights). 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:

$$\mathit{Maximize} \space \sum_{i = 1}^{I} -s \space c_i \space x_i + \sum_{j = 1}^{J} a_j w_j \\ \mathit{subject \space to} \\ a_j = \sum_{i = 1}^{I} x_i r_{ij} \space \forall j \in J \\ \sum_{i = 1}^{I} x_i c_i \leq B$$

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 $i$, $A_j$ is the amount of feature $j$ represented in in the solution, 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, $c_i$ is the cost of planning unit $i$, and $s$ is a scaling factor used to shrink the costs so that the problem will return a cheapest solution when there are multiple solutions that represent the same amount of all features within the budget.

Please note that in versions prior to 3.0.0.0, this objective function was implemented in the add_max_cover_objective but has since been renamed as add_max_utility_objective to avoid confusion with historical formulations of the maximum coverage problem.

Examples

Run this code

# NOT RUN {
# load data
data(sim_pu_raster, sim_pu_zones_stack, sim_features, sim_features_zones)

# create problem with maximum utility objective
p1 <- problem(sim_pu_raster, sim_features) %>%
      add_max_utility_objective(5000) %>%
      add_binary_decisions() %>%
      add_default_solver(gap = 0)
# }
# NOT RUN {
# solve problem
s1 <- solve(p1)

# plot solution
plot(s1, main = "solution", axes = FALSE, box = FALSE)
# }
# NOT RUN {
# create multi-zone problem with maximum utility objective that
# has a single budget for all zones
p2 <- problem(sim_pu_zones_stack, sim_features_zones) %>%
      add_max_utility_objective(5000) %>%
      add_binary_decisions() %>%
      add_default_solver(gap = 0)
# }
# NOT RUN {
# solve problem
s2 <- solve(p2)

# plot solution
plot(category_layer(s2), main = "solution", axes = FALSE, box = FALSE)
# }
# NOT RUN {
# create multi-zone problem with maximum utility objective that
# has separate budgets for each zone
p3 <- problem(sim_pu_zones_stack, sim_features_zones) %>%
      add_max_utility_objective(c(1000, 2000, 3000)) %>%
      add_binary_decisions() %>%
      add_default_solver(gap = 0)
# }
# NOT RUN {
# solve problem
s3 <- solve(p3)

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