Set the objective of a conservation planning problem() to minimize the overall shortfall for as many targets as possible while ensuring that the cost of the solution does not exceed a budget.

add_min_shortfall_objective(x, budget)

Arguments

x

problem() (i.e. ConservationProblem) 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

Object (i.e. ConservationProblem) 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 minimum shortfall objective aims to find the set of planning units that minimize the overall (weighted sum) shortfall for the representation targets---that is, the fraction of each target that remains unmet---for as many features as possible while staying within a fixed budget (inspired by Table 1, equation IV, Arponen et al. 2005). Additionally, weights can be used to favor the representation of certain features over other features (see add_feature_weights().

The minimum shortfall objective for the reserve design problem can be expressed mathematically for a set of planning units (\(I\) indexed by \(i\)) and a set of features (\(J\) indexed by \(j\)) as:

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

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 \(j\), \(y_j\) denotes the representation shortfall for 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, \(c_i\) is the cost of planning unit \(i\). Note that \(y_j\) is a continuous variable bounded between zero and infinity, and denotes the shortfall for target \(j\).

References

Arponen A, Heikkinen RK, Thomas CD, and Moilanen A (2005) The value of biodiversity in reserve selection: representation, species weighting, and benefit functions. Conservation Biology, 19: 2009--2014.

See also

Examples

# load data data(sim_pu_raster, sim_pu_zones_stack, sim_features, sim_features_zones) # create problem with minimum shortfall objective p1 <- problem(sim_pu_raster, sim_features) %>% add_min_shortfall_objective(1800) %>% add_relative_targets(0.1) %>% add_binary_decisions() %>% add_default_solver(verbose = FALSE) # \dontrun{ # solve problem s1 <- solve(p1) # plot solution plot(s1, main = "solution", axes = FALSE, box = FALSE)
# } # create multi-zone problem with minimum shortfall objective, # with 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_pu_zones_stack, sim_features_zones) %>% add_min_shortfall_objective(3000) %>% add_relative_targets(matrix(0.1, ncol = 3, nrow = 5)) %>% add_binary_decisions() %>% add_default_solver(verbose = FALSE) # \dontrun{ # solve problem s2 <- solve(p2) # plot solution plot(category_layer(s2), main = "solution", axes = FALSE, box = FALSE)
# } # create multi-zone problem with minimum shortfall objective, # with 10% representation targets for each feature, and set # separate budgets for each management zone p3 <- problem(sim_pu_zones_stack, sim_features_zones) %>% add_min_shortfall_objective(c(3000, 3000, 3000)) %>% add_relative_targets(matrix(0.1, ncol = 3, nrow = 5)) %>% add_binary_decisions() %>% add_default_solver(verbose = FALSE) # \dontrun{ # solve problem s3 <- solve(p3) # plot solution plot(category_layer(s3), main = "solution", axes = FALSE, box = FALSE)
# }