Calculate irreplaceability scores for planning units selected in a solution using the replacement cost method (Cabeza and Moilanen 2006).

# S4 method for ConservationProblem,numeric
replacement_cost(x, solution, rescale, run_checks, force, threads, ...)

# S4 method for ConservationProblem,matrix
replacement_cost(x, solution, rescale, run_checks, force, threads, ...)

# S4 method for ConservationProblem,data.frame
replacement_cost(x, solution, rescale, run_checks, force, threads, ...)

# S4 method for ConservationProblem,Spatial
replacement_cost(x, solution, rescale, run_checks, force, threads, ...)

# S4 method for ConservationProblem,Raster
replacement_cost(x, solution, rescale, run_checks, force, threads, ...)

Arguments

x

ConservationProblem-class object.

solution

numeric, matrix, data.frame, Raster-class, or Spatial-class object. See the Details section for more information.

rescale

logical flag indicating if replacement cost values---excepting infinite (Inf) and zero values---should be rescaled to range between 0.01 and 1. Defaults to TRUE.

run_checks

logical flag indicating whether presolve checks should be run prior solving the problem. These checks are performed using the presolve_check function. Defaults to TRUE. Skipping these checks may reduce run time for large problems.

force

logical flag indicating if an attempt to should be made to solve the problem even if potential issues were detected during the presolve checks. Defaults to FALSE.

threads

integer number of threads to use for the optimization algorithm. The default value of 1 will result in only one thread being used.

...

not used.

Value

A numeric, matrix, RasterLayer-class, or Spatial-class object containing the replacement costs for each planning unit in the solution.

Details

Using this method, the score for each planning unit is calculated as the difference in the objective value of a solution when each planning selected planning units locked in. In other words, the replacement cost metric corresponds to change in solution quality incurred if a given planning unit cannot be acquired when implementing the solution and the next best planning unit (or set of planning units) will need to be considered instead. Thus planning units with a higher score are more irreplaceable. For example, when using the minimum set objective function (add_min_set_objective), the replacement cost scores correspond to the additional costs needed to meet targets when each planning unit is locked out. When using the maximum utility objective function (add_max_utility_objective, the replacement cost scores correspond to the reduction in the utility when each planning unit is locked out. Infinite values mean that no feasible solution exists when planning units are locked out---they are absolutely essential for obtaining a solution (e.g. they contain rare species that are not found in any other planning units or were locked in). Zeros values mean that planning units can swapped with other planning units and this will have no effect on the performance of the solution at all (e.g. because they were only selected due to spatial fragmentation penalties). Since these calculations can take a long time to complete, we recommend calculating these scores without additional penalties (e.g. add_boundary_penalties) or constraints (e.g. link{add_contiguity_constraints}). They can be sped up further by using proportion-type decisions when calculating the scores (see below for an example).

Note that all arguments to solution must correspond to the planning unit data in the argument to x in terms of data representation, dimensionality, and spatial attributes (if applicable). This means that if the planning unit data in x is a numeric vector then the argument to solution must be a numeric vector with the same number of elements, if the planning unit data in x is a RasterLayer-class then the argument to solution must also be a RasterLayer-class with the same number of rows and columns and the same resolution, extent, and coordinate reference system, if the planning unit data in x is a Spatial-class object then the argument to solution must also be a Spatial-class object and have the same number of spatial features (e.g. polygons) and have the same coordinate reference system, if the planning units in x are a data.frame then the argument to solution must also be a data.frame with each column correspond to a different zone and each row correspond to a different planning unit, and values correspond to the allocations (e.g. values of zero or one).

Solutions must have planning unit statuses set to missing (NA) values for planning units that have missing (NA) cost data. For problems with multiple zones, this means that planning units must have missing (NA) allocation values in zones where they have missing (NA) cost data. In other words, planning units that have missing (NA) cost values in x should always have a missing (NA) value the argument to solution. If an argument is supplied to solution where this is not the case, then an error will be thrown.

References

Cabeza M and Moilanen (2006) Replacement cost: A practical measure of site value for cost-effective reserve planning. Biological Conservation, 132: 336--342.

See also

Examples

# seed seed for reproducibility set.seed(600) # load data data(sim_pu_raster, sim_features, sim_pu_zones_stack, sim_features_zones) # create minimal problem with binary decisions p1 <- problem(sim_pu_raster, sim_features) %>% add_min_set_objective() %>% add_relative_targets(0.1) %>% add_binary_decisions() %>% add_default_solver(gap = 0, verbose = FALSE)
# solve problem s1 <- solve(p1) # print solution print(s1)
#> class : RasterLayer #> dimensions : 10, 10, 100 (nrow, ncol, ncell) #> resolution : 0.1, 0.1 (x, y) #> extent : 0, 1, 0, 1 (xmin, xmax, ymin, ymax) #> crs : NA #> source : memory #> names : layer #> values : 0, 1 (min, max) #>
# plot solution plot(s1, main = "solution", axes = FALSE, box = FALSE)
# calculate irreplaceability scores rc1 <- replacement_cost(p1, s1) # print irreplaceability scores print(rc1)
#> class : RasterLayer #> dimensions : 10, 10, 100 (nrow, ncol, ncell) #> resolution : 0.1, 0.1 (x, y) #> extent : 0, 1, 0, 1 (xmin, xmax, ymin, ymax) #> crs : NA #> source : memory #> names : layer #> values : 0, 1 (min, max) #>
# plot irreplaceability scores plot(rc1, main = "replacement cost", axes = FALSE, box = FALSE)
# since replacement cost scores can take a long time to calculate with # binary decisions, we can calculate them using proportion-type # decision variables. Note we are still calculating the scores for our # previous solution (s1), we are just using a different optimization # problem when calculating the scores. p2 <- problem(sim_pu_raster, sim_features) %>% add_min_set_objective() %>% add_relative_targets(0.1) %>% add_proportion_decisions() %>% add_default_solver(gap = 0, verbose = FALSE) # calculate irreplaceability scores using proportion type decisions
rc2 <- replacement_cost(p2, s1) # print irreplaceability scores based on proportion type decisions print(rc2)
#> class : RasterLayer #> dimensions : 10, 10, 100 (nrow, ncol, ncell) #> resolution : 0.1, 0.1 (x, y) #> extent : 0, 1, 0, 1 (xmin, xmax, ymin, ymax) #> crs : NA #> source : memory #> names : layer #> values : 0, 1 (min, max) #>
# plot irreplacability scores based on proportion type decisions # we can see that the irreplaceability values in rc1 and rc2 are similar, # and this confirms that the proportion type decisions are a good # approximation plot(rc2, main = "replacement cost", axes = FALSE, box = FALSE)
# build multi-zone conservation problem with binary decisions p3 <- problem(sim_pu_zones_stack, sim_features_zones) %>% add_min_set_objective() %>% add_relative_targets(matrix(runif(15, 0.1, 0.2), nrow = 5, ncol = 3)) %>% add_binary_decisions() %>% add_default_solver(gap = 0, verbose = FALSE)
# solve the problem s3 <- solve(p3) # print solution print(s3)
#> class : RasterStack #> dimensions : 10, 10, 100, 3 (nrow, ncol, ncell, nlayers) #> resolution : 0.1, 0.1 (x, y) #> extent : 0, 1, 0, 1 (xmin, xmax, ymin, ymax) #> crs : NA #> names : layer.1.1, layer.1.2, layer.1.3 #> min values : 0, 0, 0 #> max values : 1, 1, 1 #>
# plot solution # each panel corresponds to a different zone, and data show the # status of each planning unit in a given zone plot(s3, main = paste0("zone ", seq_len(nlayers(s3))), axes = FALSE, box = FALSE)
# calculate irreplaceability scores rc3 <- replacement_cost(p3, s3) # plot irreplaceability # each panel corresponds to a different zone, and data show the # irreplaceability of each planning unit in a given zone plot(rc3, main = paste0("zone ", seq_len(nlayers(s3))), axes = FALSE, box = FALSE)