Calculate the total cost of a solution to a conservation planning
problem().
For example, if the planning unit cost data describe land acquisition costs
(USD), then the total cost would be net cost (USD) needed to acquire
all planning units selected within the solution.
eval_cost_summary(x, solution)
# S3 method for default
eval_cost_summary(x, solution)
# S3 method for ConservationProblem
eval_cost_summary(x, solution)Arguments
| x |
|
|---|---|
| solution |
|
Value
tibble::tibble() object containing the solution cost.
It contains the following columns:
- summary
characterdescription of the summary statistic. The statistic associated with the"overall"value in this column is calculated using the entire solution (including all management zones if there are multiple zones). If multiple management zones are present, then summary statistics are also provided for each zone separately (indicated using zone names).- cost
numericcost value. Greater values correspond to solutions that are more costly to implement. Thus conservation planning exercises typically prefer solutions with smaller values, because they are cheaper to implement (assuming all other relevant factors, such as feature representation, are equal).
Details
This metric is equivalent to the Cost metric reported by the
Marxan software (Ball et al. 2009).
Specifically, the cost of a solution is defined as the sum of the cost
values, supplied when creating a problem() object
(e.g. using the cost_column argument),
weighted by the status of each planning unit in the solution.
Solution format
Broadly speaking, the argument to solution must be in the same format as
the planning unit data in the argument to x.
Further details on the correct format are listed separately
for each of the different planning unit data formats:
xhasnumericplanning unitsThe argument to
solutionmust be anumericvector with each element corresponding to a different planning unit. It should have the same number of planning units as those in the argument tox. Additionally, any planning units missing cost (NA) values should also have missing (NA) values in the argument tosolution.xhasmatrixplanning unitsThe argument to
solutionmust be amatrixvector with each row corresponding to a different planning unit, and each column correspond to a different management zone. It should have the same number of planning units and zones as those in the argument tox. Additionally, any planning units missing cost (NA) values for a particular zone should also have a missing (NA) values in the argument tosolution.xhasRasterplanning unitsThe argument to
solutionbe aRasterobject where different grid cells (pixels) correspond to different planning units and layers correspond to a different management zones. It should have the same dimensionality (rows, columns, layers), resolution, extent, and coordinate reference system as the planning units in the argument tox. Additionally, any planning units missing cost (NA) values for a particular zone should also have missing (NA) values in the argument tosolution.xhasdata.frameplanning unitsThe argument to
solutionmust be adata.framewith each column corresponding to a different zone, each row corresponding to a different planning unit, and cell values corresponding to the solution value. This means that if adata.frameobject containing the solution also contains additional columns, then these columns will need to be subsetted prior to using this function (see below for example withsf::sf()data). Additionally, any planning units missing cost (NA) values for a particular zone should also have missing (NA) values in the argument tosolution.xhasSpatialplanning unitsThe argument to
solutionmust be aSpatialobject with each column corresponding to a different zone, each row corresponding to a different planning unit, and cell values corresponding to the solution value. This means that if theSpatialobject containing the solution also contains additional columns, then these columns will need to be subsetted prior to using this function (see below for example withsf::sf()data). Additionally, the argument tosolutionmust also have the same coordinate reference system as the planning unit data. Furthermore, any planning units missing cost (NA) values for a particular zone should also have missing (NA) values in the argument tosolution.xhassf::sf()planning unitsThe argument to
solutionmust be asf::sf()object with each column corresponding to a different zone, each row corresponding to a different planning unit, and cell values corresponding to the solution value. This means that if thesf::sf()object containing the solution also contains additional columns, then these columns will need to be subsetted prior to using this function (see below for example). Additionally, the argument tosolutionmust also have the same coordinate reference system as the planning unit data. Furthermore, any planning units missing cost (NA) values for a particular zone should also have missing (NA) values in the argument tosolution.
References
Ball IR, Possingham HP, and Watts M (2009) Marxan and relatives: Software for spatial conservation prioritisation in Spatial conservation prioritisation: Quantitative methods and computational tools. Eds Moilanen A, Wilson KA, and Possingham HP. Oxford University Press, Oxford, UK.
See also
Examples
# \dontrun{
# set seed for reproducibility
set.seed(500)
# load data
data(sim_pu_raster, sim_pu_sf, sim_features,
sim_pu_zones_sf, sim_features_zones)
# build minimal conservation problem with raster data
p1 <- problem(sim_pu_raster, sim_features) %>%
add_min_set_objective() %>%
add_relative_targets(0.1) %>%
add_binary_decisions() %>%
add_default_solver(verbose = FALSE)
# solve the 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 cost of the solution
r1 <- eval_cost_summary(p1, s1)
print(r1)
#> # A tibble: 1 x 2
#> summary cost
#> <chr> <dbl>
#> 1 overall 1987.
# build minimal conservation problem with polygon (sf) data
p2 <- problem(sim_pu_sf, sim_features, cost_column = "cost") %>%
add_min_set_objective() %>%
add_relative_targets(0.1) %>%
add_binary_decisions() %>%
add_default_solver(verbose = FALSE)
# solve the problem
s2 <- solve(p2)
# plot solution
plot(s2[, "solution_1"])
# print first six rows of the attribute table
print(head(s2))
#> Simple feature collection with 6 features and 4 fields
#> Geometry type: POLYGON
#> Dimension: XY
#> Bounding box: xmin: 0 ymin: 0.9 xmax: 0.6 ymax: 1
#> CRS: NA
#> cost locked_in locked_out solution_1 geometry
#> 1 215.8638 FALSE FALSE 0 POLYGON ((0 1, 0.1 1, 0.1 0...
#> 2 212.7823 FALSE FALSE 0 POLYGON ((0.1 1, 0.2 1, 0.2...
#> 3 207.4962 FALSE FALSE 0 POLYGON ((0.2 1, 0.3 1, 0.3...
#> 4 208.9322 FALSE TRUE 0 POLYGON ((0.3 1, 0.4 1, 0.4...
#> 5 214.0419 FALSE FALSE 0 POLYGON ((0.4 1, 0.5 1, 0.5...
#> 6 213.7636 FALSE FALSE 0 POLYGON ((0.5 1, 0.6 1, 0.6...
# calculate cost of the solution
r2 <- eval_cost_summary(p2, s2[, "solution_1"])
print(r2)
#> # A tibble: 1 x 2
#> summary cost
#> <chr> <dbl>
#> 1 overall 1793.
# manually calculate cost of the solution
r2_manual <- sum(s2$solution * sim_pu_sf$cost, na.rm = TRUE)
print(r2_manual)
#> [1] 1792.535
# build multi-zone conservation problem with polygon (sf) data
p3 <- problem(sim_pu_zones_sf, sim_features_zones,
cost_column = c("cost_1", "cost_2", "cost_3")) %>%
add_min_set_objective() %>%
add_relative_targets(matrix(runif(15, 0.1, 0.2), nrow = 5,
ncol = 3)) %>%
add_binary_decisions() %>%
add_default_solver(verbose = FALSE)
# solve the problem
s3 <- solve(p3)
# print first six rows of the attribute table
print(head(s3))
#> Simple feature collection with 6 features and 9 fields
#> Geometry type: POLYGON
#> Dimension: XY
#> Bounding box: xmin: 0 ymin: 0.9 xmax: 0.6 ymax: 1
#> CRS: NA
#> cost_1 cost_2 cost_3 locked_1 locked_2 locked_3 solution_1_zone_1
#> 1 215.8638 183.3344 205.4113 FALSE FALSE FALSE 0
#> 2 212.7823 189.4978 209.6404 FALSE FALSE FALSE 0
#> 3 207.4962 193.6007 215.4212 TRUE FALSE FALSE 0
#> 4 208.9322 197.5897 218.5241 FALSE FALSE FALSE 0
#> 5 214.0419 199.8033 220.7100 FALSE FALSE FALSE 0
#> 6 213.7636 203.1867 224.6809 FALSE FALSE FALSE 0
#> solution_1_zone_2 solution_1_zone_3 geometry
#> 1 0 1 POLYGON ((0 1, 0.1 1, 0.1 0...
#> 2 0 0 POLYGON ((0.1 1, 0.2 1, 0.2...
#> 3 0 0 POLYGON ((0.2 1, 0.3 1, 0.3...
#> 4 0 0 POLYGON ((0.3 1, 0.4 1, 0.4...
#> 5 0 0 POLYGON ((0.4 1, 0.5 1, 0.5...
#> 6 1 0 POLYGON ((0.5 1, 0.6 1, 0.6...
# create new column representing the zone id that each planning unit
# was allocated to in the solution
s3$solution <- category_vector(
s3[, c("solution_1_zone_1", "solution_1_zone_2", "solution_1_zone_3")])
s3$solution <- factor(s3$solution)
# plot solution
plot(s3[, "solution"])
# calculate cost of the solution
r3 <- eval_cost_summary(
p3, s3[, c("solution_1_zone_1", "solution_1_zone_2", "solution_1_zone_3")])
print(r3)
#> # A tibble: 4 x 2
#> summary cost
#> <chr> <dbl>
#> 1 overall 10160.
#> 2 zone_1 3395.
#> 3 zone_2 3322.
#> 4 zone_3 3443.
# }