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)
x 


solution 

tibble::tibble()
object containing the solution cost.
It contains the following columns:
character
description 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).
numeric
cost 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).
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.
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:
x
has numeric
planning unitsThe argument to solution
must be a
numeric
vector with each element corresponding to a different planning
unit. It should have the same number of planning units as those
in the argument to x
. Additionally, any planning units missing
cost (NA
) values should also have missing (NA
) values in the
argument to solution
.
x
has matrix
planning unitsThe argument to solution
must be a
matrix
vector 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 to x
. Additionally, any planning units
missing cost (NA
) values for a particular zone should also have a
missing (NA
) values in the argument to solution
.
x
has Raster
planning unitsThe argument to solution
be a Raster
object 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 to x
. Additionally,
any planning units missing cost (NA
) values for a particular zone
should also have missing (NA
) values in the argument to solution
.
x
has data.frame
planning unitsThe argument to solution
must
be a data.frame
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 a data.frame
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 with sf::sf()
data).
Additionally, any planning units missing cost
(NA
) values for a particular zone should also have missing (NA
)
values in the argument to solution
.
x
has Spatial
planning unitsThe argument to solution
must be a Spatial
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 the
Spatial
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 with sf::sf()
data).
Additionally, the argument to solution
must 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 to solution
.
x
has sf::sf()
planning unitsThe argument to solution
must be
a sf::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 the
sf::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 to solution
must 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 to solution
.
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.
# \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 multizone 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.
# }