Evaluate cost of solution

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.

Usage

eval_cost_summary(x, solution)

# S3 method for class 'ConservationProblem'
eval_cost_summary(x, solution)

# S3 method for class 'MultiConservationProblem'
eval_cost_summary(x, solution)

Arguments

x: problem() or multi_problem() object.
solution: numeric, matrix, data.frame, terra::rast(), or sf::sf() object. Note that solution must have the same format as the planning unit data in x. See the Solution format section for more information.

Value

A tibble::tibble() object describing the solution cost. It contains the following columns.

problem: character name of problem. Note that this column is only present if x is a multi_problem() object.
summary: 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 x has multiple zones). If x has multiple management zones, then summary statistics are also provided for each zone separately (indicated using zone names).
cost: 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 else is 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., per cost_column), weighted by the status of each planning unit in the solution.

Solution format

Broadly speaking, solution must be in the same format as the planning unit data in x. Further details on the correct format are listed separately for each of the different planning unit data formats.

x has numeric planning units: Here 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 x. Additionally, any planning units with missing cost (NA) values should also have missing (NA) values in the solution.
x has matrix planning units: Here 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 x. Additionally, any planning units with missing cost (NA) values for a particular zone should also have a missing (NA) values in solution.
x has terra::rast() planning units: Here solution be a terra::rast() object where different cells 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 x. Additionally, any planning units with missing cost (NA) values for a particular zone should also have missing (NA) values in solution.
x has data.frame planning units: Here 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 with missing cost (NA) values for a particular zone should also have missing (NA) values in solution.
x has sf::sf() planning units: Here 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, solution must also have the same coordinate reference system as the planning unit data. Furthermore, any planning units with missing cost (NA) values for a particular zone should also have missing (NA) values in solution.

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.

Examples

# set seed for reproducibility
set.seed(500)

# load data
sim_pu_raster <- get_sim_pu_raster()
sim_pu_polygons <- get_sim_pu_polygons()
sim_features <- get_sim_features()
sim_zones_pu_polygons <- get_sim_zones_pu_polygons()
sim_zones_features <- get_sim_zones_features()

# 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       : SpatRaster
#> size        : 10, 10, 1  (nrow, ncol, nlyr)
#> resolution  : 0.1, 0.1  (x, y)
#> extent      : 0, 1, 0, 1  (xmin, xmax, ymin, ymax)
#> coord. ref. : WGS 84 / Pseudo-Mercator (EPSG:3857)
#> source(s)   : memory
#> varname     : sim_pu_raster
#> name        : layer
#> min value   :     0
#> max value   :     1

# plot solution
plot(s1, main = "solution", axes = FALSE)


# calculate cost of the solution
r1 <- eval_cost_summary(p1, s1)
print(r1)
#> # A tibble: 1 × 2
#>   summary  cost
#>   <chr>   <dbl>
#> 1 overall 1987.

# build minimal conservation problem with polygon data
p2 <-
  problem(sim_pu_polygons, 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 solution
print(s2)
#> Simple feature collection with 90 features and 4 fields
#> Geometry type: POLYGON
#> Dimension:     XY
#> Bounding box:  xmin: 0 ymin: 0 xmax: 1 ymax: 1
#> Projected CRS: WGS 84 / Pseudo-Mercator
#> # A tibble: 90 × 5
#>     cost locked_in locked_out solution_1                                geometry
#>  * <dbl> <lgl>     <lgl>           <dbl>                           <POLYGON [m]>
#>  1  216. FALSE     FALSE               0     ((0 1, 0.1 1, 0.1 0.9, 0 0.9, 0 1))
#>  2  213. FALSE     FALSE               0 ((0.1 1, 0.2 1, 0.2 0.9, 0.1 0.9, 0.1 …
#>  3  207. FALSE     FALSE               0 ((0.2 1, 0.3 1, 0.3 0.9, 0.2 0.9, 0.2 …
#>  4  209. FALSE     TRUE                0 ((0.3 1, 0.4 1, 0.4 0.9, 0.3 0.9, 0.3 …
#>  5  214. FALSE     FALSE               0 ((0.4 1, 0.5 1, 0.5 0.9, 0.4 0.9, 0.4 …
#>  6  214. FALSE     FALSE               0 ((0.5 1, 0.6 1, 0.6 0.9, 0.5 0.9, 0.5 …
#>  7  210. FALSE     FALSE               0 ((0.6 1, 0.7 1, 0.7 0.9, 0.6 0.9, 0.6 …
#>  8  211. FALSE     TRUE                0 ((0.7 1, 0.8 1, 0.8 0.9, 0.7 0.9, 0.7 …
#>  9  210. FALSE     FALSE               0 ((0.8 1, 0.9 1, 0.9 0.9, 0.8 0.9, 0.8 …
#> 10  204. FALSE     FALSE               0   ((0.9 1, 1 1, 1 0.9, 0.9 0.9, 0.9 1))
#> # ℹ 80 more rows

# calculate cost of the solution
r2 <- eval_cost_summary(p2, s2[, "solution_1"])
print(r2)
#> # A tibble: 1 × 2
#>   summary  cost
#>   <chr>   <dbl>
#> 1 overall 1793.

# manually calculate cost of the solution
r2_manual <- sum(s2$solution_1 * sim_pu_polygons$cost, na.rm = TRUE)
print(r2_manual)
#> [1] 1792.535

# build multi-zone conservation problem with polygon data
p3 <-
  problem(
    sim_zones_pu_polygons, sim_zones_features,
    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 solution
print(s3)
#> Simple feature collection with 90 features and 9 fields
#> Geometry type: POLYGON
#> Dimension:     XY
#> Bounding box:  xmin: 0 ymin: 0 xmax: 1 ymax: 1
#> Projected CRS: WGS 84 / Pseudo-Mercator
#> # A tibble: 90 × 10
#>    cost_1 cost_2 cost_3 locked_1 locked_2 locked_3 solution_1_zone_1
#>  *  <dbl>  <dbl>  <dbl> <lgl>    <lgl>    <lgl>                <dbl>
#>  1   216.   183.   205. FALSE    FALSE    FALSE                    0
#>  2   213.   189.   210. FALSE    FALSE    FALSE                    0
#>  3   207.   194.   215. TRUE     FALSE    FALSE                    0
#>  4   209.   198.   219. FALSE    FALSE    FALSE                    0
#>  5   214.   200.   221. FALSE    FALSE    FALSE                    0
#>  6   214.   203.   225. FALSE    FALSE    FALSE                    0
#>  7   211.   209.   223. FALSE    FALSE    FALSE                    0
#>  8   210.   212.   222. TRUE     FALSE    FALSE                    0
#>  9   204.   218.   214. FALSE    FALSE    FALSE                    0
#> 10   213.   183.   206. FALSE    FALSE    FALSE                    0
#> # ℹ 80 more rows
#> # ℹ 3 more variables: solution_1_zone_2 <dbl>, solution_1_zone_3 <dbl>,
#> #   geometry <POLYGON [m]>

# 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 × 2
#>   summary   cost
#>   <chr>    <dbl>
#> 1 overall 10452.
#> 2 zone_1   3432.
#> 3 zone_2   3354.
#> 4 zone_3   3666.