Project cost effectiveness

Source: R/project_cost_effectiveness.R

Calculate the individual cost-effectiveness of each conservation project in a project prioritization problem() (Joseph, Maloney & Possingham 2009).

project_cost_effectiveness(x)

Arguments

x

project prioritization problem().

Value

A tibble::tibble() table containing the following columns:

"project"

character name of each project

"cost"

numeric cost of each project.

"benefit"

numeric benefit for each project. For a given project, this is calculated as the difference between (i) the objective value for a solution containing all of the management actions associated with the project and all zero cost actions, and (ii) the objective value for a solution containing the baseline project.

"ce"

numeric cost-effectiveness of each project. For a given project, this is calculated as the difference between the the benefit for the project and the benefit for the baseline project, divided by the cost of the project. Note that the baseline project will have a NaN value because it has a zero cost.

"rank"

numeric rank for each project according to is cost-effectiveness value. The project with a rank of one is the most cost-effective project. Ties are accommodated using averages.

Details

Note that project cost-effectiveness cannot be calculated for problems with minimum set objectives because the objective function for these problems is to minimize cost and not maximize some measure of biodiversity persistence.

References

Joseph LN, Maloney RF & Possingham HP (2009) Optimal allocation of resources among threatened species: A project prioritization protocol. Conservation Biology, 23, 328--338.

Examples

# load data
data(sim_projects, sim_features, sim_actions)

# print project data
print(sim_projects)
#> # A tibble: 6 × 13
#>   name       success     F1     F2      F3     F4     F5 F1_ac…¹ F2_ac…² F3_ac…³
#>   <chr>        <dbl>  <dbl>  <dbl>   <dbl>  <dbl>  <dbl> <lgl>   <lgl>   <lgl>  
#> 1 F1_project   0.919  0.791 NA     NA      NA     NA     TRUE    FALSE   FALSE  
#> 2 F2_project   0.923 NA      0.888 NA      NA     NA     FALSE   TRUE    FALSE  
#> 3 F3_project   0.829 NA     NA      0.502  NA     NA     FALSE   FALSE   TRUE   
#> 4 F4_project   0.848 NA     NA     NA       0.690 NA     FALSE   FALSE   FALSE  
#> 5 F5_project   0.814 NA     NA     NA      NA      0.617 FALSE   FALSE   FALSE  
#> 6 baseline_…   1      0.298  0.250  0.0865  0.249  0.182 FALSE   FALSE   FALSE  
#> # … with 3 more variables: F4_action <lgl>, F5_action <lgl>,
#> #   baseline_action <lgl>, and abbreviated variable names ¹​F1_action,
#> #   ²​F2_action, ³​F3_action

# print action data
print(sim_features)
#> # A tibble: 5 × 2
#>   name  weight
#>   <chr>  <dbl>
#> 1 F1     0.211
#> 2 F2     0.211
#> 3 F3     0.221
#> 4 F4     0.630
#> 5 F5     1.59 

# print feature data
print(sim_actions)
#> # A tibble: 6 × 4
#>   name             cost locked_in locked_out
#>   <chr>           <dbl> <lgl>     <lgl>     
#> 1 F1_action        94.4 FALSE     FALSE     
#> 2 F2_action       101.  FALSE     FALSE     
#> 3 F3_action       103.  TRUE      FALSE     
#> 4 F4_action        99.2 FALSE     FALSE     
#> 5 F5_action        99.9 FALSE     TRUE      
#> 6 baseline_action   0   FALSE     FALSE     

# build problem
p <- problem(sim_projects, sim_actions, sim_features,
             "name", "success", "name", "cost", "name") %>%
     add_max_richness_objective(budget = 400) %>%
     add_feature_weights("weight") %>%
     add_binary_decisions()

# print problem
print(p)
#> Project Prioritization Problem
#>   actions          F1_action, F2_action, F3_action, ... (6 actions)
#>   projects         F1_project, F2_project, F3_project, ... (6 projects)
#>   features         F1, F2, F3, ... (5 features)
#>   action costs:    min: 0, max: 103.22583
#>   project success: min: 0.81379, max: 1
#>   objective:       Maximum richness objective [budget (400)]
#>   targets:         none
#>   weights:         min: 0.21136, max: 1.59167
#>   decisions        Binary decision 
#>   constraints:     <none>
#>   solver:          default

# calculate cost-effectiveness of each project
pce <- project_cost_effectiveness(p)

# print project costs, benefits, and cost-effectiveness values
print(pce)
#> # A tibble: 6 × 6
#>   project           cost   obj benefit         ce  rank
#>   <chr>            <dbl> <dbl>   <dbl>      <dbl> <dbl>
#> 1 F1_project        94.4 0.689  0.108    0.00114      4
#> 2 F2_project       101.  0.711  0.130    0.00129      3
#> 3 F3_project       103.  0.665  0.0841   0.000815     5
#> 4 F4_project        99.2 0.858  0.277    0.00279      2
#> 5 F5_project        99.9 1.23   0.653    0.00654      1
#> 6 baseline_project   0   0.581  0      NaN            6

# plot histogram of cost-effectiveness values
hist(pce$ce, xlab = "Cost effectiveness", main = "")