Solution statistics

Calculate statistics to describe a solution to a project prioritization problem.

Usage

solution_statistics(x, solution)

Arguments

x

problem() or multi_problem() object.

solution

base::data.frame() or tibble::tibble() containing the solutions. Here, rows correspond to different solutions and columns correspond to different actions. Each column in the argument to solution should be named according to a different action in x. Cell values indicate if an action is funded in a given solution or not, and should be either zero or one. Arguments to solution can contain additional columns, though they will be ignored.

Value

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

"cost"

This column contains numeric values describing the cost of each solution.

"obj"

This column contains numeric values describing the objective value for each solution. This is calculated using the objective function defined for the argument to x. Note that if x is a multi_problem() object, then an objective column will be created for each problem in x.

x$project_names()

These columns contain logical values that indicate if each project had all of its actions selected for funding or not.

x$feature_names()

These columns contain numeric values that describe the expected outcome for each feature based on the actions selected for funding.

See also

Other functions for evaluating solutions: project_cost_effectiveness(), rank_importance(), replacement_costs()

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_action F2_action
#>   <chr>            <dbl>  <dbl>  <dbl>   <dbl>  <dbl>  <dbl> <lgl>     <lgl>    
#> 1 F1_project       0.919  0.791 NA     NA      NA     NA     TRUE      FALSE    
#> 2 F2_project       0.923 NA      0.888 NA      NA     NA     FALSE     TRUE     
#> 3 F3_project       0.829 NA     NA      0.502  NA     NA     FALSE     FALSE    
#> 4 F4_project       0.848 NA     NA     NA       0.690 NA     FALSE     FALSE    
#> 5 F5_project       0.814 NA     NA     NA      NA      0.617 FALSE     FALSE    
#> 6 baseline_proj…   1      0.298  0.250  0.0865  0.249  0.182 FALSE     FALSE    
#> # ℹ 4 more variables: F3_action <lgl>, F4_action <lgl>, F5_action <lgl>,
#> #   baseline_action <lgl>

# 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_wtd_sum_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:    continuous values (between 0 and 103.226)
#> project success: proportion values (between 0.814 and 1)
#> objective:       maximum weighted sum objective
#> targets:         none specified
#> weights:         feature weights
#> constraints:     none specified
#> decisions:       binary decision
#> solver:          none specified

# create a table with some solutions
solutions <- data.frame(
  F1_action = c(0, 1, 1),
  F2_action = c(0, 1, 0),
  F3_action = c(0, 1, 1),
  F4_action = c(0, 1, 0),
  F5_action = c(0, 1, 1),
  baseline_action = c(1, 1, 1)
)

# print the solutions
# the first solution only has the baseline action funded
# the second solution has every action funded
# the third solution has only some actions funded
print(solutions)
#>   F1_action F2_action F3_action F4_action F5_action baseline_action
#> 1         0         0         0         0         0               1
#> 2         1         1         1         1         1               1
#> 3         1         0         1         0         1               1

# calculate statistics for the solutions
solution_statistics(p, solutions)
#> # A tibble: 3 × 13
#>    cost   obj F1_project F2_project F3_project F4_project F5_project
#>   <dbl> <dbl> <lgl>      <lgl>      <lgl>      <lgl>      <lgl>     
#> 1    0  0.581 FALSE      FALSE      FALSE      FALSE      FALSE     
#> 2  498. 1.83  TRUE       TRUE       TRUE       TRUE       TRUE      
#> 3  298. 1.43  TRUE       FALSE      TRUE       FALSE      TRUE      
#> # ℹ 6 more variables: baseline_project <lgl>, F1 <dbl>, F2 <dbl>, F3 <dbl>,
#> #   F4 <dbl>, F5 <dbl>