Calculate statistics describing a solution to a project prioritization
problem().
solution_statistics(x, solution)project prioritization problem().
base::data.frame() or
tibble::tibble() table 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, and they will be ignored.
A tibble::tibble() table containing the following
columns:
"cost"numeric cost of each solution.
"obj"numeric objective value for each solution.
This is calculated using the objective function defined for the
argument to x.
x$project_names()numeric column for each
project indicating if it was completely funded (with a value of 1)
or not (with a value of 0).
x$feature_names()numeric column for each
feature indicating the probability that it will persist into
the future given each solution.
# 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_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
# 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
solution_statistics(p, solutions)
#> # A tibble: 3 × 13
#> cost obj F1_project F2_project F3_project F4_project F5_project
#> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#> 1 0 0.581 0 0 0 0 0
#> 2 498. 1.83 1 1 1 1 1
#> 3 298. 1.43 1 0 1 0 1
#> # ℹ 6 more variables: baseline_project <dbl>, F1 <dbl>, F2 <dbl>, F3 <dbl>,
#> # F4 <dbl>, F5 <dbl>