Create a project prioritization problem. This function is used to
specify the underlying data used in a prioritization problem: the projects,
the management actions, and the features that need
to be conserved (e.g. species, ecosystems). After constructing this
ProjectProblem-class object,
it can be customized using objectives, targets,
weights, constraints, decisions and
solvers. After building the problem, the
solve() function can be used to identify solutions.
problem( projects, actions, features, project_name_column, project_success_column, action_name_column, action_cost_column, feature_name_column, adjust_for_baseline = TRUE )
Arguments
| projects |
|
|---|---|
| actions |
|
| features |
|
| project_name_column |
|
| project_success_column |
|
| action_name_column |
|
| action_cost_column |
|
| feature_name_column |
|
| adjust_for_baseline |
|
Value
A new ProjectProblem object.
Details
A project prioritization problem has actions, projects,
and features. Features are the biological entities that need to
be conserved (e.g. species, populations, ecosystems). Actions are
real-world management actions that can be implemented to enhance
biodiversity (e.g. habitat restoration, monitoring, pest eradication). Each
action should have a known cost, and this usually means that each
action should have a defined spatial extent and time period (though this
is not necessary). Conservation projects are groups of management actions
(they can also comprise a singular action too), and each project is
associated with a probability of success if all of its associated actions
are funded. To determine which projects should be funded, each project is
associated with an probability of persistence for the
features that they benefit. These values should indicate the
probability that each feature will persist if only that project funded
and not the additional benefit relative to the baseline project. Missing
(NA) values should be used to indicate which projects do not
enhance the probability of certain features.
The goal of a project prioritization exercise is then to identify which management actions---and as a consequence which conservation projects---should be funded. Broadly speaking, the goal of an optimization problem is to minimize (or maximize) an objective function given a set of control variables and decision variables that are subject to a series of constraints. In the context of project prioritization problems, the objective is usually some measure of utility (e.g. the net probability of each feature persisting into the future), the control variables determine which actions should be funded or not, the decision variables contain additional information needed to ensure correct calculations, and the constraints impose limits such as the total budget available for funding management actions. For more information on the mathematical formulations used in this package, please refer to the manual entries for the available objectives (listed in objectives).
See also
constraints, decisions,
objectives, solvers, targets,
weights, solution_statistics(),
plot.ProjectProblem().
Examples
#> # A tibble: 6 x 13 #> name success F1 F2 F3 F4 F5 F1_action F2_action #> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <lgl> <lgl> #> 1 F1_p~ 0.919 0.791 NA NA NA NA TRUE FALSE #> 2 F2_p~ 0.923 NA 0.888 NA NA NA FALSE TRUE #> 3 F3_p~ 0.829 NA NA 0.502 NA NA FALSE FALSE #> 4 F4_p~ 0.848 NA NA NA 0.690 NA FALSE FALSE #> 5 F5_p~ 0.814 NA NA NA NA 0.617 FALSE FALSE #> 6 base~ 1 0.298 0.250 0.0865 0.249 0.182 FALSE FALSE #> # ... with 4 more variables: F3_action <lgl>, F4_action <lgl>, F5_action <lgl>, #> # baseline_action <lgl>#> # A tibble: 5 x 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#> # A tibble: 6 x 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)#> #> #> #> #> #> #> #> #> #> #> #>#> Gurobi Optimizer version 9.0.2 build v9.0.2rc0 (linux64) #> Optimize a model with 47 rows, 47 columns and 102 nonzeros #> Model fingerprint: 0x5daec544 #> Variable types: 0 continuous, 42 integer (42 binary) #> Semi-Variable types: 5 continuous, 0 integer #> Coefficient statistics: #> Matrix range [9e-02, 1e+02] #> Objective range [2e-01, 2e+00] #> Bounds range [1e+00, 1e+00] #> RHS range [1e+00, 4e+02] #> Found heuristic solution: objective 0.6654645 #> Presolve removed 16 rows and 12 columns #> Presolve time: 0.00s #> Presolved: 31 rows, 35 columns, 64 nonzeros #> Variable types: 0 continuous, 35 integer (35 binary) #> Presolved: 31 rows, 35 columns, 64 nonzeros #> #> #> Root relaxation: objective 1.749045e+00, 11 iterations, 0.00 seconds #> #> Nodes | Current Node | Objective Bounds | Work #> Expl Unexpl | Obj Depth IntInf | Incumbent BestBd Gap | It/Node Time #> #> * 0 0 0 1.7490448 1.74904 0.00% - 0s #> #> Explored 0 nodes (11 simplex iterations) in 0.00 seconds #> Thread count was 1 (of 4 available processors) #> #> Solution count 1: 1.74904 #> #> Optimal solution found (tolerance 0.00e+00) #> Best objective 1.749044775334e+00, best bound 1.749044775334e+00, gap 0.0000%#> # A tibble: 1 x 21 #> solution status obj cost F1_action F2_action F3_action F4_action F5_action #> <int> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> #> 1 1 OPTIM~ 1.75 395. 1 1 0 1 1 #> # ... with 12 more variables: baseline_action <dbl>, F1_project <dbl>, #> # F2_project <dbl>, F3_project <dbl>, F4_project <dbl>, F5_project <dbl>, #> # baseline_project <dbl>, F1 <dbl>, F2 <dbl>, F3 <dbl>, F4 <dbl>, F5 <dbl># print which actions are funded in the solution s[, sim_actions$name, drop = FALSE]#> # A tibble: 1 x 6 #> F1_action F2_action F3_action F4_action F5_action baseline_action #> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> #> 1 1 1 0 1 1 1# print the expected probability of persistence for each feature # if the solution were implemented s[, sim_features$name, drop = FALSE]#> # A tibble: 1 x 5 #> F1 F2 F3 F4 F5 #> <dbl> <dbl> <dbl> <dbl> <dbl> #> 1 0.808 0.865 0.0865 0.688 0.592# }
