R/add_min_largest_shortfall_objective.R
add_min_largest_shortfall_objective.Rd
Set the objective of a conservation planning problem()
to
minimize the largest target shortfall while ensuring that
the cost of the solution does not exceed a budget. Note that if the
target shortfall for a single feature cannot be decreased beyond a certain
point (e.g. because all remaining planning units occupied by that feature
are too costly or are locked out), then solutions may only use a small
proportion of the specified budget.
add_min_largest_shortfall_objective(x, budget)
x 


budget 

Object (i.e. ConservationProblem
) with the objective
added to it.
A problem objective is used to specify the overall goal of the conservation planning problem. Please note that all conservation planning problems formulated in the prioritizr package require the addition of objectivesfailing to do so will return an error message when attempting to solve problem.
The minimum largest shortfall objective aims to
find the set of planning units that minimize the largest
shortfall for any of the representation targetsthat is, the fraction of
each target that remains unmetfor as many features as possible while
staying within a fixed budget. This objective is different from the
minimum shortfall objective (add_min_shortfall_objective()
) because this
objective minimizes the largest (maximum) target shortfall,
whereas the minimum shortfall objective
minimizes the total (weighted sum) of the target shortfalls.
Note that this objective function is not compatible with feature weights
(add_feature_weights()
).
The minimum largest shortfall objective for the reserve design problem can be expressed mathematically for a set of planning units (\(I\) indexed by \(i\)) and a set of features (\(J\) indexed by \(j\)) as:
$$\mathit{Minimize} \space l \\ \mathit{subject \space to} \\ \sum_{i = 1}^{I} x_i r_{ij} + y_j \geq t_j \forall j \in J \\ l \geq \frac{y_j}{t_j} \forall j \in J \\ \sum_{i = 1}^{I} x_i c_i \leq B$$
Here, \(x_i\) is the decisions variable (e.g. specifying whether planning unit \(i\) has been selected (1) or not (0)), \(r_{ij}\) is the amount of feature \(j\) in planning unit \(i\), and \(t_j\) is the representation target for feature \(j\). Additionally, \(y_j\) denotes the target shortfall for the target \(t_j\) for feature \(j\), and \(l\) denotes the largest target shortfall. Furthermore, \(B\) is the budget allocated for the solution, \(c_i\) is the cost of planning unit \(i\). Note that \(y_j\) and \(s\) are continuous variables bounded between zero and infinity.
# load data data(sim_pu_raster, sim_pu_zones_stack, sim_features, sim_features_zones) # create problem with minimum largest shortfall objective p1 < problem(sim_pu_raster, sim_features) %>% add_min_largest_shortfall_objective(1800) %>% add_relative_targets(0.1) %>% add_binary_decisions() %>% add_default_solver(verbose = FALSE) # \dontrun{ # solve problem s1 < solve(p1)#> $LogToConsole #> [1] 0 #> #> $LogFile #> [1] "" #> #> $Presolve #> [1] 2 #> #> $MIPGap #> [1] 0.1 #> #> $TimeLimit #> [1] 2147483647 #> #> $Threads #> [1] 1 #> #> $NumericFocus #> [1] 0 #># } # create multizone problem with minimum largest shortfall objective, # with 10% representation targets for each feature, and set # a budget such that the total maximum expenditure in all zones # cannot exceed 3000 p2 < problem(sim_pu_zones_stack, sim_features_zones) %>% add_min_largest_shortfall_objective(3000) %>% add_relative_targets(matrix(0.1, ncol = 3, nrow = 5)) %>% add_binary_decisions() %>% add_default_solver(verbose = FALSE) # \dontrun{ # solve problem s2 < solve(p2)#> $LogToConsole #> [1] 0 #> #> $LogFile #> [1] "" #> #> $Presolve #> [1] 2 #> #> $MIPGap #> [1] 0.1 #> #> $TimeLimit #> [1] 2147483647 #> #> $Threads #> [1] 1 #> #> $NumericFocus #> [1] 0 #># } # create multizone problem with minimum largest shortfall objective, # with 10% representation targets for each feature, and set # separate budgets for each management zone p3 < problem(sim_pu_zones_stack, sim_features_zones) %>% add_min_largest_shortfall_objective(c(3000, 3000, 3000)) %>% add_relative_targets(matrix(0.1, ncol = 3, nrow = 5)) %>% add_binary_decisions() %>% add_default_solver(verbose = FALSE) # \dontrun{ # solve problem s3 < solve(p3)#> $LogToConsole #> [1] 0 #> #> $LogFile #> [1] "" #> #> $Presolve #> [1] 2 #> #> $MIPGap #> [1] 0.1 #> #> $TimeLimit #> [1] 2147483647 #> #> $Threads #> [1] 1 #> #> $NumericFocus #> [1] 0 #># }