Add minimum largest shortfall objective
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 |
|
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 objectives—failing 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 targets—that is, the fraction of
each target that remains unmet—for 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:
Minimize l subject to sum_i^I (xi * rij) + yj >= tj for all j in J & l >= (yj / tj) for all j in J & sum_i^I (xi * ci) <= B
Here, xi is the decisions variable (e.g. specifying whether planning unit i has been selected (1) or not (0)), rij is the amount of feature j in planning unit i, and tj is the representation target for feature j. Additionally, yj denotes the target shortfall for the target tj for feature j, and l denotes the largest target shortfall. Furthermore, B is the budget allocated for the solution, ci is the cost of planning unit i. Note that yj and s are continuous variables bounded between zero and infinity.
Object (i.e. ConservationProblem
) with the objective
added to it.
# 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) ## Not run: # solve problem s1 <- solve(p1) # plot solution plot(s1, main = "solution", axes = FALSE, box = FALSE) ## End(Not run) # create multi-zone 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) ## Not run: # solve problem s2 <- solve(p2) # plot solution plot(category_layer(s2), main = "solution", axes = FALSE, box = FALSE) ## End(Not run) # create multi-zone 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) ## Not run: # solve problem s3 <- solve(p3) # plot solution plot(category_layer(s3), main = "solution", axes = FALSE, box = FALSE) ## End(Not run)
Please choose more modern alternatives, such as Google Chrome or Mozilla Firefox.