Add connectivity penalties
Add penalties to a conservation planning problem()
to favor
solutions that select planning units with high connectivity between them.
## S4 method for signature 'ConservationProblem,ANY,ANY,matrix' add_connectivity_penalties(x, penalty, zones, data) ## S4 method for signature 'ConservationProblem,ANY,ANY,Matrix' add_connectivity_penalties(x, penalty, zones, data) ## S4 method for signature 'ConservationProblem,ANY,ANY,data.frame' add_connectivity_penalties(x, penalty, zones, data) ## S4 method for signature 'ConservationProblem,ANY,ANY,dgCMatrix' add_connectivity_penalties(x, penalty, zones, data) ## S4 method for signature 'ConservationProblem,ANY,ANY,array' add_connectivity_penalties(x, penalty, zones, data)
x |
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penalty |
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zones |
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data |
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This function adds penalties to conservation planning problem to penalize solutions that have low connectivity. Specifically, it favors pair-wise connections between planning units that have high connectivity values. It was inspired by Beger et al. (2010) and can symmetric and asymmetric connectivity relationships between planning units.
The connectivity penalties are calculated using the following equations.
Let I represent the set of planning units
(indexed by i or j), Z represent the set
of management zones (indexed by z or y), and Xiz
represent the decision variable for planning unit i for in zone
z (e.g. with binary
values one indicating if planning unit is allocated or not). Also, let
p represent the argument to penalty
, D represent the
argument to data
, and W represent the argument
to zones
.
If the argument to data
is supplied as a matrix
or
Matrix
object, then the penalties are calculated as:
sum_i^I sum_j^I sum_z^Z sum_y^Z (-p * Xiz * Xjy * Dij * Wzy)
Otherwise, if the argument to data
is supplied as a
data.frame
or array
object, then the penalties are
calculated as:
sum_i^I sum_j^I sum_z^Z sum_y^Z (-p * Xiz * Xjy * Dijzy)
Note that when the problem objective is to maximize some measure of benefit and not minimize some measure of cost, the term -p is replaced with p.
Object (i.e. ConservationProblem
) with the penalties
added to it.
The argument to data
can be specified using several different formats.
These formats can be used to describe symmetric or
asymmetric relationships between planning units.
matrix
, Matrix
where rows and columns represent
different planning units and the value of each cell represents the
strength of connectivity between two different planning units. Cells
that occur along the matrix diagonal are treated as weights which
indicate that planning units are more desirable in the solution.
The argument to zones
can be used to control
the strength of connectivity between planning units in different zones.
The default argument for zones
is to treat planning units
allocated to different zones as having zero connectivity.
data.frame
containing the fields (columns)
"id1"
, "id2"
, and "boundary"
. Here, each row
denotes the connectivity between two planning units following the
Marxan format. The data can be used to denote symmetric or
asymmetric relationships between planning units. By default,
input data is assumed to be symmetric unless asymmetric data is
also included (e.g. if data is present for planning units 2 and 3, then
the same amount of connectivity is expected for planning units 3 and 2,
unless connectivity data is also provided for planning units 3 and 2).
If the argument to x
contains multiple zones, then the columns
"zone1"
and "zone2"
can optionally be provided to manually
specify the connectivity values between planning units when they are
allocated to specific zones. If the columns "zone1"
and
"zone2"
are present, then the argument to zones
must be
NULL
.
array
containing four-dimensions where cell values
indicate the strength of connectivity between planning units
when they are assigned to specific management zones. The first two
dimensions (i.e. rows and columns) indicate the strength of
connectivity between different planning units and the second two
dimensions indicate the different management zones. Thus
the data[1, 2, 3, 4]
indicates the strength of
connectivity between planning unit 1 and planning unit 2 when planning
unit 1 is assigned to zone 3 and planning unit 2 is assigned to zone 4.
Beger M, Linke S, Watts M, Game E, Treml E, Ball I, and Possingham, HP (2010) Incorporating asymmetric connectivity into spatial decision making for conservation, Conservation Letters, 3: 359–368.
# set seed for reproducibility set.seed(600) # load Matrix package for visualizing matrices require(Matrix) # load data data(sim_pu_polygons, sim_pu_zones_stack, sim_features, sim_features_zones) # define function to rescale values between zero and one so that we # can compare solutions from different connectivity matrices rescale <- function(x, to = c(0, 1), from = range(x, na.rm = TRUE)) { (x - from[1]) / diff(from) * diff(to) + to[1] } # create basic problem p1 <- problem(sim_pu_polygons, sim_features, "cost") %>% add_min_set_objective() %>% add_relative_targets(0.2) %>% add_default_solver(verbose = FALSE) # create a symmetric connectivity matrix where the connectivity between # two planning units corresponds to their shared boundary length b_matrix <- boundary_matrix(sim_pu_polygons) # standardize matrix values to lay between zero and one b_matrix[] <- rescale(b_matrix[]) # visualize connectivity matrix ## Not run: image(b_matrix) ## End(Not run) # create a symmetric connectivity matrix where the connectivity between # two planning units corresponds to their spatial proximity # i.e. planning units that are further apart share less connectivity centroids <- rgeos::gCentroid(sim_pu_polygons, byid = TRUE) d_matrix <- (1 / (as(dist(centroids@coords), "Matrix") + 1)) # standardize matrix values to lay between zero and one d_matrix[] <- rescale(d_matrix[]) # remove connections between planning units without connectivity to # reduce run-time d_matrix[d_matrix < 0.7] <- 0 # visualize connectivity matrix ## Not run: image(d_matrix) ## End(Not run) # create a symmetric connectivity matrix where the connectivity # between adjacent two planning units corresponds to their combined # value in a field in the planning unit attribute data # for example, this field could describe the extent of native vegetation in # each planning unit and we could use connectivity penalties to identify # solutions that cluster planning units together that both contain large # amounts of native vegetation c_matrix <- connectivity_matrix(sim_pu_polygons, "cost") # standardize matrix values to lay between zero and one c_matrix[] <- rescale(c_matrix[]) # visualize connectivity matrix ## Not run: image(c_matrix) ## End(Not run) # create an asymmetric connectivity matrix. Here, connectivity occurs between # adjacent planning units and, due to rivers flowing southwards # through the study area, connectivity from northern planning units to # southern planning units is ten times stronger than the reverse. ac_matrix <- matrix(0, length(sim_pu_polygons), length(sim_pu_polygons)) ac_matrix <- as(ac_matrix, "Matrix") adjacent_units <- rgeos::gIntersects(sim_pu_polygons, byid = TRUE) for (i in seq_len(length(sim_pu_polygons))) { for (j in seq_len(length(sim_pu_polygons))) { # find if planning units are adjacent if (adjacent_units[i, j]) { # find if planning units lay north and south of each other # i.e. they have the same x-coordinate if (centroids@coords[i, 1] == centroids@coords[j, 1]) { if (centroids@coords[i, 2] > centroids@coords[j, 2]) { # if i is north of j add 10 units of connectivity ac_matrix[i, j] <- ac_matrix[i, j] + 10 } else if (centroids@coords[i, 2] < centroids@coords[j, 2]) { # if i is south of j add 1 unit of connectivity ac_matrix[i, j] <- ac_matrix[i, j] + 1 } } } } } # standardize matrix values to lay between zero and one ac_matrix[] <- rescale(ac_matrix[]) # visualize asymmetric connectivity matrix ## Not run: image(ac_matrix) ## End(Not run) # create penalties penalties <- c(10, 25) # create problems using the different connectivity matrices and penalties p2 <- list(p1, p1 %>% add_connectivity_penalties(penalties[1], data = b_matrix), p1 %>% add_connectivity_penalties(penalties[2], data = b_matrix), p1 %>% add_connectivity_penalties(penalties[1], data = d_matrix), p1 %>% add_connectivity_penalties(penalties[2], data = d_matrix), p1 %>% add_connectivity_penalties(penalties[1], data = c_matrix), p1 %>% add_connectivity_penalties(penalties[2], data = c_matrix), p1 %>% add_connectivity_penalties(penalties[1], data = ac_matrix), p1 %>% add_connectivity_penalties(penalties[2], data = ac_matrix)) # assign names to the problems names(p2) <- c("basic problem", paste0("b_matrix (", penalties,")"), paste0("d_matrix (", penalties,")"), paste0("c_matrix (", penalties,")"), paste0("ac_matrix (", penalties,")")) ## Not run: # solve problems s2 <- lapply(p2, solve) # plot solutions par(mfrow = c(3, 3)) for (i in seq_along(s2)) { plot(s2[[i]], main = names(p2)[i], cex = 1.5, col = "white") plot(s2[[i]][s2[[i]]$solution_1 == 1, ], col = "darkgreen", add = TRUE) } ## End(Not run) # create minimal multi-zone problem and limit solver to one minute # to obtain solutions in a short period of time p3 <- problem(sim_pu_zones_stack, sim_features_zones) %>% add_min_set_objective() %>% add_relative_targets(matrix(0.15, nrow = 5, ncol = 3)) %>% add_binary_decisions() %>% add_default_solver(time_limit = 60, verbose = FALSE) # create matrix showing which planning units are adjacent to other units a_matrix <- adjacency_matrix(sim_pu_zones_stack) # visualize matrix ## Not run: image(a_matrix) ## End(Not run) # create a zone matrix where connectivities are only present between # planning units that are allocated to the same zone zm1 <- as(diag(3), "Matrix") # print zone matrix print(zm1) # create a zone matrix where connectivities are strongest between # planning units allocated to different zones zm2 <- matrix(1, ncol = 3, nrow = 3) diag(zm2) <- 0 zm2 <- as(zm2, "Matrix") # print zone matrix print(zm2) # create a zone matrix that indicates that connectivities between planning # units assigned to the same zone are much higher than connectivities # assigned to different zones zm3 <- matrix(0.1, ncol = 3, nrow = 3) diag(zm3) <- 1 zm3 <- as(zm3, "Matrix") # print zone matrix print(zm3) # create a zone matrix that indicates that connectivities between planning # units allocated to zone 1 are very high, connectivities between planning # units allocated to zones 1 and 2 are moderately high, and connectivities # planning units allocated to other zones are low zm4 <- matrix(0.1, ncol = 3, nrow = 3) zm4[1, 1] <- 1 zm4[1, 2] <- 0.5 zm4[2, 1] <- 0.5 zm4 <- as(zm4, "Matrix") # print zone matrix print(zm4) # create a zone matrix with strong connectivities between planning units # allocated to the same zone, moderate connectivities between planning # unit allocated to zone 1 and zone 2, and negative connectivities between # planning units allocated to zone 3 and the other two zones zm5 <- matrix(-1, ncol = 3, nrow = 3) zm5[1, 2] <- 0.5 zm5[2, 1] <- 0.5 diag(zm5) <- 1 zm5 <- as(zm5, "Matrix") # print zone matrix print(zm5) # create vector of penalties to use creating problems penalties2 <- c(5, 15) # create multi-zone problems using the adjacent connectivity matrix and # different zone matrices p4 <- list( p3, p3 %>% add_connectivity_penalties(penalties2[1], zm1, a_matrix), p3 %>% add_connectivity_penalties(penalties2[2], zm1, a_matrix), p3 %>% add_connectivity_penalties(penalties2[1], zm2, a_matrix), p3 %>% add_connectivity_penalties(penalties2[2], zm2, a_matrix), p3 %>% add_connectivity_penalties(penalties2[1], zm3, a_matrix), p3 %>% add_connectivity_penalties(penalties2[2], zm3, a_matrix), p3 %>% add_connectivity_penalties(penalties2[1], zm4, a_matrix), p3 %>% add_connectivity_penalties(penalties2[2], zm4, a_matrix), p3 %>% add_connectivity_penalties(penalties2[1], zm5, a_matrix), p3 %>% add_connectivity_penalties(penalties2[2], zm5, a_matrix)) # assign names to the problems names(p4) <- c("basic problem", paste0("zm", rep(seq_len(5), each = 2), " (", rep(penalties2, 2), ")")) ## Not run: # solve problems s4 <- lapply(p4, solve) s4 <- lapply(s4, category_layer) s4 <- stack(s4) # plot solutions plot(s4, main = names(p4), axes = FALSE, box = FALSE) ## End(Not run) # create an array to manually specify the connectivities between # each planning unit when they are allocated to each different zone # for real-world problems, these connectivities would be generated using # data - but here these connectivity values are assigned as random # ones or zeros c_array <- array(0, c(rep(ncell(sim_pu_zones_stack[[1]]), 2), 3, 3)) for (z1 in seq_len(3)) for (z2 in seq_len(3)) c_array[, , z1, z2] <- round(runif(ncell(sim_pu_zones_stack[[1]]) ^ 2, 0, 0.505)) # create a problem with the manually specified connectivity array # note that the zones argument is set to NULL because the connectivity # data is an array p5 <- list(p3, p3 %>% add_connectivity_penalties(15, zones = NULL, c_array)) # assign names to the problems names(p5) <- c("basic problem", "connectivity array") ## Not run: # solve problems s5 <- lapply(p5, solve) s5 <- lapply(s5, category_layer) s5 <- stack(s5) # plot solutions plot(s5, main = names(p5), axes = FALSE, box = FALSE) ## End(Not run)
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