FRST90015 Landscape Ecology - Assignment
Harrison White 913819
r format(Sys.time(), ‘%d %B %Y’)
#Apologies, I had finished this on Friday but had forgotten to submit it. PLease forgive meConnectivity and Conservation: Student Notebook
Exercise 1: Code and plots
#Add your Task 1.1 code here. Make sure to remember to add fig.cap = 'XXX' to other chunks in this notebook that will output figures. Correctly refer to the figures in your answers
all_binary <- list.files(path = 'data/raw',pattern = "^Binary.*.asc$", full.names = TRUE)
all_binary <- lapply(all_binary, raster)
binary_stack <- do.call(stack, all_binary)
spplot(binary_stack)
Figure 1. Spatial configuration of habitat for 9 landscapes
#Add your Task 1.2 code here
binary_curmap <- list.files(path = "data/landscapes", pattern = "*curmap.asc", full.names = TRUE)
binary_curmap <- lapply(binary_curmap, raster)
curmap_stack <- do.call(stack, binary_curmap)
spplot(curmap_stack, col=plasma(30))
Figure 2a. Cumulative current of habitat for 9 landscapes
plot(curmap_stack, col=plasma(30))
Figure 2b. Cumulative current of habitat for 9 landscapes
plot(curmap_stack)
Figure 2c. Cumulative current of habitat for 9 landscapes
#Add your Task 1.3 code here
res<-read.csv('data/resistances.csv')
head(res)## Node1 Node2 Resistance Landscape
## 1 1 2 -1 10
## 2 1 3 -1 10
## 3 1 4 -1 10
## 4 1 5 -1 10
## 5 2 3 -1 10
## 6 2 4 -1 10res_ex1<-res%>%filter(!Landscape %in% c('sp_a', 'sp_b', 'sp_c'))%>%
mutate(res_fix=ifelse(Resistance == -1, 50, Resistance), flow = paste0('N',Node1,'-N',Node2))%>%
droplevels()
head(res_ex1)## Node1 Node2 Resistance Landscape res_fix flow
## 1 1 2 -1 10 50 N1-N2
## 2 1 3 -1 10 50 N1-N3
## 3 1 4 -1 10 50 N1-N4
## 4 1 5 -1 10 50 N1-N5
## 5 2 3 -1 10 50 N2-N3
## 6 2 4 -1 10 50 N2-N4ggplot(res_ex1, aes(Landscape, res_fix)) +
stat_summary(geom = "bar", fun = mean, position = "dodge")+
stat_summary(geom = "errorbar", fun.data = mean_se, position = "dodge")+
labs(x = 'Landscape', y = 'Resistance')
Figure 3. Relationship between the proportion of the landscape that is habitat and average resistance across all comparisons
#Add your Task 1.4 code here
ggplot(data=res_ex1, aes(x=Landscape, y=res_fix, group=as.factor(flow), color = as.factor(flow))) +
geom_point()+
geom_line(lwd = 1.5)+
labs(x = 'Landscape', y = 'Resistance', color = 'Flow')
Figure 4. Relationship between the proportion of the landscape that is habitat and resistance
#Add your Task 1.5 code here
connect<-res_ex1%>%group_by(Landscape)%>%tally(Resistance>0)
ggplot(connect,(aes(Landscape, n)))+
stat_summary(geom = "bar", fun = max, position = "dodge")+
labs(x = 'Landscape', y = 'Number of connections')
Figure 5. Number of connected habitat patches as a function of proportion of the landscape that is habitat
write.csv(res_ex1, 'outputs/landscape_resistances.csv')
write.csv(connect, 'outputs/landscape_connections.csv')Exercise 1: Questions
1. How is resistance related to the proportion of the landscape that is habitat? Describe the relationship.
Resistance decreases as the proportion of landscape that is habitat increases. Figure 3 shows that at 50% of the landscape there is a significant decrease in the resistance experienced.
2. Is there a threshold at which connectivity is lost? If yes, what is this threshold?
Connectivity is entirely lost when less habitat is 30% or less.
3. Do all habitat patches respond the same? Explain why or why not.
Not all habitats respond the same as indicated by figure 2. This is due to factors such as the size of the habitat and its close proximity to other habitats. By being larger, a habitat has more potential pathways for organisms to travel to another habitat. By being closer to other habitats there is less potential for pathways to contain resistance to travel.
4. What is a key factor underpinning the response of organisms when applying a binary landscape model?
The key factor is the amount of habitat that is in the landscape
5. How do species perceive a binary landscape?
Species view a binary landscape as having patches of habitable patches within a larger landscape of inhabitable patches
6. Which meta-population model is synonymous with the binary landscape model?
Patch dynamics is a meta-population model that utilities the information acquired from a binary landscape model.
7. In this model what determines community structure and how is it measured?
In patch dynamics community structure is determined by both colonisation and extinction as these two factors influence metapopulations in the different patches. These are measured by analysing connectivity and how species disperse throughtout the landscape.
Exercise 2: Code and plots
#Add your Task 2.1 code here (loading data only)
all_species <- list.files(path = "data/raw", pattern = "^Species.*.asc$", full.names = TRUE)
all_species <- lapply(all_species, raster)
species_stack <- do.call(stack, all_species)
spplot(species_stack)
Figure 6. Raw resistances for Species A, B, & C
#Add your Task 2.2 code here
species_curmap <- list.files(path = "data/species", pattern = "*curmap.asc", full.names = TRUE)
species_curmap <- lapply(species_curmap, raster)
par(mfrow=c(1,3))
sp_A <- species_curmap[[1]]
plot(sp_A, col = plasma(30))
sp_B <- species_curmap[[2]]
plot(sp_B, col = plasma(30))
sp_C <- species_curmap[[3]]
plot(sp_C, col = plasma(30))
Figure 7a. Maps of conductance for Species A, B, & C
reclass_raster<-function(x, threshold){
z<-reclassify(x, c(-Inf,threshold,0, threshold,+Inf,1))
return(z)
}
par(mfrow=c(1,3))
sp_A_re<-reclass_raster(sp_A, 0.3)
plot(sp_A_re, col = c('white', 'black'))
sp_B_re<-reclass_raster(sp_B, 0.3)
plot(sp_B_re, col = c('white', 'black'))
sp_C_re<-reclass_raster(sp_C, 0.3)
plot(sp_C_re, col = c('white', 'black'))
Figure 7b. Maps of conductance for Species A, B, & C
#Add your Task 2.3 code here
reserves<-raster('data/raw/Reserves.asc')
#check:
plot(reserves)
Figure 8a. Reserve Map
reserves_poly<-rasterToPolygons(reserves, dissolve = T)
par(mfrow=c(1,3))
plot(sp_A_re, col = c('grey', 'black'), legend = FALSE, main = 'Species A')
plot(reserves_poly, col = 'yellow', add = T)
legend("topright", legend = c("Matrix", "Corridors", 'Reserves'), fill = c('grey', 'black', 'yellow'))
plot(sp_B_re, col = c('grey', 'black'), legend = FALSE, main = 'Species B')
plot(reserves_poly, col = 'red', add = T)
legend("topright", legend = c("Matrix", "Corridors", 'Reserves'), fill = c('grey', 'black', 'red'))
plot(sp_C_re, col = c('grey', 'black'), legend = FALSE, main = 'Species C')
plot(reserves_poly, col = 'green', add = T)
legend("topright", legend = c("Matrix", "Corridors", 'Reserves'), fill = c('grey', 'black', 'green'))
Figure 8b. Reserve and habitat map for Species A, B & C
Exercise 2: Questions
1. Compare and contrast the pairwise resistances for the three species. Which species will have a harder time moving around the landscape?
# Add any code needed to answer this question here
res_sp<-res%>%filter(Landscape %in% c('sp_a', 'sp_b', 'sp_c'))%>%
mutate(res_fix=ifelse(Resistance == -1, 50, Resistance), flow = paste0('N',Node1,'-N',Node2))%>%
droplevels()
pairwise<-pivot_wider(res_sp, names_from = c('Landscape'), values_from = c('Resistance'))%>%
group_by(flow) %>%
summarize(sp_a=max(sp_a,na.rm=TRUE),sp_b=max(sp_b,na.rm=TRUE),sp_c=max(sp_c,na.rm=TRUE))
pairwise## # A tibble: 10 × 4
## flow sp_a sp_b sp_c
## <chr> <dbl> <dbl> <dbl>
## 1 N1-N2 36.1 4.66 177.
## 2 N1-N3 39.4 4.89 203.
## 3 N1-N4 68.6 8.24 337.
## 4 N1-N5 18.3 2.32 96.5
## 5 N2-N3 44.5 5.52 224.
## 6 N2-N4 69.5 8.20 336.
## 7 N2-N5 20.4 2.61 103.
## 8 N3-N4 48.6 5.87 248.
## 9 N3-N5 27.1 3.34 141.
## 10 N4-N5 54.3 6.38 264.write.csv(pairwise, 'outputs/species_pairwise.csv')
knitr::kable(pairwise,
caption = "Table 1. Summary of resistances for different species",
digits = 3,
col.names = c('Flow',
'Species A',
'Species B',
'Species C'))| Flow | Species A | Species B | Species C |
|---|---|---|---|
| N1-N2 | 36.058 | 4.664 | 176.576 |
| N1-N3 | 39.366 | 4.890 | 202.915 |
| N1-N4 | 68.567 | 8.237 | 336.726 |
| N1-N5 | 18.344 | 2.320 | 96.453 |
| N2-N3 | 44.487 | 5.520 | 224.498 |
| N2-N4 | 69.541 | 8.200 | 336.210 |
| N2-N5 | 20.409 | 2.610 | 102.686 |
| N3-N4 | 48.588 | 5.870 | 248.336 |
| N3-N5 | 27.061 | 3.345 | 141.301 |
| N4-N5 | 54.261 | 6.384 | 264.195 |
Species C will have the hardest time moving around the landscape as it has the highest resistance values whilst Species B will have the easiest time moving around the landscape as it has the lowest resistance values. Species A sits between Species A and Species B for resistance.
2. Which habitat patches are better connected for each species? How does the connectivity between nodes differ between species?
Patch 1 and Patch 5 are the best connected patch for each species as this pathway has the lowest resistance value for all three species. Overall the connectivity between patches is experienced similarly between the different species as when each pathway is ranked according to the resistance experienced the order remains the same for all three species with the exception of Patch 1- Patch 4 and Patch 2 - Patch 4 which differ slightly across the species.
3. Do the species have the same network of habitat through the matrix? Compare and contrast.
The species have similar habitat through the matrix but it differs in how connected/fragmented each networks are. In Figure 8 it is evident that Species C lacks well connected pathways between the habitat patches causing it to have such high resistance values. For Species B the network of habitat through the matrix in seen in Figure 8 is thicker and more connected causing its lower resistance values. Species A again is similar to that of B and C but in between the two in terms of connectivity of quality habitat between the patches.
Exercise 3: Code and plots
#Add your Task 3.1 code here
sum<-sp_A+sp_B+sp_C
plot(sum)
Figure 9a. Sum of Species A, B & C rasters
corridors<-reclassify(sum, c(-Inf, 0.4,0, #Matrix becomes 0
0.4,10,1, # Corridors become 1
10,Inf,2)) #Reserves become 2
plot(corridors, col = plasma(3), legend = FALSE)
legend("topright", legend = c("Matrix", "Corridors", 'Reserves'), fill = plasma(3))
Figure 9b. Map of Reserves & Corridors for Species A, B & C
#Add your Task 3.2 code here
corridors_df<-corridors%>%as.data.frame(xy = T, na.rm = T)
corridors_df$ID<-1 #to count pixels for area calculation
reso<-res(corridors)[1] #save the raster resolution for area calculation
areas<-corridors_df%>%group_by(layer)%>%
summarise(area = sum(ID), area_ha = (area*reso^2)/10000)%>%
mutate(sumA = sum(area), per = 100*area/sumA)%>%rename(class = layer)
areas## # A tibble: 3 × 5
## class area area_ha sumA per
## <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 0 6610 6610 10000 66.1
## 2 1 2780 2780 10000 27.8
## 3 2 610 610 10000 6.1write.csv(areas, 'outputs/species_areas.csv')Exercise 3: Questions
1. How many hectares is your corridor network (do not include the area occupied by the habitat patches)?
The corridor network is 2780 hectares
2. What proportion of the landscape is within your corridor network (do not include the area in the habitat patches)?
27.8% of the landscape is within the corridor network
3. What is a key factor underpinning the response of organisms when applying a mosaic landscape model? Hint: Look in Landscape Models Lecture.
In a mosaic landscape model the key factor that underpins the response of organisms is the quality of the habitat
4. How do species perceive a mosaic landscape?
In a mosaic landscape species perceive the landscape as a gradient of different quality habitats. The quality of the habitat would be based on its ability to facilitate feeding, reproductive and transport.
5. Which meta-population model is synonymous with the mosaic landscape model?
The species sorting meta-population model is synonymous with the mosaic landscape model as in this model species perceive the landscape as mosaic of different quality habitats. The quality of habitat is influenced by local conditions which impacts on its ability to provide different resources.
6. In this model, what determines community structure and how is it measured?
Community structure is determined by environmental conditions that make the landscape a gradient of various quality of habitat. The community structure is measured based on how many different species are present at each habitat patch and how many individuals make up the metapopulation in these patches.
7. In the landscape, 50% of the area outside the reserve network is habitat. As a conservation planner you are tasked with determining the minimum amount of habitat that needs to be maintained to conserve the three species of interest. How much of the remaining habitat on the landscape must be maintained?
The 27.8% that makes up the habitat corridor needs to be maintained to allow the dispersal of the three species between the 5 habitat reserves. Maintaining these corridors is important to ensure these patches remain connected and can allow safe passage for the species so patches do not become isolated and at risk of local extinction.
8. Are any of the habitat patches more important than the others to the network?
The summary of resistance values in Table 1 indicate that the most important patches are patches 1 and 5 as there are well connected and offer the least resistance for all three species. If any prioritisation of patches is required, these two should be prioritised
9. If you lost a habitat patch in the network would it still function as a meta-community? Why or why not?
The impact of losing a habitat patch would be determined by the properties of the habitat patch. If the patch was of high quality and offered a lot of resources it would be seen as a source that has a positive impact on local recruitment. The loss of the habitat patch could potentially have detrimental impacts on the meta-community. If it is of low quality and is seen as a sink as it has a negative impact on local recruitment, the loss of the patch may have less of an impact on the meta-community. If species view the landscape as a binary landscape the loss of a habitat patch may disrupt the meta-community as it is requried to support movement to other patches. If the species view the landscape as a mosaic landscape the loss of a patch may be less detrimental for the metacommunity as it is may still have the ability to disperse to other patches of higher quality habitat.
10. Compare and contrast the connectivity results when considering the landscape using a binary model versus mosaic model. What are the implications of these approaches for conservation planning of the three species in the landscape?
In a binary model, dispersal between patches is facilitated by corridors that connect the patches. Therefore the loss of any of these corridors may cause patches to become isolated and may lead to local extinction.This model interprets the landscape as either habitat or not habitat for specues and therefore would influence conservation planning to preserve more of the landscape for the species.
In a mosaic model, dispersal between patches may not require all patches to be connected as species may have the ability to move through gradients of different quality habitat in order to reach suitable habitat patches.This model may influence conservation planning to focus of the areas of greatest habitat quality and connectivity such as habitat patches 1 and 5.
Exercise 4: Code and plots
#Add your Task 4.1 code here (table)
lbp_res<-read.csv('data/lbp_resistances.csv')
lbp_res<-lbp_res%>%mutate(threshold = 20,
connected = ifelse(resistance<=threshold, 1, 0))
lpb_res_summary<-lbp_res%>%
group_by(replicate, scenario, timestep)%>%
summarise(connected_nodes = sum(connected))## `summarise()` has grouped output by 'replicate', 'scenario'. You can override
## using the `.groups` argument.lpb_res_summary## # A tibble: 14 × 4
## # Groups: replicate, scenario [6]
## replicate scenario timestep connected_nodes
## <int> <chr> <int> <dbl>
## 1 1 BAU 0 364
## 2 1 BAU 20 405
## 3 1 BAU 100 13
## 4 1 NM 0 405
## 5 1 NM 20 405
## 6 1 NM 100 124
## 7 2 BAU 20 2
## 8 2 BAU 100 109
## 9 2 NM 20 54
## 10 2 NM 100 19
## 11 3 BAU 20 405
## 12 3 BAU 100 0
## 13 3 NM 20 405
## 14 3 NM 100 0knitr::kable(lpb_res_summary,
caption = "Table 2. Summary of connected nodes for each replicate and scenario across different timesteps for Leadbeater's Possum",
digits = 2,
col.names = c('Replicate',
'Scenario',
'Timestep',
'No. of connected nodes'))| Replicate | Scenario | Timestep | No. of connected nodes |
|---|---|---|---|
| 1 | BAU | 0 | 364 |
| 1 | BAU | 20 | 405 |
| 1 | BAU | 100 | 13 |
| 1 | NM | 0 | 405 |
| 1 | NM | 20 | 405 |
| 1 | NM | 100 | 124 |
| 2 | BAU | 20 | 2 |
| 2 | BAU | 100 | 109 |
| 2 | NM | 20 | 54 |
| 2 | NM | 100 | 19 |
| 3 | BAU | 20 | 405 |
| 3 | BAU | 100 | 0 |
| 3 | NM | 20 | 405 |
| 3 | NM | 100 | 0 |
#Add your Task 4.2 code here (bar graphs)
lpb_res_summary%>%filter(scenario == 'NM')%>%
ggplot(aes(as.factor(timestep), connected_nodes))+
stat_summary(geom = 'bar', fill = 'dodgerblue', width = 0.6)+
stat_summary(geom = 'errorbar', width = 0.1)+
labs(y = 'Connceted nodes', x = 'Time (years)', title = 'Connected Nodes - NM')+
scale_x_discrete(labels = c("0","20", '100'))+
theme_bw()## No summary function supplied, defaulting to `mean_se()`
## No summary function supplied, defaulting to `mean_se()`
Figure 10a. Graph of conneceted nodes for no management
lpb_res_summary%>%filter(scenario == 'BAU')%>%
ggplot(aes(as.factor(timestep), connected_nodes))+
stat_summary(geom = 'bar', fill = 'dodgerblue', width = 0.6)+
stat_summary(geom = 'errorbar', width = 0.1)+
labs(y = 'Connceted nodes', x = 'Time (years)', title = 'Connected Nodes - BAU')+
scale_x_discrete(labels = c("0","20", '100'))+
theme_bw()## No summary function supplied, defaulting to `mean_se()`
## No summary function supplied, defaulting to `mean_se()`
Figure 10b. Graph of connected nodes for business as usual
#Add your Task 4.3 code here
NM_curmap <- list.files(path = "data/LBP/NM", pattern = "*curmap.asc", full.names = TRUE)
NM_curmap <- lapply(NM_curmap, raster)
NM1_T0 <- NM_curmap[[1]]
NM1_T20 <- NM_curmap[[3]] #mind the alphabetical order of our list/folder (1 before 2), here we bring it in the right order
NM1_T100 <- NM_curmap[[2]]
NM2_T0 <- NM_curmap[[4]]
NM2_T20 <- NM_curmap[[6]]
NM2_T100 <- NM_curmap[[5]]
NM3_T0 <- NM_curmap[[7]]
NM3_T20 <- NM_curmap[[9]]
NM3_T100 <- NM_curmap[[8]]
# bring in right order and plot all LBP landscapes
NM_stack <- stack(NM1_T0,NM1_T20,NM1_T100, NM2_T0,NM2_T20, NM2_T100,NM3_T0, NM3_T20, NM3_T100)
plot(NM_stack) #base plot will automatically choose a 3x3 arrangement for the maps
Figure 11a. Panel graph of conductance for Leadbeater’s Possum for no management scenario
#Add your Task 4.4 code here
BAU_curmap <- list.files(path = "data/LBP/BAU", pattern = "*curmap.asc", full.names = TRUE)
BAU_curmap <- lapply(BAU_curmap, raster)
BAU1_T0 <- BAU_curmap[[1]]
BAU1_T20 <- BAU_curmap[[3]]
BAU1_T100 <- BAU_curmap[[2]]
BAU2_T0 <- BAU_curmap[[4]]
BAU2_T20 <- BAU_curmap[[6]]
BAU2_T100 <- BAU_curmap[[5]]
BAU3_T0 <- BAU_curmap[[7]]
BAU3_T20 <- BAU_curmap[[9]]
BAU3_T100 <- BAU_curmap[[8]]
BAU_stack <- stack(BAU1_T0,BAU1_T20,BAU1_T100, BAU2_T0,BAU2_T20, BAU2_T100,BAU3_T0, BAU3_T20, BAU3_T100)
plot(BAU_stack) #base plot will automatically choose a 3x3 arrangement for the maps
Figure 11b. Panel graph of conductance for Leadbeater’s Possum for business as usual scenario
#Add your Task 4.5 code here
#NM
T0<-stack(NM1_T0, NM2_T0, NM3_T0)
names(T0)<-c('T0_NM1', 'T0_NM2', 'T0_NM3')
re_T0 <- reclass_raster(T0, 5) #function from above, set threshold to 5
cellStats(re_T0, 'sum') #in ha per rep## T0_NM1 T0_NM2 T0_NM3
## 1867 1867 1867T20<-stack(NM1_T20, NM2_T20, NM3_T20)
names(T20)<-c('T20_NM1', 'T20_NM2', 'T20_NM3')
re_T20 <- reclass_raster(T20, 5)
cellStats(re_T20, 'sum')## T20_NM1 T20_NM2 T20_NM3
## 1925 1660 1925T100<-stack(NM1_T100, NM2_T100, NM3_T100)
names(T100)<-c('T100_NM1', 'T100_NM2', 'T100_NM3')
re_T100 <- reclass_raster(T100, 5)
cellStats(re_T100, 'sum')## T100_NM1 T100_NM2 T100_NM3
## 1697 1631 1655#Plot of NM Corridor Network
NM_Net_stack <- stack(re_T0[[1]],re_T20[[1]],re_T100[[1]],
re_T0[[2]],re_T20[[2]], re_T100[[2]],
re_T0[[3]],re_T20[[3]], re_T100[[3]])
plot(NM_Net_stack)
Figure 12a. Habitat corridor network for Leadbeater’s Possum for each replicate and time period under no management scenario
#Task 4.5
#BAU
T0<-stack(BAU1_T0, BAU2_T0, BAU3_T0)
names(T0)<-c('T0_BAU1', 'T0_BAU2', 'T0_BAU3')
re_T0 <- reclass_raster(T0, 5) #function from above, set threshold to 5
cellStats(re_T0, 'sum') #in ha per rep## T0_BAU1 T0_BAU2 T0_BAU3
## 1867 1867 1867T20<-stack(BAU1_T20, BAU2_T20, BAU3_T20)
names(T20)<-c('T20_BAU1', 'T20_BAU2', 'T20_BAU3')
re_T20 <- reclass_raster(T20, 5)
cellStats(re_T20, 'sum')## T20_BAU1 T20_BAU2 T20_BAU3
## 1915 1994 1914T100<-stack(BAU1_T100, BAU2_T100, BAU3_T100)
names(T100)<-c('T100_BAU1', 'T100_BAU2', 'T100_BAU3')
re_T100 <- reclass_raster(T100, 5)
cellStats(re_T100, 'sum')## T100_BAU1 T100_BAU2 T100_BAU3
## 1589 1718 1535#Plot of BAU Corridor Network
BAU_Net_stack <- stack(re_T0[[1]],re_T20[[1]],re_T100[[1]],
re_T0[[2]],re_T20[[2]], re_T100[[2]],
re_T0[[3]],re_T20[[3]], re_T100[[3]])
plot(BAU_Net_stack)
Figure 12b. Habitat corridor network for Leadbeater’s Possum for each replicate and time period under business as usual scenario
#Add your Task 4.6 code here
NM_LBP<-sum(NM_Net_stack)
plot(NM_LBP,col = plasma(9))
Figure 13a. Aggregated habitat corridor network for Leadbeater’s Possum under no management scenairo
#Task 4.6
#BAU
BAU_LBP<-sum(BAU_Net_stack)
plot(BAU_LBP,col = plasma(9))
Figure 13b. Aggregated habitat corridor network for Leadbeater’s Possum under business as usual scenairo
#Add your Task 4.7 code here
NM_LBP_nozero<-NM_LBP #duplicate layer to not overwrite it
NM_LBP_nozero[NM_LBP_nozero == 0] <- NA #replace no data values with NA
corridors_LBP_NM<-NM_LBP_nozero%>%as.data.frame(xy = T, na.rm = T) #make raster into table
corridors_LBP_NM$ID<-1 #assign unit (1 = 1ha)
areas_LBP_NM<-corridors_LBP_NM%>%group_by(layer)%>%
summarise(area = sum(ID))%>%
mutate(sumA = 6784, per = 100*area/sumA)%>%
rename(class = layer)
areas_LBP_NM## # A tibble: 9 × 4
## class area sumA per
## <dbl> <dbl> <dbl> <dbl>
## 1 1 657 6784 9.68
## 2 2 247 6784 3.64
## 3 3 420 6784 6.19
## 4 4 188 6784 2.77
## 5 5 217 6784 3.20
## 6 6 308 6784 4.54
## 7 7 159 6784 2.34
## 8 8 475 6784 7.00
## 9 9 565 6784 8.33write.csv(areas_LBP_NM, 'outputs/LBP_areas_NM.csv')#Task 4.7
#BAU
BAU_LBP_nozero<-BAU_LBP #duplicate layer to not overwrite it
BAU_LBP_nozero[BAU_LBP_nozero == 0] <- NA #replace no data values with NA
corridors_LBP_BAU<-BAU_LBP_nozero%>%as.data.frame(xy = T, na.rm = T) #make raster into table
corridors_LBP_BAU$ID<-1 #assign unit (1 = 1ha)
areas_LBP_BAU<-corridors_LBP_BAU%>%group_by(layer)%>%
summarise(area = sum(ID))%>%
mutate(sumA = 6784, per = 100*area/sumA)%>%
rename(class = layer)
areas_LBP_BAU## # A tibble: 9 × 4
## class area sumA per
## <dbl> <dbl> <dbl> <dbl>
## 1 1 682 6784 10.1
## 2 2 491 6784 7.24
## 3 3 319 6784 4.70
## 4 4 296 6784 4.36
## 5 5 372 6784 5.48
## 6 6 268 6784 3.95
## 7 7 353 6784 5.20
## 8 8 156 6784 2.30
## 9 9 586 6784 8.64write.csv(areas_LBP_BAU, 'outputs/LBP_areas_BAU.csv')Exercise 4: Questions
21. Compare and contrast the change in the meta-population network within and between scenarios. Does the meta-population network remain static over time? What do you think is influencing the connectivity between LBP colonies overtime? Integrate the maps and graphs created above into this answer. Note: this question is worth ~20% of the entire assessment for this Prac!
The 30 LBP colonies that make up the meta-population change differently over time under the two different scenarios.
In the no management scenario, Figure 10a demonstrates a decline in connected nodes over time. The conductance of the network also changes over time as it becomes less fragmented and has more low quality connections as indicated by Figure 11a. The network corridor changes in Figure 12a to become less spread out over time and concentrates in a in one large central patch of habitat. This is likely due to fire being the only major disturbance event which acts to homogenise areas due fires ability kill large stands of Mountain Ash trees.
In the business as usual scenario, Figure 10b demonstrates a similar decline in connected nodes over time as is the case in the no management scenario. Similarly to the no management scenario, the conductance changes over time in Figure 11b as the landscape becomes more connected but there is more variability in the quality of these connections due to more frequent disturbances caused by timber harvesting. The network corridor changes in Figure 12b over time but does not appear to follow any particular pattern unlike the no management scenario. This reflects the randomness of the timber harvesting disturbances
Figures 15a and 15b demonstrate how the quality of habitat is similarly distributed between the no management scenario and the business as usual scenario. By grouping the different classes in these figures into low (classes 1-3), moderate (classes 4-6) and high (classes 7-9) it gives an indication on how the landscape is configured spatially. In the no management scenario the landscape consist of 19.51% low quality, 10.5% moderate quality and 17.66% high quality. The business as usual scenario landscape consists of 21.99% low quality, 13.79 moderate quality and 16.14% high quality. Although the differences are minor, the differences in these values would be likely due to the impacts of timber harvesting in the business as usual scenario which enables the production of acacia species as they are no longing being out competed by Mountain ash. The slightly less values in high quality habitat would be due to some of this being harvested and and given the length it takes for hollows to develop in living mountain ash trees (100-200 years) it will take longer to replenish than these models consider.
The factors that are influencing the connectivity between LBP colonies over time is the quality of resources provided by mountain ash trees and acacia trees. As the LBP colonies rely on hollows in the mountain ash for refuge and acacias for foraging the frequency of both occurring is important for creating high quality habitat. The hollows in mountain ash take 100-200 years to develop in living tree. They also occur in dead trees but the quality of these hollows deteriorate and after a few decades these become unsuitable for LBP. The acacias develop following disturbances such as fire and timber harvesting as seeds in the soil have enough exposure to light to grow. After 80-90 years most of the acacia will have died due to either senescence or out competition for the mountain ash and will no longer be able to provide foraging resources to the LBP colonies. These areas therefore rely on regular disturbances over a period of 50-100 years to continue to be able to provide the suitable resources required for the LBP colonies. The disturbance of timber harvesting can simulate the effects of harvesting as it promotes the growth of acacias but it also limits the amount of hollow bearing trees and can therefore degrade the amount of high quality habitat for the LBP colonies.