1. Autocorrelation around a stationary mean

2. Autocorrelation and stationary mean a lognormal “stationary walk”
CV in N
Amount of autocorrelation
Random process error
Stationary mean in N
14 Rand autocorrel logn stationary.r, slightly modified code from Michael Wilberg
Wilberg MJ & Miller TJ (2007) Comment on “Impacts of biodiversity loss on ocean ecosystem services”. Science 316: 1285b

3. Autocorrelation and stationary mean properties for large numbers of years

4. “Random” number seeds Computers cannot generate truly random numbers
Use a variety of ingenious methods to generate pseudo-random numbers that appear random
Each requires a starting point (a random number “seed”)
Successive numbers generated from the previous number in the sequence
Sequence does not repeat for a very long time
In R: set.seed(some.positive.number) picks a sequence
Chapter 7 Press et al. (2007) Numerical Recipes. Cambridge University Press. 1235pp.

5. Different seeds, different CVs
Seed=5, CV=0.1,0.25,0.4
Seed=6, CV=0.1,0.25,0.4
Abundance
Year
Year
14 Rand autocorrel logn stationary.r

6. Application: assess validity of catch status plots
Froese R & Kesner-Reyes K (2002) Impact of fishing on the abundance of marine species. ICES paper CM 2002/L:12: 15pp
Pauly D (2008) Global fisheries: a brief review. Journal of Biological Research-Thessaloniki 9: 3-9
Froese R, Zeller D, Kleisner K & Pauly D (2012) What catch data can tell us about the status of global fisheries. Mar. Biol. 159:
Pauly D (2013) Does catch reflect abundance? Yes, it is a crucial signal. Nature 494:

7. Autocorrelated time series, fluctuating around a mean
Low variability
High variability
Relative catch
Expect to see no trend over time!
Percentage of stocks
Year
Branch et al. (2011) Conservation Biology 25:

8. Spatial models (meta-population models)

9. Readings
Hilborn R et al. (2006) Integrating marine protected areas with catch regulation. CJFAS 63:

10. Why worry about space?

11. Dutch beam trawl fleet Netherlands England
Rijnsdorp AD et al. (1998) Micro-scale distribution of beam trawl effort in the southern North Sea between 1993 and 1996 in relation to the trawling frequency of the sea bed and the impact on benthic organisms. ICES Journal of Marine Science 55:

12. UK fisheries Beam trawlers Dredgers Netters Otter trawlers Potters
All combined
Jennings S & Lee J (2012) Defining fishing grounds with vessel monitoring system data. ICES Journal of Marine Science 69:51-63

13. Ideal free distribution
Abundance
(variable)
CPUE
(constant)
Latitude (°N)
Effort
(variable)
Longitude (°W)
Swain DP & EJ Wade (2003) Spatial distribution of catch and effort in a fishery for snow crab (Chionoectes opilio): tests of predictions of the ideal free distribution. CJFAS 60:

14. MPA MPA Marine protected area (MPA) MPA MPA
Murawski SA et al. (2005) Effort distribution and catch patterns adjacent to temperate MPAs. ICES Journal of Marine Science 62:

15. Reason for “fishing the line”
Dollars per hour
CV of dollars
Hours trawled
Minimum distance to a closed area (km)
Murawski SA et al. (2005) Effort distribution and catch patterns adjacent to temperate MPAs. ICES Journal of Marine Science 62:

16. Why space matters Populations are not uniform
Exploitation is not uniform across space
Local depletion can make exploitation more costly
Unique attributes of subpopulations may be lost (genetic, behavior, migration, etc.)

17. Metapopulation models
A metapopulation is a series of discrete populations that are isolated but have limited exchange.

18. Generalized metapopulation model
Emigration: movement out of area i into area j
Numbers in area i in time t
Immigration: movement from area j into area i
Environmental conditions in area i in time t

19. Important details A model of population change (dynamics) in each area
A model of dispersal among areas
A model for uncertain events (e.g. environmental changes)
What happens at the boundaries?

20. Boundaries? Absorbing boundary Reflective boundary
Disappear/die on hitting the boundary
E.g. another country, advection into open ocean, natural range bounds
Reflective boundary
Pile up in the cell next to the boundary
E.g. mountain range, river barrier
Pac-man (circular coastline)
Reappear on the opposite side
E.g. circumpolar, islands

21. One-dimensional logistic-growth model with harvesting
Numbers in area i in time t
Logistic model
Density
dependence
Exploitation rate
Harvest
Immigration
Emigration
Cell on the left
Cell on the right
Only survivors of harvest will move
Movement rate the
same in all cells
Harvest, then movement

22. 15 spatial models animation.r

23. Diffusion scenario
21 areas, 50 time steps, migration rate m = 0.2, r = 0.2, K = 1000, reflective boundary
Diffusion scenario: starting population = K in center cell, 0 in all other cells, exploitation rate 0
15 spatial models animation.r

24. Diffusion, no harvesting, N11 = K (darker = older years, light gray = most recent)
Abundance
####FISH 458 lecture on spatial models
#Written by Trevor A. Branch starting 6 May 2012
#Completed 7 May 2012
#######################################################
###Model 1: one-D spatial model logistic growth with harvesting
###assume reflection when hits boundary
one.d.logistic <- function(ncells=101, expl.rate=rep(0,ncells), ntime.steps=20, K=1000, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,50),K,rep(0,50))) {
N.mat <- matrix(nrow=ntime.steps,ncol=ncells)
N.mat[1,] <- start.N.vec
for (year in 1:(ntime.steps-1)) {
#modelled as reflection, so no movement when hit bounds
#left-most cell
N.mat[year+1,1] <- N.mat[year,1] + r*N.mat[year,1]*(1-N.mat[year,1]/K) -
expl.rate[1]*N.mat[year,1] +
m.rate*((1-expl.rate[1+1])*N.mat[year,1+1]) -
m.rate*((1-expl.rate[1])*N.mat[year,1])
#right-most cell
N.mat[year+1,ncells] <- N.mat[year,ncells] + r*N.mat[year,ncells]*(1-N.mat[year,ncells]/K) -
expl.rate[ncells]*N.mat[year,ncells] +
m.rate*((1-expl.rate[ncells-1])*N.mat[year,ncells-1]) -
m.rate*((1-expl.rate[ncells])*N.mat[year,ncells])
#all the cells in between
for (i in 2:(ncells-1)) {
N.mat[year+1,i] <- N.mat[year,i] + r*N.mat[year,i]*(1-N.mat[year,i]/K) -
expl.rate[i]*N.mat[year,i] +
m.rate*((1-expl.rate[i-1])*N.mat[year,i-1]+(1-expl.rate[i+1])*N.mat[year,i+1]) -
2*m.rate*((1-expl.rate[i])*N.mat[year,i])
#print(N.mat[year+1,i]) }
#plot results
gray.pal <- (1:ntime.steps)/ntime.steps* #returns values between 0.2 (light gray) and 1 (black)
#print(gray.pal)
plot(x=1:ncells,y=N.mat[1,], type="l",xaxs="i",yaxs="i",las=1, col=gray(gray.pal[1]),
ylim=c(0,K*1.02), xlab="", ylab="")
for (year in 2:ntime.steps) {
par(new=T)
lines(x=1:ncells,y=N.mat[year,], col=gray(gray.pal[year]), ylim=c(0,K*1.02))
return(N.mat)
K.start <- 1000
ncells <- 21
par(oma=c(0,0,0,0), mar=c(5,5,1,1))
x <- one.d.logistic(ncells=ncells, expl.rate=rep(0,ncells), ntime.steps=10, K=K.start, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,9),K.start/100,K.start,K.start/100,rep(0,9)))
###Set up animation; all in middle cell, no harvest
#requires installation of ImageMagick, followed by machine restart
require(animation)
outdir <- "C:\\Users\\Trevor Branch\\Documents\\FISH458 in 2012\\aLectures\\Rpics"
oopts = ani.options(interval = 0.03, ani.dev="png", outdir = outdir, width=1400, height=600) #interval is wait in seconds between frames in video
par(bg = "white") #ensure the background color is white
ani.record(reset = TRUE) #clear history before recording
for (year in 2:50) {
x <- one.d.logistic(ncells=ncells, expl.rate=rep(0,ncells), ntime.steps=year, K=K.start, r=0.2, m.rate=0.1,
ani.record() #record the current frame
saveGIF(ani.replay(), img.name = "oneDmovie", movie.name="oneDmovie.gif", convert="convert",
ani.width=1400, ani.height=600, interval = 0.2) #ImageMagick
#####all at K, add harvest but MPA in middle cells
#oopts = ani.options(interval = 0.03, ani.dev="png", outdir = outdir, width=1400, height=600) #interval is wait in seconds between frames in video
x <- one.d.logistic(ncells=ncells, expl.rate=c(rep(0.2,8),0,0,0,0,0,rep(0.2,8)), ntime.steps=year, K=K.start, r=0.2, m.rate=0.1,
start.N.vec=rep(K.start,ncells))
saveGIF(ani.replay(), img.name = "MPA", movie.name="MPA.gif", convert="convert", interval=0.2,
ani.width=1400, ani.height=600) #ImageMagick
###for(i in 1:100) {dev.off()}
Cell number
15 spatial models animation.r

25. Marine protected area scenario
21 areas, 50 time steps, migration rate m = 0.2, r = 0.2, K = 1000, reflective boundary
MPA scenario: starting population = K in all cells, exploitation rate 0 in center 5 cells (numbers 9–13), exploitation rate 0.2 in all other cells
15 spatial models animation.r

26. MPA: no harvest in center cells (darker = older years, light gray = most recent)
Abundance
####FISH 458 lecture on spatial models
#Written by Trevor A. Branch starting 6 May 2012
#Completed 7 May 2012
#######################################################
###Model 1: one-D spatial model logistic growth with harvesting
###assume reflection when hits boundary
one.d.logistic <- function(ncells=101, expl.rate=rep(0,ncells), ntime.steps=20, K=1000, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,50),K,rep(0,50))) {
N.mat <- matrix(nrow=ntime.steps,ncol=ncells)
N.mat[1,] <- start.N.vec
for (year in 1:(ntime.steps-1)) {
#modelled as reflection, so no movement when hit bounds
#left-most cell
N.mat[year+1,1] <- N.mat[year,1] + r*N.mat[year,1]*(1-N.mat[year,1]/K) -
expl.rate[1]*N.mat[year,1] +
m.rate*((1-expl.rate[1+1])*N.mat[year,1+1]) -
m.rate*((1-expl.rate[1])*N.mat[year,1])
#right-most cell
N.mat[year+1,ncells] <- N.mat[year,ncells] + r*N.mat[year,ncells]*(1-N.mat[year,ncells]/K) -
expl.rate[ncells]*N.mat[year,ncells] +
m.rate*((1-expl.rate[ncells-1])*N.mat[year,ncells-1]) -
m.rate*((1-expl.rate[ncells])*N.mat[year,ncells])
#all the cells in between
for (i in 2:(ncells-1)) {
N.mat[year+1,i] <- N.mat[year,i] + r*N.mat[year,i]*(1-N.mat[year,i]/K) -
expl.rate[i]*N.mat[year,i] +
m.rate*((1-expl.rate[i-1])*N.mat[year,i-1]+(1-expl.rate[i+1])*N.mat[year,i+1]) -
2*m.rate*((1-expl.rate[i])*N.mat[year,i])
#print(N.mat[year+1,i]) }
#plot results
gray.pal <- (1:ntime.steps)/ntime.steps* #returns values between 0.2 (light gray) and 1 (black)
#print(gray.pal)
plot(x=1:ncells,y=N.mat[1,], type="l",xaxs="i",yaxs="i",las=1, col=gray(gray.pal[1]),
ylim=c(0,K*1.02), xlab="", ylab="")
for (year in 2:ntime.steps) {
par(new=T)
lines(x=1:ncells,y=N.mat[year,], col=gray(gray.pal[year]), ylim=c(0,K*1.02))
return(N.mat)
K.start <- 1000
ncells <- 21
par(oma=c(0,0,0,0), mar=c(5,5,1,1))
x <- one.d.logistic(ncells=ncells, expl.rate=rep(0,ncells), ntime.steps=10, K=K.start, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,9),K.start/100,K.start,K.start/100,rep(0,9)))
###Set up animation; all in middle cell, no harvest
#requires installation of ImageMagick, followed by machine restart
require(animation)
outdir <- "C:\\Users\\Trevor Branch\\Documents\\FISH458 in 2012\\aLectures\\Rpics"
oopts = ani.options(interval = 0.03, ani.dev="png", outdir = outdir, width=1400, height=600) #interval is wait in seconds between frames in video
par(bg = "white") #ensure the background color is white
ani.record(reset = TRUE) #clear history before recording
for (year in 2:50) {
x <- one.d.logistic(ncells=ncells, expl.rate=rep(0,ncells), ntime.steps=year, K=K.start, r=0.2, m.rate=0.1,
ani.record() #record the current frame
saveGIF(ani.replay(), img.name = "oneDmovie", movie.name="oneDmovie.gif", convert="convert",
ani.width=1400, ani.height=600, interval = 0.2) #ImageMagick
#####all at K, add harvest but MPA in middle cells
#oopts = ani.options(interval = 0.03, ani.dev="png", outdir = outdir, width=1400, height=600) #interval is wait in seconds between frames in video
x <- one.d.logistic(ncells=ncells, expl.rate=c(rep(0.2,8),0,0,0,0,0,rep(0.2,8)), ntime.steps=year, K=K.start, r=0.2, m.rate=0.1,
start.N.vec=rep(K.start,ncells))
saveGIF(ani.replay(), img.name = "MPA", movie.name="MPA.gif", convert="convert", interval=0.2,
ani.width=1400, ani.height=600) #ImageMagick
###for(i in 1:100) {dev.off()}
Cell number
15 spatial models animation.r

27. Do protected areas increase yields?
51 areas, migration rate m = 0.2, r = 0.2, K = 1000, start population = K in all areas
Run with u = 0, 0.01, 0.02, …, 0.9
After a large number of time steps (1000) the model is at equilibrium, and total yield in the final time step is the equilibrium yield for each value of u
No MPA: all areas have harvest rate = u
MPA: 5 middle areas have zero harvest rate, other have harvest rate = u
15 MPA yield.r

28. 15 MPA yield.r

29. Almost at equilibrium yield
Time series of catches
Initially the no MPA scenario has higher catches; after year 14 with high u = 0.25 the MPA has higher catches; with no MPA, the population can go extinct
Almost at equilibrium yield
Yield in each year
No MPA
u = 0.1
MPA
u = 0.25
Year
15 MPA yield.r

30. Equilibrium yield
Examine the equilibrium yield after running the model for 100 years
How does this differ with an MPA and without an MPA?
How does this differ for different harvest rates?

31. Do protected areas increase yield?
Higher yield without MPA
Yield maximized at higher harvest rate with MPA
At u > r, MPA prevents collapse (insurance policy)
No MPA
u = r = 0.2
Equilibrium yield
####FISH 458 lecture on spatial models
#Written by Trevor A. Branch starting 6 May 2012
#Completed 7 May 2012
#######################################################
###Model 1: one-D spatial model logistic growth with harvesting
###assume reflection when hits boundary
###Test to see what total yield is at different harvest rates, and with different numbers of cells closed
one.d.logistic <- function(ncells=101, expl.rate=rep(0,ncells), ntime.steps=20, K=1000, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,50),K,rep(0,50))) {
N.mat <- matrix(nrow=ntime.steps,ncol=ncells)
N.mat[1,] <- start.N.vec
yield <- vector(length=ntime.steps)
yield[] <- 0
for (year in 1:(ntime.steps-1)) {
yield[year] <- 0
for (i in 1:ncells) {
yield[year] <- yield[year]+expl.rate[i]*N.mat[year,i] }
#modelled as reflection, so no movement when hit bounds
#left-most cell
N.mat[year+1,1] <- N.mat[year,1] + r*N.mat[year,1]*(1-N.mat[year,1]/K) -
expl.rate[1]*N.mat[year,1] +
m.rate*((1-expl.rate[1+1])*N.mat[year,1+1]) -
m.rate*((1-expl.rate[1])*N.mat[year,1])
#right-most cell
N.mat[year+1,ncells] <- N.mat[year,ncells] + r*N.mat[year,ncells]*(1-N.mat[year,ncells]/K) -
expl.rate[ncells]*N.mat[year,ncells] +
m.rate*((1-expl.rate[ncells-1])*N.mat[year,ncells-1]) -
m.rate*((1-expl.rate[ncells])*N.mat[year,ncells])
#all the cells in between
for (i in 2:(ncells-1)) {
N.mat[year+1,i] <- N.mat[year,i] + r*N.mat[year,i]*(1-N.mat[year,i]/K) -
expl.rate[i]*N.mat[year,i] +
m.rate*((1-expl.rate[i-1])*N.mat[year,i-1]+(1-expl.rate[i+1])*N.mat[year,i+1]) -
2*m.rate*((1-expl.rate[i])*N.mat[year,i])
#print(N.mat[year+1,i])
#now calculate final year yield
yield[ntime.steps] <- yield[ntime.steps]+expl.rate[i]*N.mat[ntime.steps,i]
return(list(N.mat=N.mat, yield=yield))
###compare yield by harvest rates for 5 closed cells out of 51
K.start <- 1000
ncells <- 51
ntime.steps <- 1000
urate <- seq(0,0.2,0.01)
nharvest <- length(urate)
final.yield <- vector(length=nharvest)
for (i in 1:nharvest) {
x <- one.d.logistic(ncells=ncells,
expl.rate=c(rep(urate[i],(ncells-1)/2-2),0,0,0,0,0,rep(urate[i],(ncells-1)/2-2)), ntime.steps=ntime.steps,
K=K.start, r=0.2, m.rate=0.2,
start.N.vec=rep(K.start,ncells))
final.yield[i] <- x$yield[ntime.steps]
final.yield.noMPA <- vector(length=nharvest)
xnoMPA <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate[i],ncells), ntime.steps=ntime.steps,
final.yield.noMPA[i] <- xnoMPA$yield[ntime.steps]
plot(x=urate,y=final.yield, xaxs="i",yaxs="i",type="l",lwd=2,col="blue",las=1,
xlab="Harvest rate", ylab="Equilibrium yield", ylim=c(0,2600),cex.lab=1.3)
par(new=T)
plot(x=urate,y=final.yield.noMPA, xaxs="i",yaxs="i",type="l",lwd=2,col="green",las=1,
xlab="", ylab="",ylim=c(0,2600),axes=F)
###compare yield by harvest rates for 25 closed cells out of 51
urate <- seq(0,0.9,0.01)
expl.rate=c(rep(urate[i],13),rep(0,25),rep(urate[i],13)), ntime.steps=ntime.steps,
###Three scenarios that produce identical yield of around 1377 tons
###1. closed 25 areas, at u = 0.1
###2. all areas open at u = 0.032
###3. all areas open at u = 0.168
urate <- 0.1
expl.rate=c(rep(urate,13),rep(0,25),rep(urate,13)), ntime.steps=ntime.steps,
print(x$yield[ntime.steps])
urate<-0.032
xnoMPA1 <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate,ncells), ntime.steps=ntime.steps,
print(xnoMPA1$yield[ntime.steps])
urate<-0.168
xnoMPA2 <- one.d.logistic(ncells=ncells,
print(xnoMPA2$yield[ntime.steps])
plot(x=1:ncells,y=x$N.mat[ntime.steps,], ylim=c(0,1050), xlab="Cell number",cex.lab=1.3,ylab="Abundance",type="l",lwd=2,col="blue", las=1,xaxs="i",yaxs="i")
plot(x=1:ncells,y=xnoMPA1$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
plot(x=1:ncells,y=xnoMPA2$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
polygon(x=c(14,14,38,38),y=c(0,1050,1050,0),col="# ")
uMSY = r/2 = 0.1
MPA
Harvest rate (u)
15 MPA yield.r

32. Zoomed in No MPA MPA Equilibrium yield Harvest rate (u)
####FISH 458 lecture on spatial models
#Written by Trevor A. Branch starting 6 May 2012
#Completed 7 May 2012
#######################################################
###Model 1: one-D spatial model logistic growth with harvesting
###assume reflection when hits boundary
###Test to see what total yield is at different harvest rates, and with different numbers of cells closed
one.d.logistic <- function(ncells=101, expl.rate=rep(0,ncells), ntime.steps=20, K=1000, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,50),K,rep(0,50))) {
N.mat <- matrix(nrow=ntime.steps,ncol=ncells)
N.mat[1,] <- start.N.vec
yield <- vector(length=ntime.steps)
yield[] <- 0
for (year in 1:(ntime.steps-1)) {
yield[year] <- 0
for (i in 1:ncells) {
yield[year] <- yield[year]+expl.rate[i]*N.mat[year,i] }
#modelled as reflection, so no movement when hit bounds
#left-most cell
N.mat[year+1,1] <- N.mat[year,1] + r*N.mat[year,1]*(1-N.mat[year,1]/K) -
expl.rate[1]*N.mat[year,1] +
m.rate*((1-expl.rate[1+1])*N.mat[year,1+1]) -
m.rate*((1-expl.rate[1])*N.mat[year,1])
#right-most cell
N.mat[year+1,ncells] <- N.mat[year,ncells] + r*N.mat[year,ncells]*(1-N.mat[year,ncells]/K) -
expl.rate[ncells]*N.mat[year,ncells] +
m.rate*((1-expl.rate[ncells-1])*N.mat[year,ncells-1]) -
m.rate*((1-expl.rate[ncells])*N.mat[year,ncells])
#all the cells in between
for (i in 2:(ncells-1)) {
N.mat[year+1,i] <- N.mat[year,i] + r*N.mat[year,i]*(1-N.mat[year,i]/K) -
expl.rate[i]*N.mat[year,i] +
m.rate*((1-expl.rate[i-1])*N.mat[year,i-1]+(1-expl.rate[i+1])*N.mat[year,i+1]) -
2*m.rate*((1-expl.rate[i])*N.mat[year,i])
#print(N.mat[year+1,i])
#now calculate final year yield
yield[ntime.steps] <- yield[ntime.steps]+expl.rate[i]*N.mat[ntime.steps,i]
return(list(N.mat=N.mat, yield=yield))
###compare yield by harvest rates for 5 closed cells out of 51
K.start <- 1000
ncells <- 51
ntime.steps <- 1000
urate <- seq(0,0.2,0.01)
nharvest <- length(urate)
final.yield <- vector(length=nharvest)
for (i in 1:nharvest) {
x <- one.d.logistic(ncells=ncells,
expl.rate=c(rep(urate[i],(ncells-1)/2-2),0,0,0,0,0,rep(urate[i],(ncells-1)/2-2)), ntime.steps=ntime.steps,
K=K.start, r=0.2, m.rate=0.2,
start.N.vec=rep(K.start,ncells))
final.yield[i] <- x$yield[ntime.steps]
final.yield.noMPA <- vector(length=nharvest)
xnoMPA <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate[i],ncells), ntime.steps=ntime.steps,
final.yield.noMPA[i] <- xnoMPA$yield[ntime.steps]
plot(x=urate,y=final.yield, xaxs="i",yaxs="i",type="l",lwd=2,col="blue",las=1,
xlab="Harvest rate", ylab="Equilibrium yield", ylim=c(0,2600),cex.lab=1.3)
par(new=T)
plot(x=urate,y=final.yield.noMPA, xaxs="i",yaxs="i",type="l",lwd=2,col="green",las=1,
xlab="", ylab="",ylim=c(0,2600),axes=F)
###compare yield by harvest rates for 25 closed cells out of 51
urate <- seq(0,0.9,0.01)
expl.rate=c(rep(urate[i],13),rep(0,25),rep(urate[i],13)), ntime.steps=ntime.steps,
###Three scenarios that produce identical yield of around 1377 tons
###1. closed 25 areas, at u = 0.1
###2. all areas open at u = 0.032
###3. all areas open at u = 0.168
urate <- 0.1
expl.rate=c(rep(urate,13),rep(0,25),rep(urate,13)), ntime.steps=ntime.steps,
print(x$yield[ntime.steps])
urate<-0.032
xnoMPA1 <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate,ncells), ntime.steps=ntime.steps,
print(xnoMPA1$yield[ntime.steps])
urate<-0.168
xnoMPA2 <- one.d.logistic(ncells=ncells,
print(xnoMPA2$yield[ntime.steps])
plot(x=1:ncells,y=x$N.mat[ntime.steps,], ylim=c(0,1050), xlab="Cell number",cex.lab=1.3,ylab="Abundance",type="l",lwd=2,col="blue", las=1,xaxs="i",yaxs="i")
plot(x=1:ncells,y=xnoMPA1$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
plot(x=1:ncells,y=xnoMPA2$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
polygon(x=c(14,14,38,38),y=c(0,1050,1050,0),col="# ")
uMSY without MPA
Break-even point
uMSY with MPA
Harvest rate (u)
15 MPA yield.r

33. Bigger MPA (25 cells) (previously 5 cells)
Closing more areas reduces the maximum yield more
Maximum yield reduced from to 1377 in this simulation
No MPA
Equilibrium yield
MPA
####FISH 458 lecture on spatial models
#Written by Trevor A. Branch starting 6 May 2012
#Completed 7 May 2012
#######################################################
###Model 1: one-D spatial model logistic growth with harvesting
###assume reflection when hits boundary
###Test to see what total yield is at different harvest rates, and with different numbers of cells closed
one.d.logistic <- function(ncells=101, expl.rate=rep(0,ncells), ntime.steps=20, K=1000, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,50),K,rep(0,50))) {
N.mat <- matrix(nrow=ntime.steps,ncol=ncells)
N.mat[1,] <- start.N.vec
yield <- vector(length=ntime.steps)
yield[] <- 0
for (year in 1:(ntime.steps-1)) {
yield[year] <- 0
for (i in 1:ncells) {
yield[year] <- yield[year]+expl.rate[i]*N.mat[year,i] }
#modelled as reflection, so no movement when hit bounds
#left-most cell
N.mat[year+1,1] <- N.mat[year,1] + r*N.mat[year,1]*(1-N.mat[year,1]/K) -
expl.rate[1]*N.mat[year,1] +
m.rate*((1-expl.rate[1+1])*N.mat[year,1+1]) -
m.rate*((1-expl.rate[1])*N.mat[year,1])
#right-most cell
N.mat[year+1,ncells] <- N.mat[year,ncells] + r*N.mat[year,ncells]*(1-N.mat[year,ncells]/K) -
expl.rate[ncells]*N.mat[year,ncells] +
m.rate*((1-expl.rate[ncells-1])*N.mat[year,ncells-1]) -
m.rate*((1-expl.rate[ncells])*N.mat[year,ncells])
#all the cells in between
for (i in 2:(ncells-1)) {
N.mat[year+1,i] <- N.mat[year,i] + r*N.mat[year,i]*(1-N.mat[year,i]/K) -
expl.rate[i]*N.mat[year,i] +
m.rate*((1-expl.rate[i-1])*N.mat[year,i-1]+(1-expl.rate[i+1])*N.mat[year,i+1]) -
2*m.rate*((1-expl.rate[i])*N.mat[year,i])
#print(N.mat[year+1,i])
#now calculate final year yield
yield[ntime.steps] <- yield[ntime.steps]+expl.rate[i]*N.mat[ntime.steps,i]
return(list(N.mat=N.mat, yield=yield))
###compare yield by harvest rates for 5 closed cells out of 51
K.start <- 1000
ncells <- 51
ntime.steps <- 1000
urate <- seq(0,0.2,0.01)
nharvest <- length(urate)
final.yield <- vector(length=nharvest)
for (i in 1:nharvest) {
x <- one.d.logistic(ncells=ncells,
expl.rate=c(rep(urate[i],(ncells-1)/2-2),0,0,0,0,0,rep(urate[i],(ncells-1)/2-2)), ntime.steps=ntime.steps,
K=K.start, r=0.2, m.rate=0.2,
start.N.vec=rep(K.start,ncells))
final.yield[i] <- x$yield[ntime.steps]
final.yield.noMPA <- vector(length=nharvest)
xnoMPA <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate[i],ncells), ntime.steps=ntime.steps,
final.yield.noMPA[i] <- xnoMPA$yield[ntime.steps]
plot(x=urate,y=final.yield, xaxs="i",yaxs="i",type="l",lwd=2,col="blue",las=1,
xlab="Harvest rate", ylab="Equilibrium yield", ylim=c(0,2600),cex.lab=1.3)
par(new=T)
plot(x=urate,y=final.yield.noMPA, xaxs="i",yaxs="i",type="l",lwd=2,col="green",las=1,
xlab="", ylab="",ylim=c(0,2600),axes=F)
###compare yield by harvest rates for 25 closed cells out of 51
urate <- seq(0,0.9,0.01)
expl.rate=c(rep(urate[i],13),rep(0,25),rep(urate[i],13)), ntime.steps=ntime.steps,
###Three scenarios that produce identical yield of around 1377 tons
###1. closed 25 areas, at u = 0.1
###2. all areas open at u = 0.032
###3. all areas open at u = 0.168
urate <- 0.1
expl.rate=c(rep(urate,13),rep(0,25),rep(urate,13)), ntime.steps=ntime.steps,
print(x$yield[ntime.steps])
urate<-0.032
xnoMPA1 <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate,ncells), ntime.steps=ntime.steps,
print(xnoMPA1$yield[ntime.steps])
urate<-0.168
xnoMPA2 <- one.d.logistic(ncells=ncells,
print(xnoMPA2$yield[ntime.steps])
plot(x=1:ncells,y=x$N.mat[ntime.steps,], ylim=c(0,1050), xlab="Cell number",cex.lab=1.3,ylab="Abundance",type="l",lwd=2,col="blue", las=1,xaxs="i",yaxs="i")
plot(x=1:ncells,y=xnoMPA1$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
plot(x=1:ncells,y=xnoMPA2$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
polygon(x=c(14,14,38,38),y=c(0,1050,1050,0),col="# ")
Harvest rate (u)
15 MPA yield.r

34. Lessons Spatial scale and pattern matters Simple movement models
Marine protected areas: insurance vs. yield
Trade-offs between catch, profit, and biodiversity

35. For the same yield what is u?
MPA: u = produces MSY
No MPA: two values of u result in equivalent yield: u = and u = 0.168
No MPA
Equilibrium yield
MPA
####FISH 458 lecture on spatial models
#Written by Trevor A. Branch starting 6 May 2012
#Completed 7 May 2012
#######################################################
###Model 1: one-D spatial model logistic growth with harvesting
###assume reflection when hits boundary
###Test to see what total yield is at different harvest rates, and with different numbers of cells closed
one.d.logistic <- function(ncells=101, expl.rate=rep(0,ncells), ntime.steps=20, K=1000, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,50),K,rep(0,50))) {
N.mat <- matrix(nrow=ntime.steps,ncol=ncells)
N.mat[1,] <- start.N.vec
yield <- vector(length=ntime.steps)
yield[] <- 0
for (year in 1:(ntime.steps-1)) {
yield[year] <- 0
for (i in 1:ncells) {
yield[year] <- yield[year]+expl.rate[i]*N.mat[year,i] }
#modelled as reflection, so no movement when hit bounds
#left-most cell
N.mat[year+1,1] <- N.mat[year,1] + r*N.mat[year,1]*(1-N.mat[year,1]/K) -
expl.rate[1]*N.mat[year,1] +
m.rate*((1-expl.rate[1+1])*N.mat[year,1+1]) -
m.rate*((1-expl.rate[1])*N.mat[year,1])
#right-most cell
N.mat[year+1,ncells] <- N.mat[year,ncells] + r*N.mat[year,ncells]*(1-N.mat[year,ncells]/K) -
expl.rate[ncells]*N.mat[year,ncells] +
m.rate*((1-expl.rate[ncells-1])*N.mat[year,ncells-1]) -
m.rate*((1-expl.rate[ncells])*N.mat[year,ncells])
#all the cells in between
for (i in 2:(ncells-1)) {
N.mat[year+1,i] <- N.mat[year,i] + r*N.mat[year,i]*(1-N.mat[year,i]/K) -
expl.rate[i]*N.mat[year,i] +
m.rate*((1-expl.rate[i-1])*N.mat[year,i-1]+(1-expl.rate[i+1])*N.mat[year,i+1]) -
2*m.rate*((1-expl.rate[i])*N.mat[year,i])
#print(N.mat[year+1,i])
#now calculate final year yield
yield[ntime.steps] <- yield[ntime.steps]+expl.rate[i]*N.mat[ntime.steps,i]
return(list(N.mat=N.mat, yield=yield))
###compare yield by harvest rates for 5 closed cells out of 51
K.start <- 1000
ncells <- 51
ntime.steps <- 1000
urate <- seq(0,0.2,0.01)
nharvest <- length(urate)
final.yield <- vector(length=nharvest)
for (i in 1:nharvest) {
x <- one.d.logistic(ncells=ncells,
expl.rate=c(rep(urate[i],(ncells-1)/2-2),0,0,0,0,0,rep(urate[i],(ncells-1)/2-2)), ntime.steps=ntime.steps,
K=K.start, r=0.2, m.rate=0.2,
start.N.vec=rep(K.start,ncells))
final.yield[i] <- x$yield[ntime.steps]
final.yield.noMPA <- vector(length=nharvest)
xnoMPA <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate[i],ncells), ntime.steps=ntime.steps,
final.yield.noMPA[i] <- xnoMPA$yield[ntime.steps]
plot(x=urate,y=final.yield, xaxs="i",yaxs="i",type="l",lwd=2,col="blue",las=1,
xlab="Harvest rate", ylab="Equilibrium yield", ylim=c(0,2600),cex.lab=1.3)
par(new=T)
plot(x=urate,y=final.yield.noMPA, xaxs="i",yaxs="i",type="l",lwd=2,col="green",las=1,
xlab="", ylab="",ylim=c(0,2600),axes=F)
###compare yield by harvest rates for 25 closed cells out of 51
urate <- seq(0,0.9,0.01)
expl.rate=c(rep(urate[i],13),rep(0,25),rep(urate[i],13)), ntime.steps=ntime.steps,
###Three scenarios that produce identical yield of around 1377 tons
###1. closed 25 areas, at u = 0.1
###2. all areas open at u = 0.032
###3. all areas open at u = 0.168
urate <- 0.1
expl.rate=c(rep(urate,13),rep(0,25),rep(urate,13)), ntime.steps=ntime.steps,
print(x$yield[ntime.steps])
urate<-0.032
xnoMPA1 <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate,ncells), ntime.steps=ntime.steps,
print(xnoMPA1$yield[ntime.steps])
urate<-0.168
xnoMPA2 <- one.d.logistic(ncells=ncells,
print(xnoMPA2$yield[ntime.steps])
plot(x=1:ncells,y=x$N.mat[ntime.steps,], ylim=c(0,1050), xlab="Cell number",cex.lab=1.3,ylab="Abundance",type="l",lwd=2,col="blue", las=1,xaxs="i",yaxs="i")
plot(x=1:ncells,y=xnoMPA1$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
plot(x=1:ncells,y=xnoMPA2$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
polygon(x=c(14,14,38,38),y=c(0,1050,1050,0),col="# ")
Harvest rate (u)
15 MPA yield.r

36. For the same yield, what is biomass?
No MPA
u = 0.032
MPA
u = 0.1
Abundance
####FISH 458 lecture on spatial models
#Written by Trevor A. Branch starting 6 May 2012
#Completed 7 May 2012
#######################################################
###Model 1: one-D spatial model logistic growth with harvesting
###assume reflection when hits boundary
###Test to see what total yield is at different harvest rates, and with different numbers of cells closed
one.d.logistic <- function(ncells=101, expl.rate=rep(0,ncells), ntime.steps=20, K=1000, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,50),K,rep(0,50))) {
N.mat <- matrix(nrow=ntime.steps,ncol=ncells)
N.mat[1,] <- start.N.vec
yield <- vector(length=ntime.steps)
yield[] <- 0
for (year in 1:(ntime.steps-1)) {
yield[year] <- 0
for (i in 1:ncells) {
yield[year] <- yield[year]+expl.rate[i]*N.mat[year,i] }
#modelled as reflection, so no movement when hit bounds
#left-most cell
N.mat[year+1,1] <- N.mat[year,1] + r*N.mat[year,1]*(1-N.mat[year,1]/K) -
expl.rate[1]*N.mat[year,1] +
m.rate*((1-expl.rate[1+1])*N.mat[year,1+1]) -
m.rate*((1-expl.rate[1])*N.mat[year,1])
#right-most cell
N.mat[year+1,ncells] <- N.mat[year,ncells] + r*N.mat[year,ncells]*(1-N.mat[year,ncells]/K) -
expl.rate[ncells]*N.mat[year,ncells] +
m.rate*((1-expl.rate[ncells-1])*N.mat[year,ncells-1]) -
m.rate*((1-expl.rate[ncells])*N.mat[year,ncells])
#all the cells in between
for (i in 2:(ncells-1)) {
N.mat[year+1,i] <- N.mat[year,i] + r*N.mat[year,i]*(1-N.mat[year,i]/K) -
expl.rate[i]*N.mat[year,i] +
m.rate*((1-expl.rate[i-1])*N.mat[year,i-1]+(1-expl.rate[i+1])*N.mat[year,i+1]) -
2*m.rate*((1-expl.rate[i])*N.mat[year,i])
#print(N.mat[year+1,i])
#now calculate final year yield
yield[ntime.steps] <- yield[ntime.steps]+expl.rate[i]*N.mat[ntime.steps,i]
return(list(N.mat=N.mat, yield=yield))
###compare yield by harvest rates for 5 closed cells out of 51
K.start <- 1000
ncells <- 51
ntime.steps <- 1000
urate <- seq(0,0.2,0.01)
nharvest <- length(urate)
final.yield <- vector(length=nharvest)
for (i in 1:nharvest) {
x <- one.d.logistic(ncells=ncells,
expl.rate=c(rep(urate[i],(ncells-1)/2-2),0,0,0,0,0,rep(urate[i],(ncells-1)/2-2)), ntime.steps=ntime.steps,
K=K.start, r=0.2, m.rate=0.2,
start.N.vec=rep(K.start,ncells))
final.yield[i] <- x$yield[ntime.steps]
final.yield.noMPA <- vector(length=nharvest)
xnoMPA <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate[i],ncells), ntime.steps=ntime.steps,
final.yield.noMPA[i] <- xnoMPA$yield[ntime.steps]
plot(x=urate,y=final.yield, xaxs="i",yaxs="i",type="l",lwd=2,col="blue",las=1,
xlab="Harvest rate", ylab="Equilibrium yield", ylim=c(0,2600),cex.lab=1.3)
par(new=T)
plot(x=urate,y=final.yield.noMPA, xaxs="i",yaxs="i",type="l",lwd=2,col="green",las=1,
xlab="", ylab="",ylim=c(0,2600),axes=F)
###compare yield by harvest rates for 25 closed cells out of 51
urate <- seq(0,0.9,0.01)
expl.rate=c(rep(urate[i],13),rep(0,25),rep(urate[i],13)), ntime.steps=ntime.steps,
###Three scenarios that produce identical yield of around 1377 tons
###1. closed 25 areas, at u = 0.1
###2. all areas open at u = 0.032
###3. all areas open at u = 0.168
urate <- 0.1
expl.rate=c(rep(urate,13),rep(0,25),rep(urate,13)), ntime.steps=ntime.steps,
print(x$yield[ntime.steps])
urate<-0.032
xnoMPA1 <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate,ncells), ntime.steps=ntime.steps,
print(xnoMPA1$yield[ntime.steps])
urate<-0.168
xnoMPA2 <- one.d.logistic(ncells=ncells,
print(xnoMPA2$yield[ntime.steps])
plot(x=1:ncells,y=x$N.mat[ntime.steps,], ylim=c(0,1050), xlab="Cell number",cex.lab=1.3,ylab="Abundance",type="l",lwd=2,col="blue", las=1,xaxs="i",yaxs="i")
plot(x=1:ncells,y=xnoMPA1$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
plot(x=1:ncells,y=xnoMPA2$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
polygon(x=c(14,14,38,38),y=c(0,1050,1050,0),col="# ")
No MPA
u = 0.168
Cell number
15 MPA yield.r

37. For the same yield, what is CPUE
For the same yield, what is CPUE? (assume effort is proportional to harvest rate u)
No MPA
u = 0.032, yield = 1370, CPUE = 1370/0.032 = 42800
MPA
u = 0.1, yield = 1370, CPUE = 13700
No MPA
u = 0.168, yield = 1370, CPUE = 8200
####FISH 458 lecture on spatial models
#Written by Trevor A. Branch starting 6 May 2012
#Completed 7 May 2012
#######################################################
###Model 1: one-D spatial model logistic growth with harvesting
###assume reflection when hits boundary
###Test to see what total yield is at different harvest rates, and with different numbers of cells closed
one.d.logistic <- function(ncells=101, expl.rate=rep(0,ncells), ntime.steps=20, K=1000, r=0.2, m.rate=0.1,
start.N.vec=c(rep(0,50),K,rep(0,50))) {
N.mat <- matrix(nrow=ntime.steps,ncol=ncells)
N.mat[1,] <- start.N.vec
yield <- vector(length=ntime.steps)
yield[] <- 0
for (year in 1:(ntime.steps-1)) {
yield[year] <- 0
for (i in 1:ncells) {
yield[year] <- yield[year]+expl.rate[i]*N.mat[year,i] }
#modelled as reflection, so no movement when hit bounds
#left-most cell
N.mat[year+1,1] <- N.mat[year,1] + r*N.mat[year,1]*(1-N.mat[year,1]/K) -
expl.rate[1]*N.mat[year,1] +
m.rate*((1-expl.rate[1+1])*N.mat[year,1+1]) -
m.rate*((1-expl.rate[1])*N.mat[year,1])
#right-most cell
N.mat[year+1,ncells] <- N.mat[year,ncells] + r*N.mat[year,ncells]*(1-N.mat[year,ncells]/K) -
expl.rate[ncells]*N.mat[year,ncells] +
m.rate*((1-expl.rate[ncells-1])*N.mat[year,ncells-1]) -
m.rate*((1-expl.rate[ncells])*N.mat[year,ncells])
#all the cells in between
for (i in 2:(ncells-1)) {
N.mat[year+1,i] <- N.mat[year,i] + r*N.mat[year,i]*(1-N.mat[year,i]/K) -
expl.rate[i]*N.mat[year,i] +
m.rate*((1-expl.rate[i-1])*N.mat[year,i-1]+(1-expl.rate[i+1])*N.mat[year,i+1]) -
2*m.rate*((1-expl.rate[i])*N.mat[year,i])
#print(N.mat[year+1,i])
#now calculate final year yield
yield[ntime.steps] <- yield[ntime.steps]+expl.rate[i]*N.mat[ntime.steps,i]
return(list(N.mat=N.mat, yield=yield))
###compare yield by harvest rates for 5 closed cells out of 51
K.start <- 1000
ncells <- 51
ntime.steps <- 1000
urate <- seq(0,0.2,0.01)
nharvest <- length(urate)
final.yield <- vector(length=nharvest)
for (i in 1:nharvest) {
x <- one.d.logistic(ncells=ncells,
expl.rate=c(rep(urate[i],(ncells-1)/2-2),0,0,0,0,0,rep(urate[i],(ncells-1)/2-2)), ntime.steps=ntime.steps,
K=K.start, r=0.2, m.rate=0.2,
start.N.vec=rep(K.start,ncells))
final.yield[i] <- x$yield[ntime.steps]
final.yield.noMPA <- vector(length=nharvest)
xnoMPA <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate[i],ncells), ntime.steps=ntime.steps,
final.yield.noMPA[i] <- xnoMPA$yield[ntime.steps]
plot(x=urate,y=final.yield, xaxs="i",yaxs="i",type="l",lwd=2,col="blue",las=1,
xlab="Harvest rate", ylab="Equilibrium yield", ylim=c(0,2600),cex.lab=1.3)
par(new=T)
plot(x=urate,y=final.yield.noMPA, xaxs="i",yaxs="i",type="l",lwd=2,col="green",las=1,
xlab="", ylab="",ylim=c(0,2600),axes=F)
###compare yield by harvest rates for 25 closed cells out of 51
urate <- seq(0,0.9,0.01)
expl.rate=c(rep(urate[i],13),rep(0,25),rep(urate[i],13)), ntime.steps=ntime.steps,
###Three scenarios that produce identical yield of around 1377 tons
###1. closed 25 areas, at u = 0.1
###2. all areas open at u = 0.032
###3. all areas open at u = 0.168
urate <- 0.1
expl.rate=c(rep(urate,13),rep(0,25),rep(urate,13)), ntime.steps=ntime.steps,
print(x$yield[ntime.steps])
urate<-0.032
xnoMPA1 <- one.d.logistic(ncells=ncells,
expl.rate=rep(urate,ncells), ntime.steps=ntime.steps,
print(xnoMPA1$yield[ntime.steps])
urate<-0.168
xnoMPA2 <- one.d.logistic(ncells=ncells,
print(xnoMPA2$yield[ntime.steps])
plot(x=1:ncells,y=x$N.mat[ntime.steps,], ylim=c(0,1050), xlab="Cell number",cex.lab=1.3,ylab="Abundance",type="l",lwd=2,col="blue", las=1,xaxs="i",yaxs="i")
plot(x=1:ncells,y=xnoMPA1$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
plot(x=1:ncells,y=xnoMPA2$N.mat[ntime.steps,], ylim=c(0,1050), xlab="",cex.lab=1.3,ylab="",type="l",lwd=2,col="green", las=1,xaxs="i",yaxs="i",axes=F)
polygon(x=c(14,14,38,38),y=c(0,1050,1050,0),col="# ")
15 MPA yield.r

38. Lessons Closing areas reduce the maximum yield
The more areas closed, the lower the maximum yield
Closed areas provide insurance against high fishing pressure (bad management, lack of enforcement)
For every level of yield with an MPA there is an equivalent yield without an MPA which has:
higher biomass outside the MPA
lower biomass inside the MPA
lower harvest rate where fishing occurs
higher CPUE and hence greater profits
no completely protected areas for biodiversity

39. Tradeoffs of fishing Percent of maximum New target Old target
Getting to the new target may involve a lot of lost yield in the short term.
Exploitation rate
Worm B et al. (2009) Rebuilding global fisheries. Science 325: