Reproductive value (RV)
identifies age class that contributes most to the population
growth
measures relative reproductive potential of an individual of a
given age
when population increases then early offspring contribute more
to vx than older ones
v1 .. left eigenvector of the dominant eigenvalue
of transposed A
- v1 is proportional to the reproductive values
scaled to the first category
rx
x
o
x
rx
xx
x
el
eml
v −
−
∑
=
vx
age
1
0
111 vAv λ=′
∑ =
= S
i
RV
1 1
1
v
v
Sensitivity (s)
identifies which process (p, F, G) has largest effect on the
population increase (λ1)
- examines change in λ1 given small change in processes (aij)
- sensitivity is larger for survival of early, and for fertility of older
classes
Elasticity (E)
weighted measure of sensitivity
- measures relative contribution to the population increase
- impossible transitions = 0
v.w
ijij
ij
wv
s
′
=
ij
ij
ij s
a
E
1λ
=
← sum of pairwise products
to adopt means for population promotion or control
Conservation/control procedure
1. Construction of a life table
2. Estimation of the intrinsic rates
3. Sensitivity analysis - helps to decide where
conservation/control efforts should be focused
4. Development and application of management plan
5. Prediction of future
A mouse species has spread dramatically. You perform a lifehistory
study and find that it breeds continuously. So you
distinguish age classes based upon 3-months intervals. You obtain
the following data:
Compare fecundity (m) and RV for each age.
Estimate r.
Predict how the population size would change in another 10
years using initial population structure (30, 10, 5).
x lx mx
0 1 0
1 0.8 5
2 0.5 12
3 0.3 4
(0.5+0.8)/(1+0.8)
(0.5+0.3)/(0.8+0.5)
2*(5+0.72*12)/4
2*(12+0.72*4)/4
2*(4+0.72*0)/4
A<-matrix(c(6.8,7.2,2,
0.72,0,0,
0,0.62,0),nrow=3,byrow=T);A
L<-eigen(t(A));L
v<-Re(L$vectors[,1]);v
RV<-v/v[1];RV
x<-c(1,2,3)
mx<-c(5,12,4)
plot(x,mx,type="b",ylim=c(0,12))
lines(x,RV,lty=2)
L<-eigen(A); r<-log(max(Re(L$values))); r
N0<-c(30,10,5)
N1<-A%*%N0;N1
years<-10
Nt<-matrix(0,nrow=nrow(A),ncol=years+1)
Nt[,1]<-N0
for(i in 1:years) Nt[,i+1]<-A%*%Nt[,i]
matplot(0:years,t(Nt),type="l")
legend(2,2e+10,c(1:3),lty=1:3,col=1:3)
There is a butterfly species that appears to be rare. You perform a
life-history study and gain data on survival and reproduction. You
also observe which factors determine stage-specific survival.
Create transition matrix, estimate λ, and find stable stage
distribution.
Perform sensitivity analysis and identify which factor has most
dramatic effect on population change. Suggest a conservation
plan.
stage lx mx mortality
egg 1 0 frost in winter
larva 1 0.7 0 parasitoids
larva 2 0.3 0 bird predation
pupa 0.25 0 habitat destruction
adult 0.02 80
0.7/1
0.3/0.7
0.25/0.3
0.02/0.25
A<-matrix(c(
0,0,0,0,80,
0.7,0,0,0,0,
0,0.43,0,0,0,
0,0,0.83,0,0,
0,0,0,0.08,0),nrow=5,byrow=T);A
L<-eigen(A);L
L1<-max(Re(L$values))
w<-Re(L$vectors[,3]);w
scd<-w/sum(w);scd
M<-eigen(t(A));M
M1<-max(Re(M$values))
v<-Re(M$vectors[,5]);v
s<-v%*%t(w)
ss<-s/as.numeric(v%*%w)
E<-ss*(A/L1);E
Stano Pekár„Populační ekologie živočichů“
dN
= Nr
dt
model is based on the assumption that development rate is a linear
function of temperature
valid for the region of moderate temperatures (15-25°)
at low temperatures organisms die due to coldness, and at high
temperatures organisms die due to overheating
D .. development time (days)
v .. rate of development = 1/D
tmin .. lower temperature limit
.. temperature at which
development rate = 0
0
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
0.045
0 10 20 30 40
Temperature
Developmentrate
ET.. effective temperature .. developmental temperature = t - tmin
S .. degree-days .. number of days required to complete development
.. do not depend on temperature = D*ET
tmin and S can be estimated from the regression line of v = a + bt
tmin :
S :
b
a
t −=min






+=−=
b
a
tDttDS )( min
btav
D
+
==
11
bta
b
at
S
+
+
=
b
S
1
=
0min =+ bta
accumulated degree-days (S) are equal to area under
temperature curve restricted to the interval between current
temperature and tmin
for temperatures between tmin and tmax (upper threshold)
several different non-linear models (Briere, Lactin, etc.)
allow to estimate tmin, tmax and topt (optimum temperature)
easy to interpret for experiments with constant temperature
instead of using average temperature, use actual temperature
because below and above ET model is non-linear
Briere et al. (1999)
v .. rate of development (=1/D)
t .. experimental temperature
tmin .. low temperature threshold
tmax .. upper temperature threshold
a .. constant
Optimum temperature:
parameters are estimated using non-linear regression
tttttav −×−××= maxmin )(
10
1691634 maxmin
2
min
2
maxminmax tttttt
topt
−+++
=
Lactin et al. (1995)
v .. rate of development
t .. experimental temperature
tm, ∆, ρ, λ .. constants
tmax and tmin can be estimated from the formula:
topt can be estimated from the first derivative:
λ
ρ
ρ
+−= ∆
−
− )(
tt
t
t
m
m
eev
λ
ρ
ρ
+−= ∆
−
− )(
0
tt
t
t
m
m
ee
)(1
0 ∆
−
−
×
×





∆
+−=
TT
T
T
m
m
ee
ρ
ρ
ρρ
In the laboratory the development of Diprion pini was studied. Seven
temperatures were used. For each temperature the development time (D)
of the complete development were recorded:
Fit linear model to the data. Estimate the minimum development
temperature (tmin) and the degree-days (S).
Estimate on which day the development is be complete if you know
average day temperatures during two weeks: 17, 18, 21, 23, 24, 25,
23, 24, 21, 25, 22, 25, 26, 22 a 23 °C.
t (°C) D
5 -
10 49
15 22
20 16
25 12
30 9
t<-c(5,10,15,20,25,30)
D<-c(0,49,22,16,12,9)
v<-1/D
v[1]<-0
plot(t,v)
m<-lm(v~t)
m
abline(m)
-(-0.022336/0.004351)
1/0.004351
tem<-c(17,18,21,23,24,25,23,24,21,25,22,25,26,22,23)
ET<-tem-5.13
plot(cumsum(ET),type="s")
abline(229,0)
Effect of temperature on the development of Nephus includens was
studied in the laboratory using a range of temperatures.
Use Lactin‘s model
Estimate minimal, maximal and optimal temperature.
t D
18 23.5
20 18.5
22 13
25 7.3
28 5.5
30 5
32 10.9
t<-c(18,20,22,25,28,30,32)
D<-c(23.5,18.5,13,7.3,5.5,5,10.9)
v<-1/D
plot(t,v)
m1<-nls(v~exp(rho*t)-exp(rho*Tm-(Tm-t)/delta)+lambda,
start=c(rho=0,Tm=30,delta=1,lambda=0))
summary(m1)
x<-seq(15,40,0.1)
plot(t,v,xlim=c(10,35),ylim=c(0,0.25))
lines(x,predict(m1,list(t=x)))
library(rootSolve)
tminmax<-uniroot.all(function(x) exp(0.01*x)-exp(0.01*33.7-(33.7x)/0.7)-1.19,lower=0,upper=40);
tminmax
topt<-uniroot.all(function(x) 0.01*exp(0.01*x)(0.01+1/0.7)*exp(0.01*33.7-(33.7-x)/0.7),lower=0,upper=40);
topt