Life Tables
Life History Tables
•
•
Time (x) = time interval used for separating age categories
•
nx = number alive at the beginning at age x
•
Lx = proportion of individuals alive at age x
Dynamic (horizontal) cohort= cohort followed through time until all
members have died
Freq.
•
Static (vertical or current) = one census period (day, season, etc.);
only equivalent to dynamic if population does not change age
distribution; assumes constant survivorship.
20
40
60
80
100
Life History Tables
• dx = proportion of original population dying during the age interval x to x+1
•
Age (x)
0
1
2
3
4
5
6
70
60
50
40
30
20
10
0
qx = proportion of existing population dying during age interval x to x+1;
qx = dx/lx
nx
lx
dx
qx
200
1.000
0.100
0.100
180
0.900
0.025
0.028
175
0.875
0.275
0.314
120
0.600
0.350
0.583
50
0.250
0.235
0.940
3
0.015
0
0.000
lx
1.00
180
0.90
175
0.88
120
0.60
50
0.25
3
0.02
0
0.00
Life expectancy
•
•
•
Age (x)
0
1
2
3
4
5
6
nx
200
Ex = Tx / lx
Tx = average life expectancy from current time: e.g. how much living
will be done by cohort from beginning of period x:
Tx=Σ(Lx); summed from x to last x
Age (x)
0
1
2
3
4
5
6
nx
lx
dx
qx
200
1.000
0.100
0.100
Lx
0.950
Tx
3.130
ex
3.130
180
175
0.900
0.875
0.025
0.275
0.028
0.314
0.888
0.738
2.180
1.293
2.422
1.477
120
0.600
0.350
0.583
0.425
0.555
0.925
50
0.250
0.240
0.960
0.130
0.130
0.520
2
0.010
0
0.000
1
Natality
Reproductive Rate
•fx = total natality; number of
fertilized eggs produced in a given
year by individuals surviving to age
x
•
•
•
•mx = age specific natality; average
number of fertilized eggs produced
per individual surviving to beginning
of age x
R0 = rate of change in the population. If below 1.0, population is
shrinking
R0=∑
∑(lx m x)
Sum of the number of fertilized eggs produced per original individual
during each age
Age (x)
0
1
2
3
4
5
6
•Lotka proved that any pair of
unchanging lx and mx values will
eventually give rise to a population
with a stable age distribution
nx
lx
dx
qx
mx
fx
lxmx
200
180
1.000
0.900
0.100
0.025
0.100
0.028
Lx
0.950
0.888
Tx
3.130
2.180
ex
3.130
2.422
2
360.00
1.80
175
0.875
0.275
0.314
0.738
1.293
1.477
3
525.00
120
0.600
0.350
0.583
0.425
0.555
0.925
4
480.00
2.40
50
2
0.250
0.010
0.240
0.960
0.130
0.130
0.520
5
250.00
1.25
0
0.000
2.63
R0 = 8.05
time
time
age
fec und
0
1
2
3
sum
pct 0
pct 1
pct 2
pct 3
0
5
15
0
Survive
1
0.3
0.25
0.2
1
50
0
0
0
2
0
15
0
0
3
75
0
4
0
4
56
23
0
1
5
113
17
6
0
6
169
34
4
1
age
7
232
51
8
1
8
380
70
13
2
9
538
114
17
3
10
831
161
28
3
11
1,234
249
40
6
fec und
0
1
2
3
12
1,851
370
62
8
50
15
79
80
135
208
292
464
672
1,024
1,529
2,292
100
0
95
71
83
81
79
82
80
81
81
81
0
100
0
28
13
16
17
15
17
16
16
16
0
0
5
0
4
2
3
3
3
3
3
3
0
0
5.25
0
1.01
1
1.70
0
1.54
1
1.40
0
1.59
0
1.45
0
1.52
0
1.49
0
1.50
0
0.00
0
5
15
0
Survive
1
0.3
0.25
0.2
1
50
0
0
0
sum
pct 0
pct 1
pct 2
pct 3
2
0
15
0
0
50
3
75
0
4
0
4
56
23
0
1
5
113
17
6
0
6
169
34
4
1
7
232
51
8
1
8
380
70
13
2
9
538
114
17
3
10
831
161
28
3
11
1,234
249
40
6
12
1,851
370
62
8
15
79
80
135
208
292
464
672
1,024
1,529
2,292
100
0
95
71
83
81
79
82
80
81
81
81
0
100
0
28
13
16
17
15
17
16
16
16
0
0
5
0
4
2
3
3
3
3
3
3
0
0
5.25
0
1.01
1
1.70
0
1.54
1
1.40
0
1.59
0
1.45
0
1.52
0
1.49
0
1.50
0
0.00
R=1.50
time
age
fec und
0
1
2
3
sum
pct 0
pct 1
pct 2
pct 3
1
4
15
0
Survive
1
0.3
0.25
0.2
1
50
0
0
0
2
50
15
0
0
3
110
15
4
0
4
226
33
4
1
5
415
68
8
1
6
810
124
17
2
7
1,562
243
31
3
8
3,000
469
61
6
9
5,785
900
117
12
10
11,141
1,735
225
23
11
21,457
3,342
434
45
12
41,335
6,437
836
87
50
65
129
264
491
953
1,839
3,535
6,814
13,125
25,279
48,694
100
77
85
86
84
85
85
85
85
85
85
85
0
23
12
13
14
13
13
13
13
13
13
13
0
0
3
1
2
2
2
2
2
2
2
2
0
0
1.98
0
2.05
0
1.86
0
1.94
0
1.93
0
1.92
0
1.93
0
1.93
0
1.93
0
1.93
0
0.00
R=1.83
2
Reproductive Rates
Defining R
R0 is a multiplier allowing us to determine population size at
future generation.
If discrete generations, time in generations then t=T and
R0 = net reproductive rate; Net number of offspring
produced per individual. Also, a multiplier allowing us to
determine population size in the next generation. For
simplicity, assume discrete generations.
N t +1 = N t * R0
R0=∑
∑(lxmx)
120000
General formula for population size at time t is:
100000
N t = N 0 * R0 t
6
7
8
0
0
0
9
Ro
0
kx
0.04
0.01
0.16
0.37
1.22
Mx
0
2
3
4
5
0
0
0
0
0
fx
360.00
525.00
480.00
250.00
lxmx xlxmx
0.00
1.80 1.80
2.63 5.25
2.40 7.20
1.25 5.00
0.00
0.00
0.00
0.00
0.00
-rx
0.00
-0.88
-1.75
-2.63
-3.50
-4.38
-5.26
-6.13
-7.01
-7.89
e-rx
0.42
0.17
0.07
0.03
0.01
0.01
0.00
0.00
0.00
-rx
lxmxe
0.00
0.75
0.46
0.17
0.04
0.00
0.00
0.00
0.00
0.00
Lx
0.95
0.89
0.74
0.43
0.13
Tx
3.13
2.18
1.29
0.56
0.13
0.00
ex
3.13
2.42
1.48
0.93
0.52
N
175
120
50
2
qx Log10 ax Log10lx
0.10 2.30 0.00
0.03 2.26 -0.05
0.31 2.24 -0.06
0.58 2.08 -0.22
0.96 1.70 -0.60
0.30 -2.00
40000
200
180
2
3
4
5
dx
0.10
0.03
0.28
0.35
0.24
20000
0
1
lx
1.00
0.90
0.88
0.60
0.25
0.01
0
ax
∑lxmxe-rx=1
∑lxmx
8.075
∑xlxmx 19.250
0
40
stage
ax
0
1
200
180
2
3
175
120
4
5
50
2
6
7
0
0
8
9
0
0
lx
1.00
0.90
0.88
0.60
0.25
0.01
∑lxmx
8.075
∑xlxmx 19.250
Tc
2.384
r
0.876
Double 0.787
dx
0.10
0.03
0.28
0.35
0.24
qx Log 10ax
0.10 2.30
0.03 2.26
0.31 2.24
0.58 2.08
0.96 1.70
0.30
Log1 0lx
0.00
-0.05
-0.06
-0.22
-0.60
-2.00
60
80
100
Year
Exponential Growth
number of individuals
r = intrinsic rate of increase; per capita rate of increase; also
Malthusian Parameter. When r is >0 populations will increase,
when it is <0 populations will decrease; a population that is not
increasing or decreasing will have an r of 0. Units – number of new
individuals per unit time.
Ro
20
1.415
Reproductive Rates
•
80000
Predicts exponential growth
stage
60000
R – rate of population increase as a function of time.
Unitless and dimensionless.
10000
8000
6000
4000
2000
0
0
2
4
6
r = .1
8 10 12 14 16
time
r = .2
r = .3
3
Determining r…
Cohort Generation Time
Tc is the average birth-to-birth time for a
generation.
If generation time (T) is known, can determine r from
Lotka’s Equation:
1=
∑e
− rx
l x mx
∑ xl m
x
Tc =
using iteration
This is difficult, and the equation is not biologically
meaningful…
Recall that R0 =
r can be estimated by:
ln( R0 )
r=
Tc
Thus
Tc =
where Tc is generation time.
R0
∑lm
x
∑ xl m
∑l m
x
x
•
Tc = (19.25/8.075) = 2.3839
ax
200
1
2
3
180
175
120
4
5
50
2
6
7
8
0
0
0
9
Ro
lx
1.00
0.90
0.88
0.60
0.25
0.01
0
∑lxmx
8.075
∑xlxmx 19.250
Tc
2.384
r
0.876
Double 0.787
dx
0.10
0.03
0.28
0.35
0.24
qx Log 10 a x Log 10lx
0.10 2.30 0.00
0.03 2.26 -0.05
0.31 2.24 -0.06
0.58 2.08 -0.22
0.96 1.70 -0.60
0.30 -2.00
kx
0.04
0.01
0.16
0.37
1.22
Mx
0
2
3
4
5
0
0
0
0
0
fx
360.00
525.00
480.00
250.00
ax
0
1
200
180
2
3
175
120
4
5
50
2
6
7
0
0
8
9
0
0
x
x
Ro
x
lx
1.00
0.90
0.88
0.60
0.25
0.01
dx
0.10
0.03
0.28
0.35
0.24
qx Log 10ax
0.10 2.30
0.03 2.26
0.31 2.24
0.58 2.08
0.96 1.70
0.30
Log1 0lx
0.00
-0.05
-0.06
-0.22
-0.60
-2.00
∑lxmx
8.075
∑xlxmx 19.250
Tc
2.384
r
0.876
Double 0.787
1=
True value of r difficult to get:
r = ln(2.3839)/3.068 = 0.876
0
stage
How good is our estimate of r?
Lotka’s Solution to r, solve for Tc first, then r
stage
x
lxmx
xlxmx
0.00
1.80 1.80
2.63 5.25
2.40 7.20
1.25 5.00
0.00
0.00
0.00
0.00
0.00
-rx
0.00
-0.88
-1.75
-2.63
-3.50
-4.38
-5.26
-6.13
-7.01
-7.89
e-rx
0.42
0.17
0.07
0.03
0.01
0.01
0.00
0.00
0.00
lxmxe-rx
0.00
0.75
0.46
0.17
0.04
0.00
0.00
0.00
0.00
0.00
∑lxmxe-rx=1
Lx
0.95
0.89
0.74
0.43
0.13
stage
ax
0
200
1
2
3
180
175
120
4
5
50
2
6
7
8
0
0
0
9
Ro
1.415
lx
1.00
0.90
0.88
0.60
0.25
0.01
0
dx
0.10
0.03
0.28
0.35
0.24
qx Log 10 a x Log 10lx
0.10 2.30 0.00
0.03 2.26 -0.05
0.31 2.24 -0.06
0.58 2.08 -0.22
0.96 1.70 -0.60
0.30 -2.00
kx
0.04
0.01
0.16
0.37
1.22
Mx
0
2
3
4
5
0
0
0
0
0
fx
360.00
525.00
480.00
250.00
∑e
lxmx
xlxmx
0.00
1.80 1.80
2.63 5.25
2.40 7.20
1.25 5.00
0.00
0.00
0.00
0.00
0.00
∑lxmx
8.075
∑xlxmx 19.250
Tc
2.384
r
0.876
Double 0.787
•
Value > 1 means estimate of r is too small
•
Value < 1 means estimate is too large
•
In this case, actual r slightly greater than 0.876
− rx
-rx
0.00
-0.88
-1.75
-2.63
-3.50
-4.38
-5.26
-6.13
-7.01
-7.89
l x mx
e-rx
0.42
0.17
0.07
0.03
0.01
0.01
0.00
0.00
0.00
lxmxe-rx
0.00
0.75
0.46
0.17
0.04
0.00
0.00
0.00
0.00
0.00
Lx
0.95
0.89
0.74
0.43
0.13
∑lxmxe-rx=1
1.415
4
How do we use this?
•
N t + 1 = N t * R0
N1 = R * N 0
dN
= rN
dt
Nt = Rt * N0
However, r is much more useful as a continuous measure of
population growth.
Differential equation of logistic growth. Can only
tell you the rate (dN/dt) of growth, can’t project
population size.
R = er
N t = N 0 e rt
∑lm
x
x
By extension…
Assuming infinitely small time steps, switch
from the discrete to continuous form.
Integrate the first equation, and it is written in a
form where you can project population size
given a time period (t).
r = ln( R)
R0 =
R is the finite rate of population increase. It
represents the proportional change in a
population from t to t+1. Is always positive.
It is a ratio, dimensionless and with no units.
R0 is the net reproductive rate, corrected for
mortality. The units of R0 is number of offspring
per individual per lifespan.
If R0>1.0 population is increasing.
1=
∑e
r is thus the natural log of R
− rx
l x mx
True value of r, difficult to solve.
The estimate from R0 is usually
close enough.
R – rate of population increase as a function of absolute time.
Unitless, dimensionless.
R0 – rate of population increase as a function of generation time.
Thus R=R0 only if T=1
Estimating r:
ln( R0 )
r=
T
However, this is only
An estimation of r.
5
More about r
Human Population Growth
How long does it take a population to double in size??
N t = N 0 e rt
Nt=2 and No=1 t=?
Doubling Time for Population
Nt
= e rt
N0
70
60
time
50
ln 2=rt
40
ye a r
1970
1991
2000
30
20
r
0.02
0.018
0.0125
doubling tim e
35
39
55
10
t=0.69/r
0
0.01
0.02
0.03
0.04
intrinsic rate of increase
0.05
Given current growth rates, what will the world population be in 30 years??
Nt=N0ert
Nt=6,426,101,450 e0.0125(30)
9,349,922,439
Variability in r
An example
stage
0.025
0
1
2
3
4
5
0.02
r
0.015
0.01
0.005
45
35
25
40
20
20
20
20
15
30
20
20
20
00
10
20
20
95
90
85
75
05
20
20
19
19
19
70
65
60
55
80
19
19
19
19
19
19
19
50
0
•The maximum rate of increase obtainable by a population is rmax
ax
1500
300
200
100
50
10
Mx
0
2
3
6
2
1
•
Given these data answer the following
– What proportion of individuals survive to age
2? (l2)
– What proportion of 2 year olds die before
reaching age 3? (q2)
– What is the expected lifespan? (e0)
– How many total offspring are produced by the
third year class? (f3)
– Is the population growing or shrinking? (R0)
– What is the generation time? (T)
– How large will the population be in 243 years?
(r)
– How long before the population doubles? (r)
•Difference between rmax and r is due to environmental resistance
• rmax - quality of food, space, etc. are optimum, no competition or predation.
6
An example
•
An example
What proportion of individuals survive to age 2? (l2)
– lx= ax/a0 = 200/1500 =13.3%
ax
stage
0
1
2
3
4
5
1500
300
200
100
50
10
lx
1.00
0.20
0.13
0.07
0.03
0.01
stage
0
1
2
3
4
5
An example
• Lx= (ax+ax+1)/2
• ex=Sum(Lx)/ax
• e0=(900+150+150+75+30)/1500 = 1405/1500 = 0.94
0
1
2
3
4
5
ax
1500
300
200
100
50
10
lx
1.00
0.20
0.13
0.07
0.03
0.01
dx
0.80
0.07
0.07
0.03
0.03
qx
0.80
0.33
0.50
0.50
0.80
An example
– What is the expected lifespan? (e0)
stage
– What proportion of 2 year olds die before reaching age 3? (q2)
– qx= ax+1/ax = 100/200 =50%
ax
1500
300
200
100
50
10
lx
1.00
0.20
0.13
0.07
0.03
0.01
dx
0.80
0.07
0.07
0.03
0.03
qx
0.80
0.33
0.50
0.50
0.80
Lx
900.00
250.00
150.00
75.00
30.00
ex
0.94
1.68
1.28
1.05
0.60
– How many total offspring are produced by the third year class?
(f3)
– fx= mx*ax
– f3=m3*a3 = 6*100 = 600
stage
0
1
2
3
4
5
ax
1500
300
200
100
50
10
lx
1.00
0.20
0.13
0.07
0.03
0.01
dx
0.80
0.07
0.07
0.03
0.03
qx
0.80
0.33
0.50
0.50
0.80
Mx
0
2
3
6
2
1
fx
600.00
600.00
600.00
100.00
10.00
7
An example
An example
– Is the population growing or shrinking? (R0)
– R0=Sum(lxmx)
– R0=0+0.4+0.4+0.4+0.07+0.01 = 1.28
stage
0
1
2
3
4
5
ax
1500
300
200
100
50
10
lx
1.00
0.20
0.13
0.07
0.03
0.01
dx
0.80
0.07
0.07
0.03
0.03
qx
Log10a x Log10lx
0.80
3.18
0.00
0.33
2.48 -0.70
0.50
2.30 -0.88
0.50
2.00 -1.18
0.80
1.70 -1.48
1.00 -2.18
kx
0.70
0.17
0.30
0.29
0.66
– What is the generation time? (T)
– T=sum(xlxmx)/sum(l xmx)
– T=2.7/1.27 = 2.12
Mx
0
2
3
6
2
1
fx
600.00
600.00
600.00
100.00
10.00
lxmx
xlxmx
0.00
0.40
0.40
0.40
0.80
0.40
1.20
0.07
0.27
0.01
0.03
An example
–
–
–
–
–
–
–
stage
0
1
2
3
4
5
ax
0
1
2
3
4
5
ax
1500
300
200
100
50
10
lx
1.00
0.20
0.13
0.07
0.03
0.01
dx
0.80
0.07
0.07
0.03
0.03
qx
Log10a x Log10lx
0.80
3.18
0.00
0.33
2.48 -0.70
0.50
2.30 -0.88
0.50
2.00 -1.18
0.80
1.70 -1.48
1.00 -2.18
kx
0.70
0.17
0.30
0.29
0.66
Mx
0
2
3
6
2
1
fx
600.00
600.00
600.00
100.00
10.00
lxmx
xlxmx
0.00
0.40
0.40
0.40
0.80
0.40
1.20
0.07
0.27
0.01
0.03
An example
How large will the population be in 243 years? (r)
r=lnR0/T
r=ln(1.27)/2.12 = 0.114
Nt=N0ert
N243=1500 e114*243
N243 = 1500 e27.07
8.56 x 1014
1500
300
200
100
50
10
stage
lx
1.00
0.20
0.13
0.07
0.03
0.01
dx
0.80
0.07
0.07
0.03
0.03
qx
Log10a x Log10lx
0.80
3.18
0.00
0.33
2.48 -0.70
0.50
2.30 -0.88
0.50
2.00 -1.18
0.80
1.70 -1.48
1.00 -2.18
kx
0.70
0.17
0.30
0.29
0.66
Mx
0
2
3
6
2
1
– How long before the population doubles? (r)
– Double time = 0.69/r = 0.69/0.114 = 6.05 years
fx
600.00
600.00
600.00
100.00
10.00
lxmx
xlxmx
0.00
0.40
0.40
0.40
0.80
0.40
1.20
0.07
0.27
0.01
0.03
stage
0
1
2
3
4
5
ax
1500
300
200
100
50
10
lx
1.00
0.20
0.13
0.07
0.03
0.01
dx
0.80
0.07
0.07
0.03
0.03
qx
Log10a x Log10lx
0.80
3.18
0.00
0.33
2.48 -0.70
0.50
2.30 -0.88
0.50
2.00 -1.18
0.80
1.70 -1.48
1.00 -2.18
kx
0.70
0.17
0.30
0.29
0.66
Mx
0
2
3
6
2
1
fx
600.00
600.00
600.00
100.00
10.00
lxmx
xlxmx
0.00
0.40
0.40
0.40
0.80
0.40
1.20
0.07
0.27
0.01
0.03
8