Preliminary life history table. Actually, it's a two-stage matrix model using a variable stage duration following Caswell (2000). This is "the future" for demographic analysis of pelagic elasmobranchs on a global scale. That is, a 10x10 matrix can handle a species in 5 oceans with different mortality and fertility in each region in an open system (i.e. immigration and emigration can be included).
Here, I'm presenting a two-stage matrix model using juvenile and adult stage assuming seasonal parturition (birth-pulse) in combination with postbreeding census. This produces a 2x2 matrix. If exchange between 5 different populations were included, then the resulting matrix would be 10x10. If males were added for all 5 populations, we would have to deal with a manageable 20x20 matrix.
Age at maturity 18 years (Pauly 1978), duration of first stage T(1) =
17 years;
Longevity about 40 years (Pauly 1978), duration of second stage T(2) =
23 years;
Mortality, assume M = 0.10235 (S = 0.9027) based on longevity of
40 years and using Hoenig (1983) for cetaceans;
Fecundity, assume litter of 6 every third year, i.e. effective
annual females fecundity is 6/(2x3) =1
(Based on best current models for white shark and shortfin mako (Mollet
et al., 2000.)
(A litter of 6 is expected to weigh around
200 kg, which would be about 10% of the mass of the mother and falls into
the expected range of 10-15%),
( A model using pregnant and resting stages in combination with actual
fecundity would be better and produce a higher lambda, but here I wanted
to present the smallest matrix possible. The difference is substantial
for a species with a low age-at-maturity).
Sigma (i) = P (survival of an indiviudal in stage i)
Gamma (i) = P (growth from i to i+1 | survival) = growth probability for
surviving individuals
G(i) = sigma (i) gamma (i)
P(i) = sigma (i) (1-gamma (i) )
I use a variable stage duration for the juveniles with variance V(1) =
2 yr2 and a fixed stage distribution for the adults. I can
calculate gamma (1) from equation 6.l14 on p. 164 in Caswell (2000) and
gamma (2) from equation 6.103 on p. 161 and then calculate G(i) and P(i).
It's an iterative process. Assume lambda = 1, calculate G(i)'s and P(i)'s,
solve 2x2 matrix for lambda, continue until assumed lambda agrees with
calculated lamdda.
In this case, the elements of the projection matrix A are :
P(1) = 0.8837; P(2)
= 0.8952; G(1) = 0.01905;
Fertility F(2) = m(2) x P(2) = 1.0000 x 0.8952 = 0.8952. Coincidence that
F(2) = P(2) because m(2) = 6/(2x3) = 1.0
Then we decompose the A-matrix into a transition matrix T and fertility matrix F (A = T + F).
|
A
=
|
0.8837
|
0.8952
|
T
=
|
0.8837
|
0
|
F
=
|
0
|
0.8952
|
||
|
0.01905
|
0.8952
|
0.01905
|
0.8952
|
0
|
0
|
Lambda 1 = 1.0202, growth rate of stable population (r = ln (lambda 1) = 0.0199 yr-1)
(lambda 2 = 0.7584, will determine how fast stable populaton is reached)
Age structure (right eigenvector for lamda 1 of A): w1 (juveniles)= 86.8%, w2 (adults)= 13.2% ;
Reproductive values (left eigenvector for lamda 1 of A): v1 (juveniles at age 0) = 1.00, v2 (adults at age 18 yr, i.e at beginning of stage) = 7.17. (Note that the % values of individuals in an age-structured Leslie matrix or life history table can be added to obtain a approximate % value of indiviuals in a stage, but that the reproductive values are not additive.)
elasticity matrix (relative change of lamda with respect to relative change of matrix elements
i.e. d ln( lamda)/d ln (aij):
| S = | 0.4799 | 0.0729 | E= | 0.4140 | 0.0639 | ||
| 3.424 | 0.5221 | 0.0639 | 0.4581 |
The elasticity (0.0639) of the fertility matrix element (F2 = 0.8052)
is much smaller than the elasticities (0.414 and 0.458) of the in-stage
survival matrix elements (P1 = 0.8837 and P2 = 0.8952, respectively).
Changing fecundity (m2) and thus fertility (F2) by, say a factor of
2, has a much smaller effect on lamda than changing P by the same amount.
For example:
a) Doubling fecundity (size of litter) from 6 to 12 will produce
lambda 1 = 1.0544;
b) Using M = 0.10235/2 = 0.05118yr (Survival S = 0.9501 instead
of 0.9027, an increase of only 5% will produce lambda 1 = 1.0737.
(Pauly 2001, 1997 Sabah Proceedings in press, reported M
= 0.07 yr-1 for adults).
Solution using life cycle graphs: Here I selected the width of the arrows to be proportional to the elasticity of the matrix elements and the size ot the nodes to be proportional to a) the age structure (W-vector), b) the reproductive values (V-vector). Note that the width difference of the arrows cannot be seen on a computer screen unless the differences are large.
a) 
b)
Note that the green arc on node 1 (representing P1) should be situated below node 1, similar to the arc drawn for P2. As drawn which it might imply that it represents a F1 (reproduction in juventile stage).
|
N
=
|
8.60
|
0
|
|
1.56
|
9.55
|
The matrix elements give the mean time spent in each stage. Thus a basking shark spends, on average 8.60 years as an immature and 1.56 years as a reproductive adult. A mature adult, in contrast, spends an average of 9-10 years in that stage. The sum of the columns give the mean time to death (= life expectancy; 10.2 for juveniles, 9.6 yr for the adults).
|
R
=
|
1.4078
|
8.6210
|
|
0
|
0
|
A juvenile basking shark can expected to produce 1.41 female offspring during its life. The generation time T is ln Ro/ln lambda = 16.82. Mu1 can be calculated from the matrix (F N2) and is 18.14 yr. I have not yet figured out how to calculate Abar using matrix notation. However, since T ~ (Abar + mu1)/2, Abar is ~ 15.50 yr.
Following Coale, A. J. (1972). The growth and structure of human populations.
Princeton University Press, Princeton NJ, 227p:
The generation time T is defined by e(rT) = Ro (net reproductive
rate). It is the time T required for the population to increase by a
factor of Ro. The concept is the same when one calculates
a doubling time, the time T1/2 required for the population
to increase by a factor of 2 or a half-life T1/2 in the case of a decreasing
(=decaying) population. However, T is not, somewhat surprisingly,
equal to any of the several mean ages of childbearing that can be defined..
There are three mean ages of fertility of interest in this context:
1) the mean age (m bar) of the fertility schedule (mean age of childbearing
in a cohort subject to no mortality),
2) the mean age (mu1) of the net fertility schedule (mean age of childbearing
in a cohort),
3) the mean age (A bar) of childbearing in the stable populaton.
(See Coale (1972) or Caswell (2000) for equations).
What is missing? I should provide a confidence interval for lambda, the growth rate of the stable population. In this particular case and most other demographic analyses for sharks, only a Monte Carlo uncertainty analysis is possible because available demographic data is based on estimates of fertility and little data on survival is available. We cannot use a series approximation (sometime called the "delta method") because the variance of the parameters is not known. We cannot use a bootstrap method because the resampling procedure requires the history of individuals to be known.
See Caswell (2000) and Coale and Demeny (1966) how survivorship can be estimated from other species using model life tables. The life tables of other species are re-scaled according to age-at-first reproduction. The lemon shark would be the first species to be included for the construction of a model life table. Perhaps life tables of marine mammals (e.g. whales) should be included. We cannot use a stage-based matrix model to do this, we need to use an age-based matrix model (= Leslie matrix).