|Age at maturity (alpha) 18 years (Pauly 1978);
Longevity (omega) about 40 years (Pauly 1978),;
Mortality, assume M = 0.1151 yr-1 (S = 0.8913) based on longevity
Fecundity, assume litter of 6 every third year, i.e. effective annual females fecundity of 6/(2x3) =1
(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 fertility (6/2 = 3) would be better and produce a higher population growth rate)
|Solution using a Life History Table (LHT) or the corresponding 40x40
Lambda = 1.0030, growth rate of stable population (r = ln (lambda) = 0.003 yr-1);
Net reproductive rate Ro = 1.076,
Generation "times": Abar = 24.35 yr (an age), T = ln(Ro)/r = 24.40 yr, mu1 = 24.44 yr (an age), Abar/alpha = 1.353.
(Abar/alpha = 1.35)
|Elasticities from LHT or the corresponding 40x40
E(fertility) = E(m) = E1 = 1/Abar = 0.04107
E(juvenile survival) = E(JS) = E2 = alpha *E1 = 0.7392;
E(adult survival) = E(AS) = E3 = 1 - E2 = 0.2608;
(E1 + E2 + E3 = 1.04107; normalized elasticities in % are E1 = 3.94%, E2 = 71.01%; E3 = 25.05% (check sum = 100%). E3/E2 = 0.353 (=Abar/alpha -1)
|Estimated elasticities from age-at-first reproduction alone!
Abar is assumed to be the same a mu1 = 24.44 yr. As lambda is close to 1.0, Abar is almost the same as mu1 and the E-pattern based on mu1, which can be estimaed withouht having to solve the Euler equation, is almost the same as the one given above.