### Preliminary conclusion: We need better mortality and fertility data and a better estimate of age-at-first-maturity. Note: The population growth rate reported below is a purely analytical projections assuming that the environment is constant and that density effects are unimportant (Caswell 2001).

 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 Leslie matrix: 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 Leslie matrix: 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.

Created March 2001, revised November 2002, modified slightly April 2014. Back to previous page.