Leslie Matrix for Barndoor Skate Dipturus laevis Mitchill 1817

Based on IUCN Red List Assessment document by N. K. Dulvy: "Criticically Endangered (A2bcd)".

Input parameters: Age-at-first-reproduction 9 y; longevity 15 y;
Mortality (M, y-1) = - ln(0.01)/15 = 0.3070, Survival(S) = exp(-M) = 0.7356;
Fertility (m, assuming 50% females) = 47 eggcases per year/2 = 23.5, F (= discounted fertility) = m Sa = 23.5 x 0.7356 = 17.2876 (using a post-breeding census).
Note 1:
As I use the same survival rate for all age classes, the projetion matrix A looks exactly the same for pre- and post-breeding census. I use colors to indicate how the construction of the projection matrix is completely different.
Note 2: The discounted fertility F depends on both fertility m and the survival rates of the aduts (in the case of a post-breeding census). This is most important when calculatig the E-patternbecause the discounted fertilities contribute to the elasticity of the adults (E3). In the E-matrix below I show this by using dual entries (when there is reallly only one) and color them differently to show how the additions for the E-pattern has to be done to get the correct E-pattern. If we use a Life History Table (LHT) instead of a Leslie matrix, the problem does not arise becasue the LHT only contains fertilities m, no discounted fertilities F = m Sa.

Age at-first-reproduction may have to increased and fertility may have to be decreased in analysis because:
"Egg cases of barndoor skates are the largest among species in the Northwest Atlantic and may require as much as two years before hatching. Because their egg cases are large, are deposited on the sea floor and take so long to hatch, they may also be vulnerable to impacts from fishing gear as well as mortality when they are caught in gear and discarded." http://www.mcbi.org/bdoor/skatest.html.

Life Cycle Graph:

 Projection Matrix (A) using pre-breeding census Age class J1 J2 J3 J4 J5 J6 J7 J8 A9 A10 A11 A12 A13 A14 A15 J1 0 0 0 0 0 0 0 0 17.2876 17.2876 17.2876 17.2876 17.2876 17.2876 17.2876 J2 0.7356 0 0 0 0 0 0 0 0 0 0 0 0 0 0 J3 0 0.7356 0 0 0 0 0 0 0 0 0 0 0 0 0 J4 0 0 0.7356 0 0 0 0 0 0 0 0 0 0 0 0 J5 0 0 0 0.7356 0 0 0 0 0 0 0 0 0 0 0 J6 0 0 0 0 0.7356 0 0 0 0 0 0 0 0 0 0 J7 0 0 0 0 0 0.7356 0 0 0 0 0 0 0 0 0 J8 0 0 0 0 0 0 0.7356 0 0 0 0 0 0 0 0 A09 0 0 0 0 0 0 0 0.7356 0 0 0 0 0 0 0 A10 0 0 0 0 0 0 0 0 0.7356 0 0 0 0 0 0 A11 0 0 0 0 0 0 0 0 0 0.7356 0 0 0 0 0 A12 0 0 0 0 0 0 0 0 0 0 0.7356 0 0 0 0 A13 0 0 0 0 0 0 0 0 0 0 0 0.7356 0 0 0 A14 0 0 0 0 0 0 0 0 0 0 0 0 0.7356 0 0 A15 0 0 0 0 0 0 0 0 0 0 0 0 0 0.7356 0 0.7356* Eigenvalues Eigenvectors (R&L) * Not part of Projection matrix, need to calculate F15 Real Imaginary Age struct Reprod val Abs. value rho Add age structure and reproductive values setting value of first age class to 1.000? Note that age structure and reproductive values are the solutions which go with lambda 1. 1.1624 0.0000 0.3675 0.0035 1.1624 0.9058 0.6401 0.2326 0.0056 1.1092 1.048 Approximate recovery time is ln(10)/rho = 49 yr 0.9058 -0.6401 0.1472 0.0089 1.1092 1.048 0.4558 0.5728 0.0932 0.0140 0.7321 0.4558 -0.5728 0.0590 0.0221 0.7321 0.2584 -1.0137 0.0373 0.0350 1.0461 0.2584 1.0137 0.0236 0.0552 1.0461 -0.1574 0.7151 0.0149 0.0873 0.7323 -0.1574 -0.7151 0.0095 0.1379 0.7323 -0.4681 -0.8873 0.0060 0.1345 1.0032 -0.4681 0.8873 0.0038 0.1291 1.0032 -0.6660 0.3284 0.0024 0.1207 0.7426 -0.6660 -0.3284 0.0015 0.1073 0.7426 -0.9098 -0.3373 0.0010 0.0862 0.9703 -0.9098 0.3373 0.0006 0.0528 0.9703 r 0.1505 (rate of increase, y-1) Adding F = 0.15 y-1 to M = 0.31 y-1 will produce stationary population (r = 0.0, lambda = 1.0). F = 0.15 y-1 can be related to catch if we know/assume a mean N for the virgin population in the mid-1960's. Then one could determine if observed or estimated catch was larger than F = 0.15 y-1, which would produce a declining population as implied by 96% decline in catch rate since the mid-1960's. Ro 4.9549 (expected number of replacements) T 10.63 (generation time - time for increase of Ro, yr) mu1 10.86 (mean age of parents of offspring of a cohort, yr) Abar 10.43 (mean age of parents of population at the stable age distribution, yr) N (fundamental matrix) 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.7356 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0.5412 0.7356 1 0 0 0 0 0 0 0 0 0 0 0 0 0.3981 0.5412 0.7356 1 0 0 0 0 0 0 0 0 0 0 0 0.2929 0.3981 0.5412 0.7356 1 0 0 0 0 0 0 0 0 0 0 0.2154 0.2929 0.3981 0.5412 0.7356 1 0 0 0 0 0 0 0 0 0 0.1585 0.2154 0.2929 0.3981 0.5412 0.7356 1 0 0 0 0 0 0 0 0 0.1166 0.1585 0.2154 0.2929 0.3981 0.5412 0.7356 1 0 0 0 0 0 0 0 0.0858 0.1166 0.1585 0.2154 0.2929 0.3981 0.5412 0.7356 1 0 0 0 0 0 0 0.0631 0.0858 0.1166 0.1585 0.2154 0.2929 0.3981 0.5412 0.7356 1 0 0 0 0 0 0.0464 0.0631 0.0858 0.1166 0.1585 0.2154 0.2929 0.3981 0.5412 0.7356 1 0 0 0 0 0.0341 0.0464 0.0631 0.0858 0.1166 0.1585 0.2154 0.2929 0.3981 0.5412 0.7356 1 0 0 0 0.0251 0.0341 0.0464 0.0631 0.0858 0.1166 0.1585 0.2154 0.2929 0.3981 0.5412 0.7356 1 0 0 0.0185 0.0251 0.0341 0.0464 0.0631 0.0858 0.1166 0.1585 0.2154 0.2929 0.3981 0.5412 0.7356 1 0 0.0136 0.0185 0.0251 0.0341 0.0464 0.0631 0.0858 0.1166 0.1585 0.2154 0.2929 0.3981 0.5412 0.7356 1 Sum 3.7449 3.7313 3.7129 3.6877 3.6536 3.6072 3.5441 3.4583 3.3417 3.1832 2.9678 2.6749 2.2768 1.7356 1.0000 R (expected lifetime production) 4.95 6.74 9.16 12.45 16.92 23.00 31.26 42.50 57.77 55.03 51.31 46.24 39.36 30.01 17.29 Deleted 14 rows with 0 entries Elasticity Matrix J8 A9 A15 0 0 0 0 0 0 0 0 0.0367 0.0232 0.0147 0.0093 0.0059 0.0037 0.0024 0.0367 0.0232 0.0147 0.0093 0.0059 0.0037 0.0024 0.0959 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0959 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0959 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0959 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0959 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0959 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0959 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0959 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0592 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0360 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0213 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0120 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0061 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0.0024 0 E(fertility) = E1 0.0959 Same as elasticity of each juvenile age class E1_norm 0.0875 E(juvenile survival) = E2 0.8631 ER2 = E2/E1 = 9.00 (alpha) E1_norm 0.7876 E(adult survival) = E3 0.1369 ER3 = E3/E1 = 1.43 [(1/E1) - alpha] E3_norm 0.1249 Sum = E1 + E2 + E3 1.0959 In pre- and post-breeding cycle the sum is 1 + E1 = 1.0959 Sum 1.000

Created February April 2014, revised April 2014. Back to previous page.
Please send comments or corrections to henry@elasmollet.org