Population Dynamics and the Leslie Matrix
Date: 11/04/2000 at 02:01:08 From: Tristan Smith Subject: Matrix, Leslie Design and describe: 1) an age-specific breeding and survival rates scenario that is consistent with the data contained in the matrix: [ 0 1.2 1.5 0.7 0.1 ] [ 0.65 0 0 0 0 ] A = [ 0 0.7 0 0 0 ] [ 0 0 0.75 0 0 ] [ 0 0 0 0.3 0 ] 2) an age-specific population scenario consistent with the matrix: [ 1567 ] [ 794 ] B = [ 428 ] [ 32 ] [ 8 ] 3) Now find the culling factor and hence the Leslie matrix that will stabilize the population.
Date: 11/04/2000 at 04:36:38 From: Doctor Mitteldorf Subject: Re: Matrix, Leslie Dear Tristan, I don't know how much you already understand about the subject of age-specific population dynamics in general, or the definitions of the terms, or the meanings of this matrix and vector. The math in this question is really quite simple once you understand all the definitions and what is being asked of you. First, what does the vector B mean? It is an age histogram of the population of animals. In other words, 1567 of the animals are between ages 0 and 1, 794 are between the ages 1 and 2, etc. Second, if we know the age histogram this year, what will it be next year? That is what the Leslie matrix is about. If you multiply the Leslie matrix by the age histogram vector this year, the result is the age histogram for next year. Let's see what the meaning is of the numbers in A. The top row 0 1.2 1.5 0.7 0.1 gets multiplied by all ages and the sum is the top number, the age = 0 entry, in next year's histogram. So the top row of the Leslie matrix is about fertility. It says that from age 0 to age 1, the animals haven't reached sexual maturity yet, so they're not breeding. From age 1 to age 2 they reach maturity and start to breed: the average individual produces 1.2 offspring in that year. The next year, fertility peaks at 1.5 offspring. After that, reproductive aging takes its toll and the number of offspring in the fourth year is down to 0.7. By the fifth year, fertility is only 0.1 offspring per individual on average. So we see that the top row of the matrix A tells you about fertility. Below that matrix A is all zeros except for the subdiagonal elements. The sub-diagonal of the matrix tells you about survival: each individual becomes one year older the next year providing that it survives: so the population of 3-year-olds next year is the same as the population of 2-year-olds this year, multiplied by a survival factor. In this case, the survival factor is 0.75, meaning that 75% of the individuals between ages 2 and 3 survive to become next year's population of individuals 3 to 4 years old. (Of course, the matrix and the vector should really go on forever, to account for the number of 6- and 7- and 8-year-old individuals, etc. But in practice, there are so few of these, and their fertility is so low, that they can be ignored.) The next reasonable thing to do is to multiply the matrix A by the vector B. If what you get back is exactly the vector B again, then the population is in steady-state. Next year's population histogram will be the same as this year's. If the population is not in equilibrium, often the population will APPROACH equilibrium over time. In other words, keep multiplying A*B, then A*A*B, then A*A*A*B to get the 2nd and 3rd and 4th year population histograms. The histograms may begin to look more and more alike over time, in which case the population is approaching a steady- state age distribution. The "hard" question in this science is to find the steady-state age distribution B when you know A. This is a vector which, when multiplied by A, gives itself back again. Such a vector is called an "eigenvector" of A. The procedure for finding the eigenvector of a matrix is long and complicated, but it can be performed as a "cookbook" exercise. Alternatively, you can start with a guess, multiply by A over and over again, and after 10 or 20 multiplications you'll get close enough to an eigenvector. Let me know if this explanation helps you. I think that once you understand what this is about, the things you're asked to do here are not difficult. - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/
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