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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   
Associated Topics:
College Linear Algebra

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