How to Model Animal Breeding Populations with Leslie Matrices: A 5‑Year Forecast

Breeding Farm Problem

Problem

The farm raises three age groups of animals for up to three years; after three years they are sold. Age groups: 1‑year (young, no reproduction), 2‑year and 3‑year (reproductive). The 2‑year group produces on average 4 offspring per year, the 3‑year group 3 offspring. Survival rates from younger to older groups are 0.5 for 1→2 and 0.25 for 2→3. Initially each group has 1,000 animals.

Modeling

(1) From the statement, define birth‑rate and survival‑rate vectors and let the numbers of the young, second‑year, and third‑year groups at year n be denoted. Using the given rates, a difference‑equation model is built and written in matrix form as a Leslie matrix.

Solution

Using the initial conditions, the numbers of each age group over five years are obtained as shown below.

Year  Young  Second  Third
1     7000   500     250
2     2750   3500    125
3     14375  1375    875
4     8125   7188    344
5     29781  4063    1797

Analysis of Population Growth

(2) To estimate the dynamic trend, we study the eigenvalues and eigenvectors of the Leslie matrix. Theorem 1: The Leslie matrix has a unique positive eigenvalue, which is simple, and its associated eigenvector has all positive components. Theorem 2: If the first row of the Leslie matrix contains two adjacent non‑zero elements, the positive eigenvalue is strictly dominant.

Definition 1 introduces the concept of a dominant (strictly dominant) eigenvalue. Assuming the matrix is diagonalizable with eigenvalues \(\lambda_i\) and corresponding linearly independent eigenvectors, the population vector can be expressed as a linear combination of these eigenvectors. As time grows, the term associated with the dominant eigenvalue \(\lambda_{max}\) governs the behavior.

Consequently, for sufficiently large time: if \(\lambda_{max}>1\) the population grows; if \(\lambda_{max}=1\) it remains stable; if \(\lambda_{max}<1\) it declines. For the given Leslie matrix the largest eigenvalue is \(\lambda_{max}=\) (value from calculation) with eigenvector (vector shown). Hence the population tends to infinity when \(\lambda_{max}>1\).

Market Supply Considerations

(3) If an average number of animals is sold each year, the distribution vector changes accordingly. To keep all age groups present, every component of the vector must stay positive. Using a Python program to test different parameters, we find that choosing a sale quantity of (specific value) guarantees that no age group drops to zero.

Reference

Python Mathematical Experiment and Modeling / Si Shoukui, Sun Xijing, Science Press