LAB 10: AGE-SPECIFIC POPULATION GROWTH AND MATRIX EIGENVALUES
Mathematics:
Demographic
predictions of population growth are based on a number of mathematical models.
This project exploits the Leslie matrix model, where the female
population is divided into age classes of equal duration. As time progresses,
the number of females, within each class, changes because of birth, death, and
aging.
Suppose
we study n age classes of females over maximum age in T years
and denote the number of females in the j-th class at the k-th
time instance by xj(k). Each age
class is T/n years in duration. It makes sense
therefore to study changes of the number of females xj(k)
between two observation times in T/n years. All females
of the j-th class at the k-th time pass into the (j+1)-th
class at the (k+1)-th time, except for those who died during the
time interval. Denote bj be the fraction of females
that survives in the j-th class, i.e.0 bj 1 and
xj+1(k+1)
= bj xj(k),
j = 1,2,…,n-1
The
first class at the (k+1)-th time has the number of females x1(k+1)
that was born during the time interval by females of all other classes.
Denote aj be the average number of daughters born by
each female in the j-th class during the time inteval, i.e. aj
0 and
x1(k+1)
= a1 x1(k) + a2 x2(k)
+ … + an xn(k)
The following birth and death parameters were registered from the year 1965 for Canadian females:
Age |
[0,5) |
[5,10) |
[10,15) |
[15,20) |
[20,25) |
[25,30) |
[30,35) |
[35,40) |
[40,45) |
[45,50) |
aj |
0.00 |
0.00024 |
0.05861 |
0.28608 |
0.44791 |
0.36399 |
0.22259 |
0.10457 |
0.02826 |
0.00240 |
bj |
0.99651 |
0.99820 |
0.99802 |
0.99729 |
0.99694 |
0.99621 |
0.99460 |
0.99184 |
0.98700 |
--- |
In
this table, the age classes over T = 50 years are neglected since
they rarely have children, i.e. aj = 0 , j > n, where n = 10
age classes with duration in T/n = 5 years.
Introducing
matrix notations, the Leslie model can be written as the discrete dynamical
system x(k+1) = L x(k), where x(k)
is age distribution vector at the k-th time and L is
the Leslie matrix:
L =
The
Leslie matrix L is supposed to have at least two non-zero
successive entries aj and aj+1.
The matrix L has n eigenvalues 1,…,n with n eigenvectors z1,…,zn,
such that Lzj =j zj. It can be shown that
there is unique positive eigenvalue + which is dominant, i.e. all other
negative and complex eigenvalues have smaller absolute value: |j | < +. Let z+ be the
eigenvector for the eigenvalue +. Then, the long-term behaviour of the age
distribution of the female population is:
x(k) = c +k z+
The
population increases if + > 1, decreases if + < 1 and stabilizes if + = 1.
Objectives:
·
Understand
the long-term behaviour of the female population within the Leslie model
·
Exploit
dynamical modeling of the population growth within the Leslie model
·
Exploit
the power method for finding the dominant eigenvalue and its eigenvector of the
matrix L
If Canadian women continued to reproduce and die as they did in 1965, the dynamical growth of female population in the ten age classes every five years would follow this graph:
Every
five years, the increase in the number of females at each age class, defined as
yj(k+1) = (xj(k+1) – xj(k))/xj(k),
will approach to a uniform ratio for large k, which is
approximately 0.07622 or 7.622%. In the year of 2040,
the histogram of population is distributed according to the graph:
If
the vector x(k) is normalized such that x1(k)
= 1, the vector x(k) approaches
approximately for large k:
Age class |
[0,5) |
[5,10) |
[10,15) |
[15,20) |
[20,25) |
[25,30) |
[30,35) |
[35,40) |
[40,45) |
[45,50) |
Proportion |
1.000 |
0.9259 |
0.8588 |
0.7964 |
0.7380 |
0.6836 |
0.6328 |
0.5848 |
0.5390 |
0.4943 |
Steps
in writing the MATLAB script:
Suppose
that the Leslie matrix L is diagonalizable, i.e. the eigenvectors
zj can be
combined into matrix P:
P = [ z1, z2,
…, zn ]
and
eigenvalues j
are combined into matrix D
= diag[1, 2 , …, n] such that L = P D P-1.
Then, the age distribution vector x(k) at the k-th
time is:
x(k) = Lk
x(0) = P Dk P-1
x(0)
If
the unique positive eigenvalue+ is dominant, i.e. all other eigenvalues satisfy: |j
| < +, then the age distribution vector x(k) has the limiting behavior in the
limit k :
x(k) c z+
where
z+ is the eigenvector for + and c is a constant.
The
dominant eigenvalue + and its eigenvector z+
can be found in the iterative power method based on the asymptotic limit
above. Construct a sequence:
x(k)
x(k+1) = L x(k)
where
ck = maxj |
L x(k) |j, i.e. the largest entry in the vector x(k+1)
is normalized by 1.
Then, the sequence converges to the eigenvector z+, while
the normalization constant ck converges to +. The dominant eigenvalue for the problem
is + = 1.07622 and the eigenvector z+
is given in the table:
Age class |
[0,5) |
[5,10) |
[10,15) |
[15,20) |
[20,25) |
[25,30) |
[30,35) |
[35,40) |
[40,45) |
[45,50) |
Eigenvector |
1.000 |
0.9259 |
0.8588 |
0.7964 |
0.7380 |
0.6836 |
0.6328 |
0.5848 |
0.5390 |
0.4943 |
Steps
in writing the MATLAB script:
Exploiting
the MATLAB script:
1) The value of the dominant eigenvalue+ is determined by the net reproduction rate of the population:
R
= a1 + a2 b1 + a3 b1 b2
+ … + an b1 b2 … bn-1
If R
> 1, the population increases. If R < 1, the
population decreases. If R = 1, the population stabilizes. Write
a function "[R,status] = NetRate(a,b)" that computes the
scalar R from the vectors a and b and
states in the string status whether the population increases,
decreases or stabilizes.
2)
Consider
the birth-death parameters for the animal population:
Age |
[0,5) |
[5,10) |
[10,15) |
aj |
0 |
4 |
3 |
bj |
0.5 |
0.25 |
--- |
Find
the net population growth for the animal population and state whether it
increases, decreases, or stabilizes. Compute the dynamics of the animal population
in time and confirm the prediction.