Summary of week 6
As always we started off the class with a discussion of the talking points. For the first problem
we determined that the determinant of the matrix was equal to 12 and that elementary row operations
turned out to be the simplest method to obtain the solution without using a calculator. The second
0 2 0 1
.
problem had a Leslie matrix of [ 6 0 0 0 ], a characteristic polynomial of 𝑝() = 4 − 1.22 −
0 .6 0 0
0 0 .8 0
. 5167
.288, no dominant eigenvalue since  = ±1.18532 …, and an eigenvector of [. 2616] in percentages.
. 1324
. 0894
Since problem 2 did not have a dominant eigenvalue when we used Excel to create the population there
were some interesting phenomena that occurred. First as you can see in the Excel cells below the
population only regenerates every other year, and there are either only newborns and 2 year olds or 1
year olds and 3 years olds, never 2 and 3 years olds alive at the same time. All of these phenomena
culminate in the population totals column where it can be seen that the growth rates of consecutive
years bounce between 2.193 and 0.641 every other year. All of which is a direct affect of the
characteristic polynomial not having a dominant eigenvalue.
n newborns 1 yr olds
0
100
0
1
0
60
2
120
0
3
0
72
4
172.8
0
5
0
103.68
6
241.92
0
7
0 145.152
8 340.0704
0
9
0 204.0422
2 yr olds 3 yr olds
totals
0
0
100
0
0
60
36
0
156
0
28.8
100.8
43.2
0
216
0
34.56
138.24
62.208
0 304.128
0 49.7664 194.9184
87.0912
0 427.1616
0 69.67296 273.7152
Excerpt from the Excel spreadsheet
showing the population in problem
#2 without a dominant eigenvalue.
After our discussion of the talking points we began looking at age clusters, which in effect
combine different ages into a single group just like the class intervals in relative frequency tables from
statistics. In this light we set up a human population such that each cluster encompasses 20 years.
Age cluster in years
𝑥2 +1 =
19
1
𝑥2 + 𝑥1
20
20
𝑥3 +1 =
19
1
𝑥3 + 𝑥2
20
20
𝑥4 +1 =
19
1
𝑥4 + 𝑥3
20
20
𝑥1 +1 =
1
19
𝑥2 + 𝑥1
20
20
𝑥1
𝑥2
𝑥3
𝑥4
0-19
20-39
40-59
60 +
The problem proposed was to determine the matrix that would reflect
the annual progress of a population if 95% of each age cluster
remained in the age group, 5% moved up to the next age group, and
each female has one girl during their 20’s or 30’s. Using this
1
information we created the equations shown to the left where 20
represents the 5% of the previous age group that is moving up and the
superscript +1 on the 𝑥𝑖 ’s indicates what the age group will look like
the next year. These equations become the Lefkovitch matrix 𝐴 =
. 95 . 05
0 0
.
05
.
95
0 0 ]
[
0 . 05 . 95 0
0
0 . 05 . 95
This problem can also be represented by state diagrams that show the movement between the age
groups through circles that represent the age clusters and arrows that show the movement of the
population within the clusters. The orange arrows represent the 95% of the age group that remains in
the age group, the green arrows represent the 5% of each age group that moves up to the next age
group, and the purple arrow represents the 5% increase the youngest age group when a woman in her
20’s or 30’s has a child.
0-19
20-39
40-59
60 +
Another way to represent this situation is using uneven clusters, where you combine clusters together
such that the new clusters do not have the equal distributions. For example, using the above model if
you let 𝑥0 represent the youth of the population, ages 0-19, and 𝑥1 , 𝑥2 , 𝑥3 represent the adults of the
population, ages 20 + you create uneven clusters where 3/4ths of the population are in the adult cluster.
0 . 730
This creates the matrix [
].
. 6 . 631
The final problem that we discussed in class was about harvesting where the main question is
whether or not the population will survive when a fixed number of adults are removed from the
population at specific intervals. The two vectors and one matrix that represent the situation where you
𝑥
only have youth and adults are 𝑥⃑ = [𝑦] where x represents the youth and y represents the adults of the
0 𝑏1
population. 𝐴 = [
] where 𝑏1 represents the birth rate of adults, and 𝑠1 and 𝑠2 represent the
𝑠1 𝑠2
⃑⃑ = [ 0 ] where the H represents the
survival rates of the youth and adult populations respectively, and ℎ
𝐻
fixed number of the adult population that is harvested at the end of a cycle. To determine the next
⃑⃑ and vectors and matrix delineated
year’s projected population you use this formula 𝑥⃑𝑛+1 = 𝐴𝑥⃑𝑛 − ℎ
above, which amounts to the projected youth 𝑥𝑛+1 = 𝑏1 𝑦𝑛 and the projected adults 𝑦𝑛+1 = 𝑠1 𝑥𝑛 +
𝑠2 𝑦𝑛 − 𝐻. In Excel using 𝑠1 = .7, 𝑠2 = .9, 𝑏1 = 3, 𝑎𝑛𝑑 𝐻 = 20 choosing the starting values of x = 59 and
y = 0 you can see the population increase and eventually survive from both the table and the graph.
x-youth
y-adults
0
59
0
1
0
21.3
2
63.9
-0.83
3
-2.49
23.983
4
71.949
-0.1583
5
-0.4749 30.22183
6 90.66549 6.867217
7 20.60165 49.64634
8 148.939 39.10286
9 117.3086 119.4499
200
150
100
50
0
-100
0
100
200
300
400
-50
The next question that arose from this example was whether or not there was a survival region, whether
you could locate starting values that would ensure the survival of the population. To determine the
answer to this question becomes a matter of locating the fixed point where 𝑥⃑𝑛+1 = 𝑥⃑𝑛 which becomes
⃑⃑ so you can solve for 𝑥⃑𝑛 if A-I is a non-singular matrix to determine the survival region of
(𝐴 − 𝐼)𝑥⃑𝑛 = ℎ
given birth and survival rates. For our specific example this amounted to the equations
3𝑦 = 𝑥 𝑎𝑛𝑑 .7𝑥 + .9𝑦 − 20 = 𝑦 which produced the values y = 10 and x= 30. This insinuates that if a
population started off with 10 youth and 30 adults the population would survive and that there might
exist a line through (30, 10) that may delineate a survival region where any point above the line would
guarantee that the population would survive and any point below the line would guarantee that the
population would die off. Though we never proved this fact it seems reasonable to conclude that this
line of survival may exist.