259 Lecture 7 Spring 2013

24
259 Lecture 7 Spring 2013 Population Models in Excel

description

259 Lecture 7 Spring 2013. Population Models in Excel. Toads Again!. Let’s look at the toad data again, but this time let n be the number of years after 1939 and x(n) be the area covered by toads at year n. Using Excel, we find that the best-fit exponential function for this data is - PowerPoint PPT Presentation

Transcript of 259 Lecture 7 Spring 2013

Page 1: 259  Lecture  7  Spring 2013

259 Lecture 7 Spring 2013

Population Models in Excel

Page 2: 259  Lecture  7  Spring 2013

2

Toads Again! Let’s look at the toad data

again, but this time let n be the number of years after 1939 and x(n) be the area covered by toads at year n.

Using Excel, we find that the best-fit exponential function for this data is

x(n) = 36449e0.0779n for n≥0. We can think of this function

as a recurrence relation withx(0) = 36449x(n) = f(x(n-1)) for n≥1,for some function f(x)!

Years after 1939 Area(km^2)

0 32800

5 55800

10 73600

15 138000

20 202000

25 257000

30 301000

35 584000

Page 3: 259  Lecture  7  Spring 2013

3

Toads Again! (cont.) Let’s find f(x). To do so, look at x(n) – x(n-1): x(n) – x(n-1)

= 36449e0.0779n - 36449e0.0779(n-1) = 36449e0.0779(n-1)(e0.0779 – 1)= (e0.0779 – 1)*x(n-1)

Solving for x(n), we see that x(n) = x(n-1)+(e0.0779 -1)*x(n-1) =

= e0.0779 *x(n-1), so our function is f(x) = e0.0779 *x!!

Page 4: 259  Lecture  7  Spring 2013

4

Toads Again! (cont.) Thus, the toad growth can be

modeled with the recurrence relationx(0) = 36449x(n) = e0.0779 *x(n-1) for n ≥ 1.

The closed form solution is given by our original model!

For this model, the growth of the toad population is exponential (no surprise…)

Page 5: 259  Lecture  7  Spring 2013

5

Toads Again! (cont.) So how realistic is an exponential growth

model for the toad population? For such a model, the population grows

without bound, with no limitations built in. Realistically, there should some way to limit

the growth of a population due to available space, food, or other factors.

Page 6: 259  Lecture  7  Spring 2013

6

The Logistic Model As a population increases, available resources must

be shared between more and more members of the population.

Assuming these resources are limited, here are some “reasonable” assumptions one can make how a population should grow: The population’s growth rate should eventually

decrease as the population levels increase beyond some point.

There should be a maximum allowed population level, which we will call a carrying capacity.

For population levels near the carrying capacity, the growth rate is near zero.

For population levels near zero, the growth rate should be the greatest.

Page 7: 259  Lecture  7  Spring 2013

7

The Logistic Model (cont.) The simplest model that takes these assumptions into

account is the logistic model: x(0) = x0

x(n) = x(n-1)*(R(1-x(n-1)/K)+1) for n ≥ 1 Here, x0 is the initial population size,

R is the intrinsic growth rate (i.e. growth rate without any limitations on growth),and K is the carrying capacity.

Notice that when x(n-1) is close to zero, the growth is exponential.

Also, when x(n-1) is close to K, the population stays near the constant value of K (so growth rate is close to zero).

Page 8: 259  Lecture  7  Spring 2013

8

Example 1 Use Excel to study the long-term

behavior of a population that grows logistically, with carrying capacity K = 100 and growth rate R = 0.5 (members/year).

Use x0 = 0, 25, 50, 75, 100, 125, and 150.

Page 9: 259  Lecture  7  Spring 2013

9

Example 1 (cont.)

Page 10: 259  Lecture  7  Spring 2013

10

Example 1 (cont.) Notice that X = 100 and X = 0 are

fixed points of the logistic recurrence relation.

X = 100 is stable. What about X = 0? For fun, even though this doesn’t

make sense in the real world for a population, try x0 = -1 and x0 = -10.

What happens?

Page 11: 259  Lecture  7  Spring 2013

11

Example 1 (cont.)

Page 12: 259  Lecture  7  Spring 2013

12

Example 1 (cont.) Fixed point X = 0 is unstable! In general, for the logistic equation,

the fixed points turn out to be X = 0 and X = K.

This can be shown by solving the equation X = X*(R(1-X/K)+1) for X.

Page 13: 259  Lecture  7  Spring 2013

13

Two or More Populations If two or more populations interact,

we can use a system of recurrence equations to model the population growth!

Typical examples include predator-prey, host-parasite, competitive hunters and arms races.

Page 14: 259  Lecture  7  Spring 2013

14

Predator-Prey Model As an example, let’s consider two populations that interact –

foxes (predator) and rabbits (prey). Assume no other species interact with the foxes or rabbits.

Assume the following: There is always enough food and space for the rabbits. In the absence of foxes the rabbit population grows

exponentially. In the absence of rabbits, the fox population decays

exponentially. The number of rabbits killed by foxes is proportional to the

number of encounters between the two species. This in turn is proportional to the product of the two

populations (this assumption implies fewer kills when the number of foxes or rabbits is small).

These assumptions can be modeled with the following system:

Page 15: 259  Lecture  7  Spring 2013

15

Predator-Prey Model (cont.) Let R(n) be the number of rabbits at time n

and F(n) be the number of foxes at time n. R(0) = R0

F(0) = F0

R(n) = R(n-1)+a*R(n-1) – b*R(n-1)*F(n-1)F(n) = F(n-1)-c*F(n-1) + d*R(n-1)*F(n-1) for n≥1,where a, b, c, and d are all greater than zero.

Page 16: 259  Lecture  7  Spring 2013

16

Example 2 As an example, let’s try the Rabbit-Fox

Population model with a = 0.15, b = 0.004, c = 0.1, and d = 0.001.

Assume that initially there are 200 rabbits and 50 foxes, i.e. R0 = 200 and F0 = 50.

Plot R(n) and F(n) vs. n, for 200 years. Repeat with F(n) vs. R(n), for 200 years.

Page 17: 259  Lecture  7  Spring 2013

17

Example 2 (cont.)

Page 18: 259  Lecture  7  Spring 2013

18

Example 2 (cont.)Fox and Rabbit Populations Over Time

0100200300400500600700800

0 50 100 150 200 250

Years

Num

ber

R(n)

F(n)

Page 19: 259  Lecture  7  Spring 2013

19

Example 2 (cont.)Fox Population vs. Rabbit Population

0

50

100

150

200

250

300

0 100 200 300 400 500 600 700 800

Number of Rabbits

Num

ber o

f Fox

es

Page 20: 259  Lecture  7  Spring 2013

20

Revised Predator-Prey Model (cont.) A more realistic model takes into

account the fact that there may be limits to the space available for the foxes and rabbits.

This can be modeled via a logistic growth model, in the absence of the other species!

This amounts to the following:

Page 21: 259  Lecture  7  Spring 2013

21

Revised Predator-Prey Model Let R(n) be the number of rabbits at time n

and F(n) be the number of foxes at time n. R(0) = R0

F(0) = F0

R(n) = R(n-1)+a*R(n-1) – b*R(n-1)*F(n-1) – e*R(n-1)*R(n-1)F(n) = F(n-1)-c*F(n-1) + d*R(n-1)*F(n-1) – f*F(n-1)*F(n-1) for n≥1,where a, b, c, d, e, and f are all greater than zero.

Page 22: 259  Lecture  7  Spring 2013

22

Example 3 Revise our model from Example 2

with e = 0.00015 and f = 0.00001. Keep all other parameters the same.

Page 23: 259  Lecture  7  Spring 2013

23

Example 3 (cont.)

Fox and Rabbit Populations Over Time

0

50

100

150

200

250

0 50 100 150 200 250

Years

Num

ber

R(n)

F(n)

Fox Population vs. Rabbit Population

01020304050607080

0 50 100 150 200 250

Number of Rabbits

Num

ber o

f Fox

es

Page 24: 259  Lecture  7  Spring 2013

24

References A Course in Mathematical Modeling by

Douglas Mooney and Randall Swift An Introduction to Mathematical

Models in the Social and Life Sciences by Michael Olinick