259 Lecture 7 Spring 2013

1. 259 Lecture 7 Spring 2013 Population Models in Excel

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 with x(0) = 36449 x(n) = f(x(n-1)) for n≥1, for some function f(x)!

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!!

4. Toads Again! (cont.) • Thus, the toad growth can be modeled with the recurrence relation x(0) = 36449 x(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…)

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.

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.

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).

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.

9. Example 1 (cont.)

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?

11. Example 1 (cont.)

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.

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.

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:

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.

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.

17. Example 2 (cont.)

18. Example 2 (cont.)

19. Example 2 (cont.)

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:

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.

22. Example 3 • Revise our model from Example 2 with e = 0.00015 and f = 0.00001. • Keep all other parameters the same.

23. Example 3 (cont.)

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