1. 259 Lecture 7 Spring 2015 Population Models in Excel

2. 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) = 36449e 0.0779n 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)! Years after 1939Area(km^2) 032800 555800 1073600 15138000 20202000 25257000 30301000 35584000

3. 3 Toads Again! (cont.)  Let’s find f(x).  To do so, look at x(n) – x(n-1):  x(n) – x(n-1) = 36449e 0.0779n - 36449e 0.0779(n-1) = 36449e 0.0779(n-1) (e 0.0779 – 1) = (e 0.0779 – 1)*x(n-1)  Solving for x(n), we see that  x(n) = x(n-1)+(e 0.0779 -1)*x(n-1) = = e 0.0779 *x(n-1), so our function is f(x) = e 0.0779 *x!!

4. 4 Toads Again! (cont.)  Thus, the toad growth can be modeled with the recurrence relation x(0) = 36449 x(n) = e 0.0779 *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. 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. 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. 7 The Logistic Model (cont.)  The simplest model that takes these assumptions into account is the logistic model:  x(0) = x 0 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. 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. 9 Example 1 (cont.)

10. 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. 11 Example 1 (cont.)

12. 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. 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. 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. 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) = R 0 F(0) = F 0 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. 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. R 0 = 200 and F 0 = 50.  Plot R(n) and F(n) vs. n, for 200 years.  Repeat with F(n) vs. R(n), for 200 years.

17. 17 Example 2 (cont.)

18. 18 Example 2 (cont.)

19. 19 Example 2 (cont.)

20. 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. 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) = R 0 F(0) = F 0 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. 22 Example 3  Revise our model from Example 2 with e = 0.00015 and f = 0.00001.  Keep all other parameters the same.

23. 23 Example 3 (cont.)

24. 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