1. Math Modeling Project
December 3,2007

2. The Problem
-The mountain reedbuck is a small deer common in the nature reserves of southern Africa.
-The owner of a private game reserve in the Karoo has to cull a number of his stock of reedbuck to prevent overgrazing.
-He is, however, not sure when and how many reedbuck must be removed, and is furthermore worried about the long-term effect of such a culling on his stock.

3. The Population Year Population 1988 260 1989 370 1990 500 1991 680
1992
950

4. Malthus Model N’[t]=kN[t] (Solved using Laplace Transforms)
sn(s)-n(0)=kn(s) n(0)=alpha
sn(s)-260=kn(s) alpha=260
sn(s)-kn(s)=260
n(s)(s-k)=260
n(s)=260/(s-k)
n(s)=260*InverseLaplaceTransfrom[1/(s-k)]
From Table on Page 149
N(t)=260e^(kt) or N(t)=Alpha*e^(kt)

5. Solving for the Constants
N(t)=alpha*e^(kt) alpha=260 n(1)=370
370=260*e^(kt)
(370/260)=e^(k)
ln[370/260]=k =k

6. The Model’s Results Year Population Model Predictions 1988 260 1989
370
1990
500
527
1991
680
749
1992
950
1066

7. Model Predictions

8. Culling Population using Model
Introduce a minus E(t) into model.
Table[NDSolve[{n'[t]= *n[t]-e,n[0]=a},n[t],{t,0,25}]]

9. E(t)= E(t)=92

10. Introducing E(t) as a Linear Function of Time.
Model Minus E(t)*t
E(t)=91*t E(t)=32*t

11. Conclusions about the Model
The model first of all seems to overestimate the population.
The model doesn’t seem to have a clear number of reedbucks to cull each year(91 grows the population exponentially while 92 kills the population off around 18 years).
Another method needs to be used to find a better estimation of population of culling of reedbuck.

12. Logistic Model N’[t]=k*n[t]-s*n[t]^2
after a long derivation and simplifications
the result given is as follows
N[t]=[k/((k/alpha)-s)*e^(-kt)+s)]

13. Solving for the Constants
N(1)=[k/(((k/260)-s)e^(-kt)+s))]
370*((k/260)-s)e^(-k)+s)=k
(370k/260)e^(-k)-370se^(-k)+370s=k
s( e^(-k))=k-(370k/260)e^(-k)
Equation 1 s=((k-(370k/260)e^(-k))/( e^(-k))
Plug N(2) into equation and plug Equation 2 in for s and solve for k using mathematica Find Root Option k= s=

14. The Model’s Results Year Population Model Predictions 1988 260 1989
370
1990
500
1991
680
638
1992
950
767

15. Model’s Prediction

16. Culling Population using Model
Introduced a minus E(t) into model
Used following equation
NDSolve[{n’[t]==( n[t] n[t]^2)-e ,n[0]==a},n[t],{t,0,25}]

17. Results of Model
No Culling E(t)=92

18. Culling of Population
E(t)= E(t)=98

19. Introducing E(t) as a Linear Function of Time.
Model Minus E(t)*t
E(t)=98*t E(t)=1*t

20. Conclusions about Model
The model underestimates population and eventually limits the population around 1250 without culling.
The model kills the population in 18 year if the culling is 98 or more per year.
Overall the model seems reasonable in the short-term but long-term effects seem to be more questionable.

21. Overall Conclusions
Both Models seem to closely agree with the culling amount for each year the models limits range from (as long as culling remains constant and not a function of time).
The culling could be affected by many other factors as well and the models could have some error based on constants but the best solution and long-term answer seem to be a culling of around 95 reedbuck per year. By taking an average of both models.
The logistic model with the culling constant seems to have the least long-term effect on the culling of the population.