Math Modeling Project December 3,2007. The Problem -The mountain reedbuck is a small deer common in the nature reserves of southern Africa. -The mountain.

Math Modeling ProjectMath Modeling Project

December 3,2007December 3,2007

The ProblemThe Problem

-The mountain reedbuck is a small deer -The mountain reedbuck is a small deer common in the nature reserves of southern common in the nature reserves of southern Africa.Africa.

-The owner of a private game reserve in the -The owner of a private game reserve in the Karoo has to cull a number of his stock of Karoo has to cull a number of his stock of reedbuck to prevent overgrazing.reedbuck to prevent overgrazing.

-He is, however, not sure when and how many -He is, however, not sure when and how many reedbuck must be removed, and is reedbuck must be removed, and is furthermore worried about the long-term furthermore worried about the long-term effect of such a culling on his stock.effect of such a culling on his stock.

The PopulationThe Population

YearYear PopulationPopulation

19881988 260260

19891989 370370

19901990 500500

19911991 680680

19921992 950950

Malthus Model Malthus Model

N’[t]=kN[t] N’[t]=kN[t] (Solved using Laplace Transforms)(Solved using Laplace Transforms)

sn(s)-n(0)=kn(s) n(0)=alphasn(s)-n(0)=kn(s) n(0)=alphasn(s)-260=kn(s) alpha=260sn(s)-260=kn(s) alpha=260

sn(s)-kn(s)=260sn(s)-kn(s)=260

n(s)(s-k)=260n(s)(s-k)=260

n(s)=260/(s-k)n(s)=260/(s-k)

n(s)=260*InverseLaplaceTransfrom[1/(s-k)]n(s)=260*InverseLaplaceTransfrom[1/(s-k)]

From Table on Page 149From Table on Page 149

N(t)=260e^(kt) or N(t)=Alpha*e^(kt)N(t)=260e^(kt) or N(t)=Alpha*e^(kt)

Solving for the ConstantsSolving for the Constants

N(t)=alpha*e^(kt) alpha=260 N(t)=alpha*e^(kt) alpha=260 n(1)=370n(1)=370

370=260*e^(kt)370=260*e^(kt)

(370/260)=e^(k)(370/260)=e^(k)

ln[370/260]=kln[370/260]=k

.352821=k.352821=k

The Model’s ResultsThe Model’s Results

YearYear PopulationPopulation Model Model PredictionsPredictions

19881988 260260 260260

19891989 370370 370370

19901990 500500 527527

19911991 680680 749749

19921992 950950 10661066

Model PredictionsModel Predictions

1 2 3 4

400

600

800

1000

Culling Population using Culling Population using ModelModel

Introduce a minus E(t) into model.Introduce a minus E(t) into model.

Table[NDSolve[{n'[t]=0.352821*n[t]-Table[NDSolve[{n'[t]=0.352821*n[t]-e,n[0]=a},n[t],{t,0,25}]] e,n[0]=a},n[t],{t,0,25}]]

E(t)=91 E(t)=92E(t)=91 E(t)=92

5 10 15 20 25

2000

4000

6000

8000

5 10 15 20 25

-2500

-2000

-1500

-1000

-500

Introducing E(t) as a Linear Introducing E(t) as a Linear Function of Time.Function of Time.

Model Minus E(t)*tModel Minus E(t)*t

E(t)=91*t E(t)=32*tE(t)=91*t E(t)=32*t

5 10 15 20

1000

2000

3000

4000

5000

1 2 3 4 5

-300

-200

-100

100

200

300

Conclusions about the Conclusions about the ModelModel

• The model first of all seems to The model first of all seems to overestimate the population.overestimate the population.

• The model doesn’t seem to have a clear The model doesn’t seem to have a clear number of reedbucks to cull each year(91 number of reedbucks to cull each year(91 grows the population exponentially while grows the population exponentially while 92 kills the population off around 18 years).92 kills the population off around 18 years).

• Another method needs to be used to find a Another method needs to be used to find a better estimation of population of culling of better estimation of population of culling of reedbuck.reedbuck.

Logistic ModelLogistic Model

N’[t]=k*n[t]-s*n[t]^2N’[t]=k*n[t]-s*n[t]^2

after a long derivation and after a long derivation and simplifications simplifications

the result given is as followsthe result given is as follows

N[t]=[k/((k/alpha)-s)*e^(-kt)+s)]N[t]=[k/((k/alpha)-s)*e^(-kt)+s)]

Solving for the ConstantsSolving for the Constants

N(1)=[k/(((k/260)-s)e^(-kt)+s))]N(1)=[k/(((k/260)-s)e^(-kt)+s))]370*((k/260)-s)e^(-k)+s)=k370*((k/260)-s)e^(-k)+s)=k(370k/260)e^(-k)-370se^(-k)+370s=k(370k/260)e^(-k)-370se^(-k)+370s=ks(370-370e^(-k))=k-(370k/260)e^(-k)s(370-370e^(-k))=k-(370k/260)e^(-k)Equation 1 s=((k-(370k/260)e^(-k))/(370-370e^(-k))Equation 1 s=((k-(370k/260)e^(-k))/(370-370e^(-k))Plug N(2) into equation and plug Equation 2 in for s Plug N(2) into equation and plug Equation 2 in for s

and solve for k using mathematica Find Root and solve for k using mathematica Find Root Option Option

k=0.486579k=0.486579s=0.000428s=0.000428

The Model’s ResultsThe Model’s Results

YearYear PopulationPopulation Model Model PredictionsPredictions

19881988 260260 260260

19891989 370370 370370

19901990 500500 500500

19911991 680680 638638

19921992 950950 767767

Model’s PredictionModel’s Prediction

1 2 3 4

300

400

500

600

700

Culling Population using Culling Population using ModelModel

Introduced a minus E(t) into modelIntroduced a minus E(t) into model

Used following equationUsed following equation

NDSolve[{n’[t]==(0.486579n[t]-NDSolve[{n’[t]==(0.486579n[t]-0.000428n[t]^2)-e ,n[0]==a},n[t],0.000428n[t]^2)-e ,n[0]==a},n[t],{t,0,25}]{t,0,25}]

Results of ModelResults of Model

No Culling E(t)=92No Culling E(t)=92

2 4 6 8 10 12 14

400

600

800

1000

10 20 30 40 50

300

400

500

600

700

800

900

Culling of PopulationCulling of Population

E(t)=97 E(t)=98E(t)=97 E(t)=98

10 20 30 40 50

300

400

500

600

700

800

5 10 15 20

-300

-200

-100

100

200

Introducing E(t) as a Linear Introducing E(t) as a Linear Function of Time.Function of Time.

Model Minus E(t)*tModel Minus E(t)*t

E(t)=98*t E(t)=1*tE(t)=98*t E(t)=1*t

1 2 3 4 5

-200

-100

100

200

300

25 50 75 100 125 150

200

400

600

800

1000

Conclusions about ModelConclusions about Model

• The model underestimates population The model underestimates population and eventually limits the population and eventually limits the population around 1250 without culling.around 1250 without culling.

• The model kills the population in 18 The model kills the population in 18 year if the culling is 98 or more per year if the culling is 98 or more per year. year.

• Overall the model seems reasonable Overall the model seems reasonable in the short-term but long-term effects in the short-term but long-term effects seem to be more questionable.seem to be more questionable.

Overall ConclusionsOverall Conclusions

• Both Models seem to closely agree with the Both Models seem to closely agree with the culling amount for each year the models limits culling amount for each year the models limits range from 92-98 (as long as culling remains range from 92-98 (as long as culling remains constant and not a function of time).constant and not a function of time).

• The culling could be affected by many other The culling could be affected by many other factors as well and the models could have some factors as well and the models could have some error based on constants but the best solution error based on constants but the best solution and long-term answer seem to be a culling of and long-term answer seem to be a culling of around 95 reedbuck per year. By taking an around 95 reedbuck per year. By taking an average of both models. average of both models.

• The logistic model with the culling constant The logistic model with the culling constant seems to have the least long-term effect on the seems to have the least long-term effect on the culling of the population.culling of the population.