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
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
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.