Consider the discrete time chicken and fox population model discussed in class w

Question

Consider the discrete time chicken and fox population model discussed in class with the following differences. For foxes, during 1 year period, 1/2 of the fox population dies but a number of new foxes are born (due to chicken consumption) equal to 1/4 of the chicken population. For chickens, during a 1 year period, the chicken population decreases by 1/2 of the fox population (due to consumption by foxes) and increases by 1/4 of the current chicken population (the factor is therefore 1 + 1/4 = 5/4). The initial populations are 2000 for foxes and 3000 for chickens. Formulate the model in the form x(t + 1) = Ax(t) where A is a 2 by 2 matrix then solve for the populations as functions of time using the eigenvalues and eigenvectors of A Sketch the graphs of the populations as functions of time. What are the long run populations?

Explanation / Answer

Let C = chicken population

and F = fox population

fox population in next year = F -F/2 +C/4 =F/2+C/4

Chicken population in next year = C -F/2 +C/4 =-F/2 + C/5

x(t+1) = A x(t)

A = 1/2 1/4
-1/2 1/5

F(t+1) = F(t)/2 + C(t)/4

C(t+1) =-F(t)/2 + C(t)/5

For fox : F(t+1) = F(t)/2 + C(t)/4

F(t+2) = F(t+1)/2 + C(t+1)/4 = [F(t)/2 + C(t)/4]/2 + [-F(t)/2 +C(t)/5]/4

= F(t)/4 +C(t)/8 -F(t)/8 +C(t)/20

F(t+2)=F(t)/8 +7C(t)/20

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