Fibonacci's rabbits variation

Question

In this variation on Fibonacci rabbits the growth of mature rabbits in a period has to be less than 10%.

1. After one period the rabbits are called "young rabbits", after two periods the rabbits will be mature rabbits.
2. Each pair of mature rabbits will give birth to a pair of newborn rabbits.
3. Only 80% of the rabbits survive at birth.
4. After the mature rabbits gave birth, hunters have to kill $p \cdot 100$% of the rabbits such that the growth of mature rabbits in a period is less than 10%.

What must the value $p$ at least be to meet this criterium? Can I solve this using a recurrence relation just like in the normal version of Fibonacci's rabbits?

Answer 1

Let $x_n$ be the number of mature rabbits at time $n$.

Let $y_n$ be the number of young rabbits at time $n$.

Let $z_n$ be the number of newborn rabbits at time $n$.

The number of newborn rabbits is equal to the number of mature rabbits, so $z_{n}=x_n$

80% of newborn rabbits survive to become young rabbits. This is equivalent to $y_{n+1}=0.8z_n$

With no intervention from hunters, all young rabbits would become mature rabbits and all mature rabbits would stay alive as mature rabbits. This is equivalent to $x_{n+1}=x_n+y_n$

Killing a proportion of the mature rabbits means that $x_{n+1}=(1-p)x_n+y_n$

We could create a recurrence relationship using the above, but we are told that the hunters kill enough rabbits so that in fact $x_{n+1}=1.1x_n$

This has the straightforward solution $x_n=(1.1)^nx_0$

To find out how many need to be killed to achieve this rate of growth, let's look again at the earlier equations.

$z_{n}=x_n \Rightarrow z_{n}=(1.1)^nx_0 $

$y_{n+1}=0.8z_n \Rightarrow y_{n+1}=0.8 \times (1.1)^{n}x_0 \Rightarrow y_{n}=0.8 \times (1.1)^{n-1}x_0$

Since we have both $x_{n+1}=(1-p)x_n+y_n$ and $x_{n+1}=1.1x_n$,

$1.1x_n=(1-p)x_n+y_n \Rightarrow 1.1(1.1)^nx_0=(1-p)(1.1)^nx_0+0.8 \times (1.1)^{n-1}x_0$

$(1.1)^{n+1}x_0=(1-p)(1.1)^nx_0+0.8 \times (1.1)^{n-1}x_0$

$(1.1)^{n+2}x_0=(1-p)(1.1)^{n+1}x_0+0.8 \times (1.1)^{n}x_0$

$(1.1)^2=(1-p)(1.1)+0.8$

$1.21=1.1-1.1p+0.8$

$1.1p=0.69$

$p=0.627$

So you need to kill about 63% of the mature rabbits.

Answer 2

As tomi—whose notation I shall adopt—shows in his answer, $$ y_{n+1}=0.8x_n\ , $$ and \begin{align} x_{n+1}&=(1-p)x_n+y_n\\ &=(1-p)x_n + 0.8x_{n-1}\ . \end{align} The solution of this recurrence relation is given by $$ x_n= \left(\frac{x_1-\lambda_-x_0}{\lambda_+-\lambda_-} \right)\lambda_+^n- \left(\frac{x_1-\lambda_+x_0}{\lambda_+-\lambda_-} \right)\lambda_-^n\ , $$ where $$ \lambda_\pm=\frac{1-p\pm\sqrt{(1-p)^2+3.2}}{2}\ . $$ Since $\ -1<\lambda_- < 0\ $, the term $\ \left(\frac{x_1-\lambda_+x_0}{\lambda_+-\lambda_-} \right)\lambda_-^n\ $ in the expression for $\ x_n\ $ oscillates between positive and negative values of ever decreasing magnitude which converges to $0$ as $\ n\rightarrow\infty\ $. The asymptotic growth rate of the population per period (or the constant actual growth rate, if $\ x_1=\lambda_+x_0\ $) is therefore $\ \lambda_+\ $, which is required to be less than $\ 1.1\ $. That is \begin{align} &&\frac{1-p+\sqrt{(1-p)^2+3.2}}{2}<1.1\\ &\iff&\sqrt{(1-p)^2+3.2}<p+1.2\\ &\iff& (1-p)^2+3.2<(p+1.2)^2\\ &\iff& \frac{69}{110}< p\ . \end{align}