# Puppy Power (Recursion)

• Mar 10th 2008, 05:44 AM
Tally
Puppy Power (Recursion)
The question is 10 female dogs average 16 puppies each year with 50% of those puppies being female. Every year the owners remove 60% of the female puppies and all of the male puppies.

What will the growth be of female puppies left and their mothers over 15years?

Only the original mothers and 20% of female puppies can have puppies (each year); the rest will be neutered. (Wink)

I know this is pretty easy to work out with a calculator if you keep entering the numbers, but is there a equation that will work it out?

Thanks for any help working towards finding it.
• Mar 10th 2008, 06:44 AM
Tally
Let A(n+1)=A(n)+(10X16)/2X.1
=10+(160)/2x.1
=10+(16)/2
=10+8
=18 females after first year.

A(n) being amount of present female dogs = 10

Now I need to show A+2 is right (for next year), is that correct? If so thats where I start to struggle.
• Mar 10th 2008, 06:24 PM
Tally
I understand there is away to work this out in spreadsheet, could anyone please enlighten me on what would be the easiest way.

Thanks.
• Mar 11th 2008, 02:41 PM
Tally
So would next year be A(n+2)=A(n+1)+(A(n+1)X16)/2X.1

= 18+(18X16)/2X.1
=18+288/2X.1
=18+28.8/2
=18+14.4
=32.4 round down to 32
• Mar 15th 2008, 07:58 AM
CaptainBlack
Quote:

Originally Posted by Tally
The question is 10 female dogs average 16 puppies each year with 50% of those puppies being female. Every year the owners remove 60% of the female puppies and all of the male puppies.

What will the growth be of female puppies left and their mothers over 15years?

Let there be $\displaystyle A_n$ femails in year $\displaystyle n$, then in year $\displaystyle n+1$ there are:

$\displaystyle A_{n+1}=A_n+A_n\times 8 \times 0.4=4.2A_n$

So:

$\displaystyle A_{15}=4.2A_{14}=4.2^2A_{13}= ... = 4.2^{15} A_0$

and $\displaystyle A_0=10$

RonL
• Mar 15th 2008, 08:25 AM
Tally
Thank you, things are starting to become clearer now.(Yes)