# Sustainable population - When to breed

I would like advice on an approach to solving this real world problem. I'm not certain if the solution is to solve a system of equations, or something else. The problem is stated below with the equation systems that I derived from it. I think I have it solved by using a table, but I was hoping for a system of equations

If it takes a species:

• 52 days from conception to become "useful" (and they remain "useful" until death),
• 73 days from conception to reach adulthood (ability to breed/conceive again),
• 108 days from conception until death, and
• each generation gets exactly 1 day to breed

A) What is the minimum number of living generations required to always have a useful population?

B) To maintain this minimum number of generation what day in their life cycle should they breed?

In other words, if we have the following definitions for generation, $$Gen_n$$:

$$\begin{array}{ll} c_n, & \text{day of conception} \\ b_n, & \text{day of breeding} \end{array}$$

And we have to following functions and systems:

\left\{ \begin{aligned} &Useful(c_n) = c_n + 52 \\ &Adult(c_n) = c_n + 73 \\ &Death(c_n) = c_n + 108 \\ &b_n = c_{n+2} \text{ ; breeding day is conception day} \\ &Useful(c_n) <= Death(c_{n-1}) \text{ ; Useful when prev gen dies} \\ &Useful(b_n) - Death(c_n) <= c_{n+1} \text{ ; Gen_{n+1} can survive until next gen matures} \\ &b_{n+1} >= Death(c_n) \text{ ; At most 3 generations so Gen_{n+1} can't breed until Gen_{n} has died} \\ \end{aligned} \right.

Solve for $$b_n$$

So part A's answer is 3.

But how to solve B? My brute force table shows $$b_n$$ = $$\begin{cases} 0 < b_n < 18, & \text{if n is even} \\ 0 < b_n < 21, & \text{if n is odd} \end{cases}$$ for me {0,18} is sufficient

$$\begin{array}{r|r|r|r|r|r} Range & Useful & Conception & Death & Earliest Breed & Actual Breed \\ \hline - & 52 & 0 & 108 & 73 & 73 \\ 0 & 108 & 17 & 125 & 90 & 108 \\ 21 & 125 & 73 & 181 & 146 & 146 \\ 18 & 181 & 108 & 216 & 181 & 181 \\ 21 & 216 & 146 & 254 & 219 & 219 \\ 18 & 254 & 181 & 289 & 254 & 254 \\ 21 & 289 & 219 & 327 & 292 & 292 \\ 18 & 327 & 254 & 362 & 327 & 327 \\ \end{array}$$