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I am building a financial model. I am looking for the value of Y:number of subscribers over X:month.

The model calls for the following.

  • Competing of a market with 500,000 exclusive subscribers.
  • Growth in year one will be exponential month over month with the total at the end of the year being 2500 subscribers.
  • Growth in following years is expected to be 250% but will decrease as the product saturates the market.
  • 20,000 new subscribers will be introduced into the market each year with a total of 2,000 leaving the market each year.
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I can't give you an equation for $y(x)$ but I can give you a differential equation that it must satisfy...I'm not sure if an exact (closed form) equation exists.

To derive the equation, note that if the total-subscriber population was unlimited rather than 500,000, an exponential growth model would be: $$ y'(x) = ry(x),\quad y(0) = y_0 $$ where $r>0$ is the growth factor and $y_0$ is the initial number of people subscribing to your service. The term $y'(x)$ is the change in $y$ with respect to $x$...the derivative from calculus. If you forgot your calculus, then think of it as how much $y$ changes when $x$ increases by one month (this is almost true). In this case the solution is $y(x) = y_0 e^{rx}$, which is exponential. Note that the growth is zero if you start with no one subscribing to your service! Let's assume therefore that you're able to get a small number of initial subscribers and call this number $y_0$. In fact, your assumption of 2500 year 1 subscribers and 250% growth means that $$ y_0 = 1000. $$
Your assumption of 250% yearly growth means that $$ r = \frac{\ln 2.5}{12}. $$ This model is insufficient since it doesn't take saturation into account. To add that in (still ignoring the 20,000 new - 2,000 dropped yearly subscribers) we try the model $$ y'(x) = r y(x)(1 - \frac{y}{250000}), \quad y(0) = 1000 $$ Now, when $y(x) < 250,000$ growth will be positive, and when $y>250,000$ growth with be negative. Of course $y>250,000$ is impossible, but the effect will be saturation at 250,000 (the solution will never grow beyond 250,000 unless it starts there...). One can show that the solution to this is $$ y(x) = 250000 \frac{1000 e^{rx}}{1 + 1000 e^{rx}}. $$ This is the logistic sigmoid, and this model is logistic growth, most often used in modeling populations.

We now have to take into account that the general subscriber population grows by 20,000 each year, but 2,000 people are lost. The general population growth can be modeled by replacing 250,000 with $$250,000 + 20,000\frac{x}{12}$$ Now, to deal with the lost subscribers, we note that some of those 2,000 will be our subscribers, and some will not. This complicates things significantly. We can take the approximation that the total population gains 18,000 people per year. This is approximately true when you have much less than 250,000 subscribers. While it is possible to take this complicating factor into account, I'd argue, "what's the point?" There will always be some error with your model...also, there is no way your model will simultaneously work well at this stage in time (when you have a small market share), and further down the road when you are close to I think you should just attempt a model that works well when you have a small to medium market do otherwise would be to confuse mathematical models with reality. If your investors require a model that "works" everywhere, then that can be done. In any case, our simplification means that we replace 250,000 with $$c(x) = 250,000 + 18,000\frac{x}{12}.$$ $y(x)$ then satisfies the equation $$ y' = ry(1-\frac{y}{c(x)}),\quad y(0) = 1000. $$ You can solve for $y(x)$ using e.g. MATLAB or Python or whatever. It will look like a logistic sigmoid, but the right side will slowly grow without bound.

If I wanted to complicate the model I would add some uncertainty in the growth rates using some random variables.

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This is a type of sigmoid function maybe more specifically a logistic function.

Unless you are a transhumanist, in which case the growth will obviously never stop and will instead reach a singularity like 1/0.

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