Equilibrium Problem: Landscaping Business|Accounts & Maintaining 500

[markn]

I own a landscaping business with accounts paying regularly. If I add 5 new accounts a week and on average an account stays a customer for one year before quitting, what would my equilibrium number of accounts be? How long to reach equilibrium? how many do I need to add per week to maintain at a certain number, say 500?

This is a real situation for me trying to plan business capacity in the future. A formula for the general case would be much appreciated, n accounts added, stay on for x number of days before quitting, etc.

 

[MarkFL]

Suppose you start at day zero, no accounts, and then as you progress, you have an average of $n$ new accounts per week who each stay with you for $x$ weeks. At the end of the first $x$ weeks, you then have $nx$ accounts, and thereafter, you have the same number of accounts dropping as you are adding, so your equilibrium would be $nx$. In order for this to be 500, we require:

\(\displaystyle nx=500\)

There are an infinite number of solutions...as $n$ increases, then $x$ decreases (and vice versa), as there exists an inverse relationship between the two.

 

[markn]

At the end of the first x weeks I don't think I have nx accounts. After say 50 weeks, given n = 5 it would not be 250 accounts, it must be less than 250 accounts.

If each account stays a year on average some number would have quit before 50 weeks so it must be less than 250.

My math is rusty but it's more complicated, I don't know if it's a diff eq problem, been awhile for me. Thx