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This really overcomplicates the mathematics. Steady state implies Leavers = Joiners. If 100 people join each month then you'll have steady state when there ar
by eterm 9y ago
This really overcomplicates the mathematics.
Steady state implies Leavers = Joiners.
If 100 people join each month then you'll have steady state when there are 100 leavers, i.e. when 50% of N is 100. N * 50% = 100 solves to N = 200.
In general with X% attrition and Y people joining you'll have a steady number of subscribers N using the formula
N * (1 - X) + Y = N
Which can be rearranged to N * X = Y.
- solonagathon 9y agoHow would you calculate the timeframe until steady number of subscribers is achieve d?
- codefined 9y agoWouldn't it always require an infinite amount of time? You're summing a series of: 0.5^n * x From n = 1 to n = infinity
- Sean1708 9y agoOnly if you can have a fraction of a subscriber, otherwise there will a finite number of months after which you will attain steady state.
- Sean1708 9y agoIf M is the number of months then (since you can't have a fraction of a subscriber) you will be in steady state when Y * X^(M - 1) < 1 so once M > 1 - (ln Y / ln X)
- smallnamespace 9y agoI'm constantly amazed at the lengths some coders will go to in order to avoid doing a couple of lines of algebra. Have seen people code up complicated Monte Carlo simulations, with all the attendant estimation noise, when basic stats and algebra gives the exact answer, sometimes even as a one-liner. Serious question: Is it simply a strong math phobia, or have people neglected the skill for so long that they've actually forgotten middle school level maths?
- mherrmann 9y agoI think it's that they don't have enough confidence in their mathematical ability to get to the correct answer. Programming lets you experiment and play around with the numbers more easily so you can convince yourself whether what you're doing is correct.
- pdkl95 9y agoI know I've had a few situations where the "play around with the numbers" part happened first. Usually once the actual goal was known the algebra (or similar details) would be obvious... but sometimes the inefficient/toy implementation was good enough. Sometimes half-assed is ok, if it's the the right half of the ass.[1] The wrong half becomes yet another type of technical debt to be refactored "someday". (this is not an excuse to ignore algebra or blindly assume the easiest/most-fun method is correct/useful/sane) [1] https://twitter.com/jimweirich/status/29662011323326464 https://twitter.com/jimweirich/status/29662011323326464 via https://www.youtube.com/watch?v=sCZJblyT_XM https://www.youtube.com/watch?v=sCZJblyT_XM
- dguest 9y agoI think it has as much to do with how math is taught as anything: it's easy to learn how to solve an equation, but convincing yourself that you didn't screw up without looking at the back of the textbook is another thing entirely. Of course there are a lot of softer tricks (like checking limiting cases) which would boost people's confidence, but for some reason this is only really taught in college level physics.
- fovc 9y agoI've seen both banking and consulting analysts try to solve this exact problem in excel. The ambitious ones will use circular references and let excel find the fixed point. In (good) excel, most formulas you write have a clear input => output relationship. Solving a system of equations (no matter how simple) breaks this relationship down. I suspect a similar thing happens when programming.
- Grustaf 9y ago
- patio11 9y agoI mean if you really want to nerdsnipe you could go straight to sum of an infinite geometric series.
- TimSAstro 9y agoVery Von Neuman! http://www.primepuzzle.com/leeslightest/howfar.html http://www.primepuzzle.com/leeslightest/howfar.html
- rtpg 9y agoI think the point here (that a lot of people might not even realize!) is that there's a steady state. One could assume that "well if I keep on going my customer count will keep on going up... right?" This is showing that it won't. Bonus pedantry: your proof just showed that _if_ there's a steady state, then that's the steady state. Turns out that you'll always get the steady state because of "maths" but if you replace 50% with 0% there's no fixed point.
- eterm 9y agoThat's a good point. I guess with a background in maths it's "obvious" that a system with constant in-flow and an out-flow proportional to the volume would approach equilibrium, but that perhaps isn't as obvious to others. I guess it's perhaps also worth pointing out then that the same equilibrium is reach from either side. If you start with a much higher volume of subscribers you'll reach the same equilibrium through attrition down to that number also.
- infinite8s 9y agoThat's only obvious if you remember differential calculus.
- bemmu 9y agoYes, that's the thing I wanted to show with this article. It wasn't entirely obvious to me that it wouldn't just forever increase as time goes by.