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Model of continuous asset growth

Posted on 2019-09-02

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FIRE stands for financial independence/early retirement. The point is to save and invest money, and pay yourself a salary from the interest, eventually becoming independent on other sources of inocme.

There is a relationship between:

I have a program named worthy (on Github) that tracks my net worth and models when will I be financially independent under various assumptions. Here I describe the slightly fancy math behind a more accurate model for this relationship I finished implementing today.

I am probably rediscovering Financial Mathematics 101 ¯\_(ツ)_/¯

The questions

First shot

Previously the tool’s model was very basic, and answered the two questions as follows:

Problems

Assuming infinite retirement time

If you pay yourself a monthly salary of $ \$1000 $ and your monthly interest is $ \$1000 $, your money will last forever, beyond your (likely) lifespan. If you are fine with retiring with $ \$0 $, you can pay yourself a bit more than just the $ \1000 interest.

Ignoring growth while saving

“Take how much money I need - how much I have, divide by monthly savings” ignores that the money I saved up so far also earn interest, before I’m done saving. It’s too pessimistic.

Stand aside, I know differential equations!

Let’s model the depletion of your money as a function f, which will map number of years since retirement to the amount of money. You start with some initial amount f(0). If we pretend you withdraw the salary for a year and add interest once yearly, we’d get:

f(x+1) = f(x) + i ⋅ f(x) − c

Where i is the yearly interest rate and c are the yearly costs. In the example above, i = 0.04 and c = 12000 USD.

Then:

f(x+1) − f(x) = i ⋅ f(x) − c

If we instead pretend that everything is continuous and squint, this looks like a differential equation:

f′(x) = i′ ⋅ f(x) − c′

(Where i′ plays sorta the same role as i - except it’s not equal to i. For now let’s pretend it’s some unknown variable. Its relationship to i will eventually pop out.)

Wikipedia’s Ordinary differential equations article says that if dy/dx = F(y), then the solution is \(x=\int^{y}{\frac {d\lambda }{F(\lambda )}}+C\). In our case, we have F : x ↦ ix − c′, so:

$$x = \int^{f(x)}{\frac{1}{i'\lambda-c'} d\lambda}+C \stackrel{\text{Wolfram Alpha}}{=} \frac{\log(i'f(x)-c')}{i'} + C$$

Solving for f(x):

$$ \log(i'f(x)-c') = i'(x-C) \\ i'f(x)-c' = \exp(i'(x-C)) \\ f(x) = \frac{\exp(i'(x-C)) + c'}{i'} $$

So, magic happened and I pulled the general form of f(x) out of a hat. We know what are the i and c values when we assumed interest and costs happen only once yearly.

What about i′? Let’s guess it. If we had no yearly costs (so c = c′ = 0), we wanted to have f growing at a constant rate, gaining i in interest per year:

f(x+1)/f(x) = 1 + i

Substituting in the above equation of f, we get: exp (i′(x+1−C))/exp (i′(x−C)) = 1 + i

When we simplify the fraction, we get exp (i′) = 1 + i and therefore i′ = log (1+i). So, we have now successfully guessed the right value for i′ :)

Now what’s the right value of c′?

If we set interest to i = 0, f(x) should simplify to a nice linear equation losing c per 1 unit of x.

$$x=\int^{f(x)} -\frac{1}{c'} d\lambda + C = -f(x)/c' + C$$

So: $$-f(x)/c' = x-C\\ -f(x)=c'(x-C)\\ f(x)=-c'(x-C) $$

So the right value for c′ is exactly c.

So we have: $$ f(x) = \frac{\exp(\log(1+i)(x-C)) + c}{\log(1+i)} = \frac{(1+i)^{x-C} + c}{\log(1+i)} $$

C mediates a multiplicative factor before (1+i)x. C is just some constant that makes the function work with the f(0) boundary condition. Instead of wiggling the C, we can instead wiggle C2 = (1+i)−C, which is the actual multiplicative factor, and relabel C2 as C. (It’s an abuse of notation, but an OK one. *handwave*)

$$ f(x) = C \cdot (1+i)^{x} + \frac{c}{\log(1+i)} $$ The one remaining unknown variable is C, which we will get from f(0) - which are the initial savings.

$$f(0) = C + \frac{c}{\log(1+i)}$$

So:

$$C = f_0 - \frac{c}{i'}$$

Okay this is a little bit ugly. Let’s play.

c = 12000  # yearly costs
f_0 = 100000  # initial savings
i = 0.04  # interest

from math import log, exp
i_prime = log(1+i)
print(f'i_prime={i_prime}')

C = f_0 - (c / i_prime)
print(f'C={C}')

def f(x):
  return C * (1+i) ** x + (c / i_prime)

for r in range(11):
  print("after", r, "years, got:", f(r))

And it gives:

i_prime=0.03922071315328133
C=-205960.78029234003
after 0 years, got: 100000.0
after 1 years, got: 91761.56878830638
after 2 years, got: 83193.60032814502
after 3 years, got: 74282.91312957724
after 4 years, got: 65015.79844306671
after 5 years, got: 55377.9991690958
after 6 years, got: 45354.68792416598
after 7 years, got: 34930.44422943902
after 8 years, got: 24089.23078692297
after 9 years, got: 12814.368806706276
after 10 years, got: 1088.512347280921

Cool, it seems to be giving reasonable results. But our two questions were: how much money do I need to pay myself a given salary and how long until I save up the money I need.

Let’s instead first solve another question: if I have 100 000 USD and spend 1000 USD per month, how long will it last me.

For that, we just need to invert the familiar function:

$$ f(x) = C \cdot (1+i)^{x} + \frac{c}{\log(1+i)} $$

We want to know the number of years x at which we will run out of money (so f(x) = 0) $$ 0 = C \cdot (1+i)^x + \frac{c}{\log(1+i)} \\ (1+i)^x = \frac{-c}{C \log(i+1)} \\ x = \frac{\log{\frac{-c}{C \cdot i'}}}{i'} $$

And let’s test it:

x = (log(-c / (C * i_prime))) / i_prime
print(x)
10.090871103712766

Cool, this matches what the Python f(x) predicted above - after 10 years, it was just dwindling at about 1088 USD.

Answering the how long question

To answer the question “if I now have 100 000 USD collecting 4% interest per year and put in 1000 USD per month, how long until I have 306 000 USD”, we can use the same procedure - just plug in a target f(x) = 306 000 instead of zero and set a negative c to represent savings instead of costs. Details left as homework for the curious reader.

If you’re curious about the Go code, see this commit.

Answering the how much question

As a reminder, the “how much” question asks: if I want to pay myself a salary of 1000 USD per month, how much money do I need. Previously, I solved that with saying “the interest should cover all the costs”, which resulted in an investment that would last forever (a perpetuity). But now have a function that models an investment under conditions of withdrawing (or saving) money, and we can use that to model with a finite time horizon, and get a better estimate.

Say that we know that we are 40 years old and want our money to run out on our 100th birthday. So, after x = 60 years of paying ourselves, say, 1000 USD per month (so the yearly costs c = 12000), we want to have f(x) = 0. How much initial money f(0) do we need for that stunt of precious timing?

Okay, from above, we know:

$$ f(x) = C (1+i)^{x} + \frac{c}{i'} = \left(f(0) - \frac{c}{i'}\right) \cdot (1+i)^{x} + \frac{c}{i'} $$

So:

$$ f(x) = f(0)(1+i)^x - \frac{c}{i'}(1+i)^x + \frac{c}{i'} \\ -f(0)(1+i)^x = \frac{c}{i'} -f(x) - \frac{c}{i'}(1+i)^x $$

Let’s remember that we want f(x) to be 0.

$$ -f(0)(1+i)^x = \frac{c}{i'} - \frac{c}{i'}(1+i)^x = \frac{c}{i'}(1-(1+i)^x) \\ f(0) = \frac{c}{i'}(1-(1+i)^{-x}) $$

Let’s try it out:

c = 12000  # yearly costs
x = 60  # years for the investment to survive
i = 0.04  # interest
i_prime = log(1+i)
f0 = (c/i_prime) * (1-(1+i)**(-x))
print(f0)
276876.0258210814

Cool!

Recalling the numbers in the first section, the first algorithm which assumed an infinite horizon prescribed 306 000 USD for that situation (“1000 USD per month at 4% interest rate”). This more precise estimate cut 30 000 USD from the number :)