### 2019-09-02 - Model of continuous asset growth

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:

- How much you have invested
- The interest your investment makes. (The widely cited “Trinity study” suggests 4% as a “safe withdrawal rate”.)
- The salary you pay yourself
- How long your savings last for you

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

- The
**“how much” question**:*I want to pay myself 1000 USD. My stocks grow 4% per year. How much money do I need?* - The
**“how long until” question**:*I have 100 000 USD and save 3000 USD per month. How long until I have 200 000 USD?*

# First shot

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

*I want to pay myself 1000 USD per month. My stocks grow 4% per year. How much money do I need?*Well, the 4% you get per year should cover the yearly costs, so 1000/(1.04^{1/12}− 1) ≈ 306 000 USD.*I have 100 000 USD and save 3000 USD per month. How long until I have the 306 000 USD that you said I need?*That I modelled linearly, with just $ (306000 - 100000) / (3000/) 69 $.

# 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 *d**y*/*d**x* = *F*(*y*), then the solution is $x=\int ^{y}{\frac {d\lambda }{F(\lambda )}}+C$. In our case, we have *F* : *x* ↦ *i**x* − *c*′, so:

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

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

*C*

_{2}= (1 +

*i*)

^{−}

*C*, which is the actual multiplicative factor, and relabel

*C*

_{2}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))
```

```
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 :)