Cheapest Way to Buy a Car

In Models of Car Depreciation, I promised the most FIRE version of car depreciation. We want a model of ownership cost to help smooth replacement costs over time, so the end goal is… replacement.

Autoreplacement

FIRE is all about putting things on autopilot. Save up a bunch of money, and let those savings generate an income stream to support your expenses in perpetuity. It kinda seems to me that our car replacement strategy should dovetail with the FIRE approach.

Imagine you have a pile of money, \mathbb{X}, that grows at rate r (e.g. 5%) faster than inflation every year so that after N years (e.g. 10 years), you have \mathbb{X}(1+r)^N. What if \mathbb{X} was large enough that it could cover the cost of a new car P (could even say “minus trade-in”) every N years and begin the process over again?

As usual, \mathbb{X} marks the spot.

So if I just bought a $30,000 car and want to replace it every 10 years going forward, I would need to set aside the following right now.

To check our math, observe that 47700\cdot1.05^{10} = 77700 = 47000 + 30000, that is the lump sum grows enough after 10 years to lay an egg and restart the process.

Depreciation Appreciation

Nothing above is new to FIRE folks; we’ve been doing it all along under a larger umbrella. Our major purpose in breaking it out as above is to capture the vehicle expense among our monthly or annual expenses.

On day 1 (a.k.a. year 0), that portion of your portfolio that produces the next car is \mathbb{X_0}=\frac{P}{1.05^N-1} as above. We can represent our monthly expense by dividing that dollar amount \mathbb{X} by 300 = 12 / 4\% (or annual expenses via dividing by 25 = 1 / 4\%).

What about year 5? How much of our portfolio is attributable to our autoreplacement? We have to allow the interest to grow the initial amount over time, but also inflation must be accounted for. Let’s fix those two parameters going forward. Let’s say our real rate of market return is 5% while inflation is 3%. Together, those produce a nominal rate of \approx8\% return. At year 5, we need to have the following set aside.

So your monthly expenses in year 5 is given by \mathbb{X}_5 / 300. In general, your car expense at any year n can be represented by the following.

And dividing that by 12 months allows you to use a constant amount within a year.

Example: a new $30k car every 10 years

Let’s run the numbers. A new (inflation-adjusted) $30k car every 10 years, a pretty reasonable example by any account. We’re assuming the market returns 5% on top of a 3% inflation rate.

The following table shows years 0, 3, 5, and 10 of a monthly depreciation schedule for autoreplacement and the methods in Models of Car Depreciation.

Notice

The columns in this table are surprisingly incomparable to each other. Straight light depreciation will account for the last purchase, while inflation-adjusted straight line accounts for the next purchase. Exponential decay shows what the last purchase is worth, inflation-adjusted, as a used car but says nothing about the next purchase.

Very relevant for our purposes, the exponential decay approach leads you to think your monthly vehicle expense is really small just before it isn’t.

Most relevant for our purposes, the autoreplacement column is increasing not because it’s actually costing us anything extra. It’s increasing because the market is growing your \mathbb{X}_n while you sleep. Autoreplacement is the sure FIRE cheapest way to buy a car!

Autoreplacement mechanics

Immediately after your last 30k purchase (month 1), you have 159\times\frac{12}{4\%} = 47,700 set aside to account for your vehicle replacement cost. The $159/mo expense that you attribute to this process earmarks $47,700 of your portfolio. I know I’m being sloppy with rounding the pennies off; we don’t have to overthink everything.

After 10 years and the next vehicle purchase (inflation-adjusting to 30,000\times1.03^{10}=40,318), you want to have the same amount left over (again adjusting for 3% inflation) of 47,700\times1.03^{10}=64,105 earmarked in your portfolio.

But at the end of the 10 years, just before the next purchase, you have \frac{12}{4\%}\times343=102,900 earmarked. This $102,900 very nearly equals $64,105 + $40,318 (the next round of autoreplacement, the next purchase resp).

The fact that it doesn’t equal perfectly follows from the fact that we’re using constant values within a year (i.e. instead of compounding monthly) and because we really should be using (1+ 0.0815) = (1 + 0.03)\times(1 + 0.05) instead of (1 + 0.08) = (1 + 0.03 + 0.05). We don’t need to overthink everything, and we certainly come close enough with the simplified arithmetic that it’s not worth being more precise.

Discussion and Assumptions

I originally named this concept autoreplacement because it automatically replaces an automobile. Being the cheapest way to buy a car, I thought it was clever enough name to suit the concept. However, the concept is so much more general and can be applied to any infrequent but semi-regular expense.

Have a roof to replace every 25ish years? Replace the HVAC every 15? New divorce every 7 years? Just kidding about the divorce example, autoreplacement won’t work with \mathbb{X}/2.

We’re using a 3% inflation rate throughout. It’s up to you if you think that’s reasonable, but it only really needs to reflect the price increase of the object to be autoreplaced.

We turn portfolio earmarks into monthly expenses via division by 300=\frac{12}{4\%}. Presumably, you’re using the 4% safe withdrawal rate to compute your F.I. number (required portfolio) based on your personal expenses, so this conversion works in that case. Feel free to adjust if you use a different withdrawal rate.

We’re using a 5% real rate of return assumption throughout, when elsewhere we use only 4%. The difference here is that annual withdrawals necessitate a smaller “safe withdrawal rate” to avoid sequence of return risk. Autoreplacement needs very infrequent withdrawals (e.g. every 10 years in the example above). It’s always up to you, but in the case of autoreplacement, I’m comfortable using the less conservative growth rate estimate. This is yet another reason autoreplacement is the cheapest way to buy a car.

The more flexibility a purchase has in choosing its timing, the less conservative you have to be. If the market crashes in 2035, then I can certainly wait until 2037 to replace my 12-year-old Honda.

Summary

Given expense P (priced at the time of the last replacement) every N years, the monthly expense that earmarks the portion of your portfolio that autoreplaces it can be given by the following.

This is nothing new for FIRE folks, and this only confirms what they already know. Today, however, we now have a name for the process and a mathematical framework to show it’s the cheapest way to buy a car.