I'm trying to figure out a formula for when I'll be able to retire at any given month

Question

I love the Mad Fientist laboratory that lets me see how far away I am from FI. Here is what I love about, the right side to the graph: But to motivate me a bit more, I wanted to calculate the FI date on my own after each month so I would be able to create a graph of how far away I am. To see if I progress after each month or not.

According to this website: https://www.moneycrashers.com/become-financially-independent-quickly-formula/

Basically, the Financial Independence Formula has two parts. The first part calculates your FI Number – the total amount of money required to give you a sufficient income for life:

FI Number = Yearly Spending / Safe Withdrawal Rate The second part of the formula uses your FI Number to figure out how many years it will take you to reach FI:

Years to FI = (FI Number – Amount Already Saved) / Yearly Saving

Which makes sense of course. But I would like to take into account also the withdrawal, growth and inflation rates. And that I'll be putting more money each year into the accounts to speed up the RE of the FIRE.

So I think it should be something like this:

//growthRate - inflationRate = rate (i.e.: 5% would be: 1,05)

Years to FI = log(rate) of -[((yearlySpendings/amountSaved * rate)*(1 - rate)) - 1]

This is how I imagine it:

But what I am missing is. But what I can't figure out is: how do I put into the equation the fact that each year I'll be adding some more money into the savings?

When I do numeric calculations in Excel, I get the same results as in Mad Fientist Lab. But I'd love to get pack it into one neat formula. :)

Edit: I think, that this answers the question: Future value of a compound-interest savings account when contributions are inflation linked but I do not have the Mathematica nor I am able to decipher the equation put there in the highest-awarded answer.

Answer 1

Copying a simple example from here, showing 4 deposits and 3 withdrawals.

Planning to retire in 4 months and draw monthly income of £1000 (present value) for 3 months, adjusted for inflation. APR is 8% and inflation is 4%. What should the pot be?

Calculating the monthly rates, assuming effective annual rates.

inf = 0.04
i = (1 + inf)^(1/12) - 1 = 0.00327374

apr = 0.08
m = (1 + apr)^(1/12) - 1 = 0.00643403

To illustrate the calculation, say we know at month 3 immediately after the final deposit, the pension pot p should be £3010.57

In month 4 it will have grown by (1 + m) and the inflation adjusted withdrawal will be w (1 + i)^4, where w = £1000. So the pension will decrease like so

p = 3010.57
p = p (1 + m) - w (1 + i)^4 = 2016.78
p = p (1 + m) - w (1 + i)^5 = 1013.28
p = p (1 + m) - w (1 + i)^6 = 0

This can be calculated in one go using a formula

o = 4  .  .  the month number
n = 3  .  .  the number of months
p = 3010.57

(-(1 + i)^(n + o) w + (1 + m)^n (i p - m p + (1 + i)^o w))/(i - m) = 0  (formula 1)

and more usefully, it can be expressed as a formula for p

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 3010.57  (formula 2)

So we need £3010.57 in month 3 for inflation adjusted withdrawals of £1000.

Starting with deposit d and increasing it to compensate for inflation

p = d
p = p (1 + m) + d (1 + i)^1 = 2.00971 d
p = p (1 + m) + d (1 + i)^2 = 3.0292 d
p = p (1 + m) + d (1 + i)^3 = 4.05854 d

This can also be calculated with a formula

q = 3  .  .  the final month number

p = (d ((1 + i)^(1 + q) - (1 + m)^(1 + q)))/(i - m) = 4.05854 d         (formula 3)

We know p = 3010.57

∴ d = 3010.57/4.05854 = 741.79

The above can be expressed as a formula for d

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 741.79            (formula 4)

So the first deposit will be £741.79

The next month the deposit will be £741.79 (1 + i) = £744.21 etc.

The first withdrawal will be £1000 (1 + i)^4 = £1013.16 etc.

Putting the steps together

inf = 0.04
i = (1 + inf)^(1/12) - 1 = 0.00327374

apr = 0.08
m = (1 + apr)^(1/12) - 1 = 0.00643403

o = 4  .  .  the first withdrawal month number
n = 3  .  .  the number of withdrawal months
w = 1000  .  the present value of the withdrawal amount

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 3010.57

q = 3  .  .  the final deposit month number

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 741.79

These same formulas can be used on a more realistically scaled calculation.

A more realistic calculation

For example, suppose someone at age 25 wants to withdraw $1000 per month present value from age 65 to 100. Inflation is 2% pa and interest is 3% pa (effective rates).

(65 - 25) * 12  = 480 deposit months
(100 - 65) * 12 = 420 withdrawals
(100 - 25) * 12 = 900 months overall

inf = 0.02
i = (1 + inf)^(1/12) - 1 = 0.00165158

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

o = 480  .  .  the first withdrawal month number
n = 420  .  .  the number of withdrawal months
w = 1000 .  .  the present value of the withdrawal amount

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 784011.41

q = 479  .  .  the final deposit month number

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 606.00

The plan could be achieved by making 480 deposits, starting aged 25 at $606, increasing monthly in line with inflation, i.e. $607, $608, $609 etc.

The first withdrawal at age 65 will be $2208.04: $1000 present value, i.e. w (1 + i)^480.

Using formula 3 and formula 1

Solution for a non-depleting annuity

In fact, a non-depleting inflation-linked annuity is not possible because the withdrawals will tend to infinity, as w (1 + i)^n when n -> infinity. However, as the OP's link states ...

"The study found that retirees ... can safely withdraw 4% of their starting money each year – adjusting annually for inflation – and have more left at the end of 30 years than they started with."

the calculation for capital required can be adjusted so that it is not depleted by a certain time, in this case 420 months. So, replaying the calculation above with the formula for p adjusted.

(65 - 25) * 12  = 480 deposit months
(100 - 65) * 12 = 420 withdrawals without depleting initial capital
(100 - 25) * 12 = 900 months overall

inf = 0.02
i = (1 + inf)^(1/12) - 1 = 0.00165158

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

o = 480  .  .  the first withdrawal month number
n = 420  .  .  the number of withdrawal months
w = 1000 .  .  the present value of the withdrawal amount

p = ((1 + i)^o (-(1 + i)^n + (1 + m)^n) w)/((-i + m) (-1 - m + (1 + m)^n))

  = 1217900.47                                                          (formula 5)

q = 479  .  .  the final deposit month number

d = ((i - m) p)/((1 + i)^(1 + q) - (1 + m)^(1 + q)) = 941.38

The plan could be achieved by making 480 deposits, starting aged 25 at $941.38, increasing monthly in line with inflation.

Again, the first withdrawal at age 65 will be $2208.04: $1000 present value, i.e. w (1 + i)^480.

The plot of capital can be extended by taking values of n beyond 420 months, illustrating how the inflation-linked withdrawals outpace the capital growth.

When will I reach a state, where my investment's monthly returns will be equal to my monthly expenses?

This is the OP's question, clarified in the comments. Without considering inflation the answer is straightforward. With effective APR of 3% and monthly deposit/withdrawals of $1000.

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

d = w = 1000

p = w/m = 405470.65  .  .  capital required

n = Log[1 + (m p)/(d + d m)]/Log[1 + m] = 280.898                       (formula 6)

So in 23 years and 5 months (281 months) the capital will not diminish.

With inflation

The same calculation compensating for inflation is more involved. With inflation at 2% and APR at 3%, deposits and withdrawals at $1000 present value.

inf = 0.02
i = (1 + inf)^(1/12) - 1 = 0.00165158

apr = 0.03
m = (1 + apr)^(1/12) - 1 = 0.00246627

Try 35 years of deposits

years = 35

d = 1000
q = years*12 - 1
o = q + 1
w = 1000

p = (d ((1 + i)^(1 + q) - (1 + m)^(1 + q)))/(i - m) = 999121.67

Using formula 5

solve ((1 + i)^o (-(1 + i)^n + (1 + m)^n) w)/((-i + m) (-1 - m + (1 + m)^n)) = p for n

n =  270.757

So with 35 years of deposits the capital will not diminish for 22 years and 6 months (270 months).

Rerunning the calculation for a range of deposit spans shows that for a 30 year period of withdrawals without diminished capital, 36 years and 10 months of savings would be required.

As previously mentioned, these inflation-linked calculations base all the deposits and withdrawals on present value, at the time of the first deposit. So in the examples above, if the deposits start today, all the future withdrawals will be worth $1000 in today's money.

Formulae derivations

The main formulae are derived from fairly simple recurrence equations, solved using Mathematica.

Answer 2

If Chris (poster of the other answer) is an equation guru, I'm an excel hack.

My approach is to create a spreadsheet. First, and most important, is that there are variables that are unknown. Inflation, wage raise, which I assign the same number, 3% or so. Percent of salary saved is under your control and I'd look at 15%, hopefully 10% from the worker, and 5% co match. And then, annual return. At the end of each year, one can update the present retirement account balance, and new salary.

The goal, the date of retirement, is based on when you hit your number which is unique to an individual. With my own goal, early retirement, I needed to be close to 20X, i.e. having about 20 times final income saved in order for a 4% withdrawal to replace 80% our pretax income. Why 80? Because while working, 15%+ went to retirement savings, and 7.5% to social security.

As someone gets closer to a regular retirement age, 65 or so, Social Security benefits can be considered. A 40% income replacement by SS means you need to replace only 40% more on your own, and 10X the final salary will do that.

Some time ago, I wrote an article on "The Number" which was a book that discussed the concept of that magic number in detail. The article links to the spreadsheet. You can and should change the numbers to reflect your assumptions. It was written as a guide, a framework for the analysis.