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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: 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: 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.

  • 4
    This is trivially easy to determine using a spreadsheet like Excel compared to constructing 1 mathematical formula. – BlackThorn Apr 16 '18 at 20:55
  • 2
    @BlackThorn - exactly. And the spreadsheet approach I use will allow changes along the way, e.g. "use 5% for my raises now, but in 10 years, change it to 3%." The single huge equation has a tough time with that. – JoeTaxpayer Apr 17 '18 at 9:58
  • Your rate equation is incorrect. See here: How rate of interest affects inflation rate – Chris Degnen Apr 17 '18 at 12:49
  • @BlackThorn & JoeTaxpayer Yes, trivially easy with a spreadsheet (or a bit of code) and I was able to do it myself, but I like elegant solutions, hence I was interested with one formula to take the resolve the question:). – Wiks Apr 17 '18 at 20:14
  • I hope you aren't going to make life choices based on theoretical models ;) – user45830 Apr 20 '18 at 7:32
9

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?

enter image description here

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

enter image description here

enter image description here

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.

enter image description here

enter image description here

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.

enter image description here

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

enter image description here

n =  270.757

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

enter image description here

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.

enter image description here

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.

enter image description here

  • Hi! Unfortunately I must've been not clear enough, I'm sorry. But it might be as well, that I misread this answer you've posted. Actually, here the answer concentrates on the 'how much' to accumulate and what I'm trying to find out is: when will I reach a state, where my investments' monthly returns will be equal to my monthly expenses. At Mad Fientist Lab they calculate that in 24 years I'll be able to pull out as much as I spend now without diminishing what I owe. – Wiks Apr 15 '18 at 17:23
  • Let's say I spend €1000 monthly. To withdraw €1000 from my investments I need: €1000*12/0,04 (4% withdrawal rate) = €300'000. Let's assume that I have nothing now. 1st year I save €10'000 and I earn 5% so I have €10'500. 2nd year I save €10'000 and I earn 5% on that so I have €20'500€ and the interest rates give me €21'525. 3rd year I save another €10'000 and so on. When I will reach €300'000? I can do this numerically in Excel but I'd love to have a single formula. I'm concentrating on when, but not how much :). – Wiks Apr 15 '18 at 17:28
  • So let me recap. You want to know how long, depositing €1000 monthly, it will take to build up enough capital so that every month €1000 can be withdrawn without diminishing the capital? And should the amounts include compensation for inflation? – Chris Degnen Apr 15 '18 at 18:37
  • Without inflation, the capital required p = w/m. The number of months to build it is n = Log[1 + (m p)/(d + d m)]/Log[1 + m]. – Chris Degnen Apr 15 '18 at 18:49
  • Calculating the capital required for a perpetual annuity that compensates for positive inflation is problematic because the withdrawal amount tends to infinity, i.e. w (1 + i)^infinity. – Chris Degnen Apr 16 '18 at 8:40
4

enter image description here

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.

  • Thank you for your answer. Actually I already have a spreadsheet where I put all my expenses and I wanted to entertain me with how much I am left to go to FI(RE). Like the one that Mad Fientist has in his lab (the first picture). I'll have a look at the spreadsheet from your post and I will let you know what I think about it! :) – Wiks Apr 17 '18 at 20:06
  • No problem. Mine is not to track expenses, but to literally track years to FIRE as requested. Only it’s a Year by year tracker, not a single formula. – JoeTaxpayer Apr 17 '18 at 20:11

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