# Minimum Required Nest Egg Formula?

As a follow-up to this question, what is the minimum size of a nest egg to support 30 years of \$5,583.33 monthly distributions? I want the \$5,583.33 to increase with inflation each month.

The present value of the first payment should be \$5,583.33.

``````First withdrawal will be in 20 years: \$5,583.33*(1 + 0.0033)^240 = \$12,310.86
``````

Here is what I've pieced together from this question:

``````Total withdrawals:          n = (30 years)(12 months) = 360 payments
Inflation per period:       i = 4.0% per year / 12 = 0.3333% per period)
Return per period:          m = 8.0% per year / 12 = 0.6666% per period)
Periods until 1st payment:  o = (20 years)(12 months) = 240 periods
First payment amount:       w = \$67,000 / 12 = \$5,583.33 (today's dollars)

p = ([(1 + i)^o]*[(1 + m)^-n]*((1 + i)^n - (1 + m)^n)*w)/(i - m)
p = ([(1 + 0.0033)^240]*[(1 + 0.00667)^-360]*((1 + 0.0033)^360 - (1 + 0.00667)^360)*5583.33)/(0.00333 - 0.00667)
p = \$2,594,790.06

where

n is the number of payments to be received
o is the number of the period at the end of which the first payment is received
w is the payment amount
m is the pension fund's periodic rate of return
i is the periodic inflation rate
``````

Is this the right equation? From what I could find on Google, this calculation is called the present value of a growing or graduated annuity. Is this correct?

Is it correct to say that \$2.5 million is the nest egg balance 20 years from now on the day the first withdrawal is made? And that \$2.5 million is not in today's dollars but in equivalent dollars 20 years from now?

• With `o` set to 240 you are applying 240 months of inflation to the first payment, so it will not be \$5,583.33 but `5583.33*(1 + 0.0033)^240 = 12310.86`. The present value of the first payment is \$5,583.33 but the actual first payment will be \$12,310.86. Likewise, the actual value of the pot at the end of the saving period will be \$2.5 million but its present value is \$1.2 million. I.e. \$1.2 will buy the same quantity of goods now as \$2.5 will buy in twenty years' time. – Chris Degnen Oct 26 '18 at 11:38
• There are so many variables, you can do this by just using some version of the 4% rule. 5K/month = 60k/year /.04 = 1,500,000. So somewhere in between 2 mill and 1.5 mil. – Pete B. Oct 26 '18 at 12:13
• A question is whether the OP wants his 60k/year upon retirement, or whether he means 60k/year in present value. With 2% inflation, in 20 years when he retires that will mean drawing `60k*1.02^20 = 89k/year`. – Chris Degnen Oct 26 '18 at 12:35
• @Chris you are right I would want the 60k adjusted for inflation. I mean 60k in today's dollars so 89 k. – random_dsp_guy Oct 26 '18 at 12:56
• The 4% rule assumes 2% inflation. Using the rule, `89k/0.04 = 2,225,000` is not far off the \$2,307,538 calculated here, which has the fund rate of return at 3% and increases pension payouts in line with inflation. – Chris Degnen Oct 26 '18 at 13:18

If you want the first payment amount to be \$5583.33 (unadjusted for inflation), `o` should be set to zero because `o` sets the number of inflation periods prior to the first payment received, (so that adjustment can be set from within the saving period).

To illustrate with a simple example, showing 4 deposits and 3 withdrawals.

Planning to retire in 4 months and draw monthly income of \$1000 for 3 months, adjusted for inflation starting from the first withdrawal. APR is 8% and inflation is 4%, both nominal rates, compounded monthly. What should the pot be?

Calculating the monthly rates.

``````inf = 0.04
i = inf/12 = 0.00333333

apr = 0.08
m = apr/12 = 0.00666667
``````

There are to be 3 payments received in total, at the end of periods 4, 5 & 6. The first payment should be \$1000 unadjusted for inflation. The second and third payments are to be adjusted for inflation.

Calculating the pot at the end of period 3 (using formula 2).

``````w = 1000
n = 3
o = 0

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 2970.28
``````

Checking the result

``````at the end of month 3, p = 2970.28
at the end of month 4, p = p (1 + m) - w (1 + i)^0 = 1990.59
at the end of month 5, p = p (1 + m) - w (1 + i)^1 = 1000.12
at the end of month 6, p = p (1 + m) - w (1 + i)^2 = 0
``````

So by the end of month 6 the pot is empty.

The three payment amounts are

``````w (1 + i)^0 = 1000
w (1 + i)^1 = 1003.33
w (1 + i)^2 = 1006.68
``````

``````w = 5583.33
n = 30*12 = 360
o = 0

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 1167478.60
``````

The pot should be \$1,167,478.60 at the start of the month prior to the first withdrawal, which will be \$5583.33.

With adjustment for inflation the final payment will be \$18,438.89.

``````w (1 + i)^(360 - 1) = 18438.89
``````

To illustrate what type of calculation this is, let inflation be zero. Then all the payments are \$5583.33 and the required pot is only \$760,915.72.

``````i = 0

p = ((1 + i)^o (1 + m)^-n ((1 + i)^n - (1 + m)^n) w)/(i - m) = 760915.72
``````

Demonstrating with Excel.

``````PV(0.08/12, 360, -5583.33, 0, 0)
``````

\$760,915.72

``````PMT(0.08/12, 360, 760915.72, 0, 0)
``````

-\$5,583.33

Excel correctly calculates the present value and payment amount. However, it does not have the facility to add in an inflation factor.

The Excel PMT calculation with cashflow at the end of each period uses the calculation of present value of an ordinary annuity, where the present value is `p`.

https://www.investopedia.com/retirement/calculating-present-and-future-value-of-annuities/

Derivations

The Excel PMT-type function can be derived from the sum of the present value of payments by induction.

``````∴ w = m (1 + 1/((1 + m)^n - 1)) p
``````

E.g.

``````m = 0.08/12
n = 360
p = 760915.72

w = m (1 + 1/((1 + m)^n - 1)) p = 5583.33
``````

With an inflation terms added: `i` and `o`, the sum of the present value of the payments becomes this, (formula 2).

• I want the first payment to be adjusted for inflation. I want it to be equivalent to \$5583.33 of today's dollars. – random_dsp_guy Oct 26 '18 at 17:32
• OK then. If you run the formula with `w = 5583.33`, `n = 360` and `o = 241` you will add twenty years of inflation to the calculation, as described my earlier answer. – Chris Degnen Oct 27 '18 at 9:25