Generic FV of PV Formula with equal inflation and savings rate

Question

I’m trying to arrive at a formula that allows me to determine the future value of a present savings amount (P), given the following:

Period (p): e.g., Annual (p=1), Semiannual (p=2), Quarterly (p=4), Monthly (p=12) Years of withdrawals: y Annual withdrawal amount: W Annual savings growth percent: r Periodic savings growth rate: m = (1+r)^(1/p)-1 Annual inflation percent: i

I found a formula on this forum that allows for the above, based on withdrawals at the END of the chosen period as follows:

FV = W/p*(r/m)[(1+i)^y-(1+m)^(yp)]/(r-i) + P*(1+m)^(y*p)

I was able to modify it to allow withdrawals at the beginning of the chosen period as follows:

FV = W/p*(1+1/m)[(1+i)^y-(1+m)^(yp)]/(r-i) + P*(1+m)^(y*p)

But the above formulas provide errors when i = r. I’ve been trying to arrive at a general formula that allows the inflation rate to equal the savings rate.

Can anyone help?

Thanks in advance.

Answer 1

Starting with a simple case: 3 years annual withdrawals and compounding, no inflation.

P = 1000
W = -120
r = 10%

P = 1000
P = (P + W) (1 + r)
P = (P + W) (1 + r)
P = (P + W) (1 + r) = 894.08

Same result using a formula, with n = 3

Now with inflation at i = 12% diminishing the purchasing value of P

P = 1000
P = (P + W) (1 + r)/(1 + i)
P = (P + W) (1 + r)/(1 + i)
P = (P + W) (1 + r)/(1 + i) = 600.084

Formula for future value adjusted for inflation is

FV = W (1+i)^-n (1+r) ((1+i)^n - (1+r)^n)/(i-r) + P (1+r)^n/(1+i)^n

For different compounding periods modify r, i and W accordingly, e.g. for monthly

r = (1 + 0.1)^(1/12) - 1
i = (1 + 0.12)^(1/12) - 1
W = -120/12 = -10

This assumes the withdrawal frequency follows the compounding frequency. Alas W is not adjusted for inflation here.

Continued in the light of OP's comment

The OP wants to apply inflation to the withdrawals so that they retain their purchasing power. So same simple 3 year model:-

P = 1000
W = -120
r = 0.1
i = 0.12

P = 1000;
P = (P + W (1 + i)^0) (1 + r)
P = (P + W (1 + i)^1) (1 + r)
P = (P + W (1 + i)^2) (1 + r) = 843.075

As a summation and formula with n = 3

When ì ≠ r the formula can be used:

FV = W (1+r) ((1+i)^n - (1+r)^n)/(i-r) + P (1+r)^n

and in the case where i = r a loop could be used, e.g.

subtotal = 0

For[k = 1, k <= n, k++,
 subtotal += W (1 + i)^(n - k) (1 + r)^k]

subtotal + P (1 + r)^n

843.075

Answer 2

As @GradeEhBacon said in the comments: if rate of inflation and rate of growth are equal, they cancel out in terms of real value of the money now vs. later. In the absence of other changes, you can just take the difference of the two as your rate of growth or reduction in relative buying power.

Of course both will actually vary over time, as do rates of deposit and withdrawal, and those details are hard to predict. Usual practice is to try to estimate likely scenarios, run calculations for them, see what the results would be and try to make plans based on the spread of those snapshots/samples (taking into account how likely you think they are, or just picking a "reasonable" case).

For back-of-the-envelope retirement planning purposes, a common guesstimate is that a reasonable set of investments will return about 4% more than inflation, meaning that if you will be living entirely on your retirement savings and you want them to last "forever" they should be at least 25x what you think your typical yearly spending will be, using costs at the time of retirement. Obviously other sources of income (pensions, annuities, SSI) also get figured into that.

(Unfortunately some costs do tend to go up faster than general inflation, such as medical bills. And end of life costs often surge. On the other hand few of us live forever, so it's close enough for a rough estimate of "can I safely retire yet". Personally my actual plan runs to age 110, and doesn't include selling the house, so I'm pretty confident that I am on the safe side of the balance)