I have a target amount T
USD (in today's value, before inflation) that I would like to accumulate over N
years. Every month I invest X
USD with an expected return of Y%
. Expected inflation is Z%
per year. How can I calculate whether I am investing enough? If that time is not enough, how long do I need to invest the money? How much should I invest monthly to meet the required deadline?
1 Answer
If you save x
every month, the future value (FV) is
where
n is the number of months
r is the monthly rate of return
So if Y%
return is 10% nominal interest compounded monthly
r = 0.10/12
For example, over 3 months, saving $100 per month
n = 3
x = 100
FV = (x (1 + r) ((1 + r)^n - 1))/r = 305.03
Checking the balance at the end of each month long-hand
b1 = 100 (1 + 0.1/12) = 100.83
b2 = (b1 + 100) (1 + 0.1/12) = 202.51
b3 = (b2 + 100) (1 + 0.1/12) = 305.03
The formula checks out.
So after 10 years, saving $100 per month
n = 120
x = 100
FV = (x (1 + r) ((1 + r)^n - 1))/r = 20655.20
Discounting for inflation at, say, 2% per annum
FV/(1 + 0.02)^10 = 16944.46
Your future saving of $20,655 would have the purchasing power of $16,944 today.
To work it backwards, if you want $100,000 in today's value in 10 years
FV = 100000 (1 + 0.02)^10 = 121899.44
r = 0.10/12
n = 120
x = (FV r)/((1 + r) ((1 + r)^n - 1)) = 590.16
You would need to save $590.16 each month.
Compensating for inflation
Inflation can be compensated for by increasing the monthly payments at the same rate as inflation. This makes the payments equal in 'value' terms.
Inflation is usually quoted as an effective annual rate, so with 2% (as before) the monthly rate is obtained like so
i = (1 + 0.02)^(1/12) - 1 = 0.00165158
and the 3 month long-hand calculation becomes
b1 = 100 (1 + 0.1/12) = 100.83
b2 = (b1 + 100 (1 + i)) (1 + 0.1/12) = 202.67
b3 = (b2 + 100 (1 + i)^2) (1 + 0.1/12) = 305.53
This can be expressed as a formula
Once again, to save $100,000 in today's value over ten years
FV = 100000 (1 + 0.02)^10 = 121899.44
n = 120
i = (1 + 0.02)^(1/12) - 1 = 0.00165158
r = 0.10/12
x = (FV (i - r))/((1 + r) ((1 + i)^n - (1 + r)^n)) = 542.84
The first payment is $542.84, and the payments increase like so
month 1 x (1 + i)^0 = 542.84
month 2 x (1 + i)^1 = 543.74
month 3 x (1 + i)^2 = 544.63
...
month 120 x (1 + i)^119 = 660.63
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5Note that these formulas assume deterministic returns. If returns are uncertain, you can use simulation to estimate your probability of having enough by a date in the future. Dec 3, 2019 at 16:20
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2
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2I have added a calculation to raise the payment amounts in line with inflation. Dec 4, 2019 at 10:45
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1Awesome! Now I just have to figure out what each term actually means. Should probably go back to high school... :p– MarsDec 4, 2019 at 11:20
T
in today's dollars