: How much do I need to invest monthly to accumulate a given amount? 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

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 0 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 0 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 ,655 would have the purchasing power of ,944 today.

To work it backwards, if you want 0,000 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 0.16 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 0,000 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 2.84, 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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