# How much do I need to save per month in an interest-bearing account to reach a certain account balance in a certain number of years?

How do I calculate what my monthly contribution should be to an interest bearing account, knowing how much I'll need to have in the account in a given number of years?

For Example,

I need \$8,203.03 in an account in 25 years. Money in the account garners 4% interest compounded monthly. My principle investment in the account is \$1,000. What is the formula that can tell me how much will I need to contribute monthly in order to have \$8,203.03 in the account in 25 years?

• It's doubtful that you can find any account that will guarantee you a particular interest rate for 25 years, and will let you keep making new deposits at that interest rate for all of those 25 years. If you ignore that (quite significant) practical difficulty, there are exact mathematical formulas to find, but they are perhaps more in scope for Mathematics. – hmakholm left over Monica May 3 '19 at 15:48

You can re-arrange the future value of an annuity equation to solve for payment (Pmt) needed instead of Future Value (FV):

`FV = Pmt * (((1 + r) ^ n) – 1)/r)`
`Pmt = FV / (((1 + r) ^ n) – 1)/r)`

That will get you payment needed to hit a given future value, but it won't account for the future value of the the starting balance, that should be removed from the desired future value after interest is calculated:

\$1,000 at 4% for 25 years with monthly compounding:

`n = number of periods = 12 months x 25 years = 300`
`r = rate per period = .04/12 = 0.003333`
`FV = Future Value = 8203.03`

`FV = PV(1+r)^n`
`FV = 1000 * (1+0.003333)^300`
`FV = 1000 * (1.003333)^300`
`FV = 1000 * 2.7135`
`FV = \$2,713.5`

So your \$1,000 will be worth \$2,713.5 without any additional contribution, remove that from your goal of \$8,203.03 and your target from the monthly contributions is: 5,489.53

So to solve for monthly contribution (Pmt) we'll use:

`n = number of periods = 12 months x 25 years = 300`
`r = rate per period = .04/12 = 0.003333`
`FV = Future Value = 5489.53`

`Pmt = FV / (((1 + r) ^ n) – 1)/r)`
`Pmt = 5489.53 / (((1 + .003333) ^ 300) – 1)/.003333)`
`Pmt = 5489.53 / ((1.003333^300) – 1)/.003333)`
`Pmt = 5489.53 / (1.7138)/.003333)`
`Pmt = 5489.53 / 514.19`
`Pmt = 10.68`

You need to contribute \$10.68 per month to hit your goal. May be off a few pennies due to rounding.

Even easier would be to leverage an online calculator, this one seems accurate and comes up with \$10.64. Also could solve in a spreadsheet which would enable you to handle rounding more flexibly.

You need the balance on month N to equal Bf. You make regular contributions of b to the account and get a monthly percentage yield of r (expressed as a real number multiplier. i.e., r = 1.005 is 0.5% monthly). Let B(n) be the balance at the beginning of month n; after the old balance has interest applied and the new contribution is added. Then

``````B(0) = B0
B(1) = B0r + b
B(2) = B0r^2 + br + b
...
B(n) = B0r^n + b(r^(n-1)+ ... + 1)
= B0r^n + b((r^n)-1)/(r-1)
``````

We can solve for b by using B(N) = Bf:

``````Bf = B0r^N + b(r^N-1)/(r-1)
b = (Bf - B0r^N) / [(r^N-1)/(r-1)]
``````

Your answer comes out to something like \$10.88. This gives you the contribution to reach the goal at the beginning of the last month in the 25-year period; to get the amount at the end, you can re-run the math for n = N + 1 and subtract the contribution from the N+1st month (leaving only the last month's interest). Or, you could rearrange the system and solve for end-of-month balances directly. That might be easier. The key is that understanding the derivation allows you to reason about the details of what your numbers mean.