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I've got a savings objective to achieve in 6 years and I'm trying to work out the percentage of salary I need to put towards it to achieve it.

I am trying to work out how to calculate the future value of my savings account.

Starting point

I have $10 in savings.

Contributions

Each year I contribute 10% of my earnings to my savings account.

Earnings

I will earn $1000 this year. Each year my salary increases with inflation and I'm assuming it's going to be 2% for the foreseeable future.

My savings account accumulates interest at a rate of 6% each year and all of that is re-invested.

I can do it the long manual way (excel spreadsheets), I was just wondering if there was an elegant formula I could use?

The question is:

Given the above, how much will I have in my savings account at the end of the savings period i.e. in 6 years time?

  • ha, not a homework - though if you have a textbook/online resource you could recommend that'd be handy. I'm using arbitrary numbers because I'd like to see if there's a formula I could use - rather than the massive spreadsheet that's inevitably prone to error owing to the many calculation steps. – user714852 Feb 8 '17 at 18:15
  • Not sure I understand the reason for the downvote. Is mathematical projection beyond standard compound interest not relevant in personal finance? – user714852 Feb 8 '17 at 18:17
  • Yes, I see that specificity is important - thank you for the suggestion. I've edited the question. – user714852 Feb 8 '17 at 18:20
2

Given the following

b[n] is the balance in period n
r is the periodic interest rate
i is the periodic inflation rate
d is the initial deposit made at period n = 0
x is the balance at period n = 0

b[n] = (d (1 + i)^n (1 + r) - (1 + r)^n (d + d r - i x + r x))/(i - r)

For example

starting with x = £1000
making quarterly payments beginning immediately with d = £100
earning interest at r = 2% per quarter
increasing payments by i = 1% per quarter to offset inflation

x = 1000
d = 100
r = 0.02
i = 0.01

The balance after four quarters and four deposits is

n = 4

b[n] = 1509.08

Mathematica calculation of formula

FullSimplify[
 RSolve[
  {b[n + 1] == (b[n] + d (1 + i)^n) (1 + r), b[0] == x}, b[n], n]]

{{b[n] -> (d (1 + i)^n (1 + r) - (1 + r)^n (d + d r - i x + r x))/( i - r)}}

  • You sir are a legend! I double checked manually in excel and it works! For future questioners - note that interest is applied on (x + d) - so it's assumed that the periodic contribution is applied at the start of the period. Thanks again Chris - I had resigned myself to tedious spreadsheets! – user714852 Feb 8 '17 at 21:37

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