# Retirement Plan Returns Formula

My income after tax is \$48,060 and I'm contributing 6% of my income into my retirement account.

My estimated income increase is 2.6% per year, which I hope to include in my retirement account.

The expected average rate of return is 10 percent and I'm planning to retire in 40 years.

What's the ending value of my nest egg?

If I start saving 10 years before my retirement, what's my return?

What's the ending value if my rate of return is 5%?

What's the ending value if I decide to work 5 extra years?

• The way you write makes this sound like a homework problem. Whether it is or isn't, we -- and you -- need to know whether that 6% is pre-tax or after tax, and whether it's 6% of your net or gross pay. – RonJohn Nov 6 at 4:46
• Also, given that your pay increases by 2.6% each year, I'd fire up a spreadsheet and do a set FV() calculation. – RonJohn Nov 6 at 4:48
• Lastly, "If I start saving 10 years before my retirement, whats my return?" makes no sense. The ending balance is required before calculating ROI. – RonJohn Nov 6 at 4:49
• April - I wrote a spreadsheet which does exactly as you wish. It will let you zero out any existing assumptions, such as saving from the time one starts working. And put it whatever percentages for salary increase, saving rate, saving return. – JTP - Apologise to Monica Nov 6 at 11:50

Starting with a simple demonstration saving over 3 years

With

c = initial salary contribution
i = salary increase
r = rate of return

c = 48060*0.06
i = 0.026
r = 0.1

The deposits over years grow with salary increases

d0 = c          = 2883.60
d1 = d0 (1 + i) = 2958.57
d2 = d1 (1 + i) = 3035.50

The accumulated savings with interest r are

s = d0 (1 + r)^3 + d1 (1 + r)^2 + d2 (1 + r) = 10756.99

A formula for this, for n periods, is

s = (c (1 + r) ((1 + i)^n - (1 + r)^n))/(i - r)

from the expression for the deposit in year n (starting from 0, because the deposits are made at the beginning of each year)

$d(n)=c(1+i)^n$

combined with deposit summation, with interest, to obtain (by induction)

$s=\sum_{k=1}^{n}c(1+i)^{n-k}(1+r)^k=\frac{c(1+r)((1+i)^n-(1+r)^n)}{i-r}$

So savings over 3 years are calculated thus

n = 3

∴ s = (c (1 + r) ((1 + i)^n - (1 + r)^n))/(i - r) = 10756.99

This matches the result of the long-hand calculation above, confirming the formula.

Turning to the OP's cases. He plans to retire in 40 years, and save 10 years prior to that. Salary 30 years hence will have increased, so

c = 48060*0.06 (1 + i)^(30 - 1) = 6070.28

n = 10

∴ s = (c (1 + r) ((1 + i)^n - (1 + r)^n))/(i - r) = 117404.54

If the rate of return is 5%

r = 0.05

∴ s = (c (1 + r) ((1 + i)^n - (1 + r)^n))/(i - r) = 89303.84

If 5 years extra are worked (and r remains at 5%)

n = 15

∴ s = (c (1 + r) ((1 + i)^n - (1 + r)^n))/(i - r) = 161811.96

Or if 5 years extra are worked with r at 10%

r = 0.1

∴ s = (c (1 + r) ((1 + i)^n - (1 + r)^n))/(i - r) = 244318.05

Plotting over years 30 to 45

• Including tax rate and inflation rate would give a much more useful and realistic (and depressing) result. – Ray Butterworth Nov 6 at 14:05