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Calculating Savings Rate

How much do you need to set aside each year? It comes down to three things: how fast your investments grow, how fast inflation erodes the target, and how far away the target sits. The notation below uses gross multipliers, not rates — $g = 1.08$ means 8% annual growth, $i = 1.035$ means 3.5% inflation. This is a retirement target, so it takes the spending deflator, not the real-return hurdle (section “Which Number to Plan With”):

— base annual contribution, in today’s dollars
— investment growth multiplier (e.g. $1.08$ ), assumed greater than $i$
— inflation multiplier (e.g. $1.035$ )
— number of years
— target amount, in today’s dollars

Suppose you index your contributions to inflation each year — your salary keeps pace, so you save a constant share of a growing paycheck. To reach a goal worth $T$ in today’s dollars, you must accumulate $T \times i^{n}$ in nominal dollars by year $n$ , and the portfolio you build is:

Accumulated value = c \frac{i (g^{n} - i^{n})}{g - i}

(2.1)

This closed form is the same geometric-series trick as the annuity formula in section “Loan Amortization (Annuities)”: the year- $k$ contribution is $c i^{k}$ (the base amount, indexed $k$ times — so the first year’s deposit is $c \times i$ , a detail the worked example below returns to), and it then compounds at $g$ for the remaining $n - k$ years; summing those $n$ terms collapses to the expression shown. The unindexed case two formulas down is the plain annuity: set the indexing to zero and the bracket $\frac{g^{n} - 1}{g - 1}$ is $F V_{payments}$ with $g = 1 + r$ .

Setting that equal to $T \times i^{n}$ and solving for the contribution gives:

c = T \frac{i^{n - 1} (g - i)}{g^{n} - i^{n}}

(2.2)

If contributions are not indexed to inflation — you save the same nominal amount every year — each dollar has to do more work, so the required contribution is higher:

Accumulated value = c \frac{g^{n} - 1}{g - 1}

(2.3)

c = T \frac{i^{n} (g - 1)}{g^{n} - 1}

(2.4)

Take $g = 1.08$ (8% growth), $i = 1.035$ (3.5% inflation), $n = 20$ years, and a target of $T = $ 1, 000, 000$ . With inflation-indexed contributions:

c = $ 1, 000, 000 \times \frac{1.03 5^{19} (1.08 - 1.035)}{1.0 8^{20} - 1.03 5^{20}} = $ 32, 388

Here $c$ is the base figure in today’s dollars; the schedule is indexed, so the first year’s actual contribution is $c \times i = $ 33, 522$ , rising with inflation each year thereafter. Holding the contribution flat in nominal terms instead costs more — $43,481 every year:

c = $ 1, 000, 000 \times \frac{1.03 5^{20} (1.08 - 1)}{1.0 8^{20} - 1} = $ 43, 481

For the latter case you can also use the PMT function in Excel or Google Sheets to calculate your savings rate — its arguments are (rate, number of periods, present value, future value), and spreadsheets report payments as negative numbers, money leaving your pocket, hence the leading minus:

=-PMT(0.08, 20, 0, 1000000*1.035̂20) $= $ 43, 481.28$

or for same case, but with bi-weekly contributions (assuming 26 payments per year):

=-PMT(1.08̂(1/26)-1, 20*26, 0, 1000000*1.035̂20) $= $ 1, 611.21$ per paycheck, $\times 26 = $ 41, 892$ a year

The per-paycheck rate is $1.0 8^{1 ∕ 26} - 1$ — the rate that compounds 26 times to 8% — and not $0.08 ∕ 26$ , which would overshoot, for the same reason multiyear inflation is not the sum of annual rates.

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