Suppose I want to retire at time \(T>0\).
I make constant payments to a savings account that earns continuously compounded interest \(r>0\).
I want my retirement fund to be worth \(V>0\) today (time \(0\)).
How much bigger do my payments have to be if I delay them?
Let \(X_d\) be the payments I have to make if I start saving at time \(d\in[0,T]\).
These payments form an annuity with value
$$\frac{X_d}{r}\left(1-e^{-r(T-d)}\right)$$
at time \(d\).
I want this value to equal \(Ve^{rd}\).
So my payments must equal
$$\begin{align} X_d &= \frac{r}{1-e^{-r(T-d)}}\times Ve^{rd} \\ &= \frac{rV}{e^{-rd}-e^{-rT}}. \end{align}$$
Therefore, delaying to time \(d\) increases my payments by a factor of
$$\frac{X_d}{X_0}=\frac{1-e^{-rT}}{e^{-rd}-e^{-rT}}.$$
The chart below shows how \(X_d/X_0\) grows with the proportion of time \(d/T\) I delay saving.
Part of this growth comes from having less time remaining: if my savings earn no interest, then the factor
$$\lim_{r\to0}\frac{X_d}{X_0}=\frac{T}{T-d}$$
equals the ratio of time until retirement and time spent saving.
Raising \(r\) raises \(X_d/X_0\) because I forgo more opportunities to earn interest on my interest the longer I delay.
This is especially true when I’m far from retiring (i.e., \(T\) is large).
Thanks to Michael Boskin for inspiring this post.