Tuesday, October 28, 2014

Compound Interest, Loans, and Saving for Retirement

Suppose you borrow $B at time t=0 at an annual interest rate i. If you don't make any payments, the loan compounds, and at the end of T years ballons to \(B (1 + i)^T\).

As a concrete example, if you borrow B=$10,000 at 5% (= 0.05) for 10 years, your debt expands to nearly $16,289 at the end of the loan term, if you don't make any payments.

Nobody likes to ponder sad eventualities, so lets consider the more optimistic problem of saving for retirement. In a subsequent post, we will connect these two problems: irresponsible borrowing, and prudent saving, for how to responsibly pay down a home mortgage, for example.

Consider you make constant regular payments of $D to your retirement account every year from t=1 to t=T.

If we assume that our money grows at the same constant rate of i=0.05, we want to figure out how much we end up with after T years.

The first installment has (T-1) periods to grow over. So it grows to $\(D (1+i)^{T-1}\).

Similarly the second installment has (T-2) periods to grow over; so it grows to $\(D (1+i)^{T-2}\).

So on and so forth.

Thus, the total amount of money at the end of T years is given by,\[S^+ = $D(1+i)^{T-1} + $D (1+i)^{T-2} + ... + $D.\] This is a simple geometrical series, which conveniently sums up to,\[S^+ = $D \frac{(1+i)^T-1}{i}.\] To finish off this post, and set up the stage for how to responsibly pay down loans, let me plot the increase in the loan amount as a function of time (I am going to call the number \(S^-\)), and the growth of the retirement account.

Again, using B = 10,000, i = 0.05, and T = 10:

Using D = 100, i = 0.05, and T = 10:
In this case, we end up with $1,257.5, out of which only $1,000 were contributions, and the rest was compound interest.

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