Loan Math

Let's set up the symbols for the parameters first.

• \(T\) as the initial principal (this parameter doesn't matter in the analysis)
• \(N\) as the max term in months
• \(R_1\) as the initial monthly interest rate, which is advertised as the overall "service charge" rate, or simply interest rate when talking casually.

This rate can be misleading, because the loan requires monthly repayment of the principal in the amount of \(\frac{T}{N}\), and consequently, the monthly interest rate \(R_i\) actually increases throughout the term. But by how much exactly?

The rule also says it's possible to repay the full principal prematurely after one year, say in \(n\) months, that is, \(12 \le n \le N\), and in the last month, you repay the rest principal at once.

What we want to figure out is the actual flat interest rate \(\hat R_n\) in the end, which is not to be confused with \(R_n\), the interest rate of the final month.

First, let's solve a simpler problem: given that you don't repay the principal prematurely, what is the monthly interest rate \(R_i\), where \(1 \le i \le N\)?

\[ R_i = \frac{R_1 T}{\frac{N+1 - i}{N} T} = \frac{N}{N+1 - i} \cdot R_1 \]

Intuitively, we observe the following:

• The last month's rate is a whopping \(N R_1\).

• This rate initially increases slowly (the larger \(N\), the slower), but it gets faster and faster towards the end of the term, e.g. \(R_N = 2 R_{N-1}\).

But the real question is, what is the overall (flat) interest rate, i.e. \(\hat R_N\)?

Is it just the simple average of the monthly rates?

Let's see what that average, \(\bar R_N\) looks like.

\[ \bar R_N = \frac{1}{N} \sum_{i=1}^N \frac{N}{N+1 - i} \cdot R_1 \\ = \sum_{i=1}^N \frac{1}{N+1 - i} \cdot R_1 \]

Note that the multiplier turns out to be just the harmonic number \(H_N\).

Now let's actually solve for the real deal, \(\hat R_N\).

\[ \hat R_N \sum_{i=1}^N \frac{N+1 - i}{N} \cdot T = N \cdot R_1 T \]

So we've got

\[ \hat R_N = \frac{N^2}{\sum_{i=1}^N N+1 - i} \cdot R_1 \]

After simplification, we've got

\[ \hat R_N = \frac{2N}{N+1} \cdot R_1 \]

Note that for large \(N\), we've got a doubling!

Finally, let's generalize to \(\hat R_n\), where \(1 \le n \le N\).

\[ \hat R_n \sum_{i=1}^n \frac{N+1 - i}{N} \cdot T = n \cdot R_1 T \]

This gives us

\[ \hat R_n = \frac{2nN}{2nN - n^2 + n} \cdot R_1 \]

Now let's see some real-life figures.

Let \(f(n) = \frac{\hat R_n}{R_1}\) denote the multiplier.

Note:

1. The rate of change of \(f(n)\) increases over time;
2. At any given \(n\), \(f(n)\) with the larger \(N\) always has the smaller value.

This is discrete math, but if you think of it in a continuous fashion for the sake of easier visualization, the effect of a larger \(N\) is essentially stretching the curve horizontally. Yet regardless of \(N\), those curves have the same kind of shape (hyperbolic) and they all grow from the initial value of 1 to \(2N/(N+1)\), which is close to 2 for large \(N\). Then it becomes quite intuitive that the ones with larger \(N\) are always underneath those with smaller \(N\).

Knowing this, it becomes obvious that you should always choose the option with the longest possible term. And the banks are aware of that. So to offset this clear advantage of the longer-term loans, they offer a more attractive initial monthly interest rate \(R_1\) if you go with one of the shorter-term options.

For instance, the \(N=96\) option has \(R_1 = 0.25 \%\), but if you go with \(N=60\), you get \(R_1 = 0.225 \%\).

Now because it's a lot more common to see annual instead of monthly interest rates, let's present the comparison all in APR instead.

Note that even though the 60-month option starts at a lower APR of \(2.7 \%\), as opposed to \(3.0 \%\) with the 96-month, they're getting close within the first two years, and we begin to pay interest at a lower APR just from the 3rd year on with the 96-month, and its 7th-year APR is actually lower than that of the 5th year with the 60-month. In addition, the 96-month requires a significantly lower monthly repayment of principal.

There is a clear winner here (and that's why they're shutting it off).

Oh, I forgot to respond to the question raised earlier, that is, Is \(\hat R_N\) equal to the simple average \(\bar R_N\)?

Well, now that we know the multiplier of the latter is just \(H_N\), we can simply compare that with the result in Table 1.

• \(H_{60} = 4.680\), whereas \(f_{N=60} = 1.967\)
• \(H_{96} = 5.147\), whereas \(f_{N=96} = 1.979\)

That's way off! For instance, an APR for \(N=96\) being 15.44% is just nuts. So it's clearly not the simple average of monthly rates. But that's not too much of a surprise. Intuitively, recall that the last month's rate is a whopping \(NR_1\), but at the same time, the remaining principal is also a minute \(T/N\), so the weight of that crazy rate should also be the smallest. (Remember, the equation is about an invariant on the total interest paid, not just the interest rate, therefore, each monthly remaining principal must also play a role here.)

But I don't know what the exact weights of all the monthly rates are, this might be a good exercise for later.