A Multi-Loan Engine

After I started work on my loan savings derivation, I realized that scheduled monthly payments creates an opportunity to build a sort of engine. I shall define an "engine" here as a device which cyclicly harvests the difference between conditions internal and external. A combustion engine harvests the difference in temperature and pressure between its piston bores and the exhaust pipe. The efficiency of the engine depends upon the magnitude of the difference, and if the external temperature and pressure was as great as that internal, the engine would not function at all.

We know from our previous work that a loan with consistent monthly payments can be modeled $$ P_{n} = Pr^n - \sum_{l=0}^{n-1}\$r^{l}. $$

Suppose now that instead of making the same monthly payment, $\$$, we made some different payment, $\$ + \Delta$. The difference can be positive or negative, and if it depends on when it is made we write a new model for $P_n$ where $$\begin{aligned} P_n &= Pr^n - \sum_{l=0}^{n-1}(\$ + \Delta_l)r^l\\ &= Pr^n - \sum_{l=0}^{n-1}(\$r^l + \Delta_l r^l)\\ &= Pr^n - \sum_{l=0}^{n-1}\$r^l - \sum_{l=0}^{n-1}\Delta_l r^l . \end{aligned}$$

If the sum $\sum_{l=0}^{n-1}\Delta_l r^l = 0$, then this new model will return the same results as when the monthly payment was the always the same. Let's look at a simple example to show what this means. Say you pay on average $\$$ monthly, but one month you are a bit short on cash, so you pay $\Delta$ less than $\$$. To make up for it, you earn more money so that next month you pay $\Delta r$ more than $\$$. Then the sum would be $$\begin{aligned} \sum_{l=0}^{n-1}\Delta_l r^l &= -\Delta r^l + (\Delta r)r^{l-1}\\ &= -\Delta r^l + \Delta r^{l-1+1}\\ &= -\Delta r^l + \Delta r^l\\ &= 0. \end{aligned}$$

Since you've done this, you don't fall behind on payments and are able to finish paying your loan on time. Solving $P_n$ for $n$ gives the same results as if you had never missed part of a payment.

We will now build an engine. Let's say you have two loans, each with interest rates $r_a > r_b$ and monthly payments $\$_a > \$_b$. One month you decide to take all your payment from loan $b$ and pay it towards loan $a$, which means you pay $(\$_b - \$_b)$ and $(\$_a + \$_b)$ respectively. The next month, you pay back loan $b$ using part of $\$_a$. In the same way as above, you get $$\begin{aligned} \sum_{l=0}^{n-1}\Delta_l r^l &= -\$_b r_b^l + (\$_b r_b)r_b^{l-1}\\ &= 0. \end{aligned}$$

The situation for loan $a$ is a little bit different, as seen by $$\sum_{l=0}^{n-1}\Delta_l r^l = \$_b r_{a}^{l} - (\$_b r_b + x)r_{a}^{l-1}.$$

If we set this sum to $0$ and solve for $x$, we find $$\begin{aligned} 0 &= \$_b r_{a}^{l} - (\$_b r_b + x)r_{a}^{l-1}\\ (\$_b r_b + x)r_{a}^{l-1} &= \$_b r_{a}^{l}\\ \$_b r_b + x &= \$_b r_{a}\\ x &= \$_b (r_a - r_b). \end{aligned}$$

Given that $r_a > r_b$ it is obvious that $x > 0$. How do we interpret this? Well, it means that in order to not pay loan $a$ back at a different time than expected, we have to take away $x$ more than we are paying back to loan $b$. But we know that the maximum we can take away from loan $a$'s payment is $\$_a$, so does that work? Well, if $\$_a - (\$_b r_b + x) \geq 0$, we at least break even on the transaction. We see this when $$\begin{aligned} \$_a &\geq \$_b r_b + \$_b (r_a - r_b)\\ \$_a &\geq \$_b (r_b + r_a - r_b)\\ \$_a &\geq \$_b r_a. \end{aligned}$$

So now we have the conditions to set up our engine. If we have two loans with interest rates $r_a > r_b$ and monthly payments $\$_a > \$_b$, and if $\$_a \geq \$_b r_a$, then we can run a two-payment-period engine to extract $\$_b (r_a - r_b)$ per cycle and still pay our loans off on time.

To put this in perspective, let's say you take out two loans at the same time, with principals $5000, $2000 and interest rates 5%, 3%. They both have the same term of 10 years, so your monthly payments will be $53.99, $19.65 respectively. Implementing our engine, you pay the first loan alternating between $73.64 and $34.25. The second loan payments alternate between $0 and $39.36. On each "stroke" of the engine, you pocket $0.03, and with 60 total strokes your glorious return is $1.80.

Okay, well maybe not so glorious. 18¢ per year is kind of pitiful. UNLESS YOU MULTIPLY IT BY THE MILLIONS OF PEOPLE WHO HAVE STUDENT LOANS. WHAT IF LOAN SERVICERS BUILD THESE ENGINES AS A SORT OF MAN-IN-THE-MIDDLE ATTACK BETWEEN THE DEBTORS AND CREDITORS!?!?!?

Whew, sorry, I don't know what came over me there. The point is that setting a monthly payment schedule to pay off each loan in their own separate term is not the most efficient way. The most efficient way means paying back the least amount.