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I'm trying to calculate the total loan amount when financing into the loan an interest reserve and points. An interest reserve is an amount set aside to cover interest payments for a certain period. "Points" are merely a fee that is equivalent to a % of the principal amount (so if you're charging 1 point, you could calculate that with $P·0.01$, where $P$ is the principal amount.

If you're financing into the loan amount the interest reserve and the points, both the interest reserve and the points need to increase $P$. Each subsequent increase necessitates a smaller increase until they become so small it becomes negligible.

To take an example:

  • The principal before adding the interest reserve or points, $P_0$, is $58,482.29.

  • The interest rate, $I$, is 20% annually.

  • Monthly interest can be calculated simply by dividing annual interest by 12.

  • The desired interest reserve, $R$, is 6 months.

  • The number of points, $T$, is 0.084.

If I'm not mistaken, it's possible to calculate what $P$ should be to the penny by iterating using the following pattern until $P_n-P_{n-1}<0.01$:

  • $P_0 +(P_0·\frac{I}{12}·R)+P_0·T=P_1$
  • $P_0 +(P_1·\frac{I}{12}·R)+P_1·T=P_2$
  • $P_0 +(P_2·\frac{I}{12}·R)+P_2·T=P_3$
  • $…$

Following this pattern, it takes 11 iterations until subsequent changes are less than $0.01:

  1. $58,482.29 +(58,482.29·\frac{0.2}{12}·6)+58,482.29·0.084=69,243.03$
  2. $58,482.29 +(69,243.03·\frac{0.2}{12}·6)+69,243.03·0.084=71,223.01$
  3. $58,482.29 +(71,223.01·\frac{0.2}{12}·6)+71,223.01·0.084=71,587.32$
  4. $58,482.29 +(71,587.32·\frac{0.2}{12}·6)+71,587.32·0.084=71,654.36$
  5. $58,482.29 +(71,654.36·\frac{0.2}{12}·6)+71,654.36·0.084=71,666.70$
  6. $58,482.29 +(71,666.70·\frac{0.2}{12}·6)+71,666.70·0.084=71,668.96$
  7. $58,482.29 +(71,668.96·\frac{0.2}{12}·6)+71,668.96·0.084=71,669.38$
  8. $58,482.29 +(71,669.38·\frac{0.2}{12}·6)+71,669.38·0.084=71,669.46$
  9. $58,482.29 +(71,669.46·\frac{0.2}{12}·6)+71,669.46·0.084=71,669.47$
  10. $58,482.29 +(71,669.47·\frac{0.2}{12}·6)+71,669.47·0.084=71,669.48$
  11. $58,482.29 +(71,669.48·\frac{0.2}{12}·6)+71,669.48·0.084=71,669.48$

My question is whether there's a simpler way to do this without all the iteration: is there a formula that's escaping me to calculate this? When you graph out the change, it seems to follow a $\frac{1}{x}$ type pattern, so it looks like the limit is zero. That makes me think I'm missing something much easier than what I did above.

This question seems to be of a similar vein, but I couldn't quite puzzle out how to apply the answers there to this situation.

Vincent
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1 Answers1

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You face a geometric progression. Using $i=\frac I{12}$, the general term write $$P_n=P_0\frac{ (i R +T)^{n+1}-1}{(i R +T)-1}$$ and you want to know $n$ such that $$\Delta_n= P_{n+1}-P_n\leq \epsilon$$ replacing, this means $$P_0 (i R +T)^{n+1} \leq \epsilon\implies n >\frac{\log \left(\frac{\epsilon }{P_0}\right)}{\log (iR+T)}-1$$

Using your numbers for $\epsilon=10^{-3}$, this gives $n=9.56$. Computing $$\left( \begin{array}{cc} n & \Delta_n \\ 0 & 10760.7414 \\ 1 & 1979.9764 \\ 2 & 364.3157 \\ 3 & 67.0341 \\ 4 & 12.3343 \\ 5 & 2.26951 \\ 6 & 0.4176 \\ 7 & 0.0768\\ 8 & 0.0141 \\ 9 & 0.0026 \\ 10 & 0.0005\\ 11 & 0.0001 \end{array} \right)$$

Claude Leibovici
  • 260,315
  • Thanks, this is very helpful. It seems that I can use the formula above with an arbitrarily large n (even without calculating it) such that the amount is always accurate to the penny. My follow-up question would be whether there's a way to do this (perhaps using integrals?) such that you wouldn't need to just plug in an arbitrarily large n. – Vincent Feb 01 '20 at 20:17
  • I just got the answer to my own question with the obvious realization that that would be the sum of an infinite series, and since r<1, we can use S = a_1 / (1-r), so no need for integrals. – Vincent Feb 01 '20 at 20:30