A mortgage problem (Problema sobre o prazo de amortização de uma hipoteca)

 In  a recent post  Professor Gowers  writes on the new proposal for the A-level exam on the “Use of Mathematics”.  In it  he discusses the following discrete problem which he approaches in a way that uses  continuous techniques.

Suppose for simplicity that the interest rate for an interest-only mortgage would be 5% and that this rate never changes. If I take out a repayment mortgage of £50,000 and pay £500 a month, then roughly how long will it take me to pay off the mortgage?

 

Here are two comments of mine concerning this problem.

1. “For your discrete problem about repayment mortgages I present a proposal for a direct solution. This solution, even if correct, is of course far less instructive than (*) your reasoning!
   

In general we have to find the value of constant payments A given the principal P during n monthly periods. The value of A in the period k is equivalent to the present value A/\left( 1+i\right) ^{k} monetary units, where i is the interest rate in each capitalization period. Summing in k, from 1 to n, we get the sum

\displaystyle\sum_{k=1}^{n}\dfrac{A}{\left( 1+i\right) ^{k}}

Now we have to sum a geometric progression with ratio (**) r=1/(1+i) and first term  u_{1}=A/\left( 1+i\right)

\dfrac{A}{1+i}\dfrac{\left( \dfrac{1}{1+i}\right) ^{n}-1}{\dfrac{1}{1+i}-1}=A\dfrac{\left( 1+i\right) ^{n}-1}{i\left( 1+i\right) ^{n}}=P

 or

A=P\dfrac{i\left( 1+i\right) ^{n}}{\left( 1+i\right) ^{n}-1}

For the given problem, the payments will be made during n months, with i=5/12\%=\dfrac{5}{1200} and P=50\,000. Thus

 A=50\,000\dfrac{\dfrac{5}{1200}\left( 1+\dfrac{5}{1200}\right) ^{n}}{\left( 1+\dfrac{5}{1200}\right) ^{n}-1}=500

 or

\dfrac{5}{12}\left( 1+\dfrac{5}{1200}\right) ^{n}=\left( 1+\dfrac{5}{1200}\right) ^{n}-1

Solving for n, we get

n=\dfrac{\ln 12-\ln 7}{\ln 241-\ln 240}\approx 129.63 months (10.802 years)

and, as you proved

20\ln 2\approx 13.863>10.802.

2. “Please let me only add one further interpretation of mine: the continuous effective interest rate can be derived from the 5% nominal interest rate, compounded (***) m times per year as follows:

\underset{m\rightarrow \infty }{\lim }\left( 1+\dfrac{0.05}{m}\right) ^{m}-1=\underset{m\rightarrow \infty }{\lim }\left[ \left( 1+\dfrac{0.05}{m}\right) ^{m/0.05}\right] ^{0.05}-1 =e^{0.05}-1\approx 5.127%  

 (of which approximates your formula e^{\alpha }=1.05 is an approximation for \alpha =0.05 (****)), and in the general case of an annual nominal interest rate r as

i_{E\;(m\rightarrow \infty )}=\underset{m\rightarrow \infty }{\lim }\left( 1+\dfrac{r}{m}\right) ^{m}-1  =\underset{m\rightarrow \infty }{\lim }\left[ \left( 1+\dfrac{r}{m}\right) ^{m/r}\right] ^{r}-1=e^{r}-1.

From a point of view of a purely mathematical problem the model you are discussing in the context of the “use-of-maths A’ level “ is much more interesting and worthy.”

Corrections to the original comments:

 (*) “that” in original

  (**) “rate” in original

 (***)  “compound” in original

(****) [Edited on July 28, 2009,. In original: “which approximates your formula e^{\alpha }=1.05  for \alpha =0.05“]

[Edited on July 14, 2009: First paragraph corrected and title changed]

[Edited on July 15, 2009: Last paragraph. Remark: in a comment Professor Gowers wrote “I wasn’t suggesting that the mortgage question would be a suitable one for a use-of-maths A’level” ]

Versão portuguesa

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Cálculo financeiro, Matemática, Math, Problem, Problemas com as etiquetas , , , . ligação permanente.

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