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Problems involving the amount of money required to pay off a mortgage over a fixed period of time involve the formula $$ A = \frac Pi\cdot [1 − (1 + i)^{−n}], $$ known as an ordinary annuity equation. In this equation, $A$ is the amount of the mortgage, $P$ is the amount of each payment, and $i$ is the interest rate per period for the $n$ payment periods. Suppose that a 30-year home mortgage in the amount of $\$135,000$ is needed and that the borrower can afford house payments of at most $\$1000$ per month. What is the maximal interest rate the borrower can afford to pay?

So I have:

  • $A = 135,000$
  • $n = 30$ (Or is it $360 = 30\cdot 12$ ???)
  • $P = 1000$

Next: $$ f(i) = 1000⋅(1-(1+i)^{-30})-135000i $$

But 1) Which is my point to begin with Newton method????

So tried for example with first point $= 0.28$

Result on my matlab Newton function $i = 0.0000001698$

But on the solutions of the book it says $i = 8.10\%$

Lutz Lehmann
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Isma Haro
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1 Answers1

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Since the payment is done monthly, the number of periods $n$ should be $12\times 30=360$.

So, the equation should write $$f(i) = 1000\times(1-(1+i)^{-360})-135000\times i$$ $$f'(i)=360000\times(1+i)^{-361}-135000$$ and the iterates of Newton method are given by $$i_{k+1}=i_k-\frac{f(i_k)}{f'(i_k)}$$ If you start (why ?) using $i_0=0.28$, the iterates will then be $$x_1=0.00740741$$ $$x_2=0.00676906$$ $$x_3=0.00674994$$ $$x_4=0.00674992$$ which is the solution for six significant figures. And $0.00674992\times 12 =0.080999$ which is the answer from the book.

With regard to the starting point, since $i$ is supposed to be small with respect to $1$, you could have developed $f(i)$ as a Taylor series at $i=0$ and get $$f(i)=225000 i-64980000 i^2+O\left(i^3\right)$$ and retain the non-zero solution which is $i_0=\frac{5}{1444}\approx 0.0034626$ but this would not really change the number of iterations to convergence.

Claude Leibovici
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