The interest rate on a 30 year mortgage is 12% compounded monthly. the mortgage is paid by monthly payments of 700. Suppose an additional 1000 is paid at the end of each year to payoff loan early, compute for outstanding balance at the end of 10 years.

Using the prospective method, this is what i came up with:

700 (a_240) + 1000 (a_20) = 81,619.14

a is PV annuity immediate with n=240 and n=20

i is .01

my loan is 700 (a_360) = 68,052.83

i know i'm doing something wrong coz when i use the retrospective, it doesn't equate to that. using retrospective:

68,052.83 (1.01)^120 - 700 (s_120) - 1000 (s_10) = 53,111.38

s is FV annuity immediate with n=120 and n=10

please help :s thanks!

  • 1
    $\begingroup$ I think for outstanding balance, we need to know the principal. $\endgroup$ – Ron Gordon Oct 10 '13 at 15:24
  • $\begingroup$ principal at the end of 10 yrs? $\endgroup$ – Yolo Oct 10 '13 at 16:16
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    $\begingroup$ I did the best I could analyzing this. Your notation is cryptic so I do not know how to edit into an acceptable format. My solution is fairly generic, so you can see how to attack a problem like this. $\endgroup$ – Ron Gordon Oct 10 '13 at 18:45

I'm going to formalize this a bit, but I will bring this back to earth, I promise. Assume we begin with a principal $P_0$, interest rate $i$ per compounding period (which is a month), and payment per compounding period $m$ (monthly payment). Also assume that there is another additional payment $q$ made every $\ell$ compounding periods (e.g., the additional annual payment, where $\ell=12$).

The way to express the amount owed after the first extra payment is made is to determine the present value $P_1$ of what is owed and what has been paid:

$$P_1 = \underbrace{\left [ \cdots \left [ \left [ P_0 (1+i) - m\right ] (1+i) - m \right ] (1+i) - m \cdots \right ] (1+i)}_{k \, \text{compunding periods}} - m - q$$

Collecting all the terms, we have

$$P_1 = P_0 (1+i)^k - m \sum_{j=0}^{k-1} (1+i)^j - q$$

Summing the geometric series, we finally get

$$P_1 = P_0 (1+i)^k - m \frac{(1+i)^k-1}{i} - q$$

To get the principal after $\ell$ such periods, we note that we may consider the beginning of the second period as having a principal equal to $P_1$; the beginning f the third period starts out having principal $P_2$, and so on. So we get a recurrence relation

$$P_{\ell} = P_{\ell-1} (1+i)^k - m \frac{(1+i)^k-1}{i} - q$$

This is a simple, inhomogeoneus, first-order recurrence in $P_{\ell}$ of the form

$$P_{\ell} - a P_{\ell-1} = b$$

The general solution to this equation is

$$P_{\ell} = C a^{\ell} - \frac{b}{a-1}$$


$$P_0 = C- \frac{b}{a-1}$$

Using $a=(1+i)^k$ and $b = - m \frac{(1+i)^k-1}{i} - q$, I get for the principal after $\ell$ periods:

$$P_{\ell} = P_0 (1+i)^{k \ell} - \frac{(1+i)^{k \ell}-1}{(1+i)^k-1}\frac{m [(1+i)^k-1]+q i}{i} $$

Now plug in the values $m=700$, $k=12$, $\ell=10$, $i=0.12/12=0.01$, $q=1000$; I get about

$$P_{10} \approx 3.3003869 P_0 - 179,165.88$$

($P_0$ was never specified.) If $P_{10} = 0$, then $P_0 \approx 77,884.88$.

  • $\begingroup$ the P0 would be the original loan amount? $\endgroup$ – Yolo Oct 11 '13 at 7:10
  • 2
    $\begingroup$ @Yolo: yes, that's right. $\endgroup$ – Ron Gordon Oct 11 '13 at 7:19

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