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Ian Desrosiers buys a house for \$285000. He pays \$60000 down and takes out a mortgage at 6.5% on the balance.

P.S. So, using the formula I found out the monthly payments which are \$1677.54. Then the book has the following formula for remaining payments (x) enter image description here

Where y is the remaining balance (after calculations \$112500)

R is monthly payments (\$1677.54)

i is interest (0.065/12)

n is total payments required to pay off the whole loan

x is the number of payments required to pay off half of the loan

So, I'm struggling with "x" evaluation. How do I find the x value from the given formula (if this formula is applicable for the given problem)?

WW1
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Kristina Utas
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1 Answers1

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The net present value of the future payments equals the Loan amount.

$B = P\sum_\limits{i=1}^{n} (1+y)^{-i}$

or the future value equals the future value:

$B(1+y)^n = P\sum_\limits{i=1}^{n} (1+y)^{i}$

This is a geometric series. And the sum of a geometric series equals.

$B = P \frac {1 -(1+ y)^{-n}}{y} = P\frac {(1+y)^n -1}{y(1+y)^n}$

and $P = B \frac {y(1+y)^n}{(1+y)^n -1}$

Using this value of $P,$ you want to find $k$

$P(1+y)^k\sum_\limits{i=k+1}^{n} (1+y)^{-i} = \frac 12 B$

$\frac {y(1+y)^n}{(1+y)^n -1} (1+y)^k(\frac {(1+y)^{-k} - (1+y)^{-n}}{y}) = \frac 12$

$\frac {(1+y)^n - (1+y)^k}{(1+y)^n-1}= \frac 12\\ 2(1+y)^n - 2(1+y)^{k} = ((1+y)^n - 1\\ (1+y)^n+1 = 2(1+y)^{k}$

We are going to need to take a log to find $k$

$\log ((1+y)^n+1) = \log 2 + k \log (1+y)\\ k = \frac {\log ((1+y)^n+1)-log 2}{\log (1+y)}$

One key piece of data that is missing. What is the initial term of the loan? 30 years is standard in the US. But, based on a 30 year loan, monthly payments would be $\$1,422.15$

So, I am going to guess that this is a 20 year loan.

$k = 156.44$

After the $157^{th}$ payment (of 240), the loan would be half-way to paid off.

Which shows that the initial payments are mostly going to pay interest, with very little paying down principal.

Doug M
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