**MORTGAGES**

**HOW TO COMPUTE A MONTHLY MORTGAGE PAYMENT**

If we are given the following quantities

P_{0} – loan amount to buy real estate
or the “principal”

r - monthly interest rate

n - number of monthly payments

Then we can calculate the monthly payment which completely repays the loan in the last month

M – monthly payment

At the end of each month the amount
still owed on the loan changes in two ways. In the first place, the principal,
P_{i} for i^{th} month, increases as a result of interest on
the loan. And secondly, the principal is reduced an even a greater amount by
the monthly payment. This is cumulatively expressed as

and so forth until the last month when the principal is zero having been paid off, as follows

If we let then we can solve for the summation term as

Substituting this result into the equation above gives

And we can solve directly for the required monthly payment to completely retire the loan in “n” months as

** **

**ANNUAL PERCENTAGE RATE**

When we borrow money to buy a house, lenders charge fees “C”
to originate the loan. These fees increase the loan amount beyond the actual
cost of the house “P_{0}” and effectively appear to the buyer as an increase
in the interest rate.

This new interest rate is called the Annual Percentage Rate (APR) which we represent as “A”. For convenience we will change the annual APR interest “A” to a monthly rate of “a”

To calculate the APR, we first compute the monthly payment
“M_{new}” necessary to repay the combined amount of “”
in “n” installments at the original interest rate of “r”. This is the actual
payment we will have to make.

Then for this monthly payment, we compute the effectively
increased monthly APR interest rate “a” by assuming the monthly payment, which
we just calculated of “M_{new}” was intended to repay just original principal,
that is the initial cost of the real estate ”.

We go to this trouble because we are confronted with a
monthly payment of “M_{new}” and the only thing of lasting value we
have for our money is the real estate. The original interest rate “r” has been
corrupted. Legislation requires the disclosure of the APR so lenders will not
be able to hide, much as they would like to, their incredibly over-inflated
charges which are occasionally even for non-existent services. And do not be
fooled as these charges are negotiable and frequently reduced when challenged
as they are basically insupportable. Another common fraud is “title insurance”
which, believe it or not, insures nothing.

In any event the relevant equation involving the increased effective interest rate “a”, or APR, is

Unfortunately, there is no closed form solution for this equation so we have to use the Newton-Raphson approximation to solve for “a”. This equation can be made homogeneous in the variable “a” as follows

Rearranging terms we can define the equation f(a) and calculate its derivative as follows

We then use successive approximations starting from an
initial guess “a_{0}” for the APR. For each approximation “a_{i}“
we use Newton-Raphson to get an even better approximations “a_{i+1}“ as
follows

**STARTING VALUE FOR THE APR CALCULATION**

We can get an initial guess of “a_{0}” by
considering the basic definition of APR

Then if we let and recall that

so that for

Then rewriting the above equation we have

And we can solve for “” as follows

Or finally, the improved starting value “a_{0}“ for
Newton-Raphson is

Note that this is always slightly less (typically on the order of 10%) than the final value.

**STABILITY OF THE “APR” CALCULATION FROM “a _{0}”**

Please note that the Newton-Raphson method only converges for
starting values of “a_{0}” which are reasonably close to the exact
answer. Please recall that the exact answer is that “a” which makes f(a) equal
to zero . In any event, Newton-Raphson requires that the slope of the curve
f(a) not change sign between the initial guess and the final exact solution. Otherwise
successive approximations will diverge. In our case the function f(a) has a
local maximum where its slope is zero, that is

To illustrate, a plot of a specific f(a) as a function of “a” is given below

It should be obvious that if the initial guess of “a_{0}”,
is smaller than “a_{stable}”, repeated applications of the
Newton-Raphson” iteration will give negative values and will not approach the
exact value for “a”. But if “a_{0}” is large enough so that “a_{0}
> a_{stable}”, then | f(a)-f(a_{i}) | can be made as small
as one wishes as i -> ∞.

The question arises as to whether the above estimate for the
starting value “a_{0}” is stable, which means

This can be further reduced to

And continuing

But we note from the binominal expansion by expanding the first few terms

And that

So that the equality reduces to

which for positive values of P_{0}, C, n, and r is, by
observation, always true. And so we note the starting value given above is
always in the range of stability which is to say that “a_{0} > a_{stable}”
and so starting from “a_{0}” always works.