I was recently playing around with some loan data and only happened to have the term (or length, or duration) of the loan, the amount of the recurring payment (in this case monthly) and the remaining principal owed on the loan. I figured there was an easy way to get at the interest rate, but wasn't sure how. After some badgering from my coworker +Paul, I searched the web and found a tool from CALCAmo (a site just for calculating amortizations).

Problem solved, right? Wrong. I wanted to know why; I had to go deeper. So I did a bit of math and a bit of programming and I was where I needed to be. I'll break the following down into parts before going on full steam.

- Break down the amortization schedule in terms of the variables we have and the one we want
- Determine a function we want to find zeros of
- Write some code to implement the Newton-Raphson method
- Utilize the Newton-Raphson code to find an interest rate
- Bonus: Analyze the function to make sure we are right

## Step I: Break Down the Amortization Schedule

We can do this using the series $\left\{P_i\right\}_i$ of principal owed, which varies over time and will go to zero once paid off. In this series, $P_0$ is the principal owed currently and $P_i$ is the principal owed after $i$ payments have been made. (Assuming monthly payments, this will be after $i$ months.) If the term is $T$ periods, then we have $P_T = 0$.

We have already introduced the term $T$; we also need the value of the recurring (again, usually monthly) payment $R,$ the interest rate $r$ and the initial principal owed $P_0 = P$.

#### Time-Relationship between Principal Values

If after $i$ periods, $P_i$ is owed, then after one period has elapsed, we will owe $P_i \cdot m$ where $m = m(r)$ is some multiplier based on the length of the term. For example if each period is one month, then we divide our rate by $12$ for the interest and add $1$ to note that we are adding to existing principal:

$m(r) = 1 + \frac{r}{12}.$

In addition to the interest, we will have paid off $R$ hence

$P_{i + 1} = P_i \cdot m - R.$

#### Formula for $P_i$

Using this, we can actually determine $P_i$ strictly in terms of $m, R$ and $P$. First, note that

$\begin{aligned} P_2 = P_1 \cdot m - R &= (P_0 \cdot m - R) \cdot m - R \\ &= P \cdot m^2 - R(m + 1) \end{aligned}$

since $P_0 = P$. We can show inductively that

$\displaystyle P_i = P \cdot m^i - R \cdot \sum_{j = 0}^{i - 1} m^j.$

We already have the base case $i = 1,$ by definition. Assuming it holds for $i,$ we see that

$\begin{aligned} P_{i + 1} = P_i \cdot m - R &= m \cdot \left(P \cdot m^i - R \cdot \sum_{j = 0}^{i - 1} m^j\right) - R \\ &= P \cdot m^{i + 1} - R \cdot \sum_{j = 1}^{i} m^j - R, \end{aligned}$

and our induction is complete. (We bump the index $j$ since we are multiplying each $m^j$ by $m$.) Each term in the series is related to the previous one (except $P_0,$ since time can't be negative in this case).

## Step II: Determine a Function we want to find Zeros of

Since we know $P_T = 0$ and $\displaystyle P_T = P \cdot m^T - R \cdot \sum_{j = 0}^{T - 1} m^j,$ we actually have a polynomial in place that will let us solve for $m$ and in so doing, solve for $r$.

To make our lives a tad easier, we'll do some rearranging. First, note that

$\displaystyle \sum_{j = 0}^{T - 1} m^j =m^{T - 1} + \cdots + m + 1 = \frac{m^T - 1}{m - 1}.$

We calculate this sum of a geometric series here, but I'll just refer you to the Wikipedia page instead. With this reduction we want to solve

$\begin{aligned} & 0 = P \cdot m^T - R \cdot \frac{m^T - 1}{m - 1} \\ & \Longleftrightarrow P \cdot m^T \cdot (m - 1) =R \cdot(m^T - 1). \end{aligned}$

With that, we have accomplished Step II, we have found a function (parameterized by $P, T$ and $R$ which we can use zeros from to find our interest rate:

$\begin{aligned} f_{P, T, R}(m) &= P \cdot m^T \cdot (m - 1) -R \cdot(m^T - 1) \\ &= P \cdot m^{T + 1} - (P + R) \cdot m^T + R. \end{aligned}$

## Step III: Write some code to implement the Newton-Raphson method

We use the Newton-Raphson method to get super-duper-close to a zero of the function.For in-depth coverage, see the Wikipedia page on the Newton-Raphson method, but I'll give some cursory coverage below. The methods used to show that a fixed point is found are not necessary for the intuition behind the method.

#### Intuition behind the method

For the intuition, assume we know (and can compute) a function $f,$ its derivative $f'$ at a value $x$. Assume there is some zero $y$ nearby $x$. Since they are close, we can approximate the slope of the line between the points $(x, f(x))$ and $(y, f(y))$ with the derivative nearby. Since we know $x,$ we use $f'(x)$ and intuit that

$f'(x) = \text{slope} = \frac{f(y) - f(x)}{y - x} \Rightarrow y - x = \frac{f(y) - f(x)}{f'(x)}.$

But, since we know that $y$ is a zero, $f(y) - f(x) = -f(x)$ hence

$y - x = \frac{-f(x)}{f'(x)} \Rightarrow y = x - \frac{f(x)}{f'(x)}.$

Using this method, one can start with a given value $x_0 = x$ and compute better and better approximations of a zero via the iteration above that determines $y$. We use a sequence to do so:

$x_{i + 1} = x_i - \frac{f(x_i)}{f'(x_i)}$

and stop calculating the $x_i$ either after $f(x_i)$ is below a preset threshold or after the fineness of the approximation $\left|x_i - x_{i + 1}\right|$ goes below a (likely different) preset threshold. Again, there is much that can be said about these approximations, but we are trying to accomplish things today, not theorize.

**Programming Newton-Raphson**

To perform Newton-Raphson, we'll implement a Python function that takes the initial guess $x_0$ and the functions $f$ and $f'$. We'll also (arbitrarily) stop after the value $f(x_i)$ drops below $10^{-8}$ in absolute value.

```
def newton_raphson_method(guess, f, f_prime):
def next_value(value):
return value - f(value)*1.0/f_prime(value)
current = guess
while abs(f(current)) > 10**(-8):
current = next_value(current)
return current
```

As you can see, once we have `f`

and `f_prime`

, everything else is easy
because all the work in calculating the next value (via `next_value`

)
is done by the functions.

## Step IV: Utilize the Newton-Raphson code to find an Interest Rate

We first need to implement $f_{P, T, R}(m) = P \cdot m^{T + 1} - (P + R) \cdot m^T + R$ and $f'_{P, T, R}$ in Python. Before doing so, we do a simple derivative calculation:

$f_{P, T, R}'(m) = P \cdot (T + 1) \cdot m^T - (P + R) \cdot T \cdot m^{T - 1}.$

With these formulae in
hand, we write a function which will spit out the corresponding `f`

and `f_prime`

given the parameters $P$ (`principal`

),
$T$ (`term`

) and $R$ (`payment`

):

```
def generate_polynomials(principal, term, payment):
def f(m):
return (principal*(m**(term + 1)) - (principal + payment)*(m**term) +
payment)
def f_prime(m):
return (principal*(term + 1)*(m**term) -
(principal + payment)*term*(m**(term - 1)))
return (f, f_prime)
```

Note that these functions only take a single argument (`m`

), but we are able
to use the other parameters from the parent scope beyond the life of the call
to `generate_polynomials`

due to
closure in Python.

In order to solve, we need an initial `guess`

, but we need to know the
relationship between $m$ and $r$ before
we can determine what sort of`guess`

makes sense. In addition, once a value for
$m$ is returned from Newton-Raphson, we need to be able to
turn it into an $r$ value so functions `m`

and `m_inverse`

should be implemented. For our dummy case here, we'll assume monthly
payments (and compounding):

```
def m(r):
return 1 + r/12.0
def m_inverse(m_value):
return 12.0*(m_value - 1)
```

Using these, and assuming that an interest rate of **10%** is a good
guess, we can put all the pieces together:

```
def solve_for_interest_rate(principal, term, payment, m, m_inverse):
f, f_prime = generate_polynomials(principal, term, payment)
guess_m = m(0.10) # ten percent as a decimal
m_value = newton_raphson_method(guess_m, f, f_prime)
return m_inverse(m_value)
```

To check that this makes sense, let's plug in some values. Using the bankrate.com loan calculator, if we have a 30-year loan (with $12 \cdot 30 = 360$ months of payments) of $100,000 with an interest rate of 7%, the monthly payment would be $665.30. Plugging this into our pipeline:

```
>>> principal = 100000
>>> term = 360
>>> payment = 665.30
>>> solve_for_interest_rate(principal, term, payment, m, m_inverse)
0.0699996284703
```

And we see the rate of 7% is approximated quite well!

## Bonus: Analyze the function to make sure we are right

Coming soon. We will analyze the derivative and concavity to make sure that our guess yield the correct (and unique) zero.