Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to programmatically calculate APR based on the following inputs:

  • Principal Amount
  • Number of payments (for example, 3 months loan is 3 if paying monthly, or 6 bi-weekly payments)
  • Payment each period

From this, I need the formula for APR.

Second, the customer may have an odd number of days before their first payment. I need to be able to plug this in as a variable in to calculation and get an adjusted APR.

Can someone help me with the formula for this?

share|improve this question
    
For the principal amount $a$, number of payments $n$ to be made, the payment amount $p$, are you assume that after $n$ payments the loan will be paid off? You will also need an extra input that helps distinguish between 6 monthly payments and 6 biweekly payments, as the APR will be different for these cases. –  Sasha May 23 '12 at 20:31

1 Answer 1

Let $r$ denote the APR we seek to establish, and the payments are made each $\frac{1}{n}$ of the year (i.e. $n=12$ for monthly payments, and $n=26$ for biweekly). Let $a_k$ be the principal amount of the loan at the beginning of the $k$-th period. Then $a_0 = a$, and $$ a_{k+1} = a_k (1+r)^{1/n}-p $$ Such a recurrence equation is not hard to solve using generating function technique, i.e. multiplying both sides of the equation with $x^k$ (for some indeterminate $x$) and forming $g(x) = \sum_{k=0}^\infty a_k x^k$: $$\begin{eqnarray} \sum_{k=0}^\infty a_{k+1} x^k &=& (1+r)^{1/n} \sum_{k=0}^\infty a_k x^k - p \sum_{k=0}^\infty x^k \\ \frac{f(x)-a_0}{x} &=& (1+r)^{1/n} f(x) - p \frac{1}{1-x} \\ f(x) &=& \frac{1}{(1+r)^{1/n}-1} \left( \frac{p}{1-x} - \frac{a + p - a(1+r)^{1/n}}{1 - (1+r)^{1/n} x} \right) \\ a_{k} &=& \frac{1}{(1+r)^{1/n}-1} \left( p - \left( a + p - a (1+r)^{1/n} \right) (1+r)^{k/n} \right) \end{eqnarray} $$ Requirement that after $m$ equal payments the loan is paid off gives the equation for effective annual percentage rate $r$: $$ p = \left( a + p - a (1+r)^{1/n} \right) (1+r)^{m/n} $$ As this is a non-linear equation, solving for $r$ will need to be done numerically. Here is a code in Mathematica:

In[63]:= FindAPR[a0_, p_, m_, n_] := 
 Block[{sol}, 
  sol = FindRoot[
    p == (a0 + p - a0 (1 + r)^(1/n)) ((1 + r)^(m/n)), {r, 1/2}];
  If[sol === {}, Indeterminate, r /. First[sol]]]

In[64]:= FindAPR[1000, 256, 4 (* payments *), 12 (* paid monthly *)]*100

Out[64]= 12.0876

If the time to the first payment is different, $a_0$ needs to be replaced with different effective principal amount, $a_0^\ast = a_0 \left(1 + r \right)^{d/n}$, where $d$ is the length of the no-payment period given as a fraction of payment intervals. For example, if payments are made monthly $n=12$, there are 4 payments to be made $m=4$ in the amount of $p=\$260.00$ on the loan of $\$1000.00$ with 1 month no payment period, the effective APR is

In[3]:= FindAPR2[a0_, p_, m_, n_, d_] := 
 Block[{sol}, 
  sol = FindRoot[
    p == (p - a0 (1 + r)^(d/n) ( (1 + r)^(1/n) - 1)) ((1 + r)^(m/
       n)), {r, 1/2}];
  If[sol === {}, Indeterminate, r /. First[sol]]]

In[5]:= FindAPR2[1000, 260, 4, 12, 1]*100

Out[5]= 14.4241
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.