Uniform serie present-worth factor — Calculating interest rate

In the MSE question Calculating interest rate of car financing, Juan Hynix asks:

I want a new car which costs $26.000.

But there’s an offer to finance the car: Immediate prepayment: 25% of the original price

The amount left is financed with a loan: Duration: 5 years, installment of $400 at the end of every month.

So I need to calculate the rate of interest of this loan. Do I need Excel for this exercise? Or which formula could I use for this exercise?

My answer:

You could use Excel (see below) or you could solve the equation (2) below numerically, e.g. using the secant method.

We have a so called uniform series of n=60 constant  installments m=400.

Let i be the nominal annual interest rate. The interest is compounded monthly, which means that the number of compounding periods per year is 12. Consequently, the monthly installments m are compounded at the interest rate per month i/12. The value of m in the month k is equivalent to the present value m/(1+i/12)^{k}. Summing in k, from 1 to n, we get a sum that should be equal to

P=26000-\dfrac{26000}{4}=19500.

This sum is the sum of a geometric progression of n terms, with ratio 1+i/12 and first term m/(1+i/12). So

P=\displaystyle\sum_{k=1}^{n}\dfrac{m}{\left( 1+\dfrac{i}{12}\right) ^{k}}=\dfrac{m}{1+\dfrac{i}{12}}\dfrac{\left( \dfrac{1}{1+i/12}\right) ^{n}-1}{\dfrac{1}{1+i/12}-1}=m\dfrac{\left( 1+\dfrac{i}{12}\right) ^{n}-1}{\dfrac{i}{12}\left( 1+\dfrac{i}{12}\right) ^{n}}.\qquad (1)

The ratio P/m is called the series present-worth factor (uniform series)^1.

For P=19500, m=400 and n=5\times 12=60 we have:

19500=400\dfrac{\left( 1+\dfrac{i}{12}\right) ^{60}-1}{\dfrac{i}{12}\left( 1+\dfrac{i}{12}\right) ^{60}}.\qquad (2)

I solved  numerically (2) for i using SWP and got

i\approx 0.084923\approx 8.49\%.\qquad (3)

ADDED. Computation in Excel for the principal P=19500 and interest rate i=0.084923 computed above. I used a Portuguese version, that’s why the decimal values show a comma instead of the decimal point.

– The Column k is the month (1\le k\le 60).
– The 2nd. column is the amount P_k still to be payed at the beginning of month k.
– The 3rd. column is the interest P_ki/12 due to month k.
– The 4th. column is the sum P_k+P_ki/12.
– The 5th column is the installment payed at the end of month k.

The amount P_k satisfies P_{k+1}=P_k+P_ki/12-m. We see that at the end of month k=60, P_{60}+P_{60}i/12=400=m. The last installment m=400 at the end of month k=60 balances entirely the remaining debt, which is also 400. We could find i by trial and error. Start with i=0.01  and let the spreadsheet compute the table values, until we have in the last row exactly P_{60}+P_{60}i/12=400.

^1 James Riggs, David Bedworwth and Sabah Randdhava, Engineering Economics, McGraw-Hill, 4th. ed., 1996, p.43.

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Cálculo financeiro, Matemática, Math, Mathematics Stack Exchange com as etiquetas , , , . ligação permanente.

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