Convert flat interest rate to effective interest rate

Question

My bank currently has promotion on personal loan on flat interest rate which is 3.80%. And there are some minor remarks

Terms and Conditions apply. Effective rates vary from 7.07% p.a. to 7.15% p.a. for tenures from 2 – 5 years.

Based on the internet search, flat interest rate is based on the total loan amount, while effective interest rate is based on the remaining loan amount. May I know to convert the flat rate to effective rate? For example, I loan 100k for 4 years tenures. I try to apply the formula $r_e=e^i-1=e^{0.038}-1=0.0387=3.87\%$ but it seems too far away from the correct answer. Any help is appreciated.

Edit:

For example, I loan 100,000 for 3.80% flat rate interest with 5 years tenure. Then

• Monthly installment amount will be 1,983.33
• Total payment will be 119,000.00
• Effective interest rate p.a. will be 7.07% (as the remark above)

May I know how to get the value of 7.07%? Any help is appreciated.

Answer 1

As @saulspatz answered, the solution of $$k={1-v^{60}\over i}\qquad \text{with}\qquad v={1\over1+i}\qquad \text{and}\qquad k=\frac{ 100000}{1983.33 }$$ requires some numerical method.

However, you can make some approximation using Taylor series built at $i=0$; this would give $$k=60-1830 i+37820 i^2-595665 i^3+7624512 i^4-82598880 i^5+778789440 i^6+O\left(i^{7}\right)$$

Now, using series reversion, this would give $$i=x+\frac{62 }{3}x^2+\frac{9517 }{18}x^3+\frac{1979939 }{135}x^4+\frac{686499247 }{1620} x^5+\frac{5077158734 }{405}x^6+O\left(x^7\right)$$ where $x=\frac{60-k}{1830}$.

Using your value of $k$, this gives $x=\frac{189998}{36294939}\approx 0.00523483$ and a few terms of the above expansions will very quickly give $i\approx 0.00589$ which, multiplied by $12$, gives $0.07068$ as annual rate.

The exact solution, using Newton method, would be $i=0.0058900057$

Edit

For a shortcut evaluation of $i$, we could also use Padé approximants instead of Taylor series. To stay with simple equations to solve, let us use $$k=\frac{1-\frac{1}{(i+1)^{60}}}{i}\sim \frac{60-318 i+3422 i^2}{1+\frac{126 }{5}i+\frac{1953 }{10}i^2 }$$ and solving the quadratic equation in $i$ directly leads to $i\approx 0.00589003$.

Answer 2

The effective rate of interest is the rate that makes the present value of the repayments equal to the principal. If the monthly interest rate is $i$ then if we invest one dollar for a month, at the end of month we will have $(1+i)$ dollars. The present value of one dollar a month from now is ${1\over1+i}$ dollars, since if we invest that amount for a month we will have a dollar at the end of the month.

Let the discount rate be $v={1\over1+i}$. The present value of one dollar a month from now is $v$. The present value of a dollar two months from now is $v^2,$ and so on. The present value of all the payments is $$1983.33(v+v^2+\cdots+v^{60})=1983.33{1-v^{60}\over i}$$ and we must find the value of $i$ that makes this last expression equal to $100000.$

There is no closed-form solution to this equation; it must be solved numerically. You can check that the solution is approximately $.00589$ It is customary to quote mortgage interest rates as nominal annual rates, compounded monthly; that is to say, we multiply the monthly rate by $12,$ which gives approximately $.0707$ as the annual rate.

I hadn't seen these flat rates for a long time. As I recall, automobile loans used to be quoted this way before the Truth in Lending Law in the United States, about $50$ years ago. The law required the lender to disclose the APR according to the method given above (known in law as the "actuarial method") but it didn't require that the unpaid balance of the loan be computed by the actuarial method. The result was that if you paid off the loan early, the balance would be way too high. Interest rates were much higher back then than they are today, so the effect was much more pronounced. Still, I would check carefully how the unpaid balance of the loan is to be computed before taking out a flat rate loan.