# Nominal rate of interest for interest reinvested.

Sally lends $10000$ to Tim. Tim agrees to pay back the loan over $5$ years with monthly payments at the end of each month. Sally can reinvest the the monthly payments from Tim in a savings account paying interest at $6$% compounded monthly. The yield rate earned on Sally's investment over the five-year period turned out to be $7.45$%,compounded semi-annually. What nominal rate of interest, compounded monthly, did Sally charge Tim on the loan?

To find the interest Tim pays to Sally

$$P=\frac{1000}{a_{n|j}}$$

It is this loan amount that is reinvested, so

$$P\left(\frac{1.02^{60}-1}{0.02}\right)=10000(1+0.00375)^{10}$$

$$\therefore P=126.39$$

$$a_{60|i^{(12)}/12}=79.12$$

$$\frac{1-(\frac{1+i^{(12)}}{12})^{-60}}{\frac{i^{(12)}}{12}}=79.12$$

Then I start to have problems to solve this.

• I haven´t checked if your equation is right but in general you cannot solve such equation for $i$ straightforward (algebraically). I would transform it to a polynomial (I think it has the degree of 61) and apply the approximation method of newton and raphson. – callculus May 17 '16 at 9:54
• That is what I was thinking too. But I think there is a simpler way of doing it, I am not sure of my calculation. And to be able to use Newton Raphson, I need at least an $i_0$. – Tosh May 17 '16 at 10:06
• As an initial value in context of finance $6\%$ is not a bad idea-independent from the exercise. In most cases the interest rate is between $4%$ and $12%$ Sure not in the real world, especially in these days. – callculus May 17 '16 at 10:23
• Thanks, I shall check this later and come back to this post. – Tosh May 17 '16 at 10:46

## 1 Answer

The value of the end of the 5 years of Sally’s investment is $$10000\left(1+\frac{0.0745}{2}\right)^{10}=14415.65$$

Let $P$ be the amount of the payments Sally receives from Tim. Then, $$P\, s_{\overline{60}|6\%/12}=14415.65\qquad \Longrightarrow\quad P=206.6167$$ Let $i^{(12)}$ be the nominal rate of interest, compounded monthly, which Sally charged Tim on the loan. Then, $$10000 = 206.6167\,a_{\overline{60}|i^{(12)}/12}$$ So $\frac{i^{(12)}}{12}=0.7333$ and $i^{(12)}=8.801\%$.