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$\mathbf{\text{(i) Given r= $0.12 $ annually compounded, find the monthly}}$ $\mathbf{\text{repayments on a $ 1 000 000$ Dollar loan to be repaid completely in 5 years}}$

$\mathbf{\text{Solution}}$

Finding equivalent rate of monthly compounding:

$(1+ 0.12)= (1+\frac{1_{12}}{12})^{12} \rightarrow r_{12}=0.11387$

Then

$1 000 000= C [\frac{1-(1+\frac{0.1139}{12})^{-60}}{\frac{0.1139}{12}}] \rightarrow C= 21 937.42$

$\mathbf{\text{(ii) Given r= $0.12 $ with continuous compounding, find the monthly}}$ $\mathbf{\text{ repayments on a $ 1 000 000$ Dollar loan to be repaid completely in 5 years}}$

$e ^ {-0.12}=(1+\frac{r_{12}}{12})^{12}$

Been getting $r_{12}$ as negative. Please help.

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The factor for an continuously compounding is $e^{r}$, not $e^{-r}$

There is a well known identity:

$$\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n=e^x$$

Since in your case n is not very large the equation doesn´t hold. But you can say that the left side is monthly compounding (with n=12) and the right side is continuously compounding. The x on both sides has to be different if the equation should be true.

$\left(1+\frac{r_{12}}{12}\right)^{12}=e^{0.12}$

Taking the 12-th root on both sides.

$\left(1+\frac{r_{12}}{12}\right)^{1}=e^{0.01}$

$\frac{r_{12}}{12}=e^{0.01}-1$

$r_{12}=12\cdot (e^{0.01}-1)=0.120602...$

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