这是一份nottingham诺丁汉大学ECON2012作业代写的成功案例

(a) We have $P=1000, i=0.06 / 12=0.005$, and $n=5 \times 12=60$. From equation (2.12),
$$
S=P(1+i)^{n}=1000(1.005)^{60}=\$ 1348.85
$$
The compound interest is $S-P=\$ 348.85$.
(b) We have $P=1000, i=0.15 / 12=0.0125$, and $n=30 \times 12=360$. From equation (2.12),
$$
S=1000(1.0125)^{360}=\$ 87,541.00
$$
The compound interest is $S-P=\$ 86,541.00$, which is more than 86 times the original investment of $\$ 1000$. If the investment had been at 15 per cent simple interest, the interest earned would have been only
$$
I=1000(0.15)(30)=\$ 4500
$$
This illustrates the power of compound interest. A high rate of interest for a long period of time generates far more than receiving only simple interest.

ECON2012 COURSE NOTES ：

$$
C=P N\left(d_{1}\right)-E e^{-r t} N\left(d_{2}\right)
$$
where $C=$ the price of a call option, $P=$ the current price of the underlying, $E=$ the exercise price, $e=$ the base of natural logarithms, $r=$ the continuously compounded interest rate, and $t=$ the remaining life of the call option.
$N\left(d_{1}\right)$ and $N\left(d_{2}\right)$ are the cumulative probabilities from the normal distribution of getting the values $d_{1}$ and $d_{2}$, where $d_{1}$ and $d_{2}$ are as follows:
$$
\begin{aligned}
&d_{1}=\frac{\ln (P / E)+\left(r+0.5 \sigma^{2}\right) t}{\sigma \sqrt{t}} \
&d_{2}=d_{1}-\sigma \sqrt{t}
\end{aligned}
$$
where $\sigma=$ the standard deviation of the continuously compounded rate of return on the underlying asset.