\documentclass[a4paper]{report}
\input{../../preamble.tex}
\title{Solutions 2}

\begin{document}
	\maketitle
	\begin{enumerate}
		\item First, we know that
			\[
				\textit{FV} = \$10,000
			.\]
			And only two payments will be made.
			Even though the interest is compounded quarterly.
			So:
			\begin{align*}
				\textit{PMT} & = \frac{10,000}{\left(\left(1 + \frac{0.04}{4}\right)^3 + \left(1 + \frac{0.04}{4}\right)^2\right)} \\
					     & = \$4,877.09
			\end{align*}
		\item We must compare the values of \textit{FV}:
			\[
				\textit{FV} = \frac{\textit{PMT}}{I}[(1+I)^n - 1]
			.\]
			\begin{enumerate}
				\item If $I = 0.07$ : Choose option 3.
					\begin{enumerate}
						\item Option 1:
							\begin{align*}
								\textit{FV} & = \textit{PV} \times (1 + I)^n \\
									    & = \$61M \times (1.07)^{31} \\
									    & = \$496.85M
							\end{align*}
						\item Option 2:
							\begin{align*}
								T_{10} & = \frac{\$9.5M}{0.07}[(1.07)^{10}-1] \\
								       & = \$131.26M \\
								\textit{FV} & = T_{10} \times (1.07)^{20} \\
									    & = \$507.92M
							\end{align*}
						\item Option 3:
							\begin{align*}
								\textit{FV} & = \frac{\$5.5M}{0.07}[(1.07)^{30}-1] \\
									    & = \$519.53M
							\end{align*}
					\end{enumerate}
				\item If $I=0.08$, using same working: Choose option 1.
					\begin{enumerate}
						\item Option 1: \$662.92M
						\item Option 2: \$641.45M
						\item Option 3: \$623.05M
					\end{enumerate}
				\item If $I=0.09$, using same working: Choose option 1.
					\begin{enumerate}
						\item Option 1: \$882.16M
						\item Option 2: \$808.90M
						\item Option 3: \$749.69M
					\end{enumerate}
				\item When $I$ was below a certain threshold, The annual payments yielded more money by the end of their payout, but above this threshold, they yielded significantly less.
					All options return more every year, but the regular payments will pay more than the non payout option yearly when the interest rate approaches 0.
			\end{enumerate}
		\item Use \textit{EAR}:
			\begin{enumerate}
				\item Bank A, highest \textit{EAR}.
				\item See table:
					\begin{center}
						\begin{tabular}{c|c|c}
							Bank A & Bank B & Bank C \\
							\hline
							\$5,408 & \$5,385 & \$5,367 \\
						\end{tabular}
					\end{center}
				\item See table:
					\begin{center}
						\begin{tabular}{c|c}
							Bank B & Bank C \\
							\hline
							3.96\% & 3.92\% \\
						\end{tabular}
					\end{center}
				\item See table:
					\begin{center}
						\begin{tabular}{c|c|c}
							Bank A & Bank B & Bank C \\
							\hline
							\$4,807 & \$2,476.78 & \$13.46 \\
						\end{tabular}
					\end{center}
			\end{enumerate}
		\item Use amortisation:
			\begin{enumerate}
				\item Use modified formula for $PV$ :
					\begin{align*}
						150k & = \frac{PMT}{\frac{I}{M}}\left[1-\frac{1}{\left(1+\frac{I}{M}\right)^{n\times M}}\right]\\
						     & = \frac{PMT}{\frac{1}{150}}\left[1-\frac{1}{\left(1+\frac{1}{150}\right)^{360}}\right]\\
						     & = PMT \times 136.28 \\
						PMT & = \$1,100.65
					\end{align*}
				\item Remaining balance will be: \$148,792.23
			\end{enumerate}
		\item Money required in bank account on 18th birthday: \$71,226. \\
			Amount that would be in savings just by compounding \$7.5k: \$10,037. \\
			Amount father has to make up for: \$61,189. \\
			Annual payments: \$10,854.84
	\end{enumerate}
\end{document}
