2 problems found
My bank pays \(\rho\%\) interest at the end of each year. I start with nothing in my account. Then for \(m\) years I deposit \(\pounds a\) in my account at the beginning of each year. After the end of the \(m\)th year, I neither deposit nor withdraw for \(l\) years. Show that the total amount in my account at the end of this period is \[\pounds a\frac{r^{l+1}(r^{m}-1)}{r-1}\] where \(r=1+{\displaystyle \frac{\rho}{100}}\). At the beginning of each of the \(n\) years following this period I withdraw \(\pounds b\) and this leaves my account empty after the \(n\)th withdrawal. Find an expression for \(a/b\) in terms of \(r\), \(l\), \(m\) and \(n\).
Solution: Rather than putting the deposits in the same account, imagine they are all put in separate accounts. Then for example, the first \(\pounds a\) will go on to become \(\pounds r^m \cdot r^l a\) from \(m\) years of compound interest as more money is deposited, followed by \(l\) years where no money is deposited. Therefore the total amount at the end is: \begin{align*} && S &= r^{m}r^l a + r^{m-1}r^l a + \cdots + r r^l a \\ &&&= r^l a(r^m + \cdots + r) \\ &&&= ar^l\frac{r^{m+1}-r}{r-1} \\ &&&= a \frac{r^{l+1}(r^m-1)}{r-1} \end{align*} Rather than withdrawing \(b\) each time, imagine that in the \(n\)th year we withdraw each \(b\) with the appropriate additional interest, ie \begin{align*} && \underbrace{a \frac{r^{l+1}(r^m-1)}{r-1}}_{\text{amount before \(n\) years}} \underbrace{r^{n-1}}_{\text{accounting for interest}} &= b r^{n-1} + br^{n-2} + \cdots + b \\ &&&= b \frac{r^n-1}{r-1} \\ \Rightarrow && \frac{a}{b} &= \frac{r^n-1}{r^{l+n}(r^m-1)} \end{align*}
A firm of engineers obtains the right to dig and exploit an undersea tunnel. Each day the firm borrows enough money to pay for the day's digging, which costs £\(c,\) and to pay the daily interest of \(100k\%\) on the sum already borrowed. The tunnel takes \(T\) days to build, and, once finished, earns £\(d\) a day, all of which goes to pay the daily interest and repay the debt until it is fully paid. The financial transactions take place at the end of each day's work. Show that \(S_{n},\) the total amount borrowed by the end of day \(n\), is given by \[ S_{n}=\frac{c[(1+k)^{n}-1]}{k} \] for \(n\leqslant T\). Given that \(S_{T+m}>0,\) where \(m>0,\) express \(S_{T+m}\) in terms of \(c,d,k,T\) and \(m.\) Show that, if \(d/c>(1+k)^{T}-1,\) the firm will eventually pay off the debt.
Solution: After \(n\) days they will have borrowed \(c\) for \(n-1\) days, \(c\) for \(n-2\) days, etc until \(c\) for no days. Therefore the outstanding balance will be: \begin{align*} c + (1+k)\cdot c+ (1+k)^2 \cdot c + \cdots + (1+k)^{n-1} \cdot c &= c\frac{(1+k)^n-1}{(1+k)-1} \\ &= \frac{c[(1+k)^n-1]}{k} \end{align*} At the end of \(T\) days the outstanding balance will be \(S_T = \frac{c[(1+k)^T-1]}{k}\). We can think of each payment of \(d\) during the subsequent period as being equivalent of a payment of \(d (1+k)^{m-1}\) \(m\) days later (as otherwise they would have accrued the equivalent amount in interest. Therefore after \(m\) days the amount paid back (equivalent) is: \begin{align*} (1+k)^{m-1} \cdot d + (1+k)^{m-2} \cdot d + \cdots + d &= \frac{d[(1+k)^m-1]}{k} \end{align*} Therefore the net position, \(S_{T+m}\) will be: \begin{align*} S_{T+m} &= \frac{c[(1+k)^T-1](1+k)^m-d[(1+k)^m-1]}{k} \\ &= \frac{(1+k)^m [c ((1+k)^T-1)-d]+d}{k} \end{align*} Therefore they will eventually pay back their debts if \( [c ((1+k)^T-1)-d]\) is negative. ie \(d > c((1+k)^T-1) \Rightarrow d/c > (1+k)^T-1\)