The question follows below:

My question is about (b)(ii). Given that the DPP is 18 years, shouldn't the accumulated amount be the future value of any remaining cash flows after 18 years? I.e. \(A(22)=10000\overline{S}_\overline{4|} \)

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You've got the right idea, but there are some subtleties.

I think it helps to first look at the full solution for b(i), and notice that it's not as simple as rounding up the answer for a(i).

**b(i)**

It's easy to see that the accumulated value at time \(t = 2\) is $$-80000(1.07)^{2} - 5000(1.07)^{1} = -80000(1.07)^t - 5000(1.07)^{t-1}$$

In the third year, the rent starts to come in continuously, and we're allowed to deposit it at \(6\%\) interest, so the accumulated value at time \(t = 3\) is $$ - 80000(1.07)^{3} - 5000(1.07)^{2} + 10000\ {\bar{s}_{\overline{1|}}}_{6\%} $$ $$= - 80000(1.07)^{t} - 5000(1.07)^{t-1} + 10000\ {\bar{s}_{\overline{1|}}}_{6\%}$$

Now, the repayment rate of \(7\%\) exceeds the interest rate on the bank deposit. So, at the end of the third year, we choose to repay part of the loan, rather than keep that money in the bank deposit. During the fourth year, we continuously invest the income at \(6\%\). It follows that the accumulated value at time \(t = 4 \) isĀ

$$ -80000(1.07)^4 - 5000(1.07)^{3} + 10000\ {\bar{s}_{\overline{1|}}}_{6\%}(1.07)^1 + {\bar{s}_{\overline{1|}}}_{6\%}$$

$$= -80000(1.07)^t - 5000(1.07)^{t-1} + 10000\ ({\bar{s}_{\overline{1|}}}_{6\%}){{s}_{\overline{t-2|}}}_{7\%}$$

And so we identify the pattern that, until the bank loan is repaid,

$$A(t) = -80000(1.07)^t - 5000(1.07)^{t-1} + 10000\ ({\bar{s}_{\overline{1|}}}_{6\%}){{s}_{\overline{t-2|}}}_{7\%}$$.

Solving for (the smallest) \(t\) such that \(A(t) = 0\), we get \(t = 17.9022\). Since the bank loan can only be repaid at the end of a complete year, we conclude that the \(\textrm{DPP} =18\) years.

**b(ii)**

Now, this seems like quite a lot of work, but it sets us up quite nicely for part (ii).

Using the formula from before, \(A(18) = 997.048\), and all the income that occurs after time \(18\) is deposited continuously at \(6\%\). Thus, the final accumulated amount at time \(22\) is $$997.048(1.06)^4 + 10000{\bar{s}_{\overline{4|}}}_{6\%} = 46279.3$$

PS, apologies if there are any typos!

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