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Question 2 of tut 2: Total accumulated value at any time t>0

+3 votes
90 views
asked Apr 10, 2016 in BUS 2016H - Financial Mathematics by Pandy (2,530 points)

The question states:

"The force of interest at any time \(t\), (measured in years) is given by

\(f(t) = \begin{cases}0.07-0.005t & 0 \le t < 5 \\ 0.06-0.003t & 5 \le t < 10 \\ 0.03 & t \ge 10 \end{cases}\)

What is the total accumulated in value at any time \(t > 0\) of investments of R100 at times 0, 4 and 6?"

I understand getting to first 'time period' i.e. for \(0 \le t \lt 4\) but from there on I get confused. The memo splits the answer into \(0 \le t \lt 4\), \(0 \le t \lt 5\), \(0 \le t \lt 6\), \(0 \le t \lt 10\), \(t \ge 10\). Why is it not split from 0 to 4, 4 to 5, 5 to 6, 6 to 10 and greater than 10?



commented Apr 10, 2016 by simon_rigby (4,220 points)
reshown Apr 10, 2016 by simon_rigby

I think there was a mistake in that memo hey. The way you understand it is correct.

commented Apr 10, 2016 by Pandy (2,530 points)

Okay great, thanks Simon. And for question 4 (or question 2, since they're basically identical but with different values), how do they simplify the two terms into one? I seem to get two terms, one for each payment when \(5 \le t < 6 \) but the memo has only one term.

commented Apr 10, 2016 by asilmotala (2,610 points)

From what I understand, you still kind of have separate values for each cashflow (summed together of course at correct time intervals, when the cashflows become relevant) at different times (which should be correct, if I understand your question correctly). The memo simply combines those cashflows at certain times, for example in question 4, at time 5, it will combine the accumulated value of the one rand invested at time 0, with the one rand invested at time 5, and then accumulate it in value further. I hope that answers your question.

commented Apr 10, 2016 by simon_rigby (4,220 points)

Oh ya, I had a second look and the memo is right. But it's a lot of working out! :/

commented Apr 10, 2016 by Pandy (2,530 points)

Thank you, that does make sense. However, I'm still struggling a bit with the manipulation.

In question 4, for \(5 \le t \lt 6\) I get the following by accumulating the first payment to \(t=4\), adding the payment at \(t=5\) and then accumulating further:

$$(e^{0.04\times5+0.0025\times25}+1)e^{\int_5^t0.04+0.005sds}$$

Simplifying this I get:

$$e^{0.04t+0.0025t^{2}}(1+e^{\frac {-21}{80}})$$

I notice that \(e^{0.570488}\) (from the memo) is roughly the same as my last bracket in the line above, but I am unsure as to how to simplify \((1+e^{\frac {-21}{80}})\) to  \(e^{0.570488}\).

And w.r.t question 2, I am very confused as to how they combine the intervals i.e. how \(0 \le t \lt 4\) and \(4 \le t \lt 5\) are combined into the interval \(0 \le t \lt 5\).

Thanks very much for the help so far.

1 Answer

+2 votes
answered Apr 10, 2016 by simon_rigby (4,220 points)
selected May 6, 2016 by Pandy
 
Best answer

Sorry for all the confusion. I've looked even more carefully, and the bounds are wrong. They should read \(0 \le t < 4\),    \(4 \le t < 5\),    \(5 \le t < 6\),    \(6 \le t < 10\), and   \( t \ge 10\).

e.g. If \(4 \leq t < 5\) then

\( A(t) = 100\exp[\int_{0}^{t}(0.07 - 0.005s)ds] + 100\exp[\int_{4}^{t}(0.07 - 0.005s)ds] \)

\( = 100\exp[0.07t - 0.0025t^{2}](1 + e^{-0.24}) = 178.663\exp[0.07t - 0.0025t^{2}] \)

Obviously it doesn't make sense if we substitute some \(t < 4 \) in the above. And similarly for the rest. 

Regarding Q4, you're spot on. What's \(\log(1 + e^{-21/80}) \)?

commented Apr 10, 2016 by Pandy (2,530 points)

Awesome! Thanks very much Simon.

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