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Finding interest rates and amount of investment after n years . picture below , highlighted question

+1 vote
asked Apr 12, 2017 in BUS 1003H - Introduction to Financial Risk by Mufhatutshedzwa (310 points)
edited Apr 12, 2017 by Rowan

 I have edited the question with a clearer picture:


commented Apr 12, 2017 by Pandy (2,850 points)

The picture is very unclear, please use Snipping Tool so you can highlight the specific question and post just that. Might make it easier to read. 


2 Answers

+1 vote
answered Apr 12, 2017 by Richard van Gysen (3,180 points)
Best answer

We begin by finding the nominal convertible quarterly interest rate i or effective quarterly rate i/4 either way.

We find the effective quarterly rate as (i/4)  for the first 4 years (16 quarters). The effective monthly rate is (3i/12)  = i/4 as well for the next 10 years (120 months).


\(80 000 = 20 000*(1+i/4)^{16} * (1+3i/12)^{120} \)

\(80 000 = 20 000*(1+i/4)^{16} * (1+i/4)^{120} \)

\(80 000 = 20 000*(1+i/4)^{136} \)

\(i/4  = 0.01024547\) effective quarterly

\(i = 0.04098188\) nominal convertible quarterly

Thus, after 3 years (12 quarters) we accumulate:

\(FV = 20 000*(1+i/4)^{12} = R22 602.32\)

+1 vote
answered Apr 12, 2017 by Rowan (4,010 points)

Hi Mnem

This question is a nice example of a problem which appears to be very complicated, but once you start working on it, a very elegant solution appears. That being said, I think it will be beneficial for you to find this solution on your own, however, let me give you a hint to help you on your way.

Remember that if we have an interest rate \(i\) which is convertible \(n^{th}\)ly then the accumulation factor for an amount \(A\) over \(z\) years is given by:

$$ FV = A(1 + \frac{i}{n})^{n\times z}$$

Try using the above accumulations and see what appears. If you would like further help, simply comment on this answer.