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Increasing Annuity

+4 votes
asked Aug 14, 2016 in BUS 2016H - Financial Mathematics by Pandy (2,150 points)

The question below is from a BUS2016H tut.


I'm struggling with both a) and b).

For a), I get that $$PV=0+V+4V^2+...+361V^{20}$$

But I don't know how to simplify it from there. I know it's an increasing annuity but I'm not sure how to get it into any form I can recognize.

And I'm completely unsure where to begin for b).

1 Answer

+4 votes
answered Aug 14, 2016 by DanNewton (970 points)
selected Aug 15, 2016 by Richard van Gysen
Best answer

First of all your present value is wrong. Annual payments in advance means that the first payment is at the beginning of the first year. Thus your PV would be:

$$PV = 1 + 4V + 9V^2 + ... + 400 V^(19) $$

Now, you know from first year that any quadratic has a common second difference. The trick here is to extract that into an understandable annuity.

To do this, take: $$PV - v*PV$$

Hint: This should result in an arithmetically increasing annuity, which you have learnt how to solve.

(I'm not sure why the notation for the last part of the PV messed up, but it's supposed to be 400V^(19))

commented Aug 14, 2016 by Pandy (2,150 points)

Ah right, misread the question. Thanks very much!

Would it be a similar idea for b) i.e. manipulate into something of a similar form to that in a)?

commented Aug 14, 2016 by DanNewton (970 points)

For (b) you'll have to go back to first principles. 

I'm sure you remember doing questions where the rate of payment is abnormal, and to solve this you plugged it into an integral.

So for (b) you're going to have to go back to the basics when it comes to continuous payments and set up an integral with rate of payment t^2.

If you are struggling I will post a couple hints.

commented Aug 14, 2016 by simon_rigby (4,220 points)

Thanks for contributing DanNewton :) Yes, even if the payment is made at time 0, because it's an annuity in advance, it's still in the first year. And the payment at time \((t-1)\) is in the \(t\)-th year, so the amount is \(t^2\).

If I remember, you get quite a nasty formula in this way (but it's right) and \((I\ddot{a})_\overline{n|} \) is one of the terms. (I think)

commented Aug 14, 2016 by Pandy (2,150 points)

Doing as you suggested I get $$PV=\int_0^{20}{t^2e^{-\int_0^t{\delta ds}}dt}$$

where \(\delta=ln(1+i)=ln(1+5\%)\).

Where have I gone wrong? Because evaluating this I get \(PV=1307.10\). Should I not use the payment stream of \(t^2\) and instead write out each integral for each year?

Thanks for all the help!

commented Aug 14, 2016 by DanNewton (970 points)

Oh sorry I've made a mistake. The question says that the payment is constant over every year. So you're setup is different than I thought. You would have to do a summation rather than an integral. I will have to be checked on this, but I think the sum would be:

$$ PV = \sum_{t=1}^{20} t^2 \bar{a}_\bar{1|} v^{t-1} $$

Explanation: You are paying a set amount continuously for each year (t^2), and then taking it back to time 0.

To solve for this you use the same method as question 1. Notice a common factor of the continuous annuity on the RHS.