A Tickler General Annuity Problem

jonah

Junior Member
Joined
May 15, 2008
Messages
52
Somebody tried to stump me lately with a tickler of general annuity problem that (as Sir TKHunny described it once when he noticed a blunder on my part at MathHelpForum.com last year) has a “nasty” factor – in this particular case – a nasty payment or deposit interval. I did got stumped (and tickled a little) for a few minutes and had a little drink to jumpstart or pushstart them old brain cells back into battle condition. Before I knew it, fell off the wagon, hit my head, and the King Tut persona (for most of us anyway - remember King Tut from the Adam West Batman series?) - the Jonah persona in my case, has taken over again. After a few weeks of sanity, I feel compelled somehow to rejoin my fellow knights-errant in their unspoken knightly vows and duties, i.e. to aid the clueless, guide the lost, entertain them angry lazy ingrates, etc., in their struggle against our beloved queen. Queen to some of us, the dark Lord Mathematicus for a great unenlightened many.

My apologies to my fellow knights-errant, particularly to Sir Subotosh Khan and Sir Denis for dereliction of my “sworn” knightly duties. I understand that these two fine knights-errant mentioned me twice at viewtopic.php?f=17&t=33390. From what I could gather, the indefatigable Sir Denis dispatched the perceived Lord Mathematicus dragon in that particular thread without so much as breaking a sweat.

A company deposits $800 at the end of every nine months in a bank that pays 3% interest compounded monthly. Find the amount of the company’s account at the end of 4½ years.

Accordingly, we have

\(\displaystyle S = 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{3.75 \times 12} + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{3 \times 12} + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{2.25 \times 12} + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{1.5 \times 12} + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{.75 \times 12} + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{0 \times 12}\)
\(\displaystyle \Leftrightarrow\)
\(\displaystyle S = \sum\limits_{k = 1}^6 {800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{\left( {4.5 - .75k} \right)\left( {12} \right)} } \approx {\rm{5,081}}{\rm{.12171632336}}...\)

OR (in reversing the order, we have)

\(\displaystyle S = 800 + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{.75 \times 12} + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{1.5 \times 12} + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{2.25 \times 12} + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{3 \times 12} + 800\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{3.75 \times 12}\)
\(\displaystyle \Leftrightarrow\)
\(\displaystyle S = 800 + 800\left( {1.0025} \right)^9 + 800\left( {1.0025} \right)^{18} + 800\left( {1.0025} \right)^{27} + 800\left( {1.0025} \right)^{36} + 800\left( {1.0025} \right)^{45}\)
\(\displaystyle \Leftrightarrow\)
\(\displaystyle S = 800\left[ {1 + \left( {1.0025} \right)^9 + \left( {1.0025} \right)^{18} + \left( {1.0025} \right)^{27} + \left( {1.0025} \right)^{36} + \left( {1.0025} \right)^{45} } \right]\)
\(\displaystyle \Leftrightarrow\)
\(\displaystyle S = 800\frac{{{\rm{(common ratio)(last term)}} - {\rm{(1st term)}}}}{{{\rm{(common ratio)}} - 1}}\)
\(\displaystyle \Leftrightarrow\)
\(\displaystyle S = 800\frac{{\left( {1.0025} \right)^9 \left( {1.0025} \right)^{45} - 1}}{{\left( {1.0025} \right)^9 - 1}}\)
\(\displaystyle \Leftrightarrow\)
\(\displaystyle S = 800\frac{{\left( {1.0025} \right)^{54} - 1}}{{\left( {1.0025} \right)^9 - 1}} \approx {\rm{5,081}}{\rm{.12171632336}}...\)

In using the (reduced form of the) general annuity formula, we have

\(\displaystyle S = 800\frac{{\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{4.5 \times 12} - 1}}{{\left( {1 + {\textstyle{{.03} \over {12}}}} \right)^{{\textstyle{{12} \over p}}} - 1}}\)

where p is the payment interval.
Clearly, p = 4/3.

To determine p without resorting to sigma notation, we consider the fact that if A is the present value of such an amount S at the end of 4½ years, and if r is the equivalent rate to {j = .03, m = 12} at which such a present value is to be accumulated every nine months, then

\(\displaystyle S = A\left( {1 + \frac{r}{p}} \right)^{4.5p}\)
Accordingly, there are 6 payment intervals in 4.5 years (i.e. 4.5/.75). Thus, one just needs to solve for p in:
6 = 4.5p
and we get the same value for p.
 
jonah said:
... Before I knew it, fell off the wagon, ...
Welcome back, Sir Jonah!

I'm now sober over 24 years (Apr 8/85); I tried 30 years ago, but kept falling off that wagon too...
finally managed to stay sober, after realising the first AA step meant:
"We admitted we were powerless over alcohol, and that our WIVES were unmanageable"

Remember when....
I walk in the men's tavern, a bit down;
the boys: whassamatta Denis, you look kinda depressed?
Denis: ya, the wife told me she'd leave me...
the boys: WHAT!! a nice guy like you...
Denis (suddenly feeling better): hey waiter, bring a round here...
 
jonah said:
A company deposits $800 at the end of every nine months in a bank that pays 3% interest compounded monthly.

Agree with your results if .03/12 = .0025 is used.

BUT BUT banks being sneaky, they'll say they meant 3% annual, hence .0024662697723... monthly,
for a result of 5077.176444122516....
 
Top