Fun problem for everyone - An annuity whose payments vary

jonah

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May 15, 2008
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Find the present value of an annuity of which the payments are $200 per month at the end of each month during the 1st year, $195 per month during the 2nd year, $190 per month during the 3rd year, etc. ($5 monthly will be paid at the end of each month during the 40th year and nothing thereafter.) Interest is at 12% compounded monthly. Ans. $16,090.80 (Borrowed/Lifted from a GoogleBooks preview of Schaum’s Outlines in Mathematics of Finance by P. Zima and R. Brown)
 
jonah said:
Find the present value of an annuity of which the payments are $200 per month at the end of each month during the 1st year, $195 per month during the 2nd year, $190 per month during the 3rd year, etc. ($5 monthly will be paid at the end of each month during the 40th year and nothing thereafter.) Interest is at 12% compounded monthly. Ans. $16,090.80 (Borrowed/Lifted from a GoogleBooks preview of Schaum’s Outlines in Mathematics of Finance by P. Zima and R. Brown)
What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you! :D

Eliz.
 
Funner problem is to do the problem in the reverse order : i.e. first year $5/month, second year 10/month, and so on - how long would it take to deplete $16,090.80 at a rate 12%/year (I wish) compounded monthly.
 
SUM[5(1 - v)/.01 / 1.01^(39*12) + 10(1 - v)/.01 / 1.01^(38*12) + ...
... + 195(1 - v)/.01 / 1.01^(1*12) + 200(1 - v)/.01 / 1.01^(0*12)] = 16090.788674169358....
where v = 1 / 1.01^12
....You're out .0113...cents!

Would be less wieldy this way:
v = 1.01, w = (1 - 1/v^12) / .01

SUM[5w / v^(39*12) + 10w / v^(38*12) + ..... + 195w / v^(1*12) + 200w / v^(0*12)]
 
Denis said:
SUM[5(1 - v)/.01 / 1.01^(39*12) + 10(1 - v)/.01 / 1.01^(38*12) + ...
... + 195(1 - v)/.01 / 1.01^(1*12) + 200(1 - v)/.01 / 1.01^(0*12)] = 16090.788674169358....
where v = 1 / 1.01^12

You're out .0113...cents!

Or is it 1.113 cents!!!
 
x = (1 - 1/1.01^12) / .01
w = 5
y = 40
n = 1 to 40
p = present value

p = SUM[w n x / r^(y - n)]

WELL?!

Btw, ya'll notice that .53446... cents will permit the twelve $5 withdrawals in 40 years :idea:
 
Notice that one solution is a variation of the old, and not entirely satisfactory, summation.


If we first establish that

\(\displaystyle v = \frac{1}{1.01}\),

\(\displaystyle w = v + v^{2} + ... + v^{12} = \frac{v-v^{13}}{1-v}\), and

\(\displaystyle x = v^{12}\)

We have

1) \(\displaystyle 200w + 195wx + 190wx^{2} + ... + 5wx^{39} = S\) = The desired sum

Now the standard trick, the use of which is not necessarily obvious at this point...

2) \(\displaystyle 200wx + 195wx^{2} + 190wx^{3} + ... + 5wx^{40} = Sx\)

Subtracting 2) from 1)

\(\displaystyle 200w - 5wx - 5wx^{2} - ... - 5wx^{40} = S(1-x)\)

That didn't help quite as much as may have been hoped, but after a little algebra on the left-hand side

\(\displaystyle 5w \cdot \left[40 - (x + x^{2} + ... + x^{40})\right] = S(1-x)\)

Now the usual formualtion for level payments

\(\displaystyle 5w\cdot \left[40 - \left(\frac{x-x^{41}}{1-x}\right)\right] = S(1-x)\)

And we see

\(\displaystyle S = 5w \cdot \left[\frac{40 - \left(\frac{x-x^{41}}{1-x}\right)}{1-x}\right] = 16090.7886741693\)

...establishing, yet another time, the value of the concept of "Basic Principles". It is unlikely one would want to memorize such a formula.

Perhaps more instructive would be:

\(\displaystyle S = w \cdot \left[\frac{200 - 5 \dcot \left(\frac{x-x^{41}}{1-x}\right)}{1-x}\right] = 16090.7886741693\)

I guess I'd have to take exception to that 80¢. It seems a little high. Too much rounding, no doubt.
 
Hey, bewteefull TK :!:

I was trying to get such a "formula" and getting frustrated...
now I can throw away a bundle of paper sheets...and sharpen my pencil !
 
I have nothing up my sleeves - except basic principles. :D
 
I have nothing to declare except my genius. - Oscar Wilde.
 
stapel said:
What are your thoughts? What have you tried? How far have you gotten? Where are you stuck?

Please be complete. Thank you! :D
Subhotosh Khan said:
at a rate 12%/year (I wish) compounded monthly.
I was trying to decide which was more amusing: stapel’s polite standard opening reply (whether one is asking for help or not) or Subhotosh Khan’s pining side remark. Considering that stapel’s “formula” is getting rather old and considering that the last time I saw a bank that offered a 10% interest on their depositor’s savings account was way back in the year 19 - oh wait, I guess there never was such a time. I guess Subhotosh Khan is my funny man for the day – although I do wish I could catch a rerun of Robin Williams’ antics on Letterman the other day.

Anyways, my solution for this “fun problem” consisted of blind application of the formula for a decreasing annuity with payments nW, (n-1)W, (n-2)W, …, 3W, 2W, W at the ends of the first n periods, with i as the rate per period. Accordingly, the present value, A, of these payments is given by
\(\displaystyle A = W \cdot \frac{{n - a_{\left. {\overline {\, n \,}}\! \right| i} }}{i}\)
In this case,
\(\displaystyle W = 5 \cdot s_{\left. {\overline {\, {12} \,}}\! \right| .01}\)
\(\displaystyle n = 40\)
\(\displaystyle i = \left( {1.01} \right)^{12} - 1\)
 
Yer purty funny yerself, Jonah :D

"he with nary a smile is but the skeleton of his inner child"

"laughter is to the mental what jogging is to the physical"
 
Sir Subhotosh Khan said:
Funner problem is to do the problem in the reverse order : i.e. first year $5/month, second year 10/month, and so on - how long would it take to deplete $16,090.80 at a rate 12%/year (I wish) compounded monthly.
From Dr. Howard W. Eves’ Return to Mathematical Circles (ca 1988)
143º The equality of null sets.

The man in the wilderness asked of me
How many strawberries grew in the sea.
I answered him as I thought good,
As many as red herring grow in the wood.
-Anonymous

how long would it take to deplete $16,090.80 at a rate 12%/year (I wish) compounded monthly?
I’d say the probability of my winning the powerball lottery (near zero as it is) is better than the probability of this fund getting depleted at the stated rate of withdrawal. Accordingly, the present value of such a problem is represented by
\(\displaystyle \begin{gathered} A = W \cdot \frac{{\left( {1 + i} \right) \cdot s_{\left. {\overline {\, n \,}}\! \right| i} - n}}{i} \cdot \left( {1 + i} \right)^{ - n} \hfill \\ \Leftrightarrow \hfill \\ \end{gathered}\)
\(\displaystyle \begin{gathered} A = \frac{W}{i}\left[ {\left( {1 + i} \right) \cdot a_{\left. {\overline {\, n \,}}\! \right| i} - n\left( {1 + i} \right)^{ - n} } \right] \hfill \\ \Leftrightarrow \hfill \\ A \approx \$ 4,236.430938... \hfill \\ \end{gathered}\)
where
\(\displaystyle \begin{gathered} W = \left( {5 \cdot s_{\left. {\overline {\, {12} \,}}\! \right| .01} } \right) \hfill \\ n = 40 \hfill \\ i = \left( {1.01} \right)^{12} - 1 \hfill \\ \end{gathered}\)
One could almost see that it would take exactly 40 years to deplete a present value of $4,236.43. Applying the so-called retrospective method for determining the fund balance on a certain date, we have:
\(\displaystyle A\left( {1 + i} \right)^{40} - W \cdot \frac{{\left( {1 + i} \right) \cdot s_{\left. {\overline {\, {40} \,}}\! \right| i} - 40}}{i} = 0\)
On the other hand, if the present value to be depleted is $16,090.80, it can be shown that unless the world suddenly ends for whatever reason, the fund value will never be depleted since, among other factors, the growth rate is greater than the depletion or withdrawal rate. As an example, at the end of 40 years, the fund balance (using Sir Denis’ results) would be
\(\displaystyle \left( {{\text{\$16,090}}{\text{.788674169358}}....} \right)\left( {1 + i} \right)^{40} - W \cdot \frac{{\left( {1 + i} \right) \cdot s_{\left. {\overline {\, {40} \,}}\! \right| i} - 40}}{i} \approx \$ {\text{1,406,492}}{\text{.57798946}}...\)
Sir Subhotosh Khan’s problem is therefore a type of a growing (arithmetic) perpetuity.

Incidentally, to determine the fund balance at the end of a month that is between years, say 2 3/12 years, with
\(\displaystyle A_1 = {\text{\$16,090}}{\text{.788674169358}}....\)
and
\(\displaystyle P = A_1 \left( {1 + i} \right)^2 - W \cdot \frac{{\left( {1 + i} \right) \cdot s_{\left. {\overline {\, 2 \,}}\! \right| i} - 2}}{i}\)
the fund balance would be
\(\displaystyle P\left( {1.01} \right)^3 - 15 \cdot s_{\left. {\overline {\, 3 \,}}\! \right| .01} \approx \$ {\text{20,800}}{\text{.3731827106}}...\)
 
Well, the present value for this exercise could be a variable; why $16,090.80 ?

How long to deplete $A with payments of $p monthly 1st year, $2p monthly 2nd year,
$3p monthly 3rd year ..... at rate of r% annually compounded monthly ?
To actually deplete, you'd need a condition like p > A(r/12).
 
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