My computer went down for the count and I thought that was the end of it. I got into a bad drunken habit of just unplugging it whenever it takes forever to shut down normally and I finally paid the price. Fortunately for me, my patron, the wine god Dionysus heard my prayers and decided to resuscitate my poor abused computer back to life. At any rate, I see that my friend, Sir Denis disagreed with my assessment of $11,043,966.33. Having just re-read Don Quixote lately – what with being cut off from the online community and all – and being put in an extremely humorous mood in the process, I’m somewhat inclined to challenge Sir Denis to a traditional joust or sword duel of sorts to make him “confess” that my assessment of $11,043,966.33 is the correct one. Also, with this re-reading of Don Quixote, I’m also inclined to believe that those who indulge in whatever facet of Mathematics, being the queen of the sciences – I distinctly remember her being called that – be they students, dabblers like myself, or professionals, are practically knights-errant of that ruthless and exacting queen; thus the prefix title of “Sir”. Ah! There goes that rambling drunk again, I could almost hear the silent rolling of eyes everywhere. With my amortization spreadsheet schedule confirming my theoretical calculations, I’m afraid I’ll have to tire you out once again with my poor excuse of an inductive argument, Sir Denis.
Let’s begin by letting
\(\displaystyle R_1 =\)$15,000 and \(\displaystyle \ddot A\)= $10,074,177.10.
From a pure growing annuity due perspective:
The account balance after withdrawal made on the 1st of the 290th month relative to age 26 is the same as
the account balance after withdrawal made on the 1st of the 2nd month relative to age 50. Since the 1st of the 2nd month is the end of the 1st month, this account balance is represented by
\(\displaystyle \begin{gathered} \left[ {\ddot A - R_1 \left( {1 + g} \right)^0 } \right]\left( {1 + i_{12} } \right)^1 - R_1 \left( {1 + g} \right)^1 \hfill \\ = \ddot A\left( {1 + i_{12} } \right)^1 - R_1 \left( {1 + g} \right)^0 \left( {1 + i_{12} } \right)^1 - R_1 \left( {1 + g} \right)^1 \hfill \\ = \ddot A\left( {1 + i_{12} } \right)^1 - R_1 \frac{{\left( {1 + g} \right)^1 \left( {1 + i_{12} } \right)^1 }}{{r - 1}} - R_1 \left( {1 + g} \right)^1 \hfill \\ \end{gathered}\)
= (FV of \(\displaystyle \ddot A\) at end of 1st month)-(FV value of
1st payment at end of 1st month)-(
2nd payment at 1st of 2nd month)
= $10,077,521.54
The account balance after withdrawal made on the 1st of the 291st month relative to age 26 is the same as
the account balance after withdrawal made on the 1st of the 3rd month relative to age 50. Since the 1st of the 3rd month is the end of the 2nd month, this account balance is represented by
\(\displaystyle \begin{gathered} \left[ {\ddot A\left( {1 + i_{12} } \right)^1 - R_1 \left( {1 + g} \right)^0 \left( {1 + i_{12} } \right)^1 - R_1 \left( {1 + g} \right)^1 } \right]\left( {1 + i_{12} } \right)^1 - R_1 \left( {1 + g} \right)^2 \hfill \\ = \ddot A\left( {1 + i_{12} } \right)^2 - R_1 \left( {1 + g} \right)^0 \left( {1 + i_{12} } \right)^2 - R_1 \left( {1 + g} \right)^1 \left( {1 + i_{12} } \right)^1 - R_1 \left( {1 + g} \right)^2 \hfill \\ = \ddot A\left( {1 + i_{12} } \right)^2 - \left[ {R_1 \left( {1 + g} \right)^0 \left( {1 + i_{12} } \right)^2 + R_1 \left( {1 + g} \right)^1 \left( {1 + i_{12} } \right)^1 } \right] - R_1 \left( {1 + g} \right)^2 \hfill \\ = \ddot A\left( {1 + i_{12} } \right)^2 - R_1 \frac{{\left( {1 + g} \right)^2 - \left( {1 + i_{12} } \right)^2 }}{{r - 1}} - R_1 \left( {1 + g} \right)^2 \hfill \\ \end{gathered}\)
= (FV of \(\displaystyle \ddot A\) at end of 2nd month)-(FV value of
1st payment +
2nd payment at end of 2nd month)
-(
3rd payment at 1st of 3rd month)
= $10,095,851.55
\(\displaystyle \begin{gathered} \bullet \hfill \\ \bullet \hfill \\ \bullet \hfill \\ \end{gathered}\)
The account balance after withdrawal made on the 1st of the 600th month relative to age 26 is the same as
the account balance after withdrawal made on the 1st of the 312th month relative to age 50. Since the 1st of the 312th month is the end of the 311th month, this account balance is represented by
\(\displaystyle \ddot A\left( {1 + i_{12} } \right)^{311}\)
\(\displaystyle - [R_1 \left( {1 + g} \right)^0 \left( {1 + i_{12} } \right)^{311} + R_1 \left( {1 + g} \right)^1 \left( {1 + i_{12} } \right)^{310} + R_1 \left( {1 + g} \right)^2 \left( {1 + i_{12} } \right)^{309} + \cdot \cdot \cdot\)
\(\displaystyle + R_1 \left( {1 + g} \right)^{310} \left( {1 + i_{12} } \right)^1 ]\) \(\displaystyle - R_1 \left( {1 + g} \right)^{311}\)
\(\displaystyle = \ddot A\left( {1 + i_{12} } \right)^{311} - R_1 \frac{{\left( {1 + g} \right)^{311} - \left( {1 + i_{12} } \right)^{311} }}{{r - 1}} - R_1 \left( {1 + g} \right)^{311}\)
= (FV of \(\displaystyle \ddot A\) at end of 311th month)
-(FV value of
1st payment +
2nd payment +
3rd payment + · · · +
311th payment at end of 311th month)
-(
312th payment at 1st of 312th month)
= $11,043,966.33
\(\displaystyle \begin{gathered} \bullet \hfill \\ \bullet \hfill \\ \bullet \hfill \\ \end{gathered}\)
The account balance after withdrawal made on the 1st of the 708th month relative to age 26 is the same as
the account balance after withdrawal made on the 1st of the 420th month relative to age 50. Since the 1st of the 420th month is the end of the 419th month, this account balance is represented by
\(\displaystyle \ddot A\left( {1 + i_{12} } \right)^{419}\)
\(\displaystyle - [R_1 \left( {1 + g} \right)^0 \left( {1 + i_{12} } \right)^{419} + R_1 \left( {1 + g} \right)^1 \left( {1 + i_{12} } \right)^{418} + R_1 \left( {1 + g} \right)^2 \left( {1 + i_{12} } \right)^{417} +\)
\(\displaystyle \begin{gathered} \cdot \cdot \cdot + R_1 \left( {1 + g} \right)^{418} \left( {1 + i_{12} } \right)^1 ] - R_1 \left( {1 + g} \right)^{419} \hfill \\ = \ddot A\left( {1 + i_{12} } \right)^{419} - R_1 \frac{{\left( {1 + g} \right)^{419} - \left( {1 + i_{12} } \right)^{419} }}{{r - 1}} - R_1 \left( {1 + g} \right)^{419} \hfill \\ \end{gathered}\)
= (FV of \(\displaystyle \ddot A\) at end of 419th month)
-(FV value of
1st payment +
2nd payment +
3rd payment + · · · +
419th payment at end of 419th month)
-(
420th payment at 1st of 420th month)
= $3,810,123.26
which is, as you pointed out in an earlier post, “the required amount in account to accommodate Sharon's 60 payments, a month before her 1st payment.” Relative to your simplified analysis on your last post, this point of time is on Nov.30/59.
\(\displaystyle \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet\)
From an ordinary growing annuity perspective:
With \(\displaystyle A =\)$10,074,177.10 - $15,000 = $10,059,177.10
And \(\displaystyle R_2 = 15,000(1 + g) = 15,075\)
The account balance after withdrawal made on the 1st of the 600th month relative to age 26 is the same as
the account balance after withdrawal made on the 1st of the 312th month relative to age 50. Since the 1st of the 312th month is the end of the 311th month, this account balance is represented by
\(\displaystyle A\left( {1 + i_{12} } \right)^{311}\)
\(\displaystyle - [R_2 \left( {1 + g} \right)^0 \left( {1 + i_{12} } \right)^{310} + R_2 \left( {1 + g} \right)^1 \left( {1 + i_{12} } \right)^{309} + R_2 \left( {1 + g} \right)^2 \left( {1 + i_{12} } \right)^{308} +\)
\(\displaystyle \cdot \cdot \cdot + R_2 \left( {1 + g} \right)^{309} \left( {1 + i_{12} } \right)^1 + R_2 \left( {1 + g} \right)^{310} \left( {1 + i_{12} } \right)^0 ]\)
\(\displaystyle = A\left( {1 + i_{12} } \right)^{311} - R_2 \frac{{\left( {1 + i_{12} } \right)^{311} - \left( {1 + g} \right)^{311} }}{{i_{12} - g}}\)
= (FV of \(\displaystyle A\)at end of 311th month)
-(FV value of
1st payment +
2nd payment +
3rd payment + · · · +
310th payment +
311th payment at end of
311th month relative to age 50)
= $11,043,966.33
Denis said:
Monthly withdrawals will then be made from the account until Dec.31/59, the 1st being $15,075.
These withdrawals will increase by 1/2 % each month.
I’d say this should be
“Monthly withdrawals will then be made from the account until
Nov.30/59, the 1st being $15,075. These withdrawals will increase by 1/2 % each month.”
Denis said:
The next withdrawal will be 1/2 of the Dec.31/59 withdrawal.
I’d say this should be
“The next withdrawal will be 1/2 of the
Nov.30/59 withdrawal.”
Denis said:
Withdrawals will continue until Dec.31/64, again increasing by 1/2% each month.
After the Dec.31/64 withdrawal, the account will be exactly ZERO.
I’d say this should be
“Withdrawals will continue until
Nov.30/64, again increasing by 1/2% each month.
After the
Nov.30/64 withdrawal, the account will be exactly ZERO.”
Denis said:
BUT teachers feel a need to "embellish", thus creating confusion :evil:
Even to the point that sometimes it takes a student longer to "decipher" the story than it does solving the problem:
very unfortunate if a test is timed...
I feel your pain Sir Knight. At least teachers these days no longer require students to use tables like what I had to go through despite the availability of scientific calculators. Or do they?