## Dual Compounding Interest Equations

Topics that don't fit anywhere else.
GreenLantern
Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm
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### Dual Compounding Interest Equations

This is from my money and banking class, I'm missing some critical know how for this specific problem and was hoping it could be answered here. It's the last letter in my sixth question, so please excuse the wall of text I'll try to divide and edit it nicely.

Base Scenario: Some friends of yours have just had a child. Thinking ahead, and realizing the power of compound interest, they are considering investing for their child’s college education, which will begin in 18 years. Assume that the cost of a college education today is $125,000; there is no inflation; and there are no taxes in interest income that is used to pay college tuition and expenses. A) If the interest rate is 5 percent, how much will your friends need to put into their savings account today to have$125,000 in 18 years?
D) Return to part A, the case with 5 percent interest rate and no inflation. Assume that your friends don’t have enough financial resources to make the entire investment at the beginning. Instead, they think they will be able to split their investment into two equal parts, one invested immediately and the second invested in five years. Describe how you would compute the required size of the two equal investments, made five years apart.

I can see the issue the question presents. The total investment in part D will be much greater than in part A, so essentially I'm looking at two separate compounding interest equations being added together. If I'm not mistaken that leads me to something like:
$125,000=P(1+\frac{.05}{1}\)^{18}+P(1+\frac{.05}{1}\)^{13}$
Solve for P and call it a day? (Embarrassing algebra will now follow.)
$125,000=P1.05^{18}+P1.05^{13}$
Removed some of the extra stuff...
$125,000=\frac{P1.05^{18}+P1.05^{13}}{1.05^{13}}\$
Gets me...
$\frac{125,000}{1.05^{13}}=P1.05^5+P$
And I get stuck, I did go further but it didn't seem productive to throw out 6 more wrong equations . Honestly I don't think I even set up the right starting equation.

stapel_eliz
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
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### Re: Dual Compounding Interest Equations

...If I'm not mistaken that leads me to something like:
$125,000=P(1+\frac{.05}{1}\)^{18}+P(1+\frac{.05}{1}\)^{13}$
Solve for P and call it a day? (Embarrassing algebra will now follow.)
$125,000=P1.05^{18}+P1.05^{13}$
Removed some of the extra stuff...
$125,000=\frac{P1.05^{18}+P1.05^{13}}{1.05^{13}}\$
How did you get your last line? Where did the divisor come from?

. . . . .$\mbox{}125,000\, =\, P1.05^{18}\, +\, P1.05^{13}$

...to:

. . . . .$\mbox{}125,000\, =\, P\left(1.05^{18}\, +\, 1.05^{13}\right)$

...by factoring. Then divide through and plug the result into your calculator, to find the value of the variable $P$.

GreenLantern
Posts: 23
Joined: Sat Mar 07, 2009 10:47 pm
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### Re: Dual Compounding Interest Equations

Bah! Factoring. I concede to you Liz, and thank you. It's always something, right?

Alright, divide by that...
$\frac{125,000}{(1.05^{18}+1.05^{13})} =P$
Punch it into my calculator, then profit. It should come out to be 29122.13 (rounded since I'm dealing with money), which is half the total investment.