graphical approximation problem

tclose

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I have no idea how to solve this type of problem. What formulas to use and how to apply it to a graph. Any help or suggestion would be appreciated.

Use graphical approximation techniques to answer the question. When would an ordinary annuity consisting of quarterly payments of $605.76 at 8% compounded quarterly be worth more than a principal of $6100 invested at 5% simple interest?
 

tkhunny

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Simple enough. Write some equations.

Start with the simple interest:

s(t) = 6100 + 6100*0.05*t = 6100*(1+0.05*t)

There you go. It's a nice, straight line.

Now the ordinary annuity. This one is a little trickier. It changes value every quarter, when a payment is made.

i = 0.08 / 4 = 0.02
v = 1/(1+i)

605.76 * (v + v^2 + v^3 + ... + v^n) -- Hmmm... See the problem, already? What's 'n'? It's pretty tough to calculate the value without knowing the period, otherwise, we would have called it a perpetuity.

Where does that leave us? Do we have the WHOLE problem statement?
 

soroban

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Jan 28, 2005
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Hello, tclose!

Use graphical approximation techniques to answer the question.
When would an ordinary annuity consisting of quarterly payments of $605.76 at 8% compounded quarterly
be worth more than a principal of $6100 invested at 5% simple interest?

\(\displaystyle \text{Annuity Formula: }\;A \;=\;D\,\frac{(1+i)^n - 1}{y} \;\text{ where: }\;\begin{Bmatrix}A &=& \text{final value} \\ D &=& \text{periodic deposit} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}\)

\(\displaystyle \text{We are given: }\:D \,=\, 605.76,\; i\,=\,\tfrac{0.08}{4} \,=\,0.02,\;4n\text{ periods}\)

. . \(\displaystyle \text{We have: }\:A \;=\;605.76\,\frac{(1.02)^{4n} - 1}{0.02}\) . [1]


\(\displaystyle \text{Simple Interest Formula: }\;A \;=\;P(1 + i\cdot n)\;\text{ where: }\;\begin{Bmatrix}A &=& \text{final amount} \\ P &=& \text{principal invested} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}\)

\(\displaystyle \text{We are given: }\:p = 6100,\;i = 0.05,\;n\text{ periods}\)

. . \(\displaystyle \text{We have: }\;A \;=\;6100(1 + 0.05n)\) . [2]



When is [1] > [2] ?

\(\displaystyle \text{That is: }\;605.76\,\frac{(1.02)^{4n}-1}{0.02} \;> \;6100(1 + 0.05n)\)

\(\displaystyle \text{We can}not\text{ solve this equation for }n.\)
. . \(\displaystyle \text{We can only approximate it.}\)

\(\displaystyle \text{The }n\text{ is both "inside" and "outside" an exponential function.}\)
. . \(\displaystyle \text{The equation is }transcendental.\)


\(\displaystyle \text{Using your graphing calculator, you can }see\text{ where}\)
. . \(\displaystyle \text{the Annuity graph exceeds the Simple Interest graph.}\)
 
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