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?
 
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?
 
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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