# graphical approximation problem

#### tclose

##### New member
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

##### Moderator
Staff member
Simple enough. Write some equations.

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

##### Elite Member
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?

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

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

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

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

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

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

When is [1] > [2] ?

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

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

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

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