journey through life: rant about rates and fees.

uh, i didn't get all the infos i need for constructing the numerical part of this thematic mathematical rant. so i just put in examples, and before that i calculate algebraically.

here goes!

We all know about the concept of interest if we are depositing our money into a bank account.

But know more than the concept.

Know that other than giving interest in a minuscule percentage, they also apply administration fee, though some kinds of accounts don't apply any fee. Let's take a look at those who have that kind of fee.

First, establish ${\color{black} m_{n}}$ as the money you have in your account in the ${\color{black} n}$ -th month, ${\color{black} i}$ as the interest rate per month, and ${\color{black} a}$ as the administration fee. Note that we can assume that ${\color{black} 0<i<1}$ .

Then, we get that the amount of money we get in the next month is the amount of money we have in the previous month, added with interest, subtracted with administration fee, or simply shown as
${\color{black} m_{n+1} = m_{n} + m_{n}i - a}$ (1)That's assuming that the interest is calculated before the fee is applied. If that's the case, the equation becomes
${\color{black} m_{n+1}=(m_{n}-a)+(m_{n}-a)i$ (2)Or simply,
${\color{black} m_{n+1}=(m_{n}-a)(i+1)}$

If we want our money to stay constant, then that means ${\color{black} m_{n+1}=m_{n}}$ which implies that ${\color{black} m_{n}i=a}$ (from (1)) or ${\color{black} (m_{n}-a)i=a}$ (from (2)).

Solving those equation for ${\color{black} m_{n}}$ yields ${\color{black} m_{n}= \frac{a}{i}}$ or ${\color{black} m_{n}=\frac{a(i+1)}{i}}$ . Now let's test them a bit.

Setting ${\color{black} a=\$ 1.00}$ and ${\color{black} i=0.0025 \: (0.25 \% )}$ , that gives us ${\color{black} m_{n}=\$ 400}$ or ${\color{black} m_{n}=\$ 401}$ .

Hmm, not much of a difference. And not a reaaly big amount of money either. Note that, I think the interest rate is pretty big. As ${\color{black} 0.0025 \times 12=0.03=3\%}$ interest rate per year. Onwards!

Now, let's assume that your money isn't that much. You have to add your monthly incomes also. Declare two new variables, ${\color{black} t}$ , denoting the time in month and ${\color{black} s}$ , denoting your monthly income.

Generally, what you have in your account next month consists of:

Your money from last month,
the interest from last month,
minus the administration fee.

So, forming up an equation from those while assuming that the interest is calculated before the administration fee is applied we get that:

For ${\color{black} t=1}$ ,
${\color{black} m_{1}=m_{0}(i+1)+s-a}$ For ${\color{black} t=2}$ ,
${\color{black} m_{2}=m_{1}(i+1)+s-a}$
${\color{black} m_{2}=(m_{0}(i+1)+s-a)(i+1)+s-a}$
${\color{black} m_{2}=m_{0}(i+1)^{2}+((i+1)+1)(s-a)}$ For ${\color{black} t=3}$ ,
${\color{black} m_{3}=m_{2}(i+1)+s-a}$
${\color{black} m_{3}=(m_{0}(i+1)^{2}+((i+1)+1)(s-a))(i+1)+s-a}$
${\color{black} m_{3}=m_{0}(i+1)^{3}+((i+1)^{2}+(i+1)+1)(s-a)}$
${\color{black} m_{3}=m_{0}(i+1)^{3}+((i+1)^{2}+(i+1)^{1}+(i+1)^{0})(s-a)}$
Generalizing this for ${\color{black} t=n}$ , we have
${\color{black} m_{n}=m_{0}(i+1)^{n}+(s-a)\sum_{j=0}^{n-1}(i+1)^j}$
Since the summation is a geometric series, we can simplify it to
${\color{black} m_{n}=m_{0}(i+1)^{n}+(s-a)\frac{(i+1)^{n}-1}{i}}$ Or further simplifying it, let's say that ${\color{black} S=(i+1)^{n}-1}$ , changing the equation into ${\color{black} m_{n}=m_{0}(i+1)^{n}+\frac{S(s-a)}{i}}$ .

Since we have assumed that the interest is applied before the administration fee, we can say that ${m_{n}=\frac{a}{i}}$ . Substituting, we have
${\color{black} \frac{a}{i}=m_{0}(i+1)^{n}+\frac{S(s-a)}{i}}$ Solving for ${n}$ to find out how many months needed to gain from interest,
${\color{black} a=m_{0}i(i+1)^{n}+S(s-a)}$
${\color{black} a=m_{0}i(i+1)^{n}+(i+1)^{n}(s-a)-(s-a)}$
${\color{black} a=(m_{0}i+s-a)(i+1)^{n}-s+a}$
${\color{black} s=(m_{0}i+s-a)(i+1)^{n}}$
${\color{black} \frac{s}{m_{0}i+s-a}=(i+1)^{n}}$
${\color{black} n=\log_{i+1} (\frac{s}{m_{0}i+s-a})}$
Since there's logarithm, we have to establish a constraint, that is,
${\frac{s}{m_{0}i+s-a}>0}$
${m_{0}i+s-a>0}$
${m_{0}i+s>a}$
And don't forget that ${\frac{s}{m_{0}i+s-a}>1}$ which results in ${a>m_{0}i}$ in order to avoid the negativity of ${n}$ , and that implies ${s>0}$ .

Okay, now the hellish part is finished, let's try to plug in values.

Assume that the interest rate is 1.5% yearly, your monthly salary is $350, initial money is $290, administration fee is $1.5, then,
${i=0.00125}$ ,
${s=\$350}$ ,
${m_{0}=\$290}$ ,
${a=\$1.5}$ .

Plugging in values,
${\color{black} n=\log_{0.00125+1} (\frac{350}{290\times0.00125+350-1.5})}$
${\color{black} n=\log_{1.00125} (\frac{350}{348.8625})}$
${\color{black} n=\log_{1.00125} (1.003261)}$
${\color{black} n=2.605861}$
Rounding up, we have ${n=3}$ . That means you need 3 months to progress before gaining from interest.

I hope there's nothing wrong with the calculation...

Saturday, July 10, 2010

rant about rates and fees.

1 comment:

Bulbanews, by Bulbagarden

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