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  #1  
Old September 27th, 2009, 11:59 PM
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Smile Who wants to do some maths for me?

I need to work out a fee calculator, but i cant for the life of me figure out the equation to do it. Anyone want to show off their skills?
The calculator works like this. The figures are approx ( a couple of dollars either way does not matter)
Min charge $4.95. The Fee starts at $4.95 and goes up to $29.95 an so on. The fees go up depending on the amount deposited with a fee ending at the amount of F. The deposit are up to the amount in D. So each dollar amount entered goes up in scale - which varies.

D refers to Deposit and F refers to the fee charged.
D $700 F $29.95
D $1000 F $32.95
D $2000 F $33.95
D $5000 F $55.00
D 10,000 F $70
D 25,000 F $160
D 50,000 F $280
D 75,000 F $350
D 100,000 F $380 which is the maximum fee.

So basically, I want field that you enter and amount into "D", the calculator has the formula already and spits out the answer of "F".

Who's game?
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  #2  
Old September 28th, 2009, 12:07 AM
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There is no clear relation ship between F & D in this case.

I would create an excel sheet with a look up table to determine F for any D entered into a cell.
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Old September 28th, 2009, 12:19 AM
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Hi,

I had thought of that. I'm sure their must be an equation though.

thanks
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Old September 28th, 2009, 12:25 AM
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None that I know.
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Old September 28th, 2009, 12:35 AM
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I did get this advice, but it is unfinshed and I have no way of testing it:

Polynomials would never be a good fit: for example it must go to either -infinity or +infinity, thus violating the assumptions for min or max.

For this question, line segments make more sense. You are given 10 data points (including (0, 4.95)): let's label them as (x_0, y_0), ..., (x_9, y_9), where (x_0, y_0) = (0, 4.95), etc.

The slopes between (x_(i-1), y_(i-1)) and (x_i, y_i) is

m_i = (y_i-y_(i-1))/(x_i-x_(i-1)), for i = 1, ..., 9.

Now you can always find the function of the form:

f(x) = sum_{i=0 to 9} a_i*|x-x_i| + b, subject to the slope condition

a_0 - a_1 - a_2 ... - a_9 = m_1
a_0 + a_1 - a_2 ... - a_9 = m_2
...
a_0 + a_1 +... +a_8 -a_9 = m_9
a_0 + a_1 +... +a_8 +a_9 = 0 (this last equation takes care of the slopes for x < 0, and x > x_9)

and b is chosen so that f(0) = 4.95.

This is a system of equation with 10x10 matrix of full rank, hence solvable. (in fact the coefficient determinant is 2^9 > 0)

Therefore the formula does exist.
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Old September 28th, 2009, 12:52 AM
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Quote:
Originally Posted by paullovesjo View Post
I need to work out a fee calculator, but i cant for the life of me figure out the equation to do it. Anyone want to show off their skills?
The calculator works like this. The figures are approx ( a couple of dollars either way does not matter)
Min charge $4.95. The Fee starts at $4.95 and goes up to $29.95 an so on. The fees go up depending on the amount deposited with a fee ending at the amount of F. The deposit are up to the amount in D. So each dollar amount entered goes up in scale - which varies.

D refers to Deposit and F refers to the fee charged.
D $700 F $29.95
D $1000 F $32.95
D $2000 F $33.95
D $5000 F $55.00
D 10,000 F $70
D 25,000 F $160
D 50,000 F $280
D 75,000 F $350
D 100,000 F $380 which is the maximum fee.

So basically, I want field that you enter and amount into "D", the calculator has the formula already and spits out the answer of "F".

Who's game?
Try:

F=20+500(1-e^{-0.000137P})

CB
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Old September 28th, 2009, 01:21 AM
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Looks interesting.

Now - how do I go about using that magical formula.

Do you know of an online calculator I could use?

Let say I wanted to enter $1500.00 and calculate what the fee would be. How do I use that formula?

You may have guessed, maths wasnt one of my strongest subjects.

Could you enlighten me?

Thanks
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  #8  
Old September 28th, 2009, 03:20 AM
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Originally Posted by paullovesjo View Post
Looks interesting.

Now - how do I go about using that magical formula.

Do you know of an online calculator I could use?

Let say I wanted to enter $1500.00 and calculate what the fee would be. How do I use that formula?

You may have guessed, maths wasnt one of my strongest subjects.

Could you enlighten me?

Thanks
Google will do it, enter as:

20+500*(1-exp(-0.000137*1500))

20+500*(1-exp(-0.000137*1500)) - Google Search

CB
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Old September 28th, 2009, 11:08 PM
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I got this from a maths guru. Unfortunately, he cant fathom how I dont understand it.

Does anyone know how I would enter this into a google calculator and where i would enter say $32,000 into the equation so that I get an answer to the equation?

Alright, if you solve the system of equations I mentioned earlier (either by hand or by using a calculator, say TI83 or TI84), you get the following numbers (there may be rounding error for the numbers listed below):
a_0 = 0.017857
a_1 = -0.01286
a_2 = -0.0045
a_3 = 0.003008
a_4 = -0.00201
a_5 = 0.0015
a_6 = -0.0006
a_7 = -0.001
a_8 = -0.0008
a_9 = -0.0006
Therefore the formula for the function is (denoting square root by sqrt)
f(x) = 0.017857*sqrt((x-0)^2) -0.01286*sqrt((x-700)^2) -0.0045*sqrt((x-1000)^2) +0.003008*sqrt((x-2000)^2) -0.00201*sqrt((x-5000)^2) +0.0015*sqrt((x-10000)^2) -0.0006*sqrt((x-25000)^2) -0.001*sqrt((x-50000)^2) -0.0008*sqrt((x-75000)^2) -0.0006*sqrt((x-100000)^2) + b.
To determine b, substitute 0 for x, since f(0) = 4.95, you get b = 192.475.
In conclusion the sought for formula is
f(x) = 0.017857*sqrt((x-0)^2) -0.01286*sqrt((x-700)^2) -0.0045*sqrt((x-1000)^2) +0.003008*sqrt((x-2000)^2) -0.00201*sqrt((x-5000)^2) +0.0015*sqrt((x-10000)^2) -0.0006*sqrt((x-25000)^2) -0.001*sqrt((x-50000)^2) -0.0008*sqrt((x-75000)^2) -0.0006*sqrt((x-100000)^2) + 192.475.
I have checked the formula in Excel, it works well, since the rounding error is negligible.
To avoid rounding error, you could obtain those a_i by the following formulas, which I have solved for you
a_0 = m_1/2,
a_1 = (m_2-m_1)/2
a_2 = (m_3-m_2)/2
a_3 = (m_4-m_3)/2
a_4 = (m_5-m_4)/2
a_5 = (m_6-m_5)/2
a_6 = (m_7-m_6)/2
a_7 = (m_8-m_7)/2
a_8 = (m_9-m_8)/2
a_9 = -m_9/2,
where m_i is the slope of the line joining (x_(i-1), y_(i-1)) to (x_i, y_i) for i = 1, 2, ..., 9. (For example, (x_0, y_0) = (0, 4.95), (x_1, y_1) = (700, 29.95), hence m_1 = (29.95 - 4.95)/(700 - 0), etc.)
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Old September 29th, 2009, 12:31 AM
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Quote:
Originally Posted by paullovesjo View Post
I got this from a maths guru. Unfortunately, he cant fathom how I dont understand it.

Does anyone know how I would enter this into a google calculator and where i would enter say $32,000 into the equation so that I get an answer to the equation?

Alright, if you solve the system of equations I mentioned earlier (either by hand or by using a calculator, say TI83 or TI84), you get the following numbers (there may be rounding error for the numbers listed below):
a_0 = 0.017857
a_1 = -0.01286
a_2 = -0.0045
a_3 = 0.003008
a_4 = -0.00201
a_5 = 0.0015
a_6 = -0.0006
a_7 = -0.001
a_8 = -0.0008
a_9 = -0.0006
Therefore the formula for the function is (denoting square root by sqrt)
f(x) = 0.017857*sqrt((x-0)^2) -0.01286*sqrt((x-700)^2) -0.0045*sqrt((x-1000)^2) +0.003008*sqrt((x-2000)^2) -0.00201*sqrt((x-5000)^2) +0.0015*sqrt((x-10000)^2) -0.0006*sqrt((x-25000)^2) -0.001*sqrt((x-50000)^2) -0.0008*sqrt((x-75000)^2) -0.0006*sqrt((x-100000)^2) + b.
To determine b, substitute 0 for x, since f(0) = 4.95, you get b = 192.475.
In conclusion the sought for formula is
f(x) = 0.017857*sqrt((x-0)^2) -0.01286*sqrt((x-700)^2) -0.0045*sqrt((x-1000)^2) +0.003008*sqrt((x-2000)^2) -0.00201*sqrt((x-5000)^2) +0.0015*sqrt((x-10000)^2) -0.0006*sqrt((x-25000)^2) -0.001*sqrt((x-50000)^2) -0.0008*sqrt((x-75000)^2) -0.0006*sqrt((x-100000)^2) + 192.475.
I have checked the formula in Excel, it works well, since the rounding error is negligible.
To avoid rounding error, you could obtain those a_i by the following formulas, which I have solved for you
a_0 = m_1/2,
a_1 = (m_2-m_1)/2
a_2 = (m_3-m_2)/2
a_3 = (m_4-m_3)/2
a_4 = (m_5-m_4)/2
a_5 = (m_6-m_5)/2
a_6 = (m_7-m_6)/2
a_7 = (m_8-m_7)/2
a_8 = (m_9-m_8)/2
a_9 = -m_9/2,
where m_i is the slope of the line joining (x_(i-1), y_(i-1)) to (x_i, y_i) for i = 1, 2, ..., 9. (For example, (x_0, y_0) = (0, 4.95), (x_1, y_1) = (700, 29.95), hence m_1 = (29.95 - 4.95)/(700 - 0), etc.)
Now why do I think that is linear interpolation between the tabulated values?

CB
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