Finding a Formula to Fit Data Points

Date: 06/23/2004 at 21:55:40
From: Kaylon
Subject: How to fit a formula to predetermined amounts

Dear Dr. Math, 

I need to develop a pay scale for companies that buy our product.  It
needs to be exponential and level off at a certain dollar amount (or
close to it).  For example, if the customer buys 500 items, they get
$150 credit.  If they buy 1000 items, they get $200 credit.  And, if
they buy 1500 items, they get $225 credit.  How do I figure out the
formula to make this work? 

Date: 06/23/2004 at 23:10:24
From: Doctor Peterson
Subject: Re: How to fit a formula to predetermined amounts

Hi, Kaylon.

I'm not sure what you mean by "exponential", but you probably have 
in mind something with a horizontal asymptote, like 1 - e^(-x) which 
would approach 1 from below.  A simpler type of equation would be 
1 - 1/x, which has the same characteristic.  The details, of course, 
seem to be entirely arbitrary.  Let's take your specific example.  You 
want it to look like this:

  300+
     |
     |
     |
  250+
     |
     |                       *
     |
  200+               *
     |
     |
     |
  150+       *
     |
     |
     |
  100+
     |
     |
     |
   50+
     |
     |
     |
     *-------+-------+-------+
     0      500    1000    1500

How closely can we match that with a simple equation?  It would be
nice if we knew what the asymptotic value should be; can we determine
it from the data?

If we take my proposed form

  y = 1 - 1/x

we have a little problem, since it is undefined at x = 0.  Instead, 
I'll try this:

  y = a - b/(x + c)

Now we have three parameters to determine, which requires only three 
data points.  Taking (0,0), (500,150), and (1000, 200), we have

  0 = a - b/c

  150 = a - b/(c + 500)

  200 = a - b/(c + 1000)

Subtracting the first from the other two, we eliminate a:

  150 = b(1/c - 1/(c + 500))

  200 = b(1/c - 1/(c + 1000))

These simplify to

  150c(c + 500) = 500b

  200c(c + 1000) = 1000b

Setting twice the first equal to the second, we eliminate b:

  300(c^2 + 500c) = 200(c^2 + 1000c)

which gives

  100c^2 = 50000c

  c = 500

and then

  b = 200c(c + 1000)/1000 = 500*1500/5 = 150,000

  a = b/c = 150000/500 = 300

so our formula is

  y = a - b/(x + c) = 300 - 150,000/(x + 500)

(Drum roll ...)

What do we get for x = 1500?

  y = 300 - 150,000/2000 = 300 - 75 = 225

which happens to be just what you were hoping for!  It didn't have to 
work out that nicely, but this suggests that my formula pretty well 
matches your expectations. It will approach y = 300 asymptotically, so 
that no one will ever get more than $300 back.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/