Hi Bob,

Ok, in order to determine the best Singular X and Y values we need to break
down or segregate the large data into the similar type of data sets.  After
this segregation we can have 2 or 3 sets of Singular Xs and Ys which can be
used to fulfill / justify the equation as one set of X and Y can satisfy one
set (segregation) of data set.

Let me know if you are OK with having three different Xs and Ys for this
kind of huge data because as per my observation, after seeing your data and
trends, it is very difficult to have a Singular X and Y to justify the
equation.

Best Regards,
DILIPandey




On Fri, Nov 26, 2010 at 9:23 AM, Bob <bobwo...@gmail.com> wrote:

> While the Goal Seek feature is interesting and I thank you for
> bringing it up, I don't think it answers my problem.
>
> I'm attempting to make a predictive model, so I'm looking for a method
> to determine the best X and Y value to fit the data set. So, while X
> and Y certainly change from record to record, I want to determine the
> best singular X and Y values to fit the data set at large.
>
> So, the answer to your question of how I'm determining the X and Y is
> that I don't know! That's why I'm posting here -- to come up with a
> method of determining a proper X and Y in a non-linear equation.
>
> Thanks again for your help!
>
> On Nov 24, 10:35 pm, Dilip Pandey <dilipan...@gmail.com> wrote:
> > Hi Bob,
> >
> > I have done something in the attached file.
> > I believe if the x and y are not fixed at all the time, then we can
> > use Goal Seek feature of Excel to obtain the desired results.
> > Question:- How are you determining X and Y when you already have As and
> Bs.
> >
> > Best Regards,
> > DILIPandey
> >
>  > On 11/23/10, Bob <bobwo...@gmail.com> wrote:
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > > Hello,
> >
> > > Thanks for your response!
> >
> > > Here's a random sampling of the data:
> >
> > > 169        140     10
> > > 164        197     19
> > > 50 339     31
> > > 73 326     19
> > > 204        238     34
> > > 143        201     18
> > > 59 74      3
> > > 296        216     20
> > > 31 121     9
> > > 108        148     7
> > > 139        317     24
> > > 41 263     14
> > > 233        286     35
> > > 221        121     17
> > > 111        304     25
> > > 39 325     13
> > > 201        211     37
> > > 474        102     72
> > > 28 158     7
> > > 268        230     34
> > > 246        202     19
> > > 199        208     50
> > > 144        447     48
> > > 358        138     61
> > > 300        8       27
> > > 301        255     37
> > > 197        169     24
> > > 236        205     54
> > > 28 329     14
> > > 150        159     23
> > > 244        298     52
> > > 74 108     6
> > > 202        159     37
> > > 259        127     23
> > > 19 350     16
> > > 189        131     29
> > > 40 183     17
> > > 369        255     62
> > > 122        236     3
> > > 310        201     51
> > > 95 162     10
> > > 335        162     37
> > > 91 447     55
> > > 171        179     21
> > > 110        215     9
> > > 87 242     17
> > > 328        219     49
> > > 225        182     21
> > > 245        243     42
> >
> > > The data in the first two columns (A and B for purposes explained
> > > below) are the independent variables, with the third column being the
> > > dependent (C)
> >
> > > Currently, the equation I'm using is:
> >
> > > -10.5348929531027+0.113896884380505*A+0.0893951738657916*B = C
> >
> > > As mentioned, this does not fit the data particularly well. It
> > > predicts too low a result when A and B are small, and too high a
> > > result when A and B are large. For this reason, I believe a
> > > logarithmic curve will best represent the data ... but other
> > > suggestions are also certainly welcome.
> >
> > > Hope you, or someone, can help! Thank you.
> >
> > > On Nov 23, 9:13 am, Dilip Pandey <dilipan...@gmail.com> wrote:
> > >> Hi Bob,
> >
> > >> This seems to be interesting to me.  Would you mind providing some
> > >> sample data with the the indicative result which you want to obtain
> > >> through Excel.
> >
> > >> Best Regards,
> > >> DILIPandey
> >
> > >> On 11/23/10, Bob <bobwo...@gmail.com> wrote:
> >
> > >> > Good afternoon,
> >
> > >> > I have a fairly large data set (roughly 7,000 samples) on which I
> > >> > would like to run some nonlinear regression analysis.
> >
> > >> > I'm fully capable of using the Data Analysis Toolpak and running the
> > >> > linear analysis there, but unless I'm totally missing it, there
> > >> > doesn't seem to be a method for running nonlinear analysis.
> >
> > >> > To be clear, what I want to do is take this data, which is in three
> > >> > columns, and essentially put together an equation to represent Ax +
> > >> > By
> > >> > = C.
> >
> > >> > However, this data does not seem to be all that well represented by
> a
> > >> > linear equation, and I believe it would be more precise to use an
> > >> > exponential or logarithmic model along the lines of A^x + B^y = C.
> >
> > >> > Can anyone help?
> >
> > >> > Thank you!
> >
> > >> > --
> > >> >
> ---------------------------------------------------------------------------
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> >
> > DILIP KUMAR PANDEY
> >   MBA-HR,B.Com(Hons),BCA
> > Mobile: +91 9810929744
> > dilipan...@gmail.com
> > dilipan...@yahoo.com
> > New Delhi - 62, India
> >
> >  Analysis.xls
> > 22KViewDownload
>
> --
>
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-- 
Thanks & Regards,

DILIP KUMAR PANDEY
  MBA-HR,B.Com(Hons),BCA
Mobile: +91 9810929744
dilipan...@gmail.com
dilipan...@yahoo.com
New Delhi - 62, India

-- 
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