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From: John D'Errico on 16 Jun 2010 12:24
"Adam Schirmacher" <adam.schirmacher(a)medtronic.com> wrote in message <hvat8d$mst$1(a)fred.mathworks.com>...
> "John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <hvapvc$mhi$1(a)fred.mathworks.com>...
> > Only 5 points to estimate 3 parameters is a bit low
> > in general, but I will pretend you really have more
> > data.
> >
> > Now, suppose we wanted to fit the model
> >
> > W = C + a(1)*X + a(2)*Y + a(3)*Z
> >
> > where C was a KNOWN constant? Can you fit that
> > as easily? Sure, because this just reduces to the
> > simple form
> >
> > (W - C) = a(1)*X + a(2)*Y + a(3)*Z
> >
> > Use the line of code to fit (W-C) as a linear function
> > of the variables X,Y,Z.
> >
> > The word intercept can mean different things to
> > different people however. It sounds like you wish to
> > ensure that the surface will pass through a specific
> > point. (x0,y0,z0,w0).
> >
> > An easy way to do this is to use lsqlin, from the
> > optimization toolbox. lsqlin can set an equality
> > constraint on the problem.
> >
> > The the call is simply:
> >
> > a = lse([X,Y,Z],W,[x0,y0,z0],w0);
> >
> > The call to lsqlin would be similar of course.
>
> Ah-ha! That is some very useful advice indeed. I didn't even know that function existed. I do see though that this does not return a constant term in the regression. It would just return coefficients A, B and C where w=Ax + By + Cz. What about returning another term for w=Ax + By + Cz + D?
>

Suppose your model is

W = a(1)*X + a(2)*Y + a(3)*Z + a(4)

where you don't know the value of a(4) either, but you
do know that the surface passes through a known point,
then the problem changes only slightly.

So if n is the number of data points in the model, just
do this:

a = lse([X,Y,Z,ones(n,1)],W,[x0,y0,z0,1],w0);

The model now has a constant term in it.

John
From: John D'Errico on 16 Jun 2010 12:27
"Matt J " <mattjacREMOVE(a)THISieee.spam> wrote in message <hvatho$btr$1(a)fred.mathworks.com>...
> "Adam Schirmacher" <adam.schirmacher(a)medtronic.com> wrote in message <hvat8d$mst$1(a)fred.mathworks.com>...
>
> > In reality the former sets an additional constraint, which is that the regression passes through (0, 0, 0, 0) which I do not need.
>
> Again, though, you can pre-shift your coordinate measurements so that the intercept *is* at (0,0,0,0) and perform the fit in this shifted space. If you do, however, I don't see why you would need any fancy iterative solvers. I think backslash \ should do fine

lse is NOT an iterative solver. Next time, try understanding
what you are talking about before you make a response!

Pre-shifting the measurements also changes the model,
forcing the user to use that preshift in all later predictions
with the model, or it forces the user to then transform the
coefficients as derived.

John
From: Matt J on 16 Jun 2010 13:45
"John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <hvau0r$bja$1(a)fred.mathworks.com>...

>
> lse is NOT an iterative solver. Next time, try understanding
> what you are talking about before you make a response!
===============

Very well, but lsqlin, which you also proposed, *is* an iterative solver. Moreover, you proposed it in preference to lse, making it natural to assume that lse was also iterative (since why would one prefer an iterative method over a closed-form one).


> Pre-shifting the measurements also changes the model,
> forcing the user to use that preshift in all later predictions
> with the model, or it forces the user to then transform the
> coefficients as derived.
========

Doesn't seem like a lot of work...
From: Adam Schirmacher on 16 Jun 2010 15:11
"John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <hvatr4$h7$1(a)fred.mathworks.com>...
> So if n is the number of data points in the model, just
> do this:
>
> a = lse([X,Y,Z,ones(n,1)],W,[x0,y0,z0,1],w0);
>
> The model now has a constant term in it.
>
> John

Thanks! I am learning a lot. I did get this to work for some of my data, but not all. I'm having a bit of a strange issue. Please check the following example. I'll dump some of the matrices here, and show the command I use to get the coefficients, and then show the output and how it's incorrect for some reason.

(neighbors is 8x3, outputs is 8x1)

[neighbors outputs]
1 30 60 5
1 30 210 8
1 70 60 12
1 70 210 20
2 30 60 6
2 30 210 13
2 70 60 16
2 70 210 30

lsqlin([neighbors, ones(8, 1)], outputs, [1, 30, 60, 1], 5)

This gives:

Warning: Large-scale algorithm can handle bound constraints only;
using medium-scale algorithm instead.
> In lsqlin at 286
Optimization terminated.

ans =

5.0000
0.2875
0.0533
-15.3250

Now theoretically, the extra inputs into lsqlin told it to pass through (1,30,60,5).
However, 5*1+0.2875*30+0.0533*60-15.3250 = 1.5. Rather than 5.

If I instead use lsqlin([neighbors, ones(8, 1)], outputs, [50, 30, 60, 1], 5), the resulting coefficients DO pass through (50, 30, 60, 5). So something's a little funny there. Any idea what could be causing this?
From: Matt J on 17 Jun 2010 10:52
"Adam Schirmacher" <adam.schirmacher(a)medtronic.com> wrote in message <hvb7k8$nga$1(a)fred.mathworks.com>...

> lsqlin([neighbors, ones(8, 1)], outputs, [1, 30, 60, 1], 5)
================

You need to review the syntax for lsqlin. You are specifying inequality constraints, rather than equality constraints.





> Now theoretically, the extra inputs into lsqlin told it to pass through (1,30,60,5).
> However, 5*1+0.2875*30+0.0533*60-15.3250 = 1.5. Rather than 5.
===============

When I use the non-iterative approach below, I get the following beta

beta =

3.2500
0.2437
0.0417
-8.0625

which does pass through the desired point. If you get lsqlin working, you might use it as a cross-check.

y=outputs;
x=neighbours;
x0=[1, 30, 60];
y0=5;

beta=bsxfun(@minus,X,x0)\(y-y0);
beta=[beta; y0-x0*beta],
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