solver can not solve polynomial eq. where one variable is five di.

A

Amitava

I tried to create equations from the set of data by solver
x= 90021 85248 78314 69768 59508 47975
y = 20.09 19.92 18.70 16.95 16.13 15.67

in the form of y= a + bx +cx2 + dx3;
where x2= x square ,x3 = x cube.

but could not do so.
the solver does fine when the value of x is small .
Your help will be hghly appreciated in this regard.

Thanks
 
T

tjtjjtjt

You might want to check out solver.com to find out exactly what the
limitations and weak points for the buitl-in Solver are.

tj
 
H

Harlan Grove

Amitava said:
I tried to create equations from the set of data by solver
x= 90021 85248 78314 69768 59508 47975
y = 20.09 19.92 18.70 16.95 16.13 15.67

in the form of y= a + bx +cx2 + dx3;
where x2= x square ,x3 = x cube.

but could not do so.
the solver does fine when the value of x is small .
Your help will be hghly appreciated in this regard.
Click to expand...

In Excel you're limited to 15 decimal digit precision most of the time. A
few intermediate calculations may have more if they can stay in FPU
registers. However, standard minimizing sum squared error fitting would at
some point involve squaring the x^3 values, at which point you've blown way
past machine precision.

That said, plot these X-Y values in an X-Y chart, add a polynomial trend
line or order 3, and display the trend line equation. It'll show the a, b, c
and d coefficients you seek.
 
T

Tushar Mehta

I imagine you could tweak Solver's parameters and/or scale your own
values to get Solver to work, but, in this specific instance, you
should consider using LINEST as Bernard suggested.

If you did, you would discover that none of the coefficients for an
order 3 fit are statistically significant at the 5% level. Only the
constant term is significant for a quadratic fit whereas both terms are
for a linear fit. To learn more about how to test for statistical
significance check the LINEST help Example 5.

--
Regards,

Tushar Mehta
www.tushar-mehta.com
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