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Nonlinear Regression

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Title: Nonlinear Regression


1
Nonlinear Regression
  • Chemical Engineering Majors
  • Authors Autar Kaw, Luke Snyder
  • http//numericalmethods.eng.usf.edu
  • Transforming Numerical Methods Education for STEM
    Undergraduates

2
Nonlinear Regression http//numericalmethod
s.eng.usf.edu
3
Nonlinear Regression
Some popular nonlinear regression models
1. Exponential model
2. Power model
3. Saturation growth model
4. Polynomial model
4
Nonlinear Regression
Given n data points
best fit
to the data, where
is a nonlinear function of
.

Figure. Nonlinear regression model for discrete y
vs. x data
5
RegressionExponential Model
6
Exponential Model
Given
best fit
to the data.
Figure. Exponential model of nonlinear regression
for y vs. x data
7
Finding constants of Exponential Model
The sum of the square of the residuals is defined
as
Differentiate with respect to a and b
8
Finding constants of Exponential Model
Rewriting the equations, we obtain
9
Finding constants of Exponential Model
Solving the first equation for
yields
Substituting
back into the previous equation
Nonlinear equation in terms of
The constant
can be found through numerical methods such
as the bisection method or secant method.
10
Example 1-Exponential Model
Many patients get concerned when a test involves
injection of a radioactive material. For example
for scanning a gallbladder, a few drops of
Technetium-99m isotope is used. Half of the
techritium-99m would be gone in about 6 hours.
It, however, takes about 24 hours for the
radiation levels to reach what we are exposed to
in day-to-day activities. Below is given the
relative intensity of radiation as a function of
time.
Table. Relative intensity of radiation as a
function of time.
t(hrs) 0 1 3 5 7 9
1.000 0.891 0.708 0.562 0.447 0.355
11
Example 1-Exponential Model cont.
The relative intensity is related to time by the
equation
Find
a) The value of the regression constants
and
b) The half-life of Technium-99m
c) Radiation intensity after 24 hours
12
Plot of data
13
Constants of the Model
The value of ? is found by solving the nonlinear
equation
14
Setting up the Equation in MATLAB
t (hrs) 0 1 3 5 7 9
? 1.000 0.891 0.708 0.562 0.447 0.355
15
Setting up the Equation in MATLAB
t0 1 3 5 7 9 gamma1 0.891 0.708 0.562
0.447 0.355 syms lamda sum1sum(gamma.t.exp(l
amdat)) sum2sum(gamma.exp(lamdat)) sum3sum(
exp(2lamdat)) sum4sum(t.exp(2lamdat)) fsu
m1-sum2/sum3sum4
16
Calculating the Other Constant
The value of A can now be calculated
The exponential regression model then is
17
Plot of data and regression curve
18
Relative Intensity After 24 hrs
The relative intensity of radiation after 24
hours
This result implies that only
radioactive intensity is left after 24 hours.
19
Homework
  1. What is the half-life of technetium 99m isotope?
  2. Compare the constants of this regression model
    with the one where the data is transformed.
  3. Write a program in the language of your choice to
    find the constants of the model.

20
  • THE END
  • http//numericalmethods.eng.usf.edu

21
Polynomial Model
Given
best fit
to a given data set.
Figure. Polynomial model for nonlinear regression
of y vs. x data
22
Polynomial Model cont.
The residual at each data point is given by
The sum of the square of the residuals then is
23
Polynomial Model cont.
To find the constants of the polynomial model, we
set the derivatives with respect to
where
equal to zero.
24
Polynomial Model cont.
These equations in matrix form are given by
The above equations are then solved for
25
Example 2-Polynomial Model
Below is given the FT-IR (Fourier Transform Infra
Red) data of a 11 (by weight) mixture of
ethylene carbonate (EC) and dimethyl carbonate
(DMC). Absorbance P is given as a function of
wavenumber m.
Table. Absorbance vs Wavenumber data
Wavenumber, m ( ) Absorbance, P (arbitrary unit)
804.184 0.1591
827.326 0.0439
846.611 0.0050
869.753 0.0073
889.038 0.0448
892.895 0.0649
900.609 0.1204
Figure. Absorbance vs. Wavenumber data
26
Example 2-Polynomial Model cont.
Regress the data to a second order polynomial
where
and find the absorbance at
The coefficients
are found as follows
27
Example 2-Polynomial Model cont.
The necessary summations are as follows
Table. Necessary summations for calculation of
polynomial model constants
i Wavenumber, m ( ) Absorbance, P (arbitrary unit) m2 m3
1 804.18 0.1591 6.4671105 5.2008108
2 827.33 0.0439 6.8447105 5.6628108
3 846.61 0.0050 7.1675105 6.0681108
4 869.75 0.0073 7.5647105 6.5794108
5 889.04 0.0448 7.9039105 7.0269108
6 892.90 0.0649 7.9726105 7.1187108
7 900.61 0.1204 8.1110105 7.3048108
6030.4 0.4454 5.2031106 4.4961109
28
Example 2-Polynomial Model cont.
Necessary summations continued
Table. Necessary summations for calculation of
polynomial model constants.
i m4 m P m2 P
1 4.18241011 127.95 1.0289105
2 4.68491011 36.319 3.0048104
3 5.13731011 4.233 3.583103
4 5.72251011 6.349 5.522103
5 6.24711011 39.828 3.5409104
6 6.35631011 57.948 5.1742104
7 6.57871011 108.43 9.7655104
3.89091012 381.06 3.2685105
29
Example 2-Polynomial Model cont.
Using these summations we have
Solving this system of equations we find
The regression model is then
30
Example 2-Polynomial Model cont.
With the model is given by
Figure. Polynomial model of Absorbance vs.
Wavenumber.
31
Example 2-Polynomial Model cont.
To find
where
we have
32
Linearization of Data
To find the constants of many nonlinear models,
it results in solving simultaneous nonlinear
equations. For mathematical convenience, some of
the data for such models can be linearized. For
example, the data for an exponential model can be
linearized.
As shown in the previous example, many chemical
and physical processes are governed by the
equation,
Taking the natural log of both sides yields,
Let
and
We now have a linear regression model where
(implying)
with
33
Linearization of data cont.
Using linear model regression methods,
Once
are found, the original constants of the model
are found as
34
Example 3-Linearization of data
Many patients get concerned when a test involves
injection of a radioactive material. For example
for scanning a gallbladder, a few drops of
Technetium-99m isotope is used. Half of the
technetium-99m would be gone in about 6 hours.
It, however, takes about 24 hours for the
radiation levels to reach what we are exposed to
in day-to-day activities. Below is given the
relative intensity of radiation as a function of
time.
Table. Relative intensity of radiation as a
function
of time
t(hrs) 0 1 3 5 7 9
1.000 0.891 0.708 0.562 0.447 0.355
Figure. Data points of relative radiation
intensity vs. time
35
Example 3-Linearization of data cont.
Find
a) The value of the regression constants
and
b) The half-life of Technium-99m
c) Radiation intensity after 24 hours
The relative intensity is related to time by the
equation
36
Example 3-Linearization of data cont.
Exponential model given as,
Assuming
,
and
we obtain
This is a linear relationship between
and
37
Example 3-Linearization of data cont.
Using this linear relationship, we can calculate
where
and
38
Example 3-Linearization of Data cont.
Summations for data linearization are as follows
With
Table. Summation data for linearization of data
model

1 2 3 4 5 6 0 1 3 5 7 9 1 0.891 0.708 0.562 0.447 0.355 0.00000 -0.11541 -0.34531 -0.57625 -0.80520 -1.0356 0.0000 -0.11541 -1.0359 -2.8813 -5.6364 -9.3207 0.0000 1.0000 9.0000 25.000 49.000 81.000
25.000 -2.8778 -18.990 165.00
39
Example 3-Linearization of Data cont.
Calculating
Since
also
40
Example 3-Linearization of Data cont.
Resulting model is
Figure. Relative intensity of radiation as a
function of temperature using linearization of
data model.
41
Example 3-Linearization of Data cont.
The regression formula is then
b) Half life of Technetium 99 is when
42
Example 3-Linearization of Data cont.
c) The relative intensity of radiation after 24
hours is then
This implies that only
of the radioactive
material is left after 24 hours.
43
Comparison
Comparison of exponential model with and without
data linearization
Table. Comparison for exponential model with and
without data linearization.
With data linearization (Example 3) Without data linearization (Example 1)
A 0.99974 0.99983
? -0.11505 -0.11508
Half-Life (hrs) 6.0248 6.0232
Relative intensity after 24 hrs. 6.320010-2 6.316010-2
The values are very similar so data linearization
was suitable to find the constants of the
nonlinear exponential model in this case.
44
Additional Resources
  • For all resources on this topic such as digital
    audiovisual lectures, primers, textbook chapters,
    multiple-choice tests, worksheets in MATLAB,
    MATHEMATICA, MathCad and MAPLE, blogs, related
    physical problems, please visit
  • http//numericalmethods.eng.usf.edu/topics/nonline
    ar_regression.html

45
  • THE END
  • http//numericalmethods.eng.usf.edu
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