Find constants of nonlinear regression models

8/11/2022 http://numericalmethods.eng.usf.edu 1
Nonlinear Regression
Major: All Engineering Majors
Authors: Autar Kaw, Luke Snyder
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates

)
( bx
ae
y 
)
( b
ax
y 








x
b
ax
y
Some popular nonlinear regression models:
1. Exponential model:
2. Power model:
3. Saturation growth model:
4. Polynomial model: )
( 1
0
m
mx
a
...
x
a
a
y 



3

Given n data points )
,
(
,
...
),
,
(
),
,
( 2
2
1
1 n
n y
x
y
x
y
x best fit )
(x
f
y 
to the data, where )
(x
f is a nonlinear function of x .
Figure. Nonlinear regression model for discrete y vs. x data
)
(x
f
y 
)
,
(
n
n
y
x
)
,
( 1
1
y
x
)
,
(
2
2
y
x
)
,
(
i
i
y
x
)
(
i
i
x
f
y 
4

Regression
Exponential Model
5

Exponential Model
)
,
(
,
...
),
,
(
),
,
( 2
2
1
1 n
n y
x
y
x
y
x
Given best fit
bx
ae
y  to the data.
Figure. Exponential model of nonlinear regression for y vs. x data
bx
ae
y 
)
,
(
n
n
y
x
)
,
( 1
1
y
x
)
,
(
2
2
y
x
)
,
(
i
i
y
x
i
bx
i ae
y 
6

Finding Constants of Exponential Model
 




n
i
bx
i
r
i
ae
y
S
1
2
The sum of the square of the residuals is defined as
Differentiate with respect to a and b
   0
2
1








i
i bx
n
i
bx
i
r
e
ae
y
a
S
   0
2
1








i
i bx
i
n
i
bx
i
r
e
ax
ae
y
b
S
7

Finding Constants of Exponential Model
Rewriting the equations, we obtain
0
1
2
1







n
i
bx
n
i
bx
i
i
i e
a
e
y
0
1
2
1






n
i
bx
i
n
i
bx
i
i
i
i e
x
a
e
x
y
8

Finding constants of Exponential Model
Substituting a back into the previous equation
0
1
2
1
2
1
1










n
i
bx
i
n
i
bx
bx
n
i
i
bx
i
n
i
i
i
i
i
i e
x
e
e
y
e
x
y
The constant b can be found through numerical
methods such as bisection method.





n
i
bx
n
i
bx
i
i
i
e
e
y
a
1
2
1
Solving the first equation for a yields
9

Example 1-Exponential Model
t(hrs) 0 1 3 5 7 9
1.000 0.891 0.708 0.562 0.447 0.355
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.

10

Example 1-Exponential Model cont.
Find:
a) The value of the regression constants A 
and
b) The half-life of Technetium-99m
c) Radiation intensity after 24 hours
The relative intensity is related to time by the equation
t
Ae
 
11

Plot of data
12

Constants of the Model
The value of λ is found by solving the nonlinear equation
  0
1
2
1
2
1
1











n
i
t
i
n
i
t
n
i
t
i
t
i
n
i
i
i
i
i
i e
t
e
e
e
t
f 










 n
i
t
n
i
t
i
i
i
e
e
A
1
2
1



t
Ae
 
13

Setting up the Equation in MATLAB
  0
1
2
1
2
1
1











n
i
t
i
n
i
t
n
i
t
i
t
i
n
i
i
i
i
i
i e
t
e
e
e
t
f 






t (hrs) 0 1 3 5 7 9
γ 1.000 0.891 0.708 0.562 0.447 0.355
14

Setting up the Equation in MATLAB
  0
1
2
1
2
1
1











n
i
t
i
n
i
t
n
i
t
i
t
i
n
i
i
i
i
i
i e
t
e
e
e
t
f 






t=[0 1 3 5 7 9]
gamma=[1 0.891 0.708 0.562 0.447 0.355]
syms lamda
sum1=sum(gamma.*t.*exp(lamda*t));
sum2=sum(gamma.*exp(lamda*t));
sum3=sum(exp(2*lamda*t));
sum4=sum(t.*exp(2*lamda*t));
f=sum1-sum2/sum3*sum4;
1151
.
0



15

Calculating the Other Constant
The value of A can now be calculated




 6
1
2
6
1
i
t
i
t
i
i
i
e
e
A



9998
.
0

The exponential regression model then is
t
e 1151
.
0
9998
.
0 


16

Plot of data and regression curve
t
e 1151
.
0
9998
.
0 


17

Relative Intensity After 24 hrs
The relative intensity of radiation after 24 hours
 
24
1151
.
0
9998
.
0 

 e

2
10
3160
.
6 


This result implies that only
%
317
.
6
100
9998
.
0
10
316
.
6 2


 
radioactive intensity is left after 24 hours.
18

Homework
• What is the half-life of Technetium-99m
isotope?
• Write a program in the language of your
choice to find the constants of the
model.
• Compare the constants of this regression
model with the one where the data is
transformed.
• What if the model was ?
19
t
e
 

THE END
20

Polynomial Model
)
,
(
,
...
),
,
(
),
,
( 2
2
1
1 n
n y
x
y
x
y
x
Given best fit
m
m
x
a
...
x
a
a
y 


 1
0
)
2
( 
 n
m to a given data set.
Figure. Polynomial model for nonlinear regression of y vs. x data
m
m
x
a
x
a
a
y 


 
1
0
)
,
(
n
n
y
x
)
,
( 1
1
y
x
)
,
(
2
2
y
x
)
,
(
i
i
y
x
)
(
i
i
x
f
y 
21

Polynomial Model cont.
The residual at each data point is given by
m
i
m
i
i
i x
a
x
a
a
y
E 



 .
.
.
1
0
The sum of the square of the residuals then is
 










n
i
m
i
m
i
i
n
i
i
r
x
a
x
a
a
y
E
S
1
2
1
0
1
2
.
.
.
22

To find the constants of the polynomial model, we set the derivatives
with respect to i
a where
 
 
  0
)
(
.
.
.
.
2
0
)
(
.
.
.
.
2
0
)
1
(
.
.
.
.
2
1
1
0
1
1
0
1
1
1
0
0

































m
i
n
i
m
i
m
i
i
m
r
i
n
i
m
i
m
i
i
r
n
i
m
i
m
i
i
r
x
x
a
x
a
a
y
a
S
x
x
a
x
a
a
y
a
S
x
a
x
a
a
y
a
S




,
,
1 m
i 
 equal to zero.
23

These equations in matrix form are given by





































































































































n
i
i
m
i
n
i
i
i
n
i
i
m
n
i
m
i
n
i
m
i
n
i
m
i
n
i
m
i
n
i
i
n
i
i
n
i
m
i
n
i
i
y
x
y
x
y
a
a
a
x
x
x
x
x
x
x
x
n
1
1
1
1
0
1
2
1
1
1
1
1
1
2
1
1
1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
The above equations are then solved for m
a
a
a ,
,
, 1
0 
24

Example 2-Polynomial Model
Temperature, T
(oF)
Coefficient of
thermal
expansion, α
(in/in/oF)
80 6.47×10−6
40 6.24×10−6
−40 5.72×10−6
−120 5.09×10−6
−200 4.30×10−6
−280 3.33×10−6
−340 2.45×10−6
Regress the thermal expansion coefficient vs. temperature data to
a second order polynomial.
1.00E-06
2.00E-06
3.00E-06
4.00E-06
5.00E-06
6.00E-06
7.00E-06
-400 -300 -200 -100 0 100 200
Temperature, o
F
Thermal
expansion
coefficient,
α
(in/in/
o
F)
Table. Data points for
temperature vs
Figure. Data points for thermal expansion coefficient vs
temperature.
α
25

Example 2-Polynomial Model cont.
2
2
1
0 T
a
T
a
a
α 


We are to fit the data to the polynomial regression model





















































































































n
i
i
i
n
i
i
i
n
i
i
n
i
i
n
i
i
n
i
i
n
i
i
n
i
i
n
i
i
n
i
i
n
i
i
T
T
a
a
a
T
T
T
T
T
T
T
T
n
1
2
1
1
2
1
0
1
4
1
3
1
2
1
3
1
2
1
1
2
1



The coefficients 2
1
0 , a
,a
a are found by differentiating the sum of the
square of the residuals with respect to each variable and setting the
values equal to zero to obtain
26

The necessary summations are as follows
Temperature, T
(oF)
Coefficient of
thermal expansion,
α (in/in/oF)
80 6.47×10−6
40 6.24×10−6
−40 5.72×10−6
−120 5.09×10−6
−200 4.30×10−6
−280 3.33×10−6
−340 2.45×10−6
Table. Data points for temperature vs. α 5
7
1
2
10
5580
.
2 



i
i
T
7
7
1
3
10
0472
.
7 




i
i
T
10
7
1
4
10
1363
.
2




i
i
T
5
7
1
10
3600
.
3 




i
i

3
7
1
10
6978
.
2 





i
i
i
T 
1
7
1
2
10
5013
.
8 




i
i
i
T 
27



















































1
3
5
2
1
0
10
7
5
7
5
2
5
2
10
5013
.
8
10
6978
.
2
10
3600
.
3
10
1363
.
2
10
0472
.
7
10
5800
.
2
10
0472
.
7
10
5800
.
2
10
600
.
8
10
5800
.
2
10
6000
.
8
0000
.
7
a
a
a
Using these summations, we can now calculate 2
1
0 , a
,a
a
Solving the above system of simultaneous linear equations we have




























11
9
6
2
1
0
10
2218
.
1
10
2782
.
6
10
0217
.
6
a
a
a
The polynomial regression model is then
2
11
9
6
2
2
1
0
T
10
1.2218
T
10
6.2782
10
6.0217
α











 T
a
T
a
a
28

Transformation 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 transformed. For example,
the data for an exponential model can be transformed.
As shown in the previous example, many chemical and physical processes
are governed by the equation,
bx
ae
y 
Taking the natural log of both sides yields,
bx
a
y 
 ln
ln
Let y
z ln
 and a
a ln
0 
(implying) o
a
e
a  with b
a 
1
We now have a linear regression model where x
a
a
z 1
0 

29

Transformation of data cont.
Using linear model regression methods,
_
1
_
0
1
2
1
2
1
1 1
1
x
a
z
a
x
x
n
z
x
z
x
n
a
n
i
n
i
i
i
n
i
i
n
i
n
i
i
i
i











 

 
 

 
Once 1
,a
ao are found, the original constants of the model are found as
0
1
a
e
a
a
b


30

THE END
31

Example 3-Transformation of
data
t(hrs) 0 1 3 5 7 9
1.000 0.891 0.708 0.562 0.447 0.355
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
0
0.5
1
0 5 10
Relative
intensity
of
radiation,
γ
Time t, (hours)
Figure. Data points of relative radiation intensity
vs. time
32

Example 3-Transformation of data
cont.
Find:
a) The value of the regression constants A 
and
b) The half-life of Technetium-99m
c) Radiation intensity after 24 hours
The relative intensity is related to time by the equation
t
Ae
 
33

Example 3-Transformation of data
cont.
t
Ae
 
Exponential model given as,
    t
A 
 
 ln
ln
Assuming 
ln

z ,  
A
ao ln
 and 

1
a we obtain
t
a
a
z
1
0


This is a linear relationship between z and t
34

Example 3-Transformation of data cont.
Using this linear relationship, we can calculate 1
0 ,a
a
 

 
 

 









n
i
n
i
i
n
i
i
n
i
n
i
i
i
i
t
t
n
z
t
z
t
n
a
1
2
1
2
1
1
1 1
1
and
t
a
z
a 1
0 

where
1
a


0
a
e
A 
35

Example 3-Transformation of Data
cont.
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
Summations for data transformation are as follows
Table. Summation data for Transformation of data
model
i i
t i
 i
i
z 
ln
 i
i
z
t 2
i
t

With 6

n
000
.
25
6
1



i
i
t




6
1
8778
.
2
i
i
z




6
1
990
.
18
i
i
i
z
t
00
.
165
6
1
2



i
i
t
36

cont.
Calculating 1
0 ,a
a
    
   2
1
25
00
.
165
6
8778
.
2
25
990
.
18
6





a 11505
.
0


 
6
25
11505
.
0
6
8778
.
2
0 



a
4
10
6150
.
2 



Since
 
A
a ln
0 
0
a
e
A 
4
10
6150
.
2 


 e 99974
.
0

11505
.
0
1 

 a

also
37

cont.
Resulting model is
t
e 11505
.
0
99974
.
0 



0
0.5
1
0 5 10
Time, t (hrs)
Relative
Intensity
of
Radiation,
t
e 11505
.
0
99974
.
0 



Figure. Relative intensity of radiation as a function of
temperature using transformation of data model.
38

cont.
The regression formula is then
t
e 11505
.
0
99974
.
0 



b) Half life of Technetium-99m is when
0
2
1


t


   
 
hours
.
t
.
t
.
.
e
e
.
e
.
t
.
.
t
.
0248
6
5
0
ln
11505
0
5
0
99974
0
2
1
99974
0
11508
0
0
11505
0
11505
0









39

cont.
c) The relative intensity of radiation after 24 hours is then
 
24
11505
.
0
99974
.
0 
 e

063200
.
0

This implies that only %
3216
.
6
100
99983
.
0
10
3200
.
6 2


 
of the radioactive
material is left after 24 hours.
40

Comparison
Comparison of exponential model with and without data
Transformation:
With data
Transformation
(Example 3)
Without data
Transformation
(Example 1)
A 0.99974 0.99983
λ −0.11505 −0.11508
Half-Life (hrs) 6.0248 6.0232
Relative intensity
after 24 hrs.
6.3200×10−2 6.3160×10−2
Table. Comparison for exponential model with and without data
Transformation.
41

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/nonlinear_r
egression.html

THE END

Find constants of nonlinear regression models

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