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http://numericalmethods.eng.usf.edu 1
Nonlinear Regression
Civil Engineering Majors
Authors: Autar Kaw, Luke Snyder
http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM
Undergraduates
Nonlinear Regression
http://numericalmethods.eng.usf.edu
Nonlinear Regression
)( bxaey
)( baxy
xb
axy
Some popular nonlinear regression models:
1. Exponential model:2. Power model:
3. Saturation growth model:4. Polynomial model: )( 10
mmxa...xaay
http://numericalmethods.eng.usf.edu3
Nonlinear Regression
Given n data points
),( , ... ),,(),,( 2211 nn yxyx yx best fit )(xfy
to the data, where
)(xf is a nonlinear function of
x .
Figure. Nonlinear regression model for discrete y vs. x data
)(xfy
),(nn
yx
),(11
yx
),(22
yx
),(ii
yx
)(ii
xfy
http://numericalmethods.eng.usf.edu4
RegressionExponential Model
http://numericalmethods.eng.usf.edu5
Exponential Model),( , ... ),,(),,( 2211 nn yxyx yxGive
nbest fit
bxaey to the data.
Figure. Exponential model of nonlinear regression for y vs. x data
bxaey
),(nn
yx
),(11
yx
),(22
yx
),(ii
yx
)(ii
xfy
http://numericalmethods.eng.usf.edu6
Finding constants of Exponential Model
n
i
bx
ir iaeyS
1
2
The sum of the square of the residuals is defined as
Differentiate with respect to a and b
021
ii bxn
i
bxi
r eaeya
S
021
ii bxi
n
i
bxi
r eaxaeyb
S
http://numericalmethods.eng.usf.edu7
Finding constants of Exponential Model
Rewriting the equations, we obtain
01
2
1
n
i
bxn
i
bxi
ii eaey
01
2
1
n
i
bxi
n
i
bxii
ii exaexy
http://numericalmethods.eng.usf.edu8
Finding constants of Exponential Model
Substituting
a back into the previous equation
01
2
1
2
1
1
n
i
bxin
i
bx
bxn
ii
bxi
n
ii
i
i
i
i exe
eyexy
The constant
b can be found through numerical methods suchas the bisection method or secant method.
Nonlinear equation in terms of
b
n
i
bx
n
i
bxi
i
i
e
eya
1
2
1
Solving the first equation for
a yields
http://numericalmethods.eng.usf.edu9
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 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.
http://numericalmethods.eng.usf.edu10
Example 1-Exponential Model cont.
Find: a) The value of the regression
constants A an
db) The half-life of Technium-99m
c) Radiation intensity after 24 hours
The relative intensity is related to time by the equationtAe
http://numericalmethods.eng.usf.edu11
Plot of data
http://numericalmethods.eng.usf.edu12
Constants of the Model
The value of λ is found by solving the nonlinear equation
01
2
1
2
1
1
n
i
tin
i
t
n
i
ti
ti
n
ii
i
i
i
i ete
eetf
n
i
t
n
i
ti
i
i
e
eA
1
2
1
tAe
http://numericalmethods.eng.usf.edu13
Setting up the Equation in MATLAB
01
2
1
2
1
1
n
i
tin
i
t
n
i
ti
ti
n
ii
i
i
i
i ete
eetf
t (hrs) 0 1 3 5 7 9
γ 1.000
0.891
0.708
0.562
0.447
0.355
http://numericalmethods.eng.usf.edu14
Setting up the Equation in MATLAB
01
2
1
2
1
1
n
i
tin
i
t
n
i
ti
ti
n
ii
i
i
i
i ete
eetf
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
http://numericalmethods.eng.usf.edu15
Calculating the Other Constant
The value of A can now be calculated
6
1
2
6
1
i
t
i
ti
i
i
e
eA
9998.0
The exponential regression model then is te 1151.0 9998.0
http://numericalmethods.eng.usf.edu16
Plot of data and regression curve
te 1151.0 9998.0
http://numericalmethods.eng.usf.edu17
Relative Intensity After 24 hrs
The relative intensity of radiation after 24 hours
241151.09998.0 e2103160.6
This result implies that only
%317.61009998.0
10316.6 2
radioactive intensity is left after 24 hours.
http://numericalmethods.eng.usf.edu18
Homework1. 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.
http://numericalmethods.eng.usf.edu19
THE END
http://numericalmethods.eng.usf.edu
http://numericalmethods.eng.usf.edu20
Polynomial Model
),( , ... ),,(),,( 2211 nn yxyx yxGiven best fit
m
mxa...xaay
10
)2( nm to a given data set.
Figure. Polynomial model for nonlinear regression of y vs. x data
m
mxaxaay
10
),(nn
yx
),(11
yx
),(22
yx
),(ii
yx
)(ii
xfy
http://numericalmethods.eng.usf.edu21
Polynomial Model cont.The residual at each data point is given by
mimiii xaxaayE ...10
The sum of the square of the residuals then is
n
i
mimii
n
iir
xaxaay
ES
1
2
10
1
2
...
http://numericalmethods.eng.usf.edu22
Polynomial Model cont.To find the constants of the polynomial model, we set the derivatives with respect to ia wher
e
0)(....2
0)(....2
0)1(....2
110
110
1
110
0
mi
n
i
mimii
m
r
i
n
i
mimii
r
n
i
mimii
r
xxaxaaya
S
xxaxaaya
S
xaxaaya
S
,,1 mi equal to zero.
http://numericalmethods.eng.usf.edu23
Polynomial Model cont.These equations in matrix form are
given by
n
ii
mi
n
iii
n
ii
mn
i
mi
n
i
mi
n
i
mi
n
i
mi
n
ii
n
ii
n
i
mi
n
ii
yx
yx
y
a
a
a
xxx
xxx
xxn
1
1
1
1
0
1
2
1
1
1
1
1
1
2
1
11
......
...
...........
...
...
The above equations are then solved for
maaa ,,, 10
http://numericalmethods.eng.usf.edu24
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, oF
Th
erm
al e
xpan
sio
n c
oef
fici
ent,
α
(in
/in
/oF
)
Table. Data points for temperature vs
Figure. Data points for thermal expansion coefficient vs temperature.
α
http://numericalmethods.eng.usf.edu25
Example 2-Polynomial Model cont.
2210 TaTaaα
We are to fit the data to the polynomial regression model
n
iii
n
iii
n
ii
n
ii
n
ii
n
ii
n
ii
n
ii
n
ii
n
ii
n
ii
T
T
a
a
a
TTT
TTT
TTn
1
2
1
1
2
1
0
1
4
1
3
1
2
1
3
1
2
1
1
2
1
The coefficients
210 , a,aa are found by differentiating the sum of thesquare of the residuals with respect to each variable and
setting thevalues equal to zero to obtain
http://numericalmethods.eng.usf.edu26
Example 2-Polynomial Model cont.
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 57
1
2 105580.2 i
iT
77
1
3 100472.7 i
iT
107
1
4 101363.2
i
iT
57
1
103600.3
i
i
37
1
106978.2
i
iiT
17
1
2 105013.8
i
iiT
http://numericalmethods.eng.usf.edu27
Example 2-Polynomial Model cont.
1
3
5
2
1
0
1075
752
52
105013.8
106978.2
103600.3
101363.2100472.7105800.2
100472.7105800.210600.8
105800.2106000.80000.7
a
a
a
Using these summations, we can now calculate
210 , a,aa
Solving the above system of simultaneous linear equations we have
11
9
6
2
1
0
102218.1
102782.6
100217.6
a
a
a
The polynomial regression model is then
21196
2210
T101.2218T106.2782106.0217
α
TaTaa
http://numericalmethods.eng.usf.edu28
yz ln
Linearization of DataTo 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,
bxaey Taking the natural log of both sides yields, bxay lnln
Let and
aa ln0
(implying)
oaea with
ba 1
We now have a linear regression model where
xaaz 10
http://numericalmethods.eng.usf.edu29
Linearization of data cont.Using linear model regression methods,
_
1
_
0
1
2
1
2
11 11
xaza
xxn
zxzxna
n
i
n
iii
n
ii
n
i
n
iiii
Once 1,aao are found, the original constants of the model are found as
0
1
aea
ab
http://numericalmethods.eng.usf.edu30
Example 3-Linearization 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
Rel
ativ
e in
ten
sity
of r
adia
tio
n, γ
Time t, (hours)
Figure. Data points of relative radiation intensity
vs. time http://numericalmethods.eng.usf.edu31
Example 3-Linearization of data cont.
Find: a) The value of the regression
constants A an
db) The half-life of Technium-99m
c) Radiation intensity after 24 hours
The relative intensity is related to time by the equation tAe
http://numericalmethods.eng.usf.edu32
Example 3-Linearization of data cont.
tAe Exponential model given as,
tA lnln
Assuming
lnz , Aao ln and 1a we obtaintaaz
10
This is a linear relationship between
z and t
http://numericalmethods.eng.usf.edu33
Example 3-Linearization of data cont.
Using this linear relationship, we can calculate
10 , aa
n
i
n
ii
n
ii
n
i
n
iiii
ttn
ztztna
1
2
1
2
1
11 1
1
and
taza 10
where
1a
0a
eA
http://numericalmethods.eng.usf.edu34
Example 3-Linearization of Data cont.
123456
013579
10.8910.7080.5620.4470.355
0.00000−0.11541−0.34531−0.57625−0.80520−1.0356
0.0000−0.11541−1.0359−2.8813−5.6364−9.3207
0.00001.00009.000025.00049.00081.000
25.000 −2.8778 −18.990 165.00
Summations for data linearization are as follows
Table. Summation data for linearization of data model
i it iii
z ln iizt 2
it
With 6n
000.256
1
i
it
6
1
8778.2i
iz
6
1
990.18i
iizt
00.1656
1
2 i
it
http://numericalmethods.eng.usf.edu35
Example 3-Linearization of Data cont.
Calculating 10 ,aa
21
2500.1656
8778.225990.186
a 11505.0
6
2511505.0
6
8778.20
a
4106150.2
Since Aa ln0 0aeA
4106150.2 e 99974.0
11505.01 aalso
http://numericalmethods.eng.usf.edu36
Example 3-Linearization of Data cont.
Resulting model is
te 11505.099974.0
0
0.5
1
0 5 10
Time, t (hrs)
Relative Intensity
of Radiation,
te 11505.099974.0
Figure. Relative intensity of radiation as a function of temperature using linearization of data model.
http://numericalmethods.eng.usf.edu37
Example 3-Linearization of Data cont.
The regression formula is thente 11505.099974.0
b) Half life of Technetium 99 is when02
1
t
hours.t
.t.
.e
e.e.
t.
.t.
02486
50ln115050
50
9997402
1999740
115080
0115050115050
http://numericalmethods.eng.usf.edu38
Example 3-Linearization of Data cont.
c) The relative intensity of radiation after 24 hours is then 2411505.099974.0 e
063200.0This implies that only
%3216.610099983.0
103200.6 2
of the radioactive
material is left after 24 hours.
http://numericalmethods.eng.usf.edu39
Comparison Comparison of exponential model with and without data
linearization:
With data linearization(Example 3)
Without data linearization(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 linearization.
The values are very similar so data linearization was suitable to find the constants of the nonlinear exponential model in this case.
http://numericalmethods.eng.usf.edu40
Additional ResourcesFor 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_regression.html
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