http://numericalmethods.eng.usf.edu 1 nonlinear regression civil engineering majors authors: autar...

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Nonlinear Regression

Civil Engineering Majors

Authors: Autar Kaw, Luke Snyder

http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM

Undergraduates

http://numericalmethods.eng.usf.edu/



http://numericalmethods.eng.usf.edu




)( 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


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


RegressionExponential Model


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


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



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



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


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.


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


Plot of data


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


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


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


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


Plot of data and regression curve

te 1151.0 9998.0


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.


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.


THE END



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


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

...


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.


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


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


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



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



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


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


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


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



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



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


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



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



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.



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



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.


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.


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



THE END



http://numericalmethods.eng.usf.edu 1 nonlinear regression civil engineering majors authors: autar...

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