10/11/2015 1 Nonlinear Regression Chemical Engineering Majors Authors: Autar Kaw, Luke Snyder

04/19/23http://

numericalmethods.eng.usf.edu 1

Nonlinear Regression

Chemical 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


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

http://numericalmethods.eng.usf.edu 25

Example 2-Polynomial ModelBelow is given the FT-IR (Fourier Transform Infra Red) data of a 1:1 (by weight) mixture of ethylene carbonate (EC) and dimethyl carbonate (DMC). Absorbance P is given as a function of wavenumber m.

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

Absorbance P, as a function of Wavenumber, m

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

780 800 820 840 860 880 900 920

Wavenumber, m (cm-1)

Ab

sorb

ance

, P

(ar

bit

rary

un

it)

Table. Absorbance vs Wavenumber data

Figure. Absorbance vs. Wavenumber data

1cm


Example 2-Polynomial Model cont.

Regress the data to a second order polynomial where 2

210 mamaaP

and find the absorbance at

11000cmm

The coefficients 210 ,a,aa are found as follows

n

iii

n

iii

n

ii

n

ii

n

ii

n

ii

n

ii

n

ii

n

ii

n

ii

n

ii

Pm

Pm

P

a

a

a

mmm

mmm

mmn

1

2

1

1

2

1

0

1

4

1

3

1

2

1

3

1

2

1

1

2

1



The necessary summations are as follows

iWavenumber, m

( )Absorbance, P(arbitrary unit)

m2 m3

1 804.18 0.1591 6.4671×105 5.2008×108

2 827.33 0.0439 6.8447×105 5.6628×108

3 846.61 0.0050 7.1675×105 6.0681×108

4 869.75 0.0073 7.5647×105 6.5794×108

5 889.04 0.0448 7.9039×105 7.0269×108

6 892.90 0.0649 7.9726×105 7.1187×108

7 900.61 0.1204 8.1110×105 7.3048×108

6030.4 0.4454 5.2031×106 4.4961×109

Table. Necessary summations for calculation of polynomial model constants

1cm

7

1i



Necessary summations continued:

i m4 m × P m2 × P

1 4.1824×1011 127.95 1.0289×105

2 4.6849×1011 36.319 3.0048×104

3 5.1373×1011 4.233 3.583×103

4 5.7225×1011 6.349 5.522×103

5 6.2471×1011 39.828 3.5409×104

6 6.3563×1011 57.948 5.1742×104

7 6.5787×1011 108.43 9.7655×104

3.8909×1012 381.06 3.2685×105

Table. Necessary summations for calculation of polynomial model constants.

7

1i



Using these summations we have

52

1

0

1296

963

63

102685.3

06.381

4454.0

108909.3104961.4102031.5

104961.4102031.5100304.6

102031.5100304.60000.7

a

a

a

Solving this system of equations we find

5

2

2

1

0

102365.1

100798.2

6623.8

a

a

a

The regression model is then

252

2210

102365.1100798.26623.8 mm

mamaaP

252 102365.1100798.26623.8 mmP



With the model is given by

Figure. Polynomial model of Absorbance vs. Wavenumber.



To find P where 11000cmm we have

22970.0

1000102365.11000100798.26623.8

102365.1100798.26623.8252

252

2210

mm

mamaaP

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

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



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10/11/2015 1 Nonlinear Regression Chemical Engineering Majors Authors: Autar Kaw, Luke Snyder