07 Regresi Non Linier

8/18/2019 07 Regresi Non Linier

1/38

Nonlinear Regression

SujantokoMathematic Engineering

Ocean Engineering FTK ITS

1


2/38


3/38

Nonlinear Regression

Given n data points ),(,...),,(),,( 2211 nn y x y x y x best fit )( x f y =to the data, where )( x f is a nonlinear function of x .

),(nn

y x

)( x f y =,

22

),(ii

y x−

),(11

y x ii

. .

3


4/38

RegressionExponential Model

4


5/38

Exponential Model

),(,...),,(),,( 2211 nn y x y x y xGiven best fit bx

ae y = to the data.

,nn

),( 22 y xbxae y =),(

ii y x

)(ii

x f y −

Figure. Exponential model of nonlinear regression for y vs. x data

,11

5


6/38

n ng onstants o xponent a o e

( )∑ −= n bxir iae yS 2

e sum o e square o e res ua s s e ne as

=i 1Differentiate with respect to a and b

n

021

=−−=∂ =

ii x

i

xi

r eaea

( )( ) 021

=−−=∂∂ ∑

=

ii bxi

n

i

bxi

r eaxae yb

S

6


7/38

Finding Constants of Exponential Model

Rewriting the equations, we obtain

01

2

1

=∑+∑−== i

bx

i

bxi

ii eae y

02 =∑−∑ n bx

in bx

iiii e xae x y

== ii

7


8/38

Finding constants of Exponential Model

∑n bx

iie y

∑= =

n bx

i

iea 2

1

Substituting a back into the previous equation

=i 1

21∑

− = n bx

bxn

ii

bxn i

i

i

e y

1

1

21 ∑ ==

= i in

i

bxii i ie

he constant b can be found through numerical

methods such bisection method.8


9/38

-Many patients get concerned when a test involves injection of a

few drops of Technetium-99m isotope is used. Half of thetechritium-99m would be gone in about 6 hours. It, however,takes about 24 hours for the radiation levels to reach what weare exposed to in day-to-day activities. Below is given therelative intensity of radiation as a function of time.

t(hrs) 0 1 3 5 7 9

. .

. . . . . .

9


10/38

Example 1-Exponential Model cont.

The relative intensity is related to time by the equationt λ =

Find:

a) The value of the regression constants λ andb) The half-life of Technium-99m

c) Radiation intensity after 24 hours

10


11/38

Plot of data

11


12/38

n n f h M l t Ae λ γ =e va ue o λ is oun y so ving t e non inear equation

n t iλ

( ) 01

2

2

1

1=∑−∑=

=

=

=

n

i

t i

n t

it i

n

ii

i

i

i et et f λ λ

λ γ λ

1=in

==n

ii

i

e A 1

γ

=i

ie1

12


13/38

Setting up the Equation in MATLAB

( ) 01

2

2

1

1=∑

∑−∑=

=

=

=

n

i

t in t

i

t i

t i

n

ii

i

i

i

i et e

et f λ λ

λ γ

γ λ

1=i

rs

γ 1.000 0.891 0.708 0.562 0.447 0.35513


14/38

Settin u the E uation in MATLAB21

∑= n t

n

i

t i

t nie

λ

λ

λ γ

−1

121 ∑

−=

== i

in

it

ii

iie λ

.

t=[0 1 3 5 7 9]amma= 1 0.891 0.708 0.562 0.447 0.355

syms lamdasum1=sum(gamma.*t.*exp(lamda*t));

sum2=sum(gamma.*exp(lamda*t));sum3=sum(exp(2*lamda*t));

= .

f=sum1-sum2/sum3*sum4;14


15/38

Calculating the Other Constant

The value of A can now be calculated

∑6 t i ie λ γ

∑

==6

2 t

i

i

e

Aλ

9998.0=

=1i

The exponential regression model then is

t e 1151.09998.0 −=γ

15


16/38

Plot of data and regression curve

t e 1151.0 9998.0 −=γ

16


17/38

( )241151.09998.0 −×= eγ 2103160.6 −×=

his result implies that only

10316.6 2× −

.9998.0 =

radioactive intensity is left after 24 hours.

17


18/38

H m w rk • What is the half-life of technetium 99m

• Compare the constants of this regressionmo e wit t e one w ere t e ata is

transformed.• Write a program in the language of your

choice to find the constants of the model(fortran, matlab, matchad, maple)

18


19/38

o ynom a o e

),(,...),,(),,( 2211 nn y x y x y xGiven best fitm

m xa... xaa y +++= 10)2( −≤ nm to a given data set.

),(nn

y x

m

m xa xaa y +++= K

10

),(22

y x

),(ii

y x

),(11

y x)( ii x f y −

Figure. Polynomial model for nonlinear regression of y vs. x data

19


20/38

Polynomial Model cont.

The residual at each data point is given bymimiii xa xaa y E −−−−= ...10

T e sum o t e square o t e resi ua s t en is

∑=

=n

iir E S

1

2

( )∑=

−−−−=n

i

mimii xa xaa y

1

2

10 ...

20


21/38


To find the constants of the polynomial model, we set the derivativeswith respect to ia where∂ nS

,,1 mi K= equal to zero.

( ) 0)(....2

....

10

110

0

=−−−−−=∂∂

=−−−−−=∂

∑

=

i

nmimii

r

iimii

x xa xaa yaS

xaaa ya

n

MMMM

0)(....2110 =−−−−−=∂ ∑=

mi

i

mimii

m

r

x xa xaa ya

21


22/38


These equations in matrix form are given by⎤⎡

⎥

⎤

⎢

⎡

nn

⎥

⎥

⎥

⎥

⎢

⎢

⎢

⎢

=⎥⎥

⎥

⎤

⎢⎢

⎢

⎡

⎥

⎥

⎥

⎢

⎢

⎢

⎟

⎞⎜

⎛ ⎟

⎞⎜

⎛ ⎟

⎞⎜

⎛

⎟

⎠⎜

⎝ ⎟

⎠⎜

⎝

∑

∑

∑∑∑

∑∑=

+

==n

ii

ii

nmi

n

i

n

i

i

mi

ii

y x

y

a

a

x x x

x xn1

1

0

12

11

......

...

⎥

⎥

⎥

⎥

⎦⎢

⎢

⎢

⎢

⎣

⎦⎣

⎥

⎥

⎥

⎥

⎦⎢

⎢

⎢

⎢

⎣

⎟

⎠ ⎞

⎜

⎝ ⎛

⎟

⎠ ⎞

⎜

⎝ ⎛

⎟

⎠ ⎞

⎜

⎝ ⎛ ∑∑∑∑

=

=

==

+

=

===

n

ii

mi

mn

i

mi

n

i

mi

n

i

mi

y x

a

x x x11

2

1

1

1

...

...

...........

The above equations are then solved for maaa ,,, 10 K

22


23/38

Example 2-Polynomial ModelRegress the thermal expansion coefficient vs. temperature data toa second order polynomial.

Temperature, T(oF)

Coefficient ofthermal

expansion, α 5.00E-06

6.00E-06

7.00E-06

o e f

f i c i e n

t , α

. temperature vs α

n n o F)

80 6.47×10 −6

40 6.24×10 −6

−40 5.72×10 −63.00E-06

4.00E-06

a l e x p a n s i o n

( i n

/ i n

/ o F )

−120 5.09×10 −6

−200 4.30×10 −6

−280 3.33×10 −61.00E-06

2.00E-06

-400 -300 -200 -100 0 100 200

o

T h e r

−340 2.45×10 − ,

Figure. Data points for thermal expansion coefficient vstemperature.

23


24/38

Example 2-Polynomial Model cont.

2210 T aT aaα ++=We are to fit the data to the polynomial regression model

210 , , square of the residuals with respect to each variable and setting thevalues equal to zero to obtain

⎥

⎥

⎥

⎤

⎢

⎢

⎢

⎡

⎥

⎤

⎢

⎡⎥

⎥

⎥

⎤

⎢

⎢

⎢

⎡

⎞⎛ ⎞⎛ ⎞⎛

⎟

⎠ ⎞

⎜

⎝ ⎛

⎟

⎠ ⎞

⎜

⎝ ⎛

∑∑∑===

n

n

ii

nnn

n

ii

n

ii

aT T n

10

32

1

2

1α

⎥⎥

⎥

⎦⎢⎢

⎢

⎣

⎥⎦⎢⎣

⎥

⎥⎥

⎦⎢

⎢⎢

⎣

⎟

⎠ ⎞

⎜

⎝ ⎛

⎟

⎠ ⎞

⎜

⎝ ⎛

⎟

⎠ ⎞

⎜

⎝ ⎛ ⎠⎝ ⎠⎝ ⎠⎝ ∑∑∑∑

=

=

===

===n

iii

iii

n

ii

n

ii

n

ii

ii

ii

ii

T a

T T T 1

2

12

1

1

4

1

3

1

2

111

α

24


25/38


The necessary summations are as followsTable. Data points for temperature vs. α 5

72 105580.2 ×=∑ iT

empera ure,(oF)

oe c en othermal expansion,

α (in/in/ oF)

80 6.47×10 −6

40 6.24×10 −6

1=i

77

1

3 100472.7 ×−=∑=i

iT

−40 5.72×10 −6

−120 5.09×10 −6

−200 4.30×10 −6

10

1

4 101363.2∑=

×=i

iT

57

103600.3 −×=α −280 3.33×10 −6

−340 2.45×10 −61=i

37

1

106978.2 −=

×−=∑i

iiT α

17

1

2 105013.8 −

=

×=∑i

iiT α

25


26/38


⎤⎡ ×⎤⎡⎤⎡ ××− −5052 103600.3105800.2106000.80000.7 a

Using these summations, we can now calculate 210 , a ,aa

⎥

⎥

⎦⎢

⎢

⎣ ×

×−=⎥

⎥

⎦⎢

⎢

⎣⎥

⎥

⎦⎢

⎢

⎣ ××−×

×−××−−

−

1

3

2

11075

752

105013.8

106978.2

101363.2100472.7105800.2

100472.7105800.210600.8

a

a

Solving the above system of simultaneous linear equations we have

⎥

⎥

⎥

⎤

⎢

⎢

⎢

⎡

×−××

=⎥

⎥

⎤

⎢

⎢

⎡

−

−

−

11

9

6

1

0

102218.1

102782.6

100217.6

a

a

a

The polynomial regression model is then2

210α ++= T aT aa21196 T101.2218T106.2782106.0217 −−− ×−×+×=

26


27/38

Linearization of Data

To find the constants of many nonlinear models, it results in solvingsimultaneous nonlinear equations. For mathematical convenience,some of the data for such models can be linearized. For exam le thedata for an exponential model can be linearized.

As shown in the previous example, many chemical and physical processesare overned b the e uation

bxae y =Taking the natural log of both sides yields,

Let y z ln= and aa ln0 =We now have a linear regression model where xaa z 10 +=(implying) oaea = with ba =1

27


28/38

Linearization of data cont.

Using linear model regression methods, z x z xn

n

i

n n

iii − ∑∑ ∑

1

2

1

2

1

x xn

an

i

n

i

ii ⎟

⎠

⎞⎜

⎝

⎛ −

=

∑ ∑= =

== =

_

1

_

0 xa z a −=

Once 1, aa o are found, the original constants of the model are found as

0

1

aea =

28


29/38

Example 3-Linearization of dataMany 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 whatwe are exposed to in day-to-day activities. Below is given the relative intensityof radiation as a function of time .

Table. Relative intensity of radiation as a function 1 , γ

t(hrs) 0 1 3 5 7 9

1.000 0.891 0.708 0.562 0.447 0.355γ

of time

0.5

n s i

t y o

f r a d

i a t i o n

0

0 5 10

R e l a t

i v e i n t

,

Figure. Data points of relative radiation intensity

vs. time29


30/38

Example 3-Linearization of data cont.

Find:a) The value of the regression constants A λ and

T e a - i e o Tec nium-99m

c) Radiation intensity after 24 hours

The relative intensity is related to time by the equationt Aeλ γ =

30


31/38


t Ae λ γ =Exponential model given as,

γ += nn Assuming γ ln= z , ( ) Aa o ln= and λ =1a we obtain

t aa z 10 +=This is a linear relationship between z and t

31


32/38


Using this linear relationship, we can calculate 10 , aa∑∑ ∑−n

i

n n

iii z t z t n

where

∑ ∑= =

== =

⎟

⎠

⎞⎜

⎝

⎛ −=

n

i

n

ii

t t n

a

1

2

1

2

1

1

t a z a 10 −=

1a=λ

0a

e=

32


33/38

Example 3-Linearization of Data cont.

Summations for data linearization are as followsTable. Summation data for linearization of data model With 6=n

1 0 1 0.00000− 0.0000− 0.0000

i it iγ ii z γ ln= ii z t 2

it 000.25

1

∑=

=i

it

−=6

8778.2 z 3456

3579

.0.7080.5620.4470.355

.−0.34531−0.57625−0.80520−1.0356

.−1.0359−2.8813−5.6364−9.3207

.9.000025.00049.00081.000

=1ii

∑=

−=6

1

990.18i

ii z t

25.000 −2.8778 −18.990 165.00∑ 00.1656

1

2 =∑=i it

33


34/38


Calculating 10, aa

( ) ( )( )218778.225990.186 −−−=a 11505.0−=

. −

( )25

11505.08778.2

0 −−−

=a 4

106150.2 −

×−=

Since( ) Aa ln0 =

e=4106150.2 −×−= e 99974.0=

also11505.01 −== aλ

34


35/38

Example 3-Linearization of Data cont.Resulting model is t e 11505.099974.0

−×=γ

1t 11505.0−

0.5

Relative

Intensity

e .. ×=γ

oRadiation,

0

0 5 10

Time, t (hrs)

Figure. Relative intensity of radiation as a function oftemperature using linearization of data model.

35


36/38


The regression formula is thent e 11505.099974.0 −×=γ 1=

02 =t

( ) ( )e.e. .t . 9997402

1999740 0115050115050 =× −−

( )hours.t

.t .

.e .

02486

50ln115050

50

==−

=−

36


37/38

Example 3-Linearization of Data cont.c) The relative intensity of radiation after 24 hours is then

( )2411505.099974.0 −= eγ .=

This implies that only %3216.610099983.0

103200.6 2 =×× −

of the radioactive

material is left after 24 hours.

37


38/38

Comparison

Table. Comparison for exponential model with and without datalinearization.

With data linearization(Example 3)

Without data linearization(Example 1)

A 0.99974 0.99983

− . − .

Half-Life (hrs) 6.0248 6.0232

Relative intensityafter 24 hrs.

6.3200×10 −2 6.3160×10 −2

The values are very similar so data linearization was suitable tofind the constants of the nonlinear exponential model in this

.

38

07 Regresi Non Linier

Documents

Transcript of 07 Regresi Non Linier