Download Linear Regression - Holistic Numerical Methods

Linear Regression
Industrial Engineering Majors
Authors: Autar Kaw, Luke Snyder
http://numericalmethods.eng.usf.edu
Transforming Numerical Methods Education for STEM
Undergraduates
8/11/2017
http://numericalmethods.eng.usf.edu
1
Linear Regression
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What is Regression?
What is regression? Given n data points ( x1, y1), ( x 2, y 2), ... , ( xn, yn )
best fit y  f (x ) to the data. The best fit is generally based on
minimizing the sum of the square of the residuals,
Sr.
Residual at a point is
( xn, yn )
i  yi  f ( xi )
y  f (x)
Sum of the square of the residuals
n
Sr   ( yi  f ( xi ))
i 1
3
2
( x1, y1)
Figure. Basic model for regression
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Linear Regression-Criterion#1
Given n data points ( x1, y1), ( x 2, y 2), ... , ( xn, yn) best fit y  a0  a1 x
to the data.
y
x,y
i
i
 i  yi  a0  a1 xi
x2 , y2
x ,y
n
n
x ,y
3
3
 i  yi  a0  a1 xi
x1 , y1
x
Figure. Linear regression of y vs. x data showing residuals at a typical point, xi .
Does minimizing
n

i 1
4
i
work as a criterion, where i  yi  (a 0  a1 xi )
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Example for Criterion#1
Example: Given the data points (2,4), (3,6), (2,6) and (3,8), best fit
the data to a straight line using Criterion#1
Table. Data Points
x
10
8
y
6
4.0
3.0
6.0
4
2.0
6.0
2
3.0
8.0
0
y
2.0
0
1
2
3
4
x
Figure. Data points for y vs. x data.
5
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Linear Regression-Criteria#1
Using y=4x-4 as the regression curve
Table. Residuals at each point for
regression model y = 4x – 4.
10
y
ypredicted
ε = y - ypredicted
8
2.0
4.0
4.0
0.0
6
3.0
6.0
8.0
-2.0
2.0
6.0
4.0
2.0
3.0
8.0
8.0
0.0
y
x
4

i 1
i
4
2
0
0
0
1
2
3
4
x
Figure. Regression curve for y=4x-4, y vs. x data
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Linear Regression-Criteria#1
Using y=6 as a regression curve
Table. Residuals at each point for y=6
x
y
ypredicted
ε = y - ypredicted
2.0
4.0
6.0
-2.0
3.0
6.0
6.0
0.0
2.0
6.0
6.0
0.0
3.0
8.0
6.0
2.0
10
8
4

i 1
i
y
6
4
2
0
0
0
1
2
3
4
x
Figure. Regression curve for y=6, y vs. x data
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Linear Regression – Criterion #1
4

i 1
i
 0 for both regression models of y=4x-4 and y=6.
The sum of the residuals is as small as possible, that is zero,
but the regression model is not unique.
Hence the above criterion of minimizing the sum of the
residuals is a bad criterion.
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Linear Regression-Criterion#2
n
Will minimizing

i 1
i
work any better?
y
x,y
i
i
 i  yi  a0  a1 xi
x2 , y2
n
n
x ,y
3
x1 , y1
x ,y
3
 i  yi  a0  a1 xi
x
Figure. Linear regression of y vs. x data showing residuals at a typical point, xi .
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Linear Regression-Criteria 2
Using y=4x-4 as the regression curve
Table. The absolute residuals
employing the y=4x-4 regression
model
y
ypredicted
|ε| = |y - ypredicted|
2.0
4.0
4.0
0.0
3.0
6.0
8.0
2.0
2.0
6.0
4.0
2.0
3.0
8.0
8.0
0.0
8
6
y
x
10
4
2
0
4

i 1
10
i
0
4
1
2
3
4
x
Figure. Regression curve for y=4x-4, y vs. x data
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Linear Regression-Criteria#2
Using y=6 as a regression curve
Table. Absolute residuals employing
the y=6 model
2.0
y
4.0
8
|ε| = |y –
ypredicted|
ypredicted
6.0
6
y
x
10
2.0
4
3.0
6.0
6.0
0.0
2.0
6.0
6.0
0.0
3.0
8.0
6.0
2.0
4

i 1
11
i
2
0
0
4
1
2
3
4
x
Figure. Regression curve for y=6, y vs. x data
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Linear Regression-Criterion#2
4

i 1
i
 4 for both regression models of y=4x-4 and y=6.
The sum of the errors has been made as small as possible, that
is 4, but the regression model is not unique.
Hence the above criterion of minimizing the sum of the absolute
value of the residuals is also a bad criterion.
4
Can you find a regression line for which
regression coefficients?
12

i 1
i
 4 and has unique
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Least Squares Criterion
The least squares criterion minimizes the sum of the square of the
residuals in the model, and also produces a unique line.
n
2
n
S r    i    yi  a0  a1 xi 
2
i 1
i 1
y
x,y
i
i
 i  yi  a0  a1 xi
x2 , y2
n
3
 i  yi  a0  a1 xi
x
13
n
x ,y
3
x1 , y1
x ,y
Figure. Linear regression of y vs. x data showing residuals at a typical point, xi .
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Finding Constants of Linear Model
n
n
2
Minimize the sum of the square of the residuals: S r    i    yi  a0  a1 xi 
To find a 0 and
a1
we minimize
Sr
with respect to
a1
i 1
2
and
i 1
a0 .
n
S r
 2  yi  a0  a1 xi  1  0
a0
i 1
n
S r
 2  yi  a0  a1 xi  xi   0
a1
i 1
giving
n
n
n
a  a x   y
i 1
0
n
1 i
i 1
n
a x  a x
i 1
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0 i
i 1
1 i
i 1
2
n
i
  yi xi
(a0  y  a1 x)
i 1
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Finding Constants of Linear Model
Solving for a 0 and
a1 
directly yields,
a1
n
n
n
i 1
i 1
i 1
n x i y i  x i  y i
n

2 
n x i   x i 
i 1
 i 1 
n
2
and
n
a0 
15
n
n
n
x y x x y
i 1
2
i
i 1
i
i 1
i
i 1
2


n xi2   xi 
i 1
 i 1 
n
n
i
i
(a0  y  a1 x)
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Example 1
As machines are used over long periods of time, the output product
can get off target. Below is the average value of how much off target a
product is getting manufactured as a function of machine use.
Table. Off target value as
a function of machine use.
16
Millimeters
Off Target,
h
30
1.10
33
1.21
34
1.25
35
1.23
39
1.30
44
1.40
45
1.42
1.4
h, (millimeters)
Hours of
Machine
Use, t
1.5
1.3
1.2
1.1
1
25
30
35
40
45
50
t,(hours)
Figure. Data points for h vs. t data
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Example 1 cont.
Regress the data to h 
will be 2mm off target.
a0  a1t
and find when the product
Table. Summation data for linear regression
With n  7
th
t
h
Hours
Millimeters
Hours2
Millimeter-Hour
30
1.10
900
33
33
1.21
1089
39.93
34
1.25
1156
42.50
35
1.23
1225
43.05
39
1.30
1521
50.70
44
1.40
1936
61.6
45
1.42
2025
63.9
260
8.91
9852
334.68
t2
a1 
7
7
7
i 1
i 1
i 1
n ti hi  ti  hi
2
7

2 
n ti   ti 
i 1
 i 1 
7334.68  260 8.91

2
79852   260 
 0.019179 mm  h
7
7

i 1
17
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Example 1 cont.
_
_
The value for a0 can then be found using a 0  h  a1 t where
7
_
h
7
h
i 1
i
 1.2729 mm
n
_
_
t
t
i 1
n
i
 37.143 hours
_
a0  h a1 t
 1.2729  0.0197937.143
 0.56050 mm  h
h  0.56050  0.019179t
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Example 1 cont.
The linear regression model is now given by h  0.56050  0.019179t
Figure. Linear regression of hours of use vs. millimeters off target.
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Example 1 cont.
Solving for when
h  2 mm yields
h  0.56050  0.19179t
2  0.56050  0.019179t
2  0.56050
t
0.019179
 75.056 hours
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Example 2
To find the longitudinal modulus of composite, the following data is
collected. Find the longitudinal modulus, E using the regression model
Table. Stress vs. Strain data
  E and the sum of the square of the
Strain
Stress
residuals.
(%)
(MPa)
21
0
0.183
306
0.36
612
0.5324
917
0.702
1223
0.867
1529
1.0244
1835
1.1774
2140
1.329
2446
1.479
2752
1.5
2767
1.56
2896
3.0E+09
Stress, σ (Pa)
0
2.0E+09
1.0E+09
0.0E+00
0
0.005
0.01
0.015
0.02
Strain, ε (m/m)
Figure. Data points for Stress vs. Strain data
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Example 2 cont.
Residual at each point is given by
 i   i  E i
The sum of the square of the residuals then is
n
S r    i2
i 1
n
   i  E i 
2
i 1
Differentiate with respect to E
n
S r
  2 i  E i ( i )  0
E
i 1
n
Therefore
E
 
i 1
n

i 1
22
i
i
2
i
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Example 2 cont.
Table. Summation data for regression model
σ
ε2
εσ
With
i
ε
1
0.0000
0.0000
0.0000
0.0000
2
1.8300×10−3
3.0600×108
3.3489×10−6
5.5998×105
3
3.6000×10−3
6.1200×108
1.2960×10−5
2.2032×106
and
4
5.3240×10−3
9.1700×108
2.8345×10−5
4.8821×106
12
5
7.0200×10−3
1.2230×109
4.9280×10−5
8.5855×106
6
8.6700×10−3
1.5290×109
7.5169×10−5
1.3256×107
7
1.0244×10−2
1.8350×109
1.0494×10−4
1.8798×107
8
1.1774×10−2
2.1400×109
1.3863×10−4
2.5196×107
9
1.3290×10−2
2.4460×109
1.7662×10−4
3.2507×107
10
1.4790×10−2
2.7520×109
2.1874×10−4
4.0702×107
11
1.5000×10−2
2.7670×109
2.2500×10−4
4.1505×107
12
1.5600×10−2
2.8960×109
2.4336×10−4
4.5178×107
1.2764×10−3
2.3337×108
12

12

i 1
2
i
 1.2764  10 3
 
i 1
i
Using
i
 2.3337  10 8
12
E
 
i 1
12
i i

i 1
2
i
2.3337 108

1.2764 103
 182.84 GPa
i 1
23
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Example 2 Results
The equation   182.84 describes the data.
Figure. Linear regression for Stress vs. Strain data
24
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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/linear_regr
ession.html
THE END
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