Leastsquare

INTEGRAL, LEAST SQUARE & CHARACTER OPERATIONS Budi Hartono, ST, MPM

integral

INTEGRAL CALCULATION

Integral Calculation

– the concept

a

standard mathematical technique  can be used to calculate the area, volume, etc  can be solved analytically or numerically  computer

 solve the integral problems numerically

expression:

Integral Calculation 

– the concept

Three numerical methods to solve integral problems   

Simpson method Rectangle Trapezoid

Similar idea: Total Area = sum of the partial area

Integral Calculation

– the concept

Analytical  y

Y = F(x)

B Area = ∫ F ( x ) dx A

A

x B

Numerical  Area = A1+A2+A3+…+An the more the better (accuration) , but time ??

Integral Calculation

– Simpson Method

 F ( X 0 ) + 4 × F ( X 0+1( p ) ) + 2 × F ( X 0+ 2( p ) )  B  P  Area = ∫ F ( x)dx = ×  + 4 × F ( X 0+ 3( p ) ) + ... +  3  A   2 × F ( X ( n − 2)( p ) ) + 4 × F ( X ( n −1)( p ) ) + F ( X ( n )( p ) ) Where: N = (B-A) / P; the number of sub intervals P : the width of sub interval X0 : A Xn : B

parabolic approach

Simpson Method – analytical vs. numerical 2

3 2 ( 4 x − 3 x + 2)dx = ??? ∫ 1

2

[

]

2

3 2 4 3 ( 4 x − 3 x + 2 ) dx = x − x + 2x 1 ∫

=12-2 = 10

1

Numerically, assumed that p=0.1

(4 • 13 − 3 • 12 + 2)    3 2 3 2 + 4(4 • 1.1 − 3 • 1.1 + 2) + 2((4 • 1.2 − 3 • 1.2 + 2)  = ( p / 3) × + ...   3 2 3 2 + 2(4 • 1.8 − 3 • 1.8 + 2) + 4(4 • 1.9 − 3 • 1.9 + 2)    3 2 + (4 • 2 − 3 • 2 + 2)  = 9.602

Simpson Method – error

Accuration limit • Ideally, the numerical calculation should result in convergence. • The more the area is divided, the more the calculation focus on a single fixed number Analytical Area

iteration #

Accuration limit

Analytical

L.C.

I.S

iteration #

• an 'accuration limit' must be determined to check the difference between the current result with the previous one • the limit is a number close to zero.

Accuration limit (2)

Integral _ Simpson − LastCalc <ε Integral _ Simpson if epsilon is nearly zero  there is no reason to continue the calculation  already close to the asymptotic line  further 'loop' will not significantly increase the accuration

Accuration limit (3)  Occasionally,

the limit is too far to reach

can not approach the asymptotic line  not convergent 

 use

additional limitation e.g. limit the loop calculation up to 20 recurrences

Simpson Method - flowchart Start

Integral_simpsoncurrentcalculation LastCalc  pervious calculation

Read F(x), A, B, ε

P width, Nnumber of sub classes

Integral_Simpson= P/3*SUM

Integral_Simpson=0 P=(B-A)/2

FOR I=1 TO 20

NO LastCalc = Integral_Simpson N=(B-A)/P

ABS [ Integral_Simpson-LastCalc)/ Integral_Simpson] < ε ??

P=P/2

Print Integral_Simpson

Next I FOR J=2,4, …, N-2 SUM = SUM + 2xF(A+JxP) +4xF(A+(J+1)xP)

Print "can' be found within 20 iterations"

Stop

Integral Calculation

– Rectangle B Area = ∫ F ( x ) dx A

y

Y = F(x)

A

x B

 F ( x0 ) + F ( x0+ p )  2   Area = ∫ F ( x)dx = P + F ( x0+ 2 p ) + ...  1 + F ( x  ) + F ( x ) ( n−2) p ( n −1) p  

Rectangle - numerical 2

[

]

2

3 2 4 3 ( 4 x − 3 x + 2 ) dx = x − x + 2x 1 ∫ 1

Numerically, assumed that p=0.1

(4 • 13 − 3 • 12 + 2)    3 2 3 2 + (4 • 1.1 − 3 • 1.1 + 2) + (4 • 1.2 − 3 • 1.2 + 2)   = ( p / 2) × + ...   3 2 3 2 + (4 • 1.8 − 3 • 1.8 + 2) + (4 • 1.9 − 3 • 1.9 + 2)   3 2 + (4 • 2 − 3 • 2 + 2)  = 9.075

Rectangle – the error Integral _ Re ct − LastCalc <ε Integral _ Re ct

Analytical Area

iteration #

Rectangle - flowchart Integral_rectcurrentcalculation

Start

LastCalc  pervious calculation P width, Nnumber of sub classes

Read F(x), A, B, ε Integral_Rect= P*SUM

Integral_rect=0 P=(B-A)/2

FOR I=1 TO 20

NO LastCalc = Integral_Rect N=(B-A)/P

ABS [ Integral_Rect-LastCalc)/ Integral_Rect] < ε ??

P=P/2

Print Integral_Rect

Next I FOR J=0 to N-1

SUM = SUM + F(A+JxP)

Print "can' be found within 20 iterations"

Stop

Integral Calculation

– Trapezoid B Area = ∫ F ( x ) dx A

y

Y = F(x)

A

x B

 F ( X 0 ) + 2 × F ( X 0+1( p ) ) + 2 × F ( X 0+ 2( p ) )   B P  + 2 × F ( X 0+ 3( p ) ) + ...  Area = ∫ F ( x)dx = ×   + 2 × F ( X ) + 2 × F ( X ) 2 ( n − 2 )( p ) ( n − 1 )( p ) A    + F ( X ( n )( p ) ) 

=12-2 = 10

Trapezoid 2

[

]

2

3 2 4 3 ( 4 x − 3 x + 2 ) dx = x − x + 2x 1 ∫ 1

Numerically, assumed that p=0.1

(4 • 13 − 3 • 12 + 2)    3 2 3 2 + 2(4 • 1.1 − 3 • 1.1 + 2) + 2((4 • 1.2 − 3 • 1.2 + 2)  = ( p / 2) × + ...   3 2 3 2 + 2(4 • 1.8 − 3 • 1.8 + 2) + 2(4 • 1.9 − 3 • 1.9 + 2)    3 2 + (4 • 2 − 3 • 2 + 2)  = 11.125

Trapezoid - flowchart Integral_trapcurrentcalculation

Start

LastCalc  pervious calculation P width, Nnumber of sub classes

Read F(x), A, B, ε Integral_Trap= P*SUM

Integral_trap=0 P=(B-A)/2

FOR I=1 TO 20

NO LastCalc = Integral_Trap N=(B-A)/P

ABS [ Integral_Trap-LastCalc)/ Integral_Trap] < ε ??

P=P/2

Print Integral_Trap

Next I FOR J=0 to N-1 SUM = SUM + F(A+JxP) +F(A+(J+1)xP)

Print "can' be found within 20 iterations"

Stop

Comparison of three methods  Generally

speaking:



simpson

 the most accurate  parabolic



rectangle  least accurate



trapezoid

 in-between

Least square

LEAST SQUARE

Least Square -

Introduction

 we

employ Regression Analysis to examine the relationship among quantitative variables.

 The

technique is used to predict the value of one variable (the dependent variable - y) based on the value of other variables (independent variables x1, x2,…xk.)

The Model Independent Variable -1 (x1) Independent Variable -2 (x2) Independent Variable -3

Dependent Variable BLACK BOX

(x3) Independent Variable -4 (x4) relationships ??

(y)

The Model  The

first order linear model or a simple regression model, y = A + Bx y = dependent variable x = independent variable Α = y-intercept Β = slope of the line ε = error variable

A and B are unknown, ytherefore, are estimated from the data.

Rise B = Rise/Run A

Run

x

Estimating the Coefficients  The   

estimates are determined by

drawing a sample from the population of interest, calculating sample statistics. producing a straight line that cuts into the data.

y

The question is: Which straight line fits best

   

      x

Least Squares Method The best line is the one that minimizes the sum of squared vertical differences between the points and the line.

2 2 2 Sum of squared differences (2 - 1) = (4 +- 2)(1.5 + - 3) (3.2 + - 4)2 = 6.89 2 2 2 Sum of squared differences (2 -2.5) = (4 +- 2.5) (1.5 + - 2.5) (3.2 +- 2.5)2 = 3.99

4 3 2.5 2

Let us compare two lines The second line is horizon

(2,4) 

(4,3.2)

(1,2)

(3,1.5)The

1

1

2

3

smaller the sum of squared differences the better the fit of the line to the data. 4

Least Square method y = A + Bx where:

(∑ xi ∑ yi ) ∑ xiyi − ∑ xiyi − nx y n ˆ βB1 = or 2 2 2 ( x ) x − n x ∑ ∑ 2 i i ∑ xi − n Aˆ0 = yYbar β − βˆ1-xB Xbar

n : # data

• Example 1 Relationship between odometer reading and a used car’s selling price. 





A car dealer wants to find the relationship between the odometer reading and the selling price of used cars. A random sample of 100 cars is selected, and the data. Find the regression line.

Car Odometer 1 37388 2 44758 3 45833 4 30862 5 31705 6 34010 . . . . . .

Price 5318 5061 5008 5795 5784 5359 . . .

Independent variable x Dependent variable

Least Square Method – flow chart START

(∑ xi ∑ yi ) ∑ xiyi − nx y n βˆB1-=# of data or N 2 2 2 ∑ xi − nx 2 (∑ xi ) ∑ xi − n βˆ0 = yYbar − βˆ1x- B Xbar A ∑ xiyi −

Read N

SUMX = 0; SUMY = 0 SUMXY=0; SUMXX=0

AVEX = SUMX /N AVEY = SUMY /N

FOR I=1 TO N

B = (SUMXY-N*AVEX*AVEY)/ (SUMXX-N*AVEX^2) A = AVEY - B*AVEX

SUMX =SUMX+X[I] SUMY =SUMY+Y[I] SUMXY =SUMXY+X[I]Y[I] SUMXX =SUMXX+X[I]X[I]

PRINT "Y=" & A &"X+" B

NEXT I STOP



Using the computer

Tools > Data analysis > Regression > [Shade the y range and the x range] > SUMMARY OUTPUT

5500 5000 4500 19000

29000

ˆ = 6533−.0312 y x

ANOVA df Regression Residual Total

Price

Regression Statistics Multiple R 0.806308 R Square 0.650132 Adjusted R Square 0.646562 Standard Error 151.5688 Observations 100

6000

1 98 99

39000 Odometer

SS MS F Significance F 4183528 4183528 182.1056 4.4435E-24 2251362 22973.09 6434890

CoefficientsStandard Error t Stat P-value Intercept 6533.383 84.51232 77.30687 1.22E-89 Odometer -0.03116 0.002309 -13.4947 4.44E-24

49000

6533

Price

6000 5500 5000 4500

0

No data

19000

29000

39000

49000

Odometer

ˆ = 6533−.0312 y x The intercept is b0 = 6533.

This is the slope of the line. For each additional mile on the odometer, the price decreases by an average of $0.031 Do not interpret the intercept as the “Price of cars that have not been driven”

character

CHARACTER OPERATIONS

character -

Inverting the sentence

first assignment

counter: I from I=1 to N

tnemngissa tsrif

counter: J from J=N to 1

inversion - flowchart start

counter: I counter: J

Read SENTENCE J=N

for I = 1 to N

INVERT[J] = SENTENCE[I] PRINT INVERT J = J-1 STOP NEXT I