Computational Modeling for Engineering MECN 6040
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Transcript of Computational Modeling for Engineering MECN 6040
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COMPUTATIONAL MODELING FOR ENGINEERINGMECN 6040
Professor: Dr. Omar E. Meza [email protected]
http://facultad.bayamon.inter.edu/omezaDepartment of Mechanical Engineering
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IMPLEMENTATION OF SOLUTION METHODSGaussian elimination, LU factorization, and others
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▪ An equation of the form ax+by+c=0 or equivalently ax+by=-c is called a linear equation in x and y variables.
▪ ax+by+cz=d is a linear equation in three variables, x, y, and z.
▪ Thus, a linear equation in n variables is
a1x1+a2x2+ … +anxn = b
▪ A solution of such an equation consists of real numbers c1, c2, c3, … , cn. If you need to work more than one linear equations, a system of linear equations must be solved simultaneously.3
INTRODUCTION
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▪ For small number of equations (n ≤ 3) linear equations can be solved readily by simple techniques such as “method of elimination.”
▪ Linear algebra provides the tools to solve such systems of linear equations.
▪ Nowadays, easy access to computers makes the solution of large sets of linear algebraic equations possible and practical.
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SOLVING SMALL NUMBER OF EQUATIONS
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▪ There are many ways to solve a system of linear equations:▪ Graphical method▪ Cramer’s rule▪ Method of elimination▪ Computer methods.
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SOLVING SMALL NUMBER OF EQUATIONS
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THE GRAPHICAL METHOD
For two equations:
Solve both equations for x2:
2222121
1212111
bxaxa
bxaxa
22
21
22
212
1212
11
12
112 intercept(slope)
a
bx
a
ax
xxa
bx
a
ax
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THE GRAPHICAL METHOD
Or equate and solve for x1
12
11
22
21
12
1
22
2
12
11
22
21
22
2
12
1
1
22
2
12
11
12
11
22
21
22
21
22
21
12
11
12
112
0
aa
aa
ab
ab
aa
aa
ab
ab
x
a
b
a
bx
a
a
a
a
a
bx
a
a
a
bx
a
ax
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THE GRAPHICAL METHOD
Ill-conditioned(Slopes are too close)
No solution Infinite solutions
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12
12
21
21
x3x3x2x
rearrange 3xx
3xx2
2x1 – x2 = 3x1 + x2 = 3
One solution
THE GRAPHICAL METHOD
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2x1 – x2 = 3
2x1 – x2 = – 1
No solution
THE GRAPHICAL METHOD
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6x1 – 3x2 = 92x1 – x2 = 3
Infinite solutions
THE GRAPHICAL METHOD
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CRAMER’S RULE▪ Compute the determinant D
▪ 2 x 2 matrix
▪ 3 x 3 matrix
211222112221
1211 aaaaaa
aaD
3231
222113
3331
232112
3332
232211
333231
232221
131211
aa
aaa
aa
aaa
aa
aaa
aaa
aaa
aaa
D
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CRAMER’S RULE▪ To find xk for the following system
▪ Replace kth column of as with bs (i.e., aik bi )14
nnnnnn
nn
nn
bxaxaxa
bxaxaxa
bxaxaxa
...
...
...
2211
22222121
11212111
)
)(
ijk D(a
matrix newDx
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CRAMER’S RULE▪ 3 x 3 matrix
333231
232221
131211
aaa
aaa
aaa
D
33231
22221
11211
33
33331
23221
13111
22
33323
23222
13121
11
1
1
1
baa
baa
baa
DD
Dx
aba
aba
aba
DD
Dx
aab
aab
aab
DD
Dx
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EXAMPLE 9.3
44.05.03.01.0
67.09.15.0
01.052.03.0
321
321
321
xxx
xxx
xxx
5.03.01.0
9.115.0
152.03.0
D
8.19
44.03.01.0
67.015.0
01.052.03.01
5.29
5.044.01.0
9.167.05.0
101.03.01
9.14
5.03.044.0
9.1167.0
152.001.01
33
22
11
DD
Dx
DD
Dx
DD
Dx
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Eliminate x2
Subtract to get
2222121
1212111
bxaxa
bxaxa
2122221212112
1222122211122
baxaaxaa
baxaaxaa
21122211
1212112
21121122
2121221
2121221211211122
aaaa
babax
aaaa
babax
babaxaaxaa
ELIMINATION METHOD
Not very practical for large number (> 4) of equations
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▪ 2 x 2 matrix
3)1(2)2(3
)18)(1()2(3
4)1(2)2(3
)2(2)18(2
2
1
x
x
22
1823
21
21
xx
xx
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nnnnnn
n
n
b
b
b
x
x
x
aaa
aaa
aaa
2
1
2
1
21
22221
11211
The system can be written in a matrix format as
NAVIE GAUSS ELIMINATION
Gauss elimination is the most important algorithm to solve systems of linear equations.
It involves by combining equations to eliminate unknowns.
It involves 2 phases:
1. Forward elimination phase: reduce the set of equations to an upper triangular system.
2. Back substitution: work from the last equation up.20
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In the first step of the forward elimination phase, x1 is eliminated from all equations except the first one.
The coefficient of x1 in the first equation is called the pivot element.
The second step is to eliminate x2 from the third equation through the nth equation.
Do the same for all variables x3 to xn-
1.
The goal is to set up upper triangular matrix
The back-substitution phase starts from the last equation up, to find the values of x1, x2, …, xn.
NAVIE GAUSS ELIMINATION
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Gauss Elimination Pseudocode
Forward elimination phase
Back substitution
NAVIE GAUSS ELIMINATION
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Use Gauss elimination to solve
Carry 6 significant figuresSolution
1. Forward elimination: eliminate x1 from equation (2):
- (0.1/3)
5617.19293333.000333.7
3.193.071.0
261667.000666667.000333333.01.0
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321
321
xx
xxx
xxx
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NAVIE GAUSS ELIMINATION
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Eliminate x1 from equation (3): - (0.3/3)
After eliminating x1 from equations (2) and (3), the system becomes
Eliminate x2 from equation (3):
+ (0.190000/7.00333)
After eliminating x2 from equation (3), the system becomes
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2. Back Substitution: find the value of x3 from equation (3):
Substitute the value of x3 in equation (2) to find the value of x2:
Substitute the values of x2 and x3 in equation (1) to find the value of x1:
0000.70120.10
0843.703 x
50000.200333.7
)0000.7(293333.05617.19
5617.19)0000.7(293333.000333.7
2
2
x
x
00000.33
)0000.7(2.0)50000.2(1.085.7
85.7)0000.7(2.0)50000.2(1.03
1
1
x
x
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0det AD
PITFALLS OF ELIMINATION
METHODS Division by zero (Partially solved by
the pivoting technique) Round-off errors (Important when
large number of equation are to be solved)
Ill-conditioned systems:
Small changes in coefficients result in large changes in the solution.
When
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0000.10000.10000.1
0001.20000.30003.0
21
21
xx
xx
TECHNIQUES FOR IMPROVING SOLUTIONS
Use of more significant figures (The simplest remedy)
Pivoting (Determine the largest available coefficient in the column below the pivot element and switch rows so that the largest element is the pivot element)
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TECHNIQUES FOR IMPROVING SOLUTIONS
▪ Scaling (Divide each row by the largest element in that row)
2
000,100000,1002
21
21
xx
xxWithout Scaling:
With Scaling:
2
100002.0
21
21
xx
xx
With Scaling and Pivoting:
100002.0
2
21
21
xx
xx
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MATLAB SCRIPT FILE: NGAUSSELIM.M
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>>Enter Matrix A > [3 -0.1 -0.2; 0.1 7 -0.3; 0.3 -0.2 10]A = 3.0000 -0.1000 -0.2000 0.1000 7.0000 -0.3000 0.3000 -0.2000 10.0000
>>Enter Solution Vector B > [7.85 -19.3 71.4]
B = 7.8500 -19.3000 71.4000
S = 3.0000 -2.5000 7.0000
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Example 9.5
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MATLAB SCRIPT FILE: GAUSSELIMPPIVOT.M
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>>Enter Matrix A > [1 0 2 3; -1 2 2 -3; 0 1 1 4; 6 2 2 4]A = 1 0 2 3 -1 2 2 -3 0 1 1 4 6 2 2 4
>>Enter Solution Vector B > [1 -1 2 1]B = 1 -1 2 1S = -0.1857 0.2286 -0.1143 0.4714
14226
24110
13221
13201
4321
4321
4321
4321
xxxx
xxxx
xxxx
xxxx
Example
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It is a variation of Gauss elimination. Both methods use row operations to eliminate variables
The major difference is that what an unknown is eliminated, it is eliminated from all other equations.
Also, all rows (equations) are normalized by dividing by their pivot elements.
The elimination phase produces an identity matrix.
It does not involve the back substitution phase.
GAUSS-JORDAN METHOD
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MATLAB SCRIPT FILE: GAUSSJORDAN.M
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>>Dame la matriz aumentada
Cuantas filas tiene la matriz: 3Cuantas columnas tiene la matriz: 4fila : 1columna : 1Numero de esta fila y columna: 3fila : 2columna : 1Numero de esta fila y columna: 0.1fila : 3columna : 1Numero de esta fila y columna: 0.3 … Continuar ingresando los valores de la matriz aumentadaa = 3.0000 -0.1000 -0.2000 7.8500 0.1000 7.0000 -0.3000 -19.3000 0.3000 -0.2000 10.0000 71.4000a = 1.0000 0 0 3.0000 0 1.0000 0 -2.5000 0 0 1.0000 7.0000
resultado
71.4 10.0 0.2- 0.3
19.3- 0.3- 7.0 0.1
7.85 0.2- 0.1- 3.0
Example 9.12
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MATLAB’s Methods
To solve the system
Enter the following commands>> A=[3 -.1 -.2; .1 7 -0.3;.3 -.2 10];>> b=[7.85;-19.3;71.4];>> inv(A)*b
ans =
3.0000 -2.5000 7.0000
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