aeb 6184 – production gauss seidel
DESCRIPTION
AEB 6184 – Production Gauss Seidel. Elluminate - 2. Basic Idea. Using a first-order Taylor series expansion. General Function. Take a general function Suppose we want to solve for. General Function. First Order Setup. Applying the first-order Taylor series expansion - PowerPoint PPT PresentationTRANSCRIPT
AEB 6184 –
PRODUCTI
ON
GAUSS SEID
EL
ELLUMIN
ATE -
2
BASIC IDEA
Using a first-order Taylor series expansion
0
0
0
1 0 1 0
1 0 1 0
1 00 1
x x
x x
x x
f xf x f x x x
x
f xf x f x x x
x
f x f xx x
f x
x
GENERAL FUNCTION
• Take a general function
• Suppose we want to solve for
2exp 0.50 0.75 2f x x x
2exp 0.50 0.75 2 1f x x x
GENERAL FUNCTION
FIRST ORDER SETUP
• Applying the first-order Taylor series expansion
• To find the solution we substitute for the desired value of the function
2exp 0.50 0.75 2 1f x x x
21.00 0.75 exp 0.50 0.75 2f x
x x xx
1
1
2
1 2 1 4.48172 2 2.6215
5.6021
x
f x fx
f x
x
ITERATIONS
X F(x) D f(x)
2.0000 4.4817 -5.60211
2.6215 1.6989 -3.17955
2.8413 1.0990 -2.29846
2.8844 1.0034 -2.14167
2.8860 1.0000 -2.13601
TWO EQUATIONS – TWO UNKNOWNS
• Building a two equation system
• Extending the Taylor expansion
2 21 2 1 1 2 2
2 21 2 1 1 2 2
, exp 0.50 0.75 0.25
, exp .75 0.35 0.45
f x x x x x x
g x x x x x x
1 1 2 2
1 1 2 2
1 2 1 20 0 0 01 2 1 2 1 1 2 2
1 2
1 2 1 20 0 0 01 2 1 2 1 1 2 2
1 2
, ,, ,
, ,, ,
x x x x
x x x x
f x x f x xf x x f x x x x x x
x x
g x x g x xg x x g x x x x x x
x x
MATRIX TAYLOR EXPANSION
• In Matrix form
1 1 2 2
1 1 2 2
1 2 1 2
0 0 01 21 2 1 11 2
0 0 01 2 1 2 1 21 2 2 2
1 2
, ,
, ,
, , ,,
x x x x
x x x x
f x x f x x
x xf x x x xf x x
g x x g x x g x xg x x x x
x x
JACOBIAN
• Jacobian
1 2
1 2
1 2 1 2
1 2 1 2
,
1.00 0.75 0.75 0.50
1.50 0.35 0.35 0.90
x
f x f x
x xf x g x
g x g x
x x
x x f x x x f x
x x g x x x g x
Iteration
1
0 0 01 1 1 21 2 1 2
0 0 01 22 2 1 2
1 2
, ,
,,
f x f xx x f x xx x f x x
g x xg x g xx x g x xx x
1
0 001 21 2 1 211
0 0 01 222 1 2
1 2
, ,
,,
f x f xf x xx x f x xxx
g x xxx g x g x g x xx x
• Suppose we want to find the solution for f(x) =0.95 and g(x) = 0.75, starting from x1 = 0.50 and x2 = 0.50.
10102
1.0000 0.1250 0.1250 0.9500 1.0000 0.7146
0.8086 0.4659 0.2224 0.7500 0.8086 0.3162
x
x
10102
0.7146 0.4282 0.3387 0.9500 0.8945 0.6464
0.3162 0.6786 0.0233 0.7500 0.7055 0.3921
x
x
10102
0.6464 0.3327 0.2726 0.9500 0.9444 0.9499
0.3921 0.6204 0.0944 0.7500 0.7454 0.7499
x
x
PARAMETERS OF THE COBB-DOUGLAS FUNCTION• Solving parameters
1 2
1 1 21 2
1
2 1 2
2
Y Ax x
w Ax xY Ax x
P x
w Ax x
P x
• X1=50, x2=65, P=3.00, w1=0.005, w2=0.004, Y=75.0
75.0 50 650
50 650.0050
3.00 500
50 650.004
3.00 65
A
A
A