see whether this gives you some ideas:
http://softwareengineering.stackexchange.com/questions/158315/what-does-value-propagation-mean
However, elimination might be more useful here. You can manipulate the equations so that this is easy.
3x+2y=4
2x-3y=7
9x+6y = 12
4x-6y = 14
13x = 26
x = 2
so y = -1
How to solve
3x+2y=4
2x-3y=7
By proporgation value of variable
2 answers
Propagation value of variable is a very useful iterative method for solving system of equations, especially if certain variables have constraints.
The given example is especially favourable example for the method because the coefficients are diagonal dominant.
We rewrite the equations as:
x=(4-2y)/3, and
y=-(7-2x)/3
Using an arbitrary initial value of (1,1), i.e. x0=1,y0=1, we get
a new set as
x1=(4-2)/3=2/3
y1=-(7-2)/3=-5/3
Repeating, we have the following sets:
n x y
1 2/3 -5/3
2 22/9 -17/9
3 70/27 -19/27
4 146/81 -49/81
5 422/243 -275/243
We see that the 5th iteration evaluates to (1.737, -1.132)
which approaches the corrected answer of (2,-1).
However, the amazing fact is that if we apply Aitken's acceleration to the 1st, 3rd and 5th sets of results, we get exactly (2,-1).
Aitken's acceleration is given by:
X=((xn*xn+2-xn+1²)/(xn+xn+2-2xn+1)
Same applies to y.
The given example is especially favourable example for the method because the coefficients are diagonal dominant.
We rewrite the equations as:
x=(4-2y)/3, and
y=-(7-2x)/3
Using an arbitrary initial value of (1,1), i.e. x0=1,y0=1, we get
a new set as
x1=(4-2)/3=2/3
y1=-(7-2)/3=-5/3
Repeating, we have the following sets:
n x y
1 2/3 -5/3
2 22/9 -17/9
3 70/27 -19/27
4 146/81 -49/81
5 422/243 -275/243
We see that the 5th iteration evaluates to (1.737, -1.132)
which approaches the corrected answer of (2,-1).
However, the amazing fact is that if we apply Aitken's acceleration to the 1st, 3rd and 5th sets of results, we get exactly (2,-1).
Aitken's acceleration is given by:
X=((xn*xn+2-xn+1²)/(xn+xn+2-2xn+1)
Same applies to y.
0 answers