A =
1 2 -1 1 1 0 0 0
2 -2 3 3 0 1 0 9
1 -1 2 -1 0 0 1 6
-4 1 -4 -3 0 0 0 -15
the pivot element is the element
1 1
1 2 -1 1 1 0 0 0
0 -6 5 1 -2 1 0 9
0 -3 3 -2 -1 0 1 6
0 9 -8 1 4 0 0 -15
here is the basic elements index
1 6 7
and here is the non-basics.....
5 2 3 4
the pivot element is the element
2 3
1.0000 0.8000 0 1.2000 0.6000 0.2000 0 1.8
0 -1.2000 1.0000 0.2000 -0.4000 0.2000 0 1.8
0 0.6000 0 -2.6000 0.2000 -0.6000 1.0000 0.6
0 -0.6000 0 2.6000 0.8000 1.6000 0 -0.6
here is the basic elements index
1 3 7
and here is the non-basics.....
5 2 6 4
the pivot element is the element
3 2
1.0000 0 0 4.6667 0.3333 1.0000 -1.3333 1
0 0 1.0 -5.0000 0 -1.0000 2.0000 3
0 1.0000 0 -4.3333 0.3333 -1.0000 1.6667 1
0 0 0 0 1.0000 1.0000 1.0000 0
here is the basic elements index
1 3 2
and here is the non-basics.....
5 7 6 4
the current soultion is optimal, the solution is not unique!.....
x = 1.0000 1.0000 3.0000 0 0 0
SECOND PAHSE II
A =
1.0 0 0 14/3 1
0 0 1.0000 -5.0000 3
0 1. 0 -13/3 1
0. 0 0 97/3 -7
the current soultion is optimal
x = 1 3 1 0
A =
5 -4 13 -2 1 1 0 20
1 -1 5 -1 1 0 1 8
-6 5 -18 3 -2 0 0 -28
the pivot element is the element
1 3
0.3846 -0.3077 1.0000 -0.1538 0.0769 0.0769 0 1.5385
-0.9231 0.5385 0 -0.2308 0.6154 -0.3846 1.00 0.3077
0.9231 -0.5385 0 0.2308 -0.6154 1.3846 0 -0.308
here is the basic elements index
3 7
and here is the non-basics.....
6 2 1 4 5
the pivot element is the element
2 2
-0.1429 0 1.0000 -0.2857 0.4286 -0.1429 0.5714 1.7143
-1.7143 1.0000 0 -0.4286 1.1429 -0.7143 1.8571 0.5714
0 0 0 0 0 1.0000 1.0000 0
here is the basic elements index
3 2
and here is the non-basics.....
6 7 1 4 5
the current soultion is optimal
the solution is not unique!.....
x = 0 0.5714 1.7143 0 0 0
7 R1+R3 -0.1429 0 1.0000 -0.2857 0.4286 1.7143
-6R2+R3 -1.7143 1.0000 0 -0.4286 1.1429 0.5714
1 6 -7 1 5 0
PhaseII
Now making the redaction elements of the basic
variables equal zero
-0.1429 0 1.0 -0.2857 0.4286 1.7143
-1.7143 1.0 0 -0.4286 1.1429 0.5714
10.2855 0 0 1.5717 1.1428 8.5717
solution is optimal
REVISED Method
First step using Phase I to find initial B.F.S then
you use the REVISED Method
But in both example the problem is solve from the first Phase.