CSC 445/545 Notes 1.2: Cycling Example

For the following linear programming problem, the pivot variable is chosen to be the one with the largest positive coefficient in the z row. After 6 pivots, the dictionary is the same as the one we started with. This results in an infinite loop.

Input file:
4 3
 10  -57   -9   -24
0 0.5 -5.5 -2.5   9
0 0.5 -1.5 -0.5   1
1 1    0    0     0 
*************** Problem     1 ***************
Phase 1: Input dictionary.
X5 =   0.00-   0.50 X1 +   5.50 X2 +   2.50 X3 -   9.00 X4 
X6 =   0.00-   0.50 X1 +   1.50 X2 +   0.50 X3 -   1.00 X4 
X7 =   1.00-   1.00 X1 +   0.00 X2 +   0.00 X3 +   0.00 X4 
-----------------------------------------------
z  =  -0.00+  10.00 X1 -  57.00 X2 -   9.00 X3 -  24.00 X4 


The initial dictionary:
X5 =   0.00-   0.50 X1 +   5.50 X2 +   2.50 X3 -   9.00 X4 
X6 =   0.00-   0.50 X1 +   1.50 X2 +   0.50 X3 -   1.00 X4 
X7 =   1.00-   1.00 X1 +   0.00 X2 +   0.00 X3 +   0.00 X4 
-----------------------------------------------
z  =  -0.00+  10.00 X1 -  57.00 X2 -   9.00 X3 -  24.00 X4 

X1  enters.   X5  leaves.  z = -0.000000 

After   1 pivot:
X1 =   0.00+  11.00 X2 +   5.00 X3 -  18.00 X4 -   2.00 X5 
X6 =   0.00-   4.00 X2 -   2.00 X3 +   8.00 X4 +   1.00 X5 
X7 =   1.00-  11.00 X2 -   5.00 X3 +  18.00 X4 +   2.00 X5 
-----------------------------------------------
z  =  -0.00+  53.00 X2 +  41.00 X3 - 204.00 X4 -  20.00 X5 

X2  enters.   X6  leaves.  z = -0.000000 

After   2 pivots:
X1 =   0.00-   0.50 X3 +   4.00 X4 +   0.75 X5 -   2.75 X6 
X2 =   0.00-   0.50 X3 +   2.00 X4 +   0.25 X5 -   0.25 X6 
X7 =   1.00+   0.50 X3 -   4.00 X4 -   0.75 X5 +   2.75 X6 
-----------------------------------------------
z  =  -0.00+  14.50 X3 -  98.00 X4 -   6.75 X5 -  13.25 X6 

X3  enters.   X1  leaves.  z = -0.000000 

After   3 pivots:
X3 =   0.00-   2.00 X1 +   8.00 X4 +   1.50 X5 -   5.50 X6 
X2 =   0.00+   1.00 X1 -   2.00 X4 -   0.50 X5 +   2.50 X6 
X7 =   1.00-   1.00 X1 +   0.00 X4 +   0.00 X5 +   0.00 X6 
-----------------------------------------------
z  =  -0.00-  29.00 X1 +  18.00 X4 +  15.00 X5 -  93.00 X6 

X4  enters.   X2  leaves.  z = -0.000000 

After   4 pivots:
X3 =   0.00+   2.00 X1 -   4.00 X2 -   0.50 X5 +   4.50 X6 
X4 =   0.00+   0.50 X1 -   0.50 X2 -   0.25 X5 +   1.25 X6 
X7 =   1.00-   1.00 X1 +   0.00 X2 +   0.00 X5 +   0.00 X6 
-----------------------------------------------
z  =  -0.00-  20.00 X1 -   9.00 X2 +  10.50 X5 -  70.50 X6 

X5  enters.   X3  leaves.  z = -0.000000 

After   5 pivots:
X5 =   0.00+   4.00 X1 -   8.00 X2 -   2.00 X3 +   9.00 X6 
X4 =   0.00-   0.50 X1 +   1.50 X2 +   0.50 X3 -   1.00 X6 
X7 =   1.00-   1.00 X1 +   0.00 X2 +   0.00 X3 +   0.00 X6 
-----------------------------------------------
z  =  -0.00+  22.00 X1 -  93.00 X2 -  21.00 X3 +  24.00 X6 

X6  enters.   X4  leaves.  z = -0.000000 

After   6 pivots:
X5 =   0.00-   0.50 X1 +   5.50 X2 +   2.50 X3 -   9.00 X4 
X6 =   0.00-   0.50 X1 +   1.50 X2 +   0.50 X3 -   1.00 X4 
X7 =   1.00-   1.00 X1 +   0.00 X2 +   0.00 X3 +   0.00 X4 
-----------------------------------------------
z  =  -0.00+  10.00 X1 -  57.00 X2 -   9.00 X3 -  24.00 X4