Example #2.)  Flip the third constraint to get initial dictionary

     z   =     0.00  +   3.00 x1  -   1.00 x2 
    ---------------------------------------------
     x3  =     8.00  -   1.00 x1              
     x4  =    10.00  -   1.00 x1  -   1.00 x2 
     x5  =     8.00  -   2.00 x1  +   1.00 x2 

Original Matrix:
   	[ 1	 0	| 1	0	0 ]
A = 	[ 1	 1	| 0	1	0 ]
	[ 2	-1	| 0	0	1 ]

scriptB = { 3, 4, 5 }

   	[ 1	0	0 ]
B =	[ 0	1	0 ]   ( equals B-inverse,
   	[ 0	0	1 ]             of course)

 ENTERING VARIABLE: x1
 LEAVING VARIABLE: x5
 NEW DICTIONARY: 

     z   =    12.00  +   0.50 x2  -   1.50 x5 
    ---------------------------------------------
     x3  =     4.00  -   0.50 x2  +   0.50 x5 
     x4  =     6.00  -   1.50 x2  +   0.50 x5 
     x1  =     4.00  +   0.50 x2  -   0.50 x5 

scriptB = { 3, 4, 1 }

   	[ 1	0	1 ]
B =	[ 0	1	1 ]
   	[ 0	0	2 ]

       	[ 1	0	-1/2 ]
B-inv =	[ 0	1	-1/2 ] (x3,x4 cols from LHS,
       	[ 0	0	 1/2 ]  x5 col pulled over from RHS)

 ENTERING VARIABLE: x2
 RATIOS (for each constraint):
 LEAVING VARIABLE: x4
 NEW DICTIONARY: 

     z   =    14.00  -   0.33 x4  -   1.33 x5 
    ---------------------------------------------
     x3  =     2.00  +   0.33 x4  +   0.33 x5 
     x2  =     4.00  -   0.67 x4  +   0.33 x5 
     x1  =     6.00  -   0.33 x4  -   0.33 x5 

scriptB = { 3, 2, 1 }

   	[ 1	 0	1 ]
B =	[ 0	 1	1 ]  (submatrix of original)
   	[ 0	-1	2 ]

       	[ 1	-1/3	-1/3 ]
B-inv =	[ 0	 2/3	-1/3 ] (x3 col from LHS,
       	[ 0	 1/3	 1/3 ]  x4,x5 cols pulled over)


 STOP:  All cj-bar <= 0 so we have reached optimality.
 
 Optimal Basis:  { 3,2,1 }
 Optimal Solution: x = ( 6, 4, 2, 0, 0 )
 Optimal Objective Value: z = 14


Example #4.) Toss in three slack variables to get initial dictionary

     z   =     0.00  +   1.00 x1  +   2.00 x2 
    ---------------------------------------------
     x3  =    40.00  -   4.00 x1  -   1.00 x2 
     x4  =    20.00  -   2.00 x1  -   1.00 x2 
     x5  =    10.00  -   1.00 x1  -   1.00 x2 

Original Matrix:
   	[ 4	 1	| 1	0	0 ]
A = 	[ 2	 1	| 0	1	0 ]
	[ 1	 1	| 0	0	1 ]

scriptB = { 3, 4, 5 }

     -1 [ 1	0	0 ]
B = B	[ 0	1	0 ]  
   	[ 0	0	1 ]   

 ENTERING VARIABLE: x1
 RATIOS (for each constraint):
 LEAVING VARIABLE: x3
 NEW DICTIONARY: 

     z   =    10.00  +   1.75 x2  -   0.25 x3 
    ---------------------------------------------
     x1  =    10.00  -   0.25 x2  -   0.25 x3 
     x4  =     0.00  -   0.50 x2  +   0.50 x3 
     x5  =     0.00  -   0.75 x2  +   0.25 x3 

scriptB = { 1, 4, 5 }

   	[ 4	0	0 ]
B =	[ 2	1	0 ]  (submatrix of original)
   	[ 1	0	1 ]

       	[ 1/4	0	0 ]
B-inv =	[-1/2	1	0 ] (x4,x5 cols from LHS,
       	[-1/4	0	1 ]  x3 col pulled over)


 ENTERING VARIABLE: x2
 RATIOS (for each constraint):
 LEAVING VARIABLE: x4
 NEW DICTIONARY:   (degenerate pivot)

     z   =    10.00  +   1.50 x3  -   3.50 x4 
    ---------------------------------------------
     x1  =    10.00  -   0.50 x3  +   0.50 x4 
     x2  =     0.00  +   1.00 x3  -   2.00 x4 
     x5  =     0.00  -   0.50 x3  +   1.50 x4 

scriptB = { 1, 2, 5 }

   	[ 4	1	0 ]
B =	[ 2	1	0 ]  (submatrix of original)
   	[ 1	1	1 ]

       	[ 1/2	-1/2	0 ]
B-inv =	[-1	2	0 ] (x5 col from LHS,
       	[ 1/2	-3/2	1 ]  x3,x4 cols pulled over)



 ENTERING VARIABLE: x3
 RATIOS (for each constraint):
 LEAVING VARIABLE: x5
 NEW DICTIONARY:  (degenerate pivot)

     z   =    10.00  +   1.00 x4  -   3.00 x5 
    ---------------------------------------------
     x1  =    10.00  -   1.00 x4  +   1.00 x5 
     x2  =     0.00  +   1.00 x4  -   2.00 x5 
     x3  =     0.00  +   3.00 x4  -   2.00 x5 

scriptB = { 1, 2, 3 }

   	[ 4	1	1 ]
B =	[ 2	1	0 ]  (submatrix of original)
   	[ 1	1	0 ]

       	[ 0	 1	-1 ]
B-inv =	[ 0	-1	 2 ] (x3 col from LHS
       	[ 1	-3	 2 ]  - basic in 3rd row -
                   x4,x5 cols pulled over from RHS)


 ENTERING VARIABLE: x4
 RATIOS (for each constraint):
 LEAVING VARIABLE: x1
 NEW DICTIONARY: 

     z   =    20.00  -   1.00 x1  -   2.00 x5 
    ---------------------------------------------
     x4  =    10.00  -   1.00 x1  +   1.00 x5 
     x2  =    10.00  -   1.00 x1  -   1.00 x5 
     x3  =    30.00  -   3.00 x1  +   1.00 x5 

scriptB = { 4, 2, 3 }

   	[ 0	1	1 ]
B =	[ 1	1	0 ]  (submatrix of original A
   	[ 0	1	0 ]  with cols indexed 4,2,3)

       	[ 0	 1	-1 ]
B-inv =	[ 0	 0	 1 ] (x3,x4 cols from LHS
       	[ 1	 0	-1 ]  - basic in 3rd row
   and 1st row, resp. - x5 col pulled over from RHS)



 STOP:  All cj-bar <= 0 so we have reached optimality.
 
 Optimal Basis:  { 4,2,3 }
 Optimal Solution: x = ( 0, 10, 30, 10, 0 )
 Optimal Objective Value: z = 20