NOTE: The initial solution has an artificial variable in row 2, but no
      column for it.


  MIN     4 X1 + 4 X2
  SUBJECT TO
         2)   X1 + X2    =    2
         3)   2 X1 + X2 >=   3
  END

 THE TABLEAU

      ROW  (BASIS)         X1        X2  SLK    3
        1 ART           4.000     4.000     0.000     0.000
        2 ART           1.000     1.000     0.000     2.000
        3 SLK    3     -2.000    -1.000     1.000    -3.000
 ART      ART          -3.000    -2.000     0.000    -5.000

       X1 ENTERS AT VALUE   2.0000     IN ROW    2 OBJ. VALUE= -8.0000


 THE TABLEAU

      ROW  (BASIS)         X1        X2  SLK    3
        1 ART           0.000     0.000     0.000    -8.000
        2       X1      1.000     1.000     0.000     2.000
        3 SLK    3      0.000     1.000     1.000     1.000
 ART      ART           0.000     0.000     0.000     0.000


 LP OPTIMUM FOUND AT STEP      1

        OBJECTIVE FUNCTION VALUE

        1)      8.000000

  VARIABLE        VALUE          REDUCED COST
        X1         2.000000          0.000000
        X2         0.000000          0.000000


       ROW   SLACK OR SURPLUS     DUAL PRICES
        2)         0.000000         -4.000000
        3)         1.000000          0.000000

 NO. ITERATIONS=       1