\Opgave 3

Primal problem

   max 8 x1 + 14 x2 + 11 x3 + 4 x4 + 12 x5 + 7 x6 + 4 x7 + 13 x8 + 9 x9

   s.t 1 x1 +  1 x2 +  1 x3                                             <= 480
                              1 x4 +  1 x5 + 1 x6                       <= 400
                                                    1 x7 +  1 x8 + 1 x9 <= 230
               1 x2                +  1 x5               +  1 x8        <= 420
                       1 x3                + 1 x6                + 1 x9 <= 250

The dual problem

  min 480 y1 + 400 y2 + 230 y3 + 420 y4 + 250 y5

  s.t     y1                                     >=  8
          y1                       + y4          >= 14
          y1                                + y5 >= 11
                   y2                            >=  4
                   y2              + y4          >= 12
                   y2                       + y5 >=  7
                            y3                   >=  4
                            y3     + y4          >= 13
                            y3              + y5 >=  9

We have the dual solution

         y1 = 8
         y2 = 5
         y3 = 6
         y4 = 7
         y5 = 3

If it should become attractive to sell fresh bellies (i.e. x4 > 0)
then constraint 4 in the dual problem must be tight. Constraint 4
however looks like

                   y2 >= 4

with y2 = 5, so the right-hand side had to be increased to 5.

If it should become attractive to sell fresh picnics (i.e. x7 > 0)
then constraint 7 in the dual problem must be tight. Constraint 7
looks like

                   y3 >= 4

with y3 = 6, so the right-hand side had to be increased to 6.