Simplex method in prime how to use the constrains maximise s.t. constraints.

[Gil]

As for last problem

[[ 1 -1 0 0 1 0 0 'E' 3 ]
[ -3 2 9 2 0 -2 0 'E' 22 ]
[ 0 1 0 0 0 -1 0 'E' 2 ]
[ 4 3 5 0 1 0 -1 'E' 4 ]
[ -1 0 3 -1 0 0 0 0 'Max' ]]
E for equal

There should have 2 distinct solutions :





Solution 1
Z-Max
: 6 :x1: 0 :x2: 2 :x3: 2 :x4: 0 :x5: 5 :x6: 0 :x7: 17 :

Solution 2
Z-Max
: 6 :x1: 5 :x2: 2 :x3: [ '11/3' 3.66666666667 ] :x4: 0 :x5: 0 :x6: 0 :x7: [ '121/3' 40.3333333333 ] :


So that solution Simplex
L×(solution1) + (1-L)×solution2

But again, here we have a special, so that the full solution should be :

{L×(solution1) + (1-L)×solution +
mu1× [ 1 1 '1/3' 0 0 1 '26/3' ] +
mu2 ×[ 0 1 0 0 1 1 4 ]}, with mu1& mu2 free.