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

[ftneek]

(11-17-2023 10:47 AM)Gil Wrote:  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 :


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.

2 solutions means infinite solutions along the edge between the two vertices of the feasible region. If you want to recover either you can substitute for t=0 and t=1.