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

[Gil]

And try this tricky case:

{
[[ 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


"{ :
S
 = \\GS(
\\Gl
i*
Sol 
i): with :0\\<=
\\Gl
i\\<=: 1 :\\GS
\\Gl
: 1 :
Main 
Sol 1
,as Zrow=
0
: in.Col\\1661\\166final.STEPS Var\\|>7\\166Steps\\|>15 Time\\|>8\\166s :


Z-Max
: 6 :x1: 0 :x2: 2 :x3: 2 :x4: 0 :x5: 5 :x6: 0 :x7: 17 :
Main 
Sol 2
,as Zrow=
0
: inCol\\{
[[ 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' ]] "{ :
S
 = \\GS(
\\Gl
i*
Sol 
i): with :0\\<=
\\Gl
i\\<=: 1 :\\GS
\\Gl
: 1 :
Main 
Sol 1
,as Zrow=
0
: in.Col\\1661\\166final.STEPS Var\\|>7\\166Steps\\|>15 Time\\|>8\\166s :


Z-Max
: 6 :x1: 0 :x2: 2 :x3: 2 :x4: 0 :x5: 5 :x6: 0 :x7: 17 :
Main 
Sol 2
,as Zrow=
0
: inCol\\1661\\166addit.STEPS Var\\|>7\\166Steps\\|>16 Time\\|>10\\166s :


Z-Max
: 6 :x1: 5 :x2: 2 :x3: [ '11/3' 3.66666666667 ] :x4: 0 :x5: 0 :x6: 0 :x7: [ '121/3' 40.3333333333 ] :
Or not? SIMPLEX 
Sol 
i: { [ 5 2 '11/3' 0 0 0 '121/3' ] [ 0 2 2 0 5 0 17 ] } :
+\\Gm
12
[
\\>=0
]
*
d
, 
d
: { [ 1 1 '1/3' 0 0 1 '26/3' ] [ 0 1 0 0 1 1 4 ] } :
\\->Mixed 
Solution
:
 
0\\<=
\\Gl
\\<=1

\\Gl
(
Sol 1
) 
+
 (
1-\\Gl
) *
(not? SIMPLEX 
Sol 2
): Plus :
\\Gm
12
[
\\>=0
]
*
d
, 
d
: { [ 1 1 '1/3' 0 0 1 '26/3' ] [ 0 1 0 0 1 1 4 ] } }" }.STEPS Var\\|>7\\166Steps\\|>16 Time\\|>10\\166s :


Z-Max
: 6 :x1: 5 :x2: 2 :x3: [ '11/3' 3.66666666667 ] :x4: 0 :x5: 0 :x6: 0 :x7: [ '121/3' 40.3333333333 ] :
Or not? SIMPLEX 
Sol 
i: { [ 5 2 '11/3' 0 0 0 '121/3' ] [ 0 2 2 0 5 0 17 ] } :
+\\Gm
12
[
\\>=0
]
*
d
, 
d
: { [ 1 1 '1/3' 0 0 1 '26/3' ] [ 0 1 0 0 1 1 4 ] } :
\\->Mixed 
Solution
:
 
0\\<=
\\Gl
\\<=1

\\Gl
(
Sol 1
) 
+
 (
1-\\Gl
) *
(not? SIMPLEX 
Sol 2
): Plus :
\\Gm
12
[
\\>=0
]
*
d
, 
d
: { [ 1 1 '1/3' 0 0 1 '26/3' ]
[ 0 1 0 0 1 1 4 ] } }" }

Here, 2 main solutions L(sol1) + (1-L)sol2+
mu1× [ 1 1 '1/3' 0 0 1 '26/3' ]
+mu2 ×[ 0 1 0 0 1 1 4 ], with mu1& mu2 free

It was for me quite a nightmare to solve these cases.