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

[ftneek]

(11-16-2023 10:50 PM)Gil Wrote:  Yes, but there are an infinity of solutions
in that special case.

My program on my HP50G EMU48 finds the following general solution:
[ '2/3' 0 '4/3' 0 0 ]
:
+µ[>0]
*
d,
with d : [ 1 1 0 0 0 ].

Try for instance with µ=10
x1: 2/3+10×1
x2: 0+ 10×1
x3: 4/3+10×0
x4: 0+10×0
x5: 0+10×0

And the minimum value remains at 14/3.

Can you share the final basic variables for the alternate solution? For example for the first solution the final basic variables are {x3,x1}. There is a 0 is c row index 2, x2 is not a basic variable, so I think there might be infinite solutions. Using same pivoting rule as before usually returns infinite solution. But there is a -1 in the column, how did you determine pivot selection? My idea was: multiply row by -1 -> negative b entry -> apply dual simplex -> arrived back at original solution...

(11-17-2023 12:19 AM)Gil Wrote:  Another case, with no max

solved
> simplex2([[2,-3,1,0,4],[5,-3,0,-1,100],[-1,-1,0,0,0]],{3,4},4,0,0)

(11-17-2023 12:25 AM)Gil Wrote:  And try this tricky case:

solved
> simplex2([[1,-1,0,0,1,0,0,1,0,0,0,3],[-3,2,9,2,0,-2,0,0,1,0,0,22],[0,1,0,0,0,-1,0,0,0,1,0,2],[4,3,5,0,1,0,-1,0,0,0,1,4],[1,0,-3,1,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,1,0]],{8,9,10,11},11,4,0)