grupp: 46
namn1: Samuel Englund
namn2: 
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Grupp 46
Samuel Englund

Eftersom tangentbordet saknar tre tangenter skriver jag inte resten pa svenska, for det blir sa fult.

Uppgift 1
optimum: z=50
basis: x2=2/3, x4=1, x6=1/3


Uppgift 2
we need to find an entry basis first. An extra variable `a` is added to constraint 2 since there is a smaller-than-sign. 
The oject of minimization is 

z`=a, which is zero when 
x4=20,x6=6, x3=10. 

Just delete row 7, where a is located and edit the
original object of minimization. Simplex some more untill we`re satisfied!

optimum: z=10
basis: x4=10, x1=10, x6=10


Uppgift 3

Kind of the same as in uppgift 3, but we`ll have two extra variables which we remove one at a time.
The answer is that we can`t solve this system.


Uppgift 4

We don`t expect dhe dual to behave well, because the original problem has no solution.
The dual is unbound.


Uppgift 5

Easy! 

optimum: z=35/2
basis: x1=0, x3=5/2, x4=1/2


Uppgift 6
We introduce a new variable, x7. Reduced cost for x7 in the basis of the optimum in Uppgift5 is -1.
The minus indicates that we`re not at optimum. 

optimum: z=26
basis: x1=5/2, x6=2/3, x7=11/6