Dual Simplex Method Example Problem I have tried to solve this Linear Program:
max z = −2*x1 −  x2 
s.t.    −2*x1 +  x2 + x3 ≤ −4
           x1 + 2x2 − x3 ≤ −6 
          x1,x2,x3≥0

Choosing -6, wouldn't work unless 0 is allowed as a ratio.
Choosing -4 works but on the next iteration, row 1 (second constraint) has a negative RHS (-4) but the coefficients for all the constraints are positive (problem)
z x1   x2   x3   s1   s2 | RHS | Basic Variables 
-------------------------|-----|---------------- 
1  0    2    1    1    0 | -4  | 
0  1 -1/2 -1/2 -1/2    0 |  2  |       x1 
0  0  5/2    0  1/2    1 | -4  |       s2

I assume that this problem is infeasible in the case however I'm not sure as I am not very confident with this method. If I am wrong in my assumption could someone demonstrate, with this example, how the dual simplex method would be applied.
 A: Math overflow answer:
0 Ratios do count for this method:
Solved:
Initial Tableau:
z x1   x2   x3   s1   s2 | RHS | Basic Variables 
-------------------------|-----|---------------- 
1  2    1    0    0    0 |  0  | 
0 -2    1    1    1    0 | -4  |       s1 
0  1    2   (-1)  0    1 | -6  |       s2       <---Leaving Variable
-------------------------|-----|----------------
x  -    -    0    x    x |Ratio|

Iteration 1:
z x1   x2   x3   s1   s2 | RHS | Basic Variables 
-------------------------|-----|---------------- 
1  2    1    0    0    0 |  0  | 
0 (-1)  3    0    1    1 | -10 |       s1       <---Leaving Variable
0 -1   -2    1    0   -1 |  6  |       x3       

x -2    -    -    x    x

Iteration 2:
z x1   x2   x3   s1   s2 | RHS | Basic Variables 
-------------------------|-----|---------------- 
1  0   -5    0   -2   -2 | -20 | 
0  1   -3    0   -1   -1 |  10 |       x1 
0  0   -5    1   -1    0 |  16 |       x3       

Complete
x1 = 10
x2 = 0
x3 = 6

Optimal Primal Solution:
z = 20

