Here is the function to be maximised: $f = x_1 + 2x_2$.
Constrains: $$-5x_1+2x_2\le 10 \\ 3x_1+2x_2\le12 \\ x_1-4x_2\le 3$$ I need to find a solution to this problem with $x_1,x_2 \in Z$ using Gomori algorithm. First I applied simplex algorithm to find the solution to the problem, where $x_1,x_2$ are not necessarily integers: Last symplex table and solution
From there, I choose to change the row of symplex table with basal variable $x_2$: $x_6 = \frac{-5}{8}+\frac{3}{16}x_3+\frac{5}{16}x_4$. Because $x_6 \ge 0 \Rightarrow 3x_3 +5x_4 \ge 10$. I express $x_3, x_4$ from LPP canonical: $$x_3 = 5x_1-2x_2+10 \\ x_4 = -3x_1-2x_2+12$$
Then $$10 \le 3(5x_1-2x_2+10) +5(-3x_1-2x_2+12) \Rightarrow x_2\le 5$$
New problem with the additional constrain rewritten in canonical form $LPP_1$: $$f=x_1+2x_2 \rightarrow \max \\ -5x_1+2x_2= 10 \\ 3x_1+2x_2=12 \\ x_1-4x_2= 3 \\ x_2+x_6 = 5$$ The last Symplex table and solution:
So, now I have $x* = (\frac{2}{3},5)^T$. The first component is non integer, so I will make changes co a Symplex table row with basal variable $x_1$:
$$x_7 = -\frac{2}{3}+\frac{1}{3}x_4 -\frac{2}{3}x_6 \Rightarrow \frac{1}{3}x_4 -\frac{2}{3}x_6 \ge \frac{2}{3}$$. Then from the $LPP_1$ I will express $x_4, x_6:$
$$x_4 = -3x_1-2x_2+12 \\ x_6 = -x_2+5$$
From there: $$2 \le (-3x_1-2x_2+12) -2(-x_2+5) \Rightarrow x_1 \le 0 \Rightarrow x_1 = 0$$ How I go from there and isn't there any mistakes in my computations?
