Quick question regarding the Primal Simplex Method:
How do you use the Primal Simplex Method when you have to start with a specific basis?
Suppose we wish to use the Primal Simplex method on the following minimization problem in standard form: $$\text{min}\quad 2x_1+2x_2+2x_3$$ $$\text{s.t. }\qquad\qquad\qquad\qquad\qquad\qquad$$ $$2x_1+4x_3+s_1=4$$ $$2x_2+2x_3+s_2=3$$ $$3x_1+x_2+x_3−s_3=8$$ $$x_i,s_i\ge0,\forall i$$ We have to start with the basic variables $BV=(x_1,s_2,s_3) $, and I have a hard time with the fact that usually I would start with the basis consisting of our slack variables, however, in this example this isn't the case. A walkthrough of understanding this would be greatly appreciated.
Consider the original standardized model (with the fixed third constraint, $a_3$ is analogous to $s_3$), $$\text{min}\quad z=2x_1+2x_2+2x_3+Ma_3$$ $$\text{s.t. }\qquad\qquad\qquad\qquad\qquad\qquad$$ $$2x_1+4x_3+s_1=4$$ $$2x_2+2x_3+s_2=3$$ $$3x_1+x_2+x_3−e_3+a_3=8$$
with the Basic Variables $=\{x_1, s_2,a_3\}$. This means $x_2=x_3=s_1=e_1=0$. Let's consider the first constraint: $$2x_1+4x_3+s_1=4$$ $$2x_1+0+0=4$$ $$x_1=2$$
Then let's consider the second constraint: $$2x_2+2x_3+s_2=3$$ $$0+0+s_2=3$$ $$s_2=3$$
The third constraint: $$3x_1+x_2+x_3-e_3+a_3=8$$ $$6+0+0-0+a_3=8$$ $$a_3=2$$
We can the put this into the Simplex tableau as such: \begin{array} {|c|c|} \hline BV & z & x_1 & x_2 & x_3 & e_3 & s_1 & s_2 & a_3 & RHS \\ \hline z & 1 & 0 & & & & & 0 & 0 & 2M+4 \\ \hline x_1 & 0 & 1 & & & & & 0 & 0 & 2 \\ s_2 & 0 & 0 & & & & & 1 & 0 & 3 \\ a_3 & 0 & 0 & & & & & 0 & 1 & 2 \\ \hline \end{array}
If we match our basic variables to our original equations like so: $$\text{min}\quad z-2x_1-2x_2-2x_3+0+0+0-Ma_3=0$$ $$x_1\longrightarrow 2x_1+0x_2+4x_3+s_1+0+0+0=4$$ $$s_2\longrightarrow 0x_1+2x_2+2x_3+0+s_2+0+0=3$$ $$a_3\longrightarrow 3x_1+x_2+x_3+0+0−e_3+a_3=8$$
We can do Gaussian Elimination from this to get a system of equations that are the reduced costs and coefficients of the basis we are trying to achieve to get the following tableau:
\begin{array} {|c|c|} \hline BV & z & x_1 & x_2 & x_3 & s_1 & e_3 & s_2 & a_3 & RHS \\ \hline z & 1 & 0 & M-2 & -5M+2 & -\frac{3}{2}M & -M & 0 & 0 & 2M+4 \\ \hline x_1 & 0 & 1 & 0 & 2 & \frac{1}{2} & 0 & 0 & 0 & 2 \\ s_2 & 0 & 0 & 2 & 2 & 0 & 0 & 1 & 0 & 3 \\ a_3 & 0 & 0 & 1 & -5 & -\frac{3}{2} & -1 & 0 & 1 & 2 \\ \hline \end{array}
From here, since $M$ is the "largest" number in $\Bbb R$, we would proceed to pivot the $x_2$ column with the $s_2$ row with the Simplex Method in tableau form.