Example about the Reduced cost in the BigM method?
The BigM method can be seen as a special case of the TwoPhase Simplex, now instead of $y_1,y_2,...,y_n$ dummy vars to reformate the base, we use a large value $M$. It is calculated pretty similarly to the TwoPhase Simplex as you can see here for the problem. Again the fundamental idea is this reduced cost formula $\bar{\bf{c'}_j}=c_j'\bf{c'}_B B^{1}A_j$. The Bertsimas has something on the page 117 but not as explicitly calculated example as here. I hope it helps!
Related information
Example
where the $15M$ should be $15M$, a typo, ie $\bar{c}_0:=\bf{c}_B'\bf{B}^{1}b=15M$. The rest is pretty much the traditional Simplex. The key is the differences with M  you consider this very large $M$ as a number, not as a limit for the positive infinite.
Hard part
I haven't yet solved this but I got infeasible base with $[x_2,x_1,x_4]$ and not optimum situation here. Then I zwitched $x_4$ and $x_3$ but the new base $[x_2,x_1,x_2]$ is infeasible and not optimum as well. I may have done some mistake somewhere  calculating the big M terms errorsome unless some thinking mistake in my simplex.
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hhh
Updated on August 01, 2022Comments

hhh over 1 year
I want to gather examples about the reduced cost in different cases, now for the BigM method. I hope this makes the methods more accesible. So
How does the BigM method work with the below?
$$\min x_12x_2+x_4$$ $$\text{ s.t. } x_1+x_2=1$$ $$x_22x_3+3x_4=10$$ $$x_1+x_3+4x_4=4$$ $$x_1,x_2,x_3,x_4\geq 0$$