Reduced cost in the Phase II of the two-phase Simplex?

My lecture slides outline how the two-phase simplex works: this table shows the end result of the phase I for the standard-form problem and the auxliary table of the phase I here. I understood until the reduced costs in the phase II (marked in red below), it needs to calculate the reduced costs again because of the changed cost function — the dummy vars will be removed at the end after reformatting the base so the reduced costs must be recalculated

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and my calculation here. I am unable to get the reduced costs $(3,3,82/7,0,0,0)$. Any help appreciated.

P.s. I know how to calculate the reduced costs in the first phase as shown here, it should be this $\bar c_j’= c_j’-c_B’B^{-1}A_j$ — perhaps misunderstanding the $c_j$ term.

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I can almost get the same result as on your slides but $\bar{c}’=[5,82/7,0,0,0]$, perhaps there is a small typo in the slide?

The basic idea is to use the formula for the reduced cost:

$$\bar c_j = c_j -\bf{c_B}B^{-1}A_j$$

where $\bar{c}_j$ means the reduced cost (not vector!) and $c_j$ denotes the terms in the minimization like $$\bf{c}’\bf{x}=c_1x_1+c_2x_2+c_3x_3+c_4x_4+c_5x_5=2x_1+3x_2+3x_3+x_4-2x_5$$

where $c_1=2$, $c_2=3$ and so on.


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