Past exam of the probability, statistics and options I course of the University of Oxford / 2009 / T / 1 / a

The entries in the two slack columns give the inverse basis matrixThus the ordered basic variables are $x_3,x_1$. Multiplying the original coefficient matrix and right-hand side by $B^{-1}$ gives the completed constraint rows. With the bottom row defined as reduced costs $c^T-c_B^T{B}^{-1}A$ and right-hand entry $-c_B^T{B}^{-1}b$, the complete tableau isThis bottom-right entry is the negative of the optimal objective under this row-operation convention. If a tableau records the positive objective value separately in that position, it is $26/3$; the completed constraint rows and reduced costs are the same.

Set the nonbasic variables $x_2,x_4,x_5$ to zero. The two constraint rows giveAll nonbasic reduced costs are negative, so increasing any of them would decrease the objective. The basic solution is feasible and therefore optimal.