max 2 = 4x X2 s.t 8×1 + 2×2 5 16 x1 + X2 5s X1, X2 20 The following Simplex tableau corresponds to a basic feasible solution to the LP problem above. Row z X1 X2 51 S2 Rhs BV 0 1 -3 0 0 1 5 Z 1 0 6 0 1 -2. 6 51 2. 0 1 1 0 1 5 X2 According to this Simplex Tableau, this basic feasible solution is x1 = ?, X2 = Si= and 52 . In this basic feasible solution, varia are basic variables, and variables are nonbasie variables. To move better adjacent basic feasible solution, variable is the entering variable (i.e entering the basis), and variable is the leaving variable (i.e., leaving from t basis). The Simplex tableau for the better adjacent basic feasible solution is: Row X1 X2 51 S2 Rhs BV Z 0 1 V V z 1 0 > > X1 2 0 X2
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