The tableau implementation (non-degenerate case) Tableau: (1) If c � 0 then STOP (we have an optimal BFS). (2) Let c

Tableau:

(1) If c � 0 then STOP (we have an optimal BFS).

(2) Let c_j < 0; write u = B⁻¹A_j. If u_i � 0 for all i then STOP ( inf c · x = −∞).

(3) Otherwise, find i with u_i > 0 and x_B(i)/u_i giving the least possible value.

(4) Apply row operations to change u to the unit vector (0, . . . , 0, 1, . . . , 0).

(5) Reduce the costs and GoTo (1).

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Example (easy to start)

Example (typical)

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Example (degenerate case)

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How this work

We have x_B = B⁻¹b in the zero column and B⁻¹A in the main body.

Let the m × m matrix V codes the row operations changing u = B⁻¹A_j to the unit vector.

• Then V B⁻¹ becomes the inverse of new basic matrix B⁻¹ and therefore V x_B = V B⁻¹b = B⁻¹b becomes the new BFS.

• Applying row operations to make cⁱ equal zero, we anyway modify the cost vector by a linear combination of the rows of A (which does not change optimality).

Why we write −c · x in the (0, 0) box of the tableau?

Recall that d_B = −B⁻¹A_j so u = −d^B.

We have the new BFS y = x + θ^∗d so we should have in that box −c · y = −c · x − θ^∗c· d.

During the row operations we get

−c · x − x_B(i)

u_i c_j = −c · x − θ^∗c_j is it is fine!