 If the minimum is positive then there is no feasible solution for the Phase I problem where the artificial variables are all zero.This implies that the feasible region for the original problem is empty, and so the original problem has no solution.Consider the linear program

This is represented by the (non-canonical) tableau

Introduce artificial variables "u" and "v" and objective function "W" = "u" + "v", giving a new tableau

The equation defining the original objective function is retained in anticipation of Phase II.