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TABLEAU FORM: THE GENERAL CASE 233

method. Further calculations with the Simplex method show that x 1 will replace
a 4 in the basic solution. The following tableau is the result of the first iteration.

Result of Iteration 1

x 1 x 2 s 1 s 2 s 3 s 4 a 4
Basis c B 50 40 0 0 0 0 M
0 0 2 1 0 0 3 3 75
s 1
0 0 1 0 1 0 0 0 20
s 2
0 0 3 0 0 1 8 8 100
s 3
50 1 1 0 0 0 1 1 25
x 1
50 50 0 0 0 50 50 1 250
z j
0 10 0 0 0 50 M 50
c j – z j

When the artificial variable a 4 ¼ 0, we have a situation in which the basic feasible
solution contained in the tableau is also a feasible solution to the real HighTech
problem. In addition, because a 4 is an artificial variable that was added simply to
obtain an initial basic feasible solution, we can now drop its associated column from
the tableau. Indeed, whenever artificial variables are used, they can be dropped from
the tableau as soon as they have been eliminated from the basic feasible solution.
The artificial variable has served its purpose in providing us with a real, basic
feasible solution and so is no longer needed.
When artificial variables are required to obtain an initial basic feasible solution,
the iterations required to eliminate the artificial variables are referred to as phase I
of the Simplex method. When all the artificial variables have been eliminated from
the basis (in more complex problems of course there may be more than one
constraint and so more than one artificial variable), phase I is complete, and a basic
feasible solution to the real problem has been obtained. So, by dropping the column
associated with a 4 from the current tableau, we obtain the following tableau at the
end of phase I.

x 1 x 2 s 1 s 2 s 3 s 4
Basis c B 50 40 0 0 0 0
0 0 2 1 0 0 3 75
s 1
0 0 1 0 1 0 0 20
s 2
0 0 3 0 0 1 8 100
s 3
50 1 1 0 0 0 1 25
x 1
z j 50 50 0 0 0 50 1 250
c j – z j 0 10 0 0 0 50

We are now ready to begin phase II of the Simplex method. This phase simply
continues the Simplex method computations after all artificial variables have been
removed. At the next iteration, variable s 4 with c j – z j ¼ 50 is entered into the
solution and variable s 3 is eliminated. The tableau after this iteration is:

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