Two-Phase Method for Solving Infeasible Linear Programming Problems, Exams of Mathematics

The two-phase method, an approach to solve infeasible linear programming problems using the simplex algorithm. The method introduces an auxiliary problem to find a feasible initial point for the original problem. The steps to write the auxiliary problem, create tableaus, and apply the simplex algorithm.

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Pre 2010

Uploaded on 08/18/2009

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8 The two-phase method.
8.1 The two-phase method.
The simplex algorithm assumes that the initial point is feasible in the primal problem. If b is
greater than or equal to zero, then the origin is feasible. If the origin is not feasible, then it is
necessary to determine some other initial point that is feasible. It is possible to introduce an
auxiliary primal problem specifically designed to help in this task. The first phase of the two-
phase method applies the simplex algorithm to the auxiliary problem. The second phase uses the
feasible point generated in the first phase as initial point in the original problem. There is
typically a need for elementary row operations to bring the tableau into the form required by the
simplex algorithm.
8.2 The auxiliary problem.
Consider the primal problem
12
12 12
12
22
max 9 8x when 3 3
,0
xx
xxx
xx
−−
−−
.
The feasible set does not include the origin. The augmented system has the two equations
121
122
22
33
xxy
xxy
−− +=
−−+=.
Write the system in the form
1 211
1222
22
33
xxyz
xxyz
++=
++=
,
using the auxiliary non-negative variables 12
,
zz. The auxiliary problem seeks to minimize 12
zz+
without violating the previous system.
8.3 Tableaux.
If the minimizer is given by 12
0zz==, then the previous system yields a solution that is
feasible in the original problem. The tableau associated with the problem to maximize 12
zz−− is
given by
12 10102
310 1013
0000 110
−−
.
Observe how there are no basic variables. This is readily fixed by adding the first and the second
row to the last row. It is convenient to extend the tableau with one more row for this particular
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8 The two-phase method.

8.1 The two-phase method.

The simplex algorithm assumes that the initial point is feasible in the primal problem. If b is greater than or equal to zero, then the origin is feasible. If the origin is not feasible, then it is necessary to determine some other initial point that is feasible. It is possible to introduce an auxiliary primal problem specifically designed to help in this task. The first phase of the two- phase method applies the simplex algorithm to the auxiliary problem. The second phase uses the feasible point generated in the first phase as initial point in the original problem. There is typically a need for elementary row operations to bring the tableau into the form required by the simplex algorithm.

8.2 The auxiliary problem.

Consider the primal problem

1 2 1 2 1 2 1 2

max 9 8x when 3 3 , 0

x x x x x x x

The feasible set does not include the origin. The augmented system has the two equations 1 2 1 1 2 2

x x y x x y

Write the system in the form 1 2 1 1 1 2 2 2

x x y z x x y z

using the auxiliary non-negative variables z (^) 1 , z (^) 2. The auxiliary problem seeks to minimize z (^) 1 + z 2 without violating the previous system.

8.3 Tableaux.

If the minimizer is given by z (^) 1 = z 2 = 0 , then the previous system yields a solution that is feasible in the original problem. The tableau associated with the problem to maximize − z 1 (^) − z 2 is given by 1 2 1 0 1 0 2 3 1 0 1 0 1 3 0 0 0 0 1 1 0

Observe how there are no basic variables. This is readily fixed by adding the first and the second row to the last row. It is convenient to extend the tableau with one more row for this particular

step. The extended tableau is given by 1 2 1 0 1 0 2 3 1 0 1 0 1 3 0 0 0 0 1 1 0 4 3 1 1 0 0 5

8.4 The auxiliary problem and the simplex method.

The simplex algorithm is now applicable and the next tableau is given by 0 5 / 3 1 1/ 3 1 1/ 3 1 1 1/ 3 0 1/ 3 0 1/ 3 1 0 5 / 3 1 1/ 3 0 4 / 3 1

The algorithm calls for one more iteration. The next tableau is given by 0 1 3 / 5 1/ 5 3 / 5 1/ 5 3 / 5 1 0 1/ 5 2 / 5 1/ 5 2 / 5 4 / 5 0 0 0 0 1 1 0

Observe how the value of the objective has increased from − 5 to − 1 and finally 0. The auxiliary variables are both non-basic and hence zero. They have served their role and produced the feasible point x (^) 1 = 4 / 5, x 2 = 3 / 5.

8.5 The original problem and the simplex method.

It is now time to bring back the original objective and delete the auxiliary variables. The tableau is given by 0 1 3 / 5 1/ 5 3 / 5 1 0 1/ 5 2 / 5 4 / 5 9 8 0 0 0

There are again no basic variables. To fix this, add eight times the first and nine times the second row to the last row. The extended tableau is given by 0 1 3 / 5 1/ 5 3 / 5 1 0 1/ 5 2 / 5 4 / 5 9 8 0 0 0 0 0 3 2 12

This is in fact a final tableau and x 1 (^) = 4 / 5, x 2 = 3 / 5 is optimal.