## Infeasible Assignment Problem

## Example 5.9 An Infeasible Problem

This is an example of the Infeasible Information Summary that is displayed when an infeasible problem is encountered. Consider the following problem:

Examination of this problem reveals that it is unsolvable. Consequently, PROC LP identifies it as infeasible. The following program attempts to solve it.

The results are shown in Output 5.9.1.

** Output 5.9.1 The Solution of an Infeasible Problem**

The LP Procedure

Max profit |

_rhs_ |

_type_ |

77.78 |

Number |

4 |

2 |

6 |

Number |

2 |

1 |

1 |

4 |

Note the information given in the Infeasible Information Summary for the infeasible row CONST1. It shows that the inequality row CONST1 with right-hand side 5 was found to be infeasible with activity 6. The summary also shows each variable that has a nonzero coefficient in that row and its activity level at the infeasibility. Examination of these model parameters might give you a clue as to the cause of infeasibility, such as an incorrectly entered coefficient or right-hand-side value.

The LP Procedure

2.5 |

2 |

0 |

0 |

0 |

0 |

5 |

0 |

2 |

1E-8 |

1.797693E308 |

100 |

100 |

99999999 |

100 |

120 |

The LP Procedure

x1 | BASIC | NON-NEG | 1 | 0.75 | 0 |

x2 | BASIC | NON-NEG | 1 | 1.75 | 0 |

x3 | NON-NEG | 1 | 0 | 0.5 | |

x4 | NON-NEG | 1 | 0 | 0 | |

const1 | BASIC | SLACK | 0 | -1 | 0 |

const2 | SLACK | 0 | 0 | 0.5 |

The LP Procedure

profit | OBJECTVE | . | 0 | 2.5 | . |

const1 | LE | 5 | 5 | 6 | 0 |

const2 | LE | 6 | 4 | 4 | -0.5 |

const3 | EQ | . | 9 | 9 | 0.5 |

1 | 0.75 | 0 | INFINITY |

3 | 1.75 | 0 | INFINITY |

2 | 0 | 0 | INFINITY |

4 | 0 | 0 | INFINITY |

__Infeasible models__

A linear program is infeasible if there exists no solution that satisfies all of the constraints -- in other words, if no *feasible* solution can be constructed. Since any real operation that you are modelling must remain within the constraints of reality, infeasibility most often indicates an error of some kind. Simplex-based LP software like lp_solve efficiently detects when no feasible solution is possible.

The source of infeasibility is often difficult to track down. It may stem from an error in specifying some of the constraints in your model, or from some wrong numbers in your data. It can be the result of a combination of factors, such as the demands at some customers being too high relative to the supplies at some warehouses.

Upon detecting infeasibility, LP codes typically show you the most recent infeasible solution that they have encountered. Sometimes this solution provides a good clue as to the source of infeasibility. If it fails to satisfy certain capacity constraints, for example, then you would do well to check whether the capacity is sufficient to meet the demand; perhaps a demand number has been mistyped, or an incorrect expression for the capacity has been used in the capacity constraint, or the model simply lacks any provision for coping with increasing demands. More often, unfortunately, LP codes respond to an infeasible problem by returning a meaninglessly infeasible solution, such as one that violates material balances. lp_solve is behaving also as such.

lp_solve currently doesn't provide analysis routines to detect infeasible constraints however that doesn't mean that it stops there.

A useful approach is to forestall meaningless infeasibilities by explicitly modelling those sources of infeasibility that you view as realistic. As a simple example, you could add a new "slack" variable on each capacity constraint, having a very high penalty cost. Then infeasibilities in your capacities would be signalled by positive values for these slacks at the optimal solution, rather than by a mysterious lack of feasibility in the linear program as a whole. Modelling approaches that use this technique are called sometimes "elastic programming" or "elastic filter".

So in practice, if a constraint is a < constraint, add a variable to the model and give it for that constraint a -1 coefficient for that variable. In the objective you give it a relative large cost. If a constraint is a > constraint, add a variable to the model and give it for that constraint a +1 coefficient for that variable. In the objective you give it a relative large cost. If a constraint is an equal constraint, add two variables to the model and give it for that constraint respectively a -1 and +1 coefficient for that variable. In the objective you give them a relative large cost. Or you only add one variable and give it an -infinite lower bound.

This will result in an automatic relaxation of the constraint(s) when needed (if that constraint would make the model infeasible). To make sure that these added variables only get non-zero values when the constraint is violating, the value in the objective must be relative large. Like that this variable gets a penalty cost and it will only become non-zero when really needed. Note that the signs of these objective coefficients must be positive when minimizing and negative when maximizing. Don't make these costs too big also because that introduces instabilities. If none of these added variables have a non-zero value then the model was initially feasible. When at least one is non-zero then the original model is infeasible. Note that the objective value will then not be very useful. However you could subtract the cost * value of all these variables from the objective to obtain the objective value of the relaxed model.

Note that a model can also become infeasible because of bounds set on variables. Above approach doesn't relax these.

#### Example:

min: x + y; c1: x >= 6; c2: y >= 6; c3: x + y <= 11;This model is clearly infeasible. Now introduce extra variables to locate the infeasibility:

min: x + y + 1000 e1 + 1000 e2 + 1000 e3; c1: x + e1 >= 6; c2: y + e2 >= 6; c3: x + y - e3 <= 11;The result of this model is:

Value of objective function: 1011 Actual values of the variables: x 5 y 6 e1 1 e2 0 e3 0With this simple example model, multiple solutions were possible. Here, the first constraint was relaxed since e1 is non-zero. Only this one constraint had to be relaxed to make the model feasible. The objective value of 1011 isn't saying very much. However if we subtract 1000 e1 + 1000 e2 + 1000 e3 from it, then it becomes 11 which is the value of the original objective function (x + y).

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