The power of linear programming: some surprising and unexpected LPs

Abstract

Linear programming has had a tremendous impact in the modeling and solution of a great diversity of applied problems, especially in the efficient allocation of resources. As a result, this methodology forms the backbone of introductory courses in operations research. What students, and others, may not appreciate is that linear programming transcends its linear nomenclature and can be applied to an even wider range of important practical problems. The objective of this article is to present a selection, and just a selection, from this range of problems that at first blush do not seem amenable to linear programming formulation. The exposition focuses on the most basic models in these selected applications, with pointers to more elaborate formulations and extensions. Thus, our intent is to expand the modeling awareness of those first encountering linear programming. In addition, we hope this article will be of interest to those who teach linear programming and to seasoned academics and practitioners, alike.

Appendices

Appendix A: On the uniqueness of solutions to linear programs

The author in Appa (2002) presents a method which examines whether an LP has a unique optimal solution, and, in the case that the solution is not unique, it provides an alternative optimal solution. In particular, this is achieved by solving an additional LP as described below.

Consider a LP in the standard form:

$$\begin{aligned}&\max {} \begin{aligned}&\sum _{j=1}^{n} c_j x_j \\ \end{aligned} \end{aligned}$$

(66)

$$\begin{aligned}&\text {subject to} \nonumber \\&\sum _{j=1}^{n} a_{ij} x_j = b_i, \quad \forall i \end{aligned}$$

(67)

$$\begin{aligned}&x_j \ge 0, \quad \forall j. \end{aligned}$$

(68)

Let $x^* = \{x^*_1, x^*_2, \ldots , x^*_n\}$ be an optimal solution to the above LP, and let T denote the set of all variables in $x^*$ that are equal to zero. We also define parameter $d_j$ to be equal to 1 if j is in T, and to equal 0 otherwise. In order to check if $x^*$ is unique, we can solve the following LP:

$$\begin{aligned}&\max {} \begin{aligned}&\sum _{j=1}^{n} d_j x_j \\ \end{aligned} \end{aligned}$$

(69)

$$\begin{aligned}&\text {subject to} \nonumber \\&\sum _{j=1}^{n} a_{ij} x_j = b_i, \quad \forall i \end{aligned}$$

(70)

$$\begin{aligned}&\sum _{j=1}^{n} c_j x_j = \sum _{j=1}^{n} c_j x^*_j \end{aligned}$$

(71)

$$\begin{aligned}&x_j \ge 0, \quad \forall j. \end{aligned}$$

(72)

Let ${\bar{x}}= \{{\bar{x}}_1, {\bar{x}}_2, \ldots , {\bar{x}}_n\}$ be an optimal solution to the above LP. If the objective value equals zero when solved optimally (i.e., $\sum _{j=1}^{n} d_j {\bar{x}}_j = 0$), then $x^*$ is a unique optimal solution to the initial LP. On the other hand, if $\sum _{j=1}^{n} d_j {\bar{x}}_j > 0$, then ${\bar{x}}$ is an alternative optimal solution of the initial LP. In other words, what the second LP does is that it forces the variables in $x^*$ that are equal to zero to take strictly positive values if it is possible (objective function (69)), while maintaining the same objective value obtained in the initial LP [constraint (71)].

Note that, if the reduced costs are available, we can modify the definitions of T and $d_j$. In particular, we can let T denote the set of all variables in $x^*$ that are equal to zero and have a reduced cost of 0. If T is empty, then there are no multiple optimal solutions. If T is not empty, set coefficient $d_j$ equal to 1 if j is in T, and equal to 0 otherwise. Then, we solve the above LP.

Appendix B: An example of a first stage LP for AHP

For the pairwise comparison matrix in Fig. 9, the first stage LP given below emerges:

$$\begin{aligned}&\min {} \begin{aligned}&\text { } z_{12} + z_{13} + z_{23} \\ \end{aligned}\\&\text {subject to} \\&x_1 - x_2 - y_{12} = ln(2), \quad x_2 - x_1 - y_{21} = ln(1/2) \\&x_1 - x_3 - y_{13} = ln(5), \quad x_3 - x_1 - y_{31} = ln(1/5) \\&x_2 - x_3 - y_{23} = ln(2), \quad x_3 - x_2 - y_{32} = ln(1/2) \\&z_{12} \ge y_{12}, \quad z_{12} \ge y_{21} \\&z_{13} \ge y_{13}, \quad z_{13} \ge y_{31} \\&z_{23} \ge y_{23}, \quad z_{23} \ge y_{32} \\&x_1 - x_2 \ge 0, \quad x_1 = 0 \\&x_1 - x_3 \ge 0, \quad z_{ij} \ge 0, \quad \forall i,j \\&x_2 - x_3 \ge 0, \quad x_i, y_{ij} \text { unrestricted,} \quad \forall i,j. \end{aligned}$$