Efstratios N. Pistikopoulos

Multi-parametric Optimization and Control


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rel="nofollow" href="#fb3_img_img_87c8d631-3846-5036-9bfc-a727003b33d1.png" alt="images"/>, images be two arbitrary feasible parameters and images be given such that images. Then there exists a path images in the mp‐LP graph images such that images.

       Proof

      Consider the line segment joining images and images, i.e.

      (2.10)equation

      Based on Theorem 2.1, images, as images is convex. Setting images in the mp‐LP problem (2.2) converts the original mp‐LP problem into a parametric linear programming (p‐LP) problem. The solution of this p‐LP problem is given by a series of line segments that are connected as the union constitutes the feasible parameter space images and is convex based on Theorem 2.1. Based on Eq. (2.6), the limits of each line segment result from the violation of a currently inactive constraint for the parametric solution of that line segment, as all the active constraints are satisfied by definition. Thus, in order to move beyond these limits, the violated constraint needs to be considered as an active constraint, i.e. a step of the dual simplex algorithm needs to be performed. This results in a new active set, and by extension in a sequence of active sets that correspond to the path images in the mp‐LP graph images.

A schematic representation of the connected-graph theorem, (a) geometric interpretation of the feasible parameter space Θf, and (b) dual pivot identified in the transition between the active sets associated with the critical regions.

       Primal degeneracy: In this case, the vertex of the optimal solution of the LP is overdefined, i.e. there exist multiple sets such that(2.11) where, based on Eq. (2.3a):(2.12)

       Dual degeneracy: If there exists more than one point, which attains the optimal objective function value, then the optimal solution is not unique. Thus, there exist multiple sets with such that(2.13) where . Note that as shown in Figure 2.4, the solution of the problem does not necessarily have to be a vertex, and thus, Eq. (2.12) does not have to hold.

Image described by caption.