Methods to Search for Configurations of Distribution Networks

Abstract

The problem of finding the complete set of admissible configurations of a distribution network is considered. When solving this problem, the tools of graph theory are used to find the limit graphs. A new method for finding the complete set of limit graphs is proposed, and a number of properties of this method and its soundness are proved. The efficiencies of various methods are compared on a qualitative level, and it is shown that our method is distinguished by a considerably higher computational speed.

APPENDIX

Proof of Assertion 1. Consider component $I$ in the output array. It is a forest, because the connectedness of each tree is accounted for by construction, and the lack of cycles is guaranteed by the cloning procedure and the lack-of-cycles check (item 2.3.1). In accordance with prime initialization (Step 0), $I$ contains all consumer vertices, and each tree in a component contains a source vertex, since it is only in this case that the component is placed in the output array. Consequently, the component $I$ and the $W$ subgraph constructed based on it are configuration graphs. It is obvious that for each limit graph one can explicitly indicate the sequence of selecting descendant vertices (item 2.2) which leads to the appearance of this graph in the output array. In accordance with the assertion proved in [8] and given above (Assertion 2), only the complete set of limit graphs will be left after the absorption check. The proof of Assertion 1 is complete. $\quad \blacksquare$

Proof of Assertion 2. Let us prove this assertion for the following case:

• – Select one consumer vertex.

• – We will construct components while ignoring the vertex selected; i.e., we will skip all actions in the method that are related to it.

• – When no active vertices are left except for the one selected, we will continue constructing the components by considering the paths from the vertex selected.

Let us show that the output arrays in this case and in the component construction method coincide. Let $M$ be the maximum length of the path to the source vertex among all consumer vertices except the one selected. Consider the graph $\varGamma$ obtained from the initial graph $H$ by the following principle: a “tail” of $M$ ordinary vertices (not switching nodes) connected in series is added to the selected consumer vertex. The component construction method will give the same, up to the “tail,” results for $\varGamma$ as for $H$. On the other hand, the sequence of selecting consumer vertices in the component construction method for $\varGamma$ will be the same as in the case described.

To generalize this reasoning, it is obvious that the output array will not change if such “tails” are added to several consumer vertices, while the length of the tails can be arbitrary. Thus, each consumer vertex can be “connected” at any step of the component method. In a similar way, it can be shown that the order in which the components are selected at Step 1 is not important. The proof of Assertion 2 is complete. $\quad \blacksquare$

Proof of Assertion 3. The component $I$ of the output array is a limit graph, because it is a forest. Let us prove the assertion by showing that under the stated conditions on the branching vertices, the $W$ subgraph constructed based on $I$ will coincide with $I$. Consider a tree $T$ in $I$ and any edge of the original graph for which one of the incident vertices belongs to $T$ (we denote it by $t$) and the other does not (we denote it by $q$). Let us show that vertex $q$ is a vertex of the set $C$. If $t\in U$, then its degree is strictly greater than two: at least two edges belong to the tree plus the edge in question; that is, $t$ is a branching vertex, and it is surrounded by switching nodes. If $t\in S$, then, by the properties of the graphs considered, $q\in C$, and $t$ cannot be a consumer vertex, because its only edge belongs to $I$. By the construction of the $W$ subgraph, these vertices of the set $C$ are removed. It follows that the $W$ subgraph constructed based on $I$ coincides with $I$. The proof of Assertion 3 is complete. $\quad \blacksquare$