The Hypergraph Network Simplex Algorithm & Railway Optimization - PowerPoint PPT Presentation

The Hypergraph Network Simplex Algorithm & Railway Optimization Ralf Borndörfer joint work with Isabel Beckenbach and Markus Reuther 4th ISM-ZIB-IMI MODAL Workshop on Mathematical Optimization and Data Analysis Tokyo, ISM, 27.03.2019 Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 1

Directed Hypergraphs Definition A directed hypergraph is a pair 𝐼 = (𝑊, 𝒝) where 𝑊 is a finite set of vertices and 𝒝 is a family of hyperarcs . A hyperarc a ∈ 𝒝 is a pair 𝑏 = (𝑢, ℎ) of disjoint sets 𝑢, ℎ ⊆ 𝑊 sets of vertices, at least one of them non-empty; 𝑢 ⊆ 𝑊 is called the tail of 𝑏 , ℎ ⊆ 𝑊 is the head . Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 2

Graph-based Hypergraphs Definition Let 𝐸 = (𝑊, 𝐵) be a simple directed draph. A directed hypergraph based on 𝐸 is a pair 𝐼 = (𝑊, 𝒝) where 𝒝 ⊆ 2 𝐵 is a set of non-empty subsets 𝑏 ⊆ 𝐵 of vertex-disjoint arcs. A directed hypergraph based on some graph 𝐸 is called graph-based . Remark A graph-based directed hypergraph is a directed hypergraph: For 𝑏 ∈ 𝒝 let 𝑢 𝑏 ≔ {𝑤 ∈ 𝑊: ∃ 𝑤, 𝑥 ∈ 𝑏} and ℎ 𝑏 ≔ 𝑥 ∈ 𝑊: ∃ 𝑤, 𝑥 ∈ 𝑏 . Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 3

The Minimum Cost Hyperflow Problem Definition Let 𝐼 = (𝑊, 𝒝) be a directed hypergraph based on a directed graph 𝐸 , 𝑑 ∈ ℝ 𝒝 a vector of costs , and 𝑐 ∈ ℝ 𝑊 of demands s.t. 𝑐 𝑈 1 = 0. The minimum cost hyperflow problem (MCH) is the linear program min ෍ 𝑑 𝑏 𝑦 𝑏 𝑏∈𝒝 ෍ 𝑦 𝑏 − ෍ 𝑦 𝑏 = 𝑐 𝑤 ∀𝑤 ∈ 𝑊 𝑏∈𝒝:𝑤∈ℎ 𝑏 𝑏∈𝒝:𝑤∈𝑢 𝑏 𝑦 ≥ 0. A vector 𝑦 ∈ ℝ 𝒝 that is feasible for this LP is a hyperflow (in 𝐼 ) (actually a circulation). Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 4

The Minimum Cost Hyperflow Problem In contrast to the graph case, there might not exist an integral ► min cost hyperfow, even if all data is integral (see example later). Finding a minimum cost integral hyperflow is NP-hard, even if the ► hyperarcs consist of at most two arcs. If the underlying digraph 𝐸 is connected and 𝐵 ⊆ 𝒝 , i.e., all arcs ► of the underlying digraph are also hyperarcs, then ෍ 𝑦 𝑏 − ෍ 𝑦 𝑏 = 𝑐 𝑤 ∀𝑤 ∈ 𝑊 𝑏∈𝒝:𝑤∈ℎ 𝑏 𝑏∈𝒝:𝑤∈𝑢 𝑏 𝑦 ≥ 0. has a solution if and only if 𝑐 𝑈 1 = 0 . Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 5

A Hypernetwork Simplex Algorithm Earlier work R. Cambini, G. Gallo, and M. G. Scutellà: Flows on hypergraphs. Mathematical Programming 78.2, p. 195-217 (1997). However We heavily use that we work on graph-based hypergraphs. ► The algorithm for our setting is simpler and closer to the original ► network simplex. Reference I. Beckenbach: Matchings and Flows in Hypergraphs. PhD thesis, Freie Universität Berlin (2019). Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 6

Characterizing the Bases Assumption The underlying digraph 𝐸 is connected and 𝐵 ⊆ 𝒝 . Let 𝑁 ∈ 0, ±1 𝑊×𝒝 be the incidence matrix of 𝐼 . The minimum ► cost hyperflow problem can then be written as min 𝑑 𝑈 𝑦, 𝑁𝑦 = 𝑐, 𝑦 ≥ 0. If 𝐶 = {𝑏 1 , … , 𝑏 𝑙 } , then let 𝑁 ⋅𝐶 ≔ (𝑁 ⋅𝑏 1 , … , 𝑁 ⋅𝑏 𝑙 ) . ► ► 𝑠𝑙 𝑁 = 𝑊 − 1 𝐶 is a basis if and only if 𝑠𝑙 𝑁 ⋅𝐶 = 𝐶 = 𝑊 − 1 . ► If 𝐶 is a basis, then 𝐼 𝐶 ∩ 𝐵 = (𝑊, 𝐶 ∩ 𝐵) is a forest that contains ► = |𝐶 ∖ 𝐵 + 1 components. 𝑊| − 𝐶 ∩ 𝐵 = 𝑊 − 𝐶 − 𝐶 ∖ 𝐵 Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 7

Characterizing the Bases Let 𝐶 ⊆ 𝒝 be s.t. 𝐶 = 𝑊 − 1, 𝐶 1 ≔ 𝐶 ∩ 𝐵 , 𝐶 2 ≔ 𝐶 ∖ 𝐵. ► 𝐼[𝐶 1 ] is a forest with 𝐶 2 + 1 components. ► For every tree of 𝐼[𝐶 1 ], choose a root 𝑠 and denote its tree by 𝑈 𝑠 . ► Let 𝑆 be the set of all such roots. ► Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 8

Characterizing the Bases If 𝐶 is a basis and 𝑠 2 ∈ 𝑆 two roots, the system ► 1 , 𝑠 𝑁 ⋅𝐶 𝑔 = −𝑓 𝑠 1 + 𝑓 𝑠 2 , 𝑦 ≥ 0 has a unique solution; we can send 1 unit of flow from 𝑠 1 to 𝑠 2 . Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 9

Characterizing the Bases If 𝐶 is a basis and 𝑠 2 ∈ 𝑆 two roots, the system ► 1 , 𝑠 𝑁 ⋅𝐶 𝑔 = −𝑓 𝑠 1 + 𝑓 𝑠 2 , 𝑔 ≥ 0 has a unique solution; we can (in a unique way) send 1 unit of flow from 𝑠 1 to 𝑠 2 in 𝐼 𝐶 ∩ 𝐵 . The unique flow of 1 unit from an arbitrary fixed root 𝑠 ∗ to some ► other other root 𝑠 ≠ 𝑠 ∗ is called elementary . We can send 1 unit of flow from 𝑠 1 to 𝑠 2 via an arbitrary ► 1 to 𝑠 ∗ to 𝑠 intermediate root 𝑠 ∗ , i.e., from 𝑠 2 . We can also (easily) send 1 unit of flow inside of a tree. ► Any flow in 𝐼 𝐶 ∩ 𝐵 is a superposition of elementary flows and ► flows on trees. Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 10

The Elementary Hyperflow Matrix 𝑠 , 𝑠 ∈ 𝑆 , the rooted trees, 𝑠 ∗ ∈ 𝑆 a fixed root. Let 𝐶 be a basis, 𝑈 ► For every 𝑠 ∈ 𝑆 ∖ 𝑠 ∗ there is a unique hyperflow 𝑔 𝑠 in 𝑊, 𝐶 that ► transports 1 unit from 𝑠 ∗ to 𝑠 . 𝑠 𝑠∈𝑆∖{𝑠 ∗ } ∈ ℝ 𝐶×𝑆∖ 𝑠 ∗ be the elementary hyperflow matrix Let 𝐺 ≔ 𝑔 ► whose 𝑠 -th column contains this flow. 𝐺 is easily reconstructed from 𝐺|𝐶 2 = 𝐺 𝐶 2 ⋅ by recomputing the ► flows on the trees (but this takes time). Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 11

The Elementary Hyperflow Matrix Example 𝑠 2 |𝐶 2 = 1/2 𝑠 ∗ = 𝑠 1 , 𝑔 1/4 Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 12

The Elementary Hyperflow Matrix Example 𝑠 3 𝐶 2 ⋅ = 1/2 1/2 𝑠 ∗ = 𝑠 1 , 𝐺 𝐶 2 ⋅ = 𝑔 𝑠 2 , 𝑔 1/4 −1/4 Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 13

The Elementary Hyperflow Matrix Example 𝐺 𝐶 2 ⋅ = 1/2 1/2 𝐹 𝑆∖ 𝑠 1 ⋅ = 1 2 𝑠 ∗ = 𝑠 1 , −2 , 1/4 −1/4 1 − 𝑊 𝑈 𝑠 2 ∩ ℎ 𝑏 ෍ 𝑔 𝑠 2 𝑏 = 1 − 𝑊 𝑈 𝑠 2 ∩ 𝑢 𝑏 𝑏∈𝐶 − 𝑊 𝑈 𝑠 3 ∩ ℎ 𝑏 ෍ 𝑔 𝑠 2 𝑏 = 0 − 𝑊 𝑈 𝑠 3 ∩ 𝑢 𝑏 𝑏∈𝐶 Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 14

Characterizing the Bases Define 𝐹 ∈ ℤ 𝑆×𝐶 2 as ► 𝐹 𝑠𝑏 ≔ 𝑊 𝑈 𝑠 ∩ ℎ 𝑏 − 𝑊 𝑈 𝑠 ∩ 𝑢 𝑏 . Lemma −1 ► 𝐹 𝑆∖ 𝑠 ∗ ⋅ = 𝐺 𝐶 2 ⋅ 𝐶 basis ⟺ 𝐼 𝐶 1 forest of 𝐶 2 + 1 components ⋀ 𝑠𝑙 𝐺 𝐶 2 ⋅ = 𝐶 2 ► ⟺ 𝐼 𝐶 1 forest of 𝐶 2 + 1 components ⋀ 𝑠𝑙 𝐹 = 𝐶 2 . Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 15

Characterizing the Bases Example −2 0 has rank 2 ⇒ 𝐶 is a basis 𝐹 = 1 2 1 −2 Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 16

Characterizing the Bases Example −1 −1 has rank 1 ⇒ 𝐶 is not a basis 𝐹 = 2 2 −1 −1 Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 17

Elementary Hyperflow and Intersection Count Matrix 𝑠 , 𝑠 ∈ 𝑆 , the rooted trees, 𝑠 ∗ ∈ 𝑆 a fixed root. Let 𝐶 be a basis, 𝑈 ► For every 𝑠 ∈ 𝑆 ∖ 𝑠 ∗ there is a unique hyperflow 𝑔 𝑠 in 𝑊, 𝐶 that ► transports 1 unit from 𝑠 ∗ to 𝑠 . 𝑠 𝑠∈𝑆∖{𝑠 ∗ } ∈ ℝ 𝐶×𝑆∖ 𝑠 ∗ be the elementary hyperflow matrix Let 𝐺 ≔ 𝑔 ► whose 𝑠 -th column contains this flow. 𝐺 is easily reconstructed from 𝐺|𝐶 2 = 𝐺 𝐶 2 ⋅ by recomputing the ► flows on the trees. Define an intersection count matrix 𝐹 ∈ ℤ 𝑆×𝐶 2 as ► 𝐹 𝑠𝑏 ≔ 𝑊 𝑈 𝑠 ∩ ℎ 𝑏 − 𝑊 𝑈 𝑠 ∩ 𝑢 𝑏 . −1 ► 𝐹 𝑆∖ 𝑠 ∗ ⋅ = 𝐺 𝐶 2 ⋅ Hypernetwork Simplex Algorithm and Railways | 4th ISM-ZIB-IMI MODAL Workshop 18