Management Science 461

Lecture 9 – Arc Routing

November 18, 2008

Changing Focus

We always assume demand and service is at the nodes …

But what if they’re along the edges?

What if the edges are more important than the nodes?

Arc Routing

Is it more important to visit every intersection, or cover every street?

What applications can you think of where arcs are more important than nodes?

What are the challenges of efficiently visiting each arc in a network?

Postman Problem

Formerly the Chinese Postman Problem, also called Route Inspection Problem

Given a network with nodes and arcs, find the shortest-distance route that traverses all arcs at least once

Don’t care which nodes to visit (or how often) Ideally, every arc is visited once (lower bound) –

is this always possible?

Graph Representation

Directed (arcs)

Non-Directed (edges)a/k/a bidirectional

Mixed (arcs and edges)

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Remember This? Try to draw the following without lifting

your pen from the paper or re-drawing any segments

The Königsberg Bridges

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Euler, L. (1736), “Solutio problematis ad geometriam situs pertinentis,” Commentarii Academiae Scientiarum Imperialis Petropolitanae 8 128-140

Euler’s answer

Calculate the degree of each node If all node degrees are

even, we can complete atour without retracingour steps

If all but two are even, canstart at one odd node, finishat the other (semi-Eulerian)

Otherwise, we have deadheading

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Formal Definition

A graph is Eulerian or unicursal if:

Undirected graph: graph is connected, all node degrees are even

Directed graph: strongly connected, degree-in equals degree-out for each node

Formal DefinitionMixed graph (combo of dir/undir)

“Every node must be incident to an even number of directed and undirected arcs. Moreover, for every subset S of all nodes V, the difference between the number of directed arcs from subset (V minus S) to subset S must be less or equal to the number of undirected arcs joining S and V minus S.”

Eiselt et al (1995), “Arc Routing Problems, Part I: The Chinese Postman Problem,” Operations Research 43(2) 231-242.

The Postman Problem

What if the graph isn’t Eulerian? Connecting two nodes of odd degree makes

them evenThere are an even number of odd-degree

nodes in any graph Algorithms to solve the Postman Problem

revolve around “matching” odd-degree nodes with minimal cost

Non-Eulerian Graphs

Guan’s algorithm (1962)

• Identify all odd-degree nodes•Find shortest paths between each pair•Select pairing of minimum cost•Augment the graph, solve Eulerian tour

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Non-Eulerian Graphs

• Identify odd degree nodes.

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Non-Eulerian Graphs

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• Pair nodes through their shortest path in the graph. Select minimum cost pair assignment.

Possible pairs:2-4 and 6-8 (3+5 = 8)2-6 and 4-8 (4+3 = 7)2-8 and 4-6 (2+6 = 8)

Non-Eulerian Graphs• Augment the graph to make it Eulerian (check node degree).• Each new arc effectively means we’reretracing our steps• Now that graph is Eulerian, we apply end-pairing to find the solution•Challenge: the matching problem is difficult and increases exponentially, making this a “hard” problem to solve.

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End-Pairing Algorithm

Hierholzer (1873)Step 1: Trace simple tour. If all edges have been

included, stop.Step 2: Find a node v on the tour that’s

connected to a node not on the tour. Form a second tour from v (do not overlap first tour)

Step 3: Merge both tours at the node v. If all edges have been traversed, stop. Otherwise go to Step 2.

End-Pairing Algorithm

Example

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End-Pairing Algorithm

Example

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Step 1: 1-2-4-1

End-Pairing Algorithm

Example

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Step 1: 1-2-4-1Step 2: v = 2 Path: 2-3-4-6-5-2Combine them 1-2-3-4-6-5-2-4-1

End-Pairing Algorithm

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Step 1: 1-2-4-1Step 2: v = 2 Path: 2-3-4-6-5-2Combine them 1-2-3-4-6-5-2-4-1Step 2: v = 6 Path: 6-5-7-6Combine them 1-2-3-4-6-5-7-6-5-2-4-1All edges visited. Stop.

Rural Postman Problem

Only a subset of the edges is mandatory Proven to be NP-Hard (Orloff, 1974) Technique: turn it into a postman problem

and solveFrederickson heuristic (1979): uses Min

Spanning Tree as a sub-problem. Sol’ns within 3% of optimal

Rural Postman Problem

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Frederickson heuristic:

1. Find a minimum spanning tree (T) to connect the components of the graph

2. Augment graph

3. Apply end-pairing

Rural Postman Problem

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4 Frederickson algorithm (1979)• Determine MST connecting connected components (T).

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Now the graph is connected… we need to:A)Make EulerianB)Find Tour

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Identify all nodes of odd degree (there are an even number of them)

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Next we find shortest paths between each pair of odd-degree nodes. We use the shortest paths to find the minimum-cost matching: match nodes to minimize total cost of the distances between them (see Excel)

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i j Paths2 3 2->36 10 6->7->108 11 8->11

12 15 12 -> 11 -> 4 -> 16 -> 1516 19 16->1917 21 17->18->21

(i, j) in Full

Rural Postman Problem

The final graph is Eulerian. Apply end-pairing algorithm.

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Review

For any arc-routing problem (bidirectional)

Step 1. If graph unconnected, connect it (min spanning tree).

Step 2. If graph is non-Eulerian, make it Eulerian (find minimum matching using shortest paths, duplicate arcs).

Step 3. Apply end-pairing algorithm to find the tour.