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The Held & Karp Relaxation of TSP

John E. Mitchell

Department of Mathematical SciencesRPI, Troy, NY 12180 USA

April 2019

Mitchell The Held & Karp Relaxation of TSP 1 / 18

A Lagrangian relaxation of the traveling salesman problem

Outline

1 A Lagrangian relaxation of the traveling salesman problem

2 Algorithm

3 Example

The traveling salesman problem

Let G = (V ,E) be the complete graph on n vertices, with edge weightswe for each e ∈ E .

A feasible solution to the traveling salesman problem satisfies threeconditions:

1 It is connected.2 It contains n edges.3 It contains exactly two edges incident to each vertex.

In the polyhedral approach to the TSP, we relaxed the first condition,connectivity. We then enforce this condition by selectively addingsubtour elimination constraints.

The Held & Karp relaxation

In the Held & Karp relaxation, the degree constraints are relaxed.

Thus, we require our solution to be connected and to contain n edges,but it might not have exactly two edges incident to each vertex.

For notational simplicity, let V = {1,2, . . . ,n}.

Our feasible solutions to our Lagrangian relaxations are required tosatisfy two criteria:

1 We have a spanning tree on nodes {2, . . . ,n}.2 We have exactly two edges incident to node 1.

Such a solution is called a 1-tree.

The Held & Karp relaxation

In the Held & Karp relaxation, the degree constraints are relaxed.

Thus, we require our solution to be connected and to contain n edges,but it might not have exactly two edges incident to each vertex.

For notational simplicity, let V = {1,2, . . . ,n}.

Our feasible solutions to our Lagrangian relaxations are required tosatisfy two criteria:

1 We have a spanning tree on nodes {2, . . . ,n}.2 We have exactly two edges incident to node 1.

Such a solution is called a 1-tree.

1-treesThus, our solution is connected, and has n edges:n − 2 from the spanning tree, plus two edges incident to node 1.

We relax the dual constraints for vertices 2, . . . ,n, adding them to theobjective with Lagrangian multipliers.

Let δ(v) denote the edges incident to vertex v .

Let x be the incidence vector of our solution. Thus, the Lagrangianrelaxation is

zHK (λ) := minx∑

e∈E wexe +∑n

j=2 λj(2−∑

e∈δ(j) xe)

subject to∑

e∈δ(1) xe = 2x gives a spanning tree on vertices 2, . . . ,nx binary

This is a relaxation of the TSP for any choice of λ.

Solving the Lagrangian dual

If the optimal solution to the relaxation is a tour then zHK (λ) is equal tothe length of the tour.

If the solution for a particular λ has the degree of a vertex v equal toone, the value of λv is increased to make it more attractive to useedges incident to vertex v .

Conversely, if a vertex v has degree greater than 2, we decrease thevalue of λv .

The Lagrangian relaxation

Note that we can rewrite the Lagrangian relaxation as follows:

zHK (λ) := 2n∑

λj + minx∑

(u,v)∈E (wuv − λu − λv ) xuv

subject to∑

e∈δ(1) xe = 2x gives a spanning tree

on vertices 2, . . . ,nx binary

Thus, as λv increases, the cost of using any edge incident to vertex vdecreases.

Algorithm

Outline

2 Algorithm

3 Example

Algorithm

The algorithm

1 Initialize: Choose initial values for λv , v = 2, . . . ,n. For example,can set λv = 0 for all vertices.

2 Solve a spanning tree problem: Use a greedy algorithm to findan optimal minimum spanning tree on vertices {2, . . . ,n}, withedge weights wuv − λu − λv .

3 Connect vertex 1: Choose the two shortest edges incident tovertex 1, with edge weights w1v − λv for v = 2, . . . ,n.

4 Terminate: If the current x constitutes a cycle, STOP with anoptimal solution to the TSP.

5 Update: Otherwise, update λ and loop back to Step 2.

Algorithm

The algorithm

Algorithm

The algorithm

Algorithm

The algorithm

Algorithm

The algorithm

Algorithm

Strength of the relaxation

Each iteration of the algorithm can run in polynomial time.

With certain update rules, it is a subgradient method.

The Lagrangian dual problem can be solved in polynomial time usingthe ellipsoid algorithm or an interior point cutting plane algorithm.

The optimal solution to the dual problem may not solve the travelingsalesman problem, as in the following example.

Algorithm

Exercise

Show that the optimal value of the Held-Karp 1-tree relaxation of thefollowing graph is 3, while the optimal value of the traveling salesmanproblem is 4.

other edge lengths equal to 15

Example

Outline

2 Algorithm

3 Example

Example

We initialize with each λv = 0, v = 2, . . . ,6. We update λ as follows:

λv ←− λv + 2 (2 − degree(v))

(Other update rules are possible.)

Example

Solution to the initial Lagrangian relaxationThe optimal solution to the first Lagrangian relaxation is as follows(spanning tree edges are red, two edges incident to vertex 1 are blue):

Vertices 2,3,5 each have degree 2, so their λ values are not changed.The values of λ4 and λ6 are updated. We now have:

λ2 = 0, λ3 = 0, λ4 = 2, λ5 = 0, λ6 = −2.

Example

Solution to the initial Lagrangian relaxationThe optimal solution to the first Lagrangian relaxation is as follows(spanning tree edges are red, two edges incident to vertex 1 are blue):

Vertices 2,3,5 each have degree 2, so their λ values are not changed.The values of λ4 and λ6 are updated. We now have:

λ2 = 0, λ3 = 0, λ4 = 2, λ5 = 0, λ6 = −2.

Example

Updated relaxationλ4 = 2, λ6 = −2.

Edge weights updated:

wuv ←− wuv − λu − λv

We solve the 1-tree relaxation on the following graph:

other edge lengths equal to 13, 15, or 17

Example

Solution to second Lagrangian relaxationThe optimal solution to the second Lagrangian relaxation is as follows(spanning tree edges are red, two edges incident to vertex 1 are blue):

We decrease λ4 and increase λ2:

λ2 = 2, λ3 = 0, λ4 = 0, λ5 = 0, λ6 = −2.

Example

Solution to second Lagrangian relaxationThe optimal solution to the second Lagrangian relaxation is as follows(spanning tree edges are red, two edges incident to vertex 1 are blue):

We decrease λ4 and increase λ2:

λ2 = 2, λ3 = 0, λ4 = 0, λ5 = 0, λ6 = −2.

Example

Next relaxationλ2 = 2, λ6 = −2.

Example

Solution to third relaxation

The optimal solution to the 1-tree relaxation is

Since the solution to the Lagrangian relaxation is a tour, it solves theoriginal problem.

Example

Solution to third relaxation

The optimal solution to the 1-tree relaxation is

Since the solution to the Lagrangian relaxation is a tour, it solves theoriginal problem.

the held & karp relaxation of tspeaton.math.rpi.edu/faculty/mitchell/courses/matp...a lagrangian...

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