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The travelling salesman problem

This guide provides a step-by-step approach to finding lower bounds for the weight of the minimum Hamiltonian cycle in the travelling salesman problem. The process involves deleting nodes and their corresponding arcs, determining the minimum connector length, and adding the two least weight arcs from the deleted node to form the lower bound. Through examples and calculations, the method is demonstrated for nodes A, B, C, D, and E, resulting in a final lower bound value.

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The travelling salesman problem

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  1. The travelling salesman problem Finding a lower bound • To find a lower bound for the weight of the minimum Hamiltonian cycle: • Choose an arbitrary node. Delete that node and all its arcs. • Find the length of the minimum connector for the remaining arcs. • Add the weights of the two least weight arcs from the deleted node to the weight of the minimum connector.

  2. 3 B A 2 4 2 5 5 5 C 4 6 6 E D The travelling salesman problem Finding a lower bound Example First, delete A and its arcs.

  3. The travelling salesman problem Finding a lower bound Weight of minimum connector = 11 Example Lower bound = 11 + 2 + 3 = 16 3 B A 2 4 2 5 5 5 C 4 6 6 E D First, delete A and its arcs. Find the weight of the minimum connector for the remaining network. Add the two least weight arcs from A to give a lower bound.

  4. 3 B A 2 4 2 5 5 5 C 4 6 6 E D The travelling salesman problem Finding a lower bound Example Now, delete B and its arcs.

  5. The travelling salesman problem Finding a lower bound Weight of minimum connector = 11 Example Lower bound = 11 + 2 + 3 = 16 3 B A 2 4 2 5 5 5 C 4 6 6 E D Now, delete B and its arcs. Find the weight of the minimum connector for the remaining network. Add the two least weight arcs from B to give a lower bound.

  6. 3 B A 2 4 2 5 5 5 C 4 6 6 E D The travelling salesman problem Finding a lower bound Example Now, delete C and its arcs.

  7. The travelling salesman problem Finding a lower bound Weight of minimum connector = 10 Example Lower bound = 10 + 2 + 4 = 16 3 B A 2 4 2 5 5 5 C 4 6 6 E D Now, delete C and its arcs. Find the weight of the minimum connector for the remaining network. Add the two least weight arcs from C to give a lower bound.

  8. 3 B A 2 4 2 5 5 5 C 4 6 6 E D The travelling salesman problem Finding a lower bound Example Now, delete D and its arcs.

  9. The travelling salesman problem Finding a lower bound Weight of minimum connector = 10 Example Lower bound = 10 + 2 + 4 = 16 3 B A 2 4 2 5 5 5 C 4 6 6 E D Now, delete D and its arcs. Find the weight of the minimum connector for the remaining network. Add the two least weight arcs from D to give a lower bound.

  10. 3 B A 2 4 2 5 5 5 C 4 6 6 E D The travelling salesman problem Finding a lower bound Example Finally, delete E and its arcs.

  11. The travelling salesman problem Finding a lower bound Weight of minimum connector = 7 Example Lower bound = 7 + 5 + 5 = 17 3 B A 2 4 2 5 5 5 C 4 6 6 E D Finally, delete E and its arcs. Find the weight of the minimum connector for the remaining network. Add the two least weight arcs from E to give a lower bound.

  12. The travelling salesperson problem Finding a lower bound The greatest lower bound is 17, by deleting vertex E. So the lower bound for the travelling salesperson problem is 17.

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