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The Capacity Scaling Algorithm Implementation

Dive into the details of implementing The Capacity Scaling Algorithm with step-by-step instructions and examples for optimal results.

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The Capacity Scaling Algorithm Implementation

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  1. 15.082 and 6.855J The Capacity Scaling Algorithm

  2. The Original Costs and Node Potentials 0 0 4 2 4 7 5 2 1 1 0 6 2 3 5 0 0

  3. The Original Capacities and Supplies/Demands 5 -2 10 2 4 30 25 23 20 1 20 20 25 3 5 -7 -19

  4. Set  = 16. This begins the -scaling phase. 5 -2 10 We send flow from nodes with excess   to nodes with deficit  . 2 4 30 25 23 20 1 20 We ignore arcs with capacity  . 20 25 3 5 -7 -19

  5. Select a supply node and find the shortest paths shortest path distance 7 8 4 2 4 7 0 5 The shortest path tree is marked in bold and blue. 2 1 1 6 2 3 5 8 6

  6. Update the Node Potentials and the Reduced Costs -7 -8 4 3 2 4 7 To update a node potential, subtract the shortest path distance. 5 0 0 2 6 1 1 1 0 0 6 2 3 5 0 -6 -8

  7. Send Flow Along a Shortest Path in G(x, 16) 5 -2 10 Send flow from node 1 to node 5. 2 4 30 25 23 20 1 1 How much flow should be sent? 20 20 3 25 5 -7 -19

  8. Update the Residual Network 5 -2 10 19 units of flow were sent from node 1 to node 5. 2 4 30 23 25 4 20 1 1 20 19 1 19 3 5 6 -19 -7 0

  9. This ends the 16-scaling phase. 5 -2 10 2 4 The -scaling phase continues whene(i)  for some i. e(j)  - for some j. There is a path from i to j. 30 25 4 20 1 1 20 19 1 19 3 5 6 0 -7

  10. This begins and ends the 8-scaling phase. 5 -2 10 2 4 The -scaling phase continues whene(i)  for some i. e(j)  - for some j. There is a path from i to j. 30 25 4 20 1 1 20 19 1 19 3 5 6 0 -7

  11. This begins 4-scaling phase. 5 -2 10 2 4 What would we do if there were arcs with capacity at least 4 and negative reduced cost? 30 25 4 20 1 1 20 19 1 19 3 5 6 0 -7

  12. Select a “large excess” node and find shortest paths. -7 -8 3 2 4 0 The shortest path tree is marked in bold and blue. 0 6 1 1 1 0 0 0 0 3 5 0 -6 -8

  13. Update the Node Potentials and the Reduced Costs -11 -7 -8 0 To update a node potential, subtract the shortest path distance. 2 4 1 0 2 1 1 0 0 Note: low capacity arcs may have a negative reduced cost 4 -4 0 3 5 0 -6 -8 -12 -10

  14. Send Flow Along a Shortest Path in G(x, 4). 5 -2 10 2 4 Send flow from node 1 to node 7 30 25 4 20 1 1 20 How much flow should be sent? 19 1 19 3 5 6 0 -7

  15. Update the Residual Network 5 -2 6 2 4 4 units of flow were sent from node 1 to node 3 4 26 0 25 4 4 20 1 1 16 19 4 1 15 3 5 10 -7 0 -3

  16. This ends the 4-scaling phase. 5 -2 6 2 4 4 26 There is no node j with e(j)  -4. 25 0 4 20 1 1 16 19 4 1 15 3 5 10 -3 0

  17. Begin the 2-scaling phase 5 -2 6 There is no node j with e(j)  -4. 2 4 4 26 25 0 4 What would we do if there were arcs with capacity at least 4 and negative reduced cost? 20 1 1 16 19 4 1 15 3 5 10 -3 0

  18. Send flow along a shortest path 5 -2 6 2 4 Send flow from node 2 to node 4 4 26 25 0 4 20 1 1 16 19 4 How much flow should be sent? 1 15 3 5 10 -3 0

  19. Update the Residual Network 3 0 5 -2 4 2 4 2 units of flow were sent from node 2 to node 4 6 26 25 0 4 20 1 1 16 19 4 1 15 3 5 10 -3 0

  20. Send Flow Along a Shortest Path 3 0 4 2 4 Send flow from node 2 to node 3 6 26 25 0 4 20 1 1 16 19 4 How much flow should be sent? 1 15 3 5 10 -3 0

  21. Update the Residual Network 0 3 0 1 2 4 3 units of flow were sent from node 2 to node 3 9 26 25 0 4 20 1 1 13 19 7 1 12 3 5 13 -3 0 0

  22. This ends the 2-scaling phase. 0 0 1 2 4 Are we optimal? 9 26 25 0 4 20 1 1 13 19 7 1 12 3 5 13 0 0

  23. Begin the 1-scaling phase. 0 0 1 2 4 Saturate any arc whose capacity is at least 1 and with negative reduced cost. 9 26 25 0 4 20 1 1 13 19 7 1 12 3 5 13 reduced cost is negative 0 0

  24. Update the Residual Network 0 0 1 2 4 Send flow from node 3 to node 1. 9 26 25 -1 4 20 1 1 13 20 Note: Node 1 is now a node with deficit 7 12 3 5 13 1 0

  25. Update the Residual Network 0 0 2 2 4 1 unit of flow was sent from node 3 to node 1. 8 27 25 0 3 20 1 1 14 20 6 Is this flow optimal? 13 3 5 12 0 0

  26. The Final Optimal Flow 5 -2 10,8 2 4 30,3 25 23 20 1 20,6 20,20 25,13 3 5 -7 -19

  27. The Final Optimal Node Potentials and the Reduced Costs -7 -11 0 2 4 Flow is at lower bound. 1 0 2 1 0 0 -4 3 5 Flow is at upper bound 0 -10 -12

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