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EMIS 8374 Flows on Cycles Updated 18 March 2008

EMIS 8374 Flows on Cycles Updated 18 March 2008. Example Max Flow Problem. (0,2). 2. 4. (0,4). (0,5). 1. 6. (0,6). (0,6). t. s. (0,6). 3. 5. (0,7). (0,5). A Feasible Flow with v = 3. (2,2). 2. 4. (0,4). (3,5). 1. 6. (5,6). (2,6). t. s. (0,6). 3. 5. (3,7). (5,5).

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EMIS 8374 Flows on Cycles Updated 18 March 2008

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  1. EMIS 8374 Flows on CyclesUpdated 18 March 2008

  2. Example Max Flow Problem (0,2) 2 4 (0,4) (0,5) 1 6 (0,6) (0,6) t s (0,6) 3 5 (0,7) (0,5)

  3. A Feasible Flow with v = 3 (2,2) 2 4 (0,4) (3,5) 1 6 (5,6) (2,6) t s (0,6) 3 5 (3,7) (5,5)

  4. Flow on a Cycle (2,2) 2 4 (0,4) (3,5) 1 6 (5,6) (2,6) t s (0,6) 3 5 (3,7) (5,5)

  5. Reduce x23 from 5 to 3 out – in = -2 (2,2) 2 4 (0,4) (3,5) 1 6 (3,6) (2,6) t s (0,6) 3 5 (3,7) (5,5) out – in = 2

  6. Reduce x35 from 5 to 3 out – in = -2 (2,2) 2 4 (0,4) (3,5) 1 6 (3,6) (2,6) t s (0,6) 3 5 (3,7) (3,5) out – in = 0 out – in = 2

  7. Reduce x54 from 2 to 0 out – in = -2 out – in = 2 (2,2) 2 4 (0,4) (3,5) 1 6 (3,6) (0,6) t s (0,6) 3 5 (3,7) (3,5) out – in = 0 out – in = 0

  8. Reduce x42 from 2 to 0 out – in = 0 out – in = 0 (0,2) 2 4 (0,4) (3,5) 1 6 (3,6) (0,6) t s (0,6) 3 5 (3,7) (3,5) out – in = 0 out – in = 0

  9. Balanced Flow with v = 3 (0,2) 2 4 (0,4) (3,5) 1 6 (3,6) (0,6) t s (0,6) 3 5 (3,7) (3,5)

  10. General Case • Let j be a node in a cycle and consider a feasible flow x. • The cycle contains exactly one arc (i, j) going into j and exactly one arc (j, k) going out of j. • Reducing xij and xjk by the same amount leaves node j balanced.

  11. Example 2 1 3

  12. Reducing the flow on the cycle by  Claim: y is a feasible flow

  13. Flow Balance Constraints for Nodes in the Cycle

  14. Rewrite Flow Balance Constraints for Nodes in the Cycle

  15. Substitute y for x • xij = yij if (i, j) not in cycle • xij = yij +δif (i, j) in cycle

  16. Substitute y for x • xij = yij if (i, j) not in cycle • xij = yij +δif (i, j) in cycle

  17. Flow is Balanced Around the Cycle Since yij = xij for all arcs not in the cycle, y also satisfies the flow balance constraints for the nodes that are not in the cycle.

  18. Conclusion • Reducing flow around a cycle yields another feasible flow • If the cycle doesn’t contain the source or the sink, then the value (v) of the flow isn’t changed • There always an acyclic maximum flow • An undirected edge {i, j} with net capacity uij can be represented by directed arcs (i, j) and (j, i) each with capacity uij

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