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EMIS 8374 Maximum Concurrent Flow Updated 3 April 2008. Maximum Concurrent Flow Problem (MCFP). Input Undirected graph G = ( V , E ) with capacity u ij for each edge { i , j } in E Upper Triangular demand matrix D where d ij is the demand for flow between vertex i and vertex j
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Maximum Concurrent Flow Problem (MCFP) • Input • Undirected graph G = (V, E) with capacity uij for each edge {i, j} in E • Upper Triangular demand matrix D where dij is the demand for flow between vertex i and vertex j • Optimization Problem • Find a feasible flow and throughput value z such that • Each vertex pair (i, j) receives zdij units of flow • The throughput z is maximized
Edge-Path Formulation: Notation • Notation: • Pij denotes the set of paths between i and j • Ep denotes the set of edges in path p • Uij denotes the set of paths that use edge {i, j} • Decision variables • fp denotes the amount of flow on path p • z denotes the value of the concurrent of flow (throughput)
Example Graph G 1 2 4 2 1 1 1 6 1 2 1 3 5
MCFP Example 1 • Example graph G with demand matrix D =
Edge-Path Formulation for Example Problem • P16 = {1, 2, 3, 4} where • E1={{1, 2}, {2, 4}, {4, 6}} • E2={{1, 2}, {2, 5}, {5, 6}} • E3= {{1, 3}, {3, 5}, {5, 6}} • E4= {{1, 3}, {3, 5}, {2, 5}, {2, 4}, {4, 6}} • U1,2 = {1, 2}, U1,3 = {3, 4} • U2,4 = {1, 4}, U2,5 = {2, 4} • U3,5 = {3, 4}, U4,6 = {1, 4}, U5,6 = {2, 3}
Edge-Path Formulation: LP Optimal Solution: Slide 8
MCFP Example 2 Example graph G with demand matrix D = Slide 9
Edge-Path Formulation for MCFP Ex. 2 P16 = {1, 2, 3, 4} where • E1={{1, 2}, {2, 4}, {4, 6}} • E2={{1, 2}, {2, 5}, {5, 6}} • E3= {{1, 3}, {3, 5}, {5, 6}} • E4= {{1, 3}, {3, 5}, {2, 5}, {2, 4}, {4, 6}} P25 = {5, 6 7} where • E5={{2, 5}} • E6={{1, 2}, {1, 3}, {3, 5}} • E7= {{2, 4}, {4, 6}, {5, 6}} Slide 10
Edge-Path Formulation for MCFP Ex. 2 U1,2 = {1, 2, 6} U1,3 = {3, 4, 6} U2,4 = {1, 4, 7} U2,5 = {2, 4, 5} U3,5 = {3, 4, 6} U4,6 = {1, 4, 7} U5,6 = {2, 3, 7} Slide 11
Edge-Path LP for MCFP Example 2 Slide 12
Upper Bounds on z d16 = 3 d25 = 2 1 2 4 2 1 1 1 6 1 2 3z 3 1 3 5 z 1 Slide 13
Upper Bounds on z d16 = 3 d25 = 2 1 2 4 2 1 1 1 6 1 2 3z + 2z 3 1 3 5 z 0.6 Slide 14
Optimal Solution for MCFP Ex. 2 • Each pair gets 60% of its demand • 1.8 units between 1 and 6 • 1.2 units between 2 and 5 Slide 15
Optimal Solution for MCFP Ex. 2 1 2 4 0.6 0.4 2 1 0.4 0.6 1 1 0.6 0.4 1 6 1 1 2 0.6 1 0.4 0.4 1 3 5 0.6 0.6 0.4 Slide 16
MCFP Example 3: Uniform Case Example graph G with uij = 1 for all edges and demand matrix D = Slide 17
Upper Bounds on z dij = 1 uij = 1 2 4 1 6 (1)(5)z 2 3 5 z 0.4 Slide 18
Upper Bounds on z dij = 1 uij = 1 2 4 1 6 (2)(4)z 3 3 5 z 0.375 Slide 19
Upper Bounds on z dij = 1 uij = 1 2 4 1 6 (4)(2)z 2 3 5 z 0.25 Slide 20
Optimal Solution for MCFP Ex. 3 0.25 2 4 1 6 3 5 Slide 21
Optimal Solution for MCFP Ex. 3 0.25 2 4 1 6 3 5 Slide 22
Optimal Solution for MCFP Ex. 3 0.25 2 4 1 6 3 5 Slide 23
Optimal Solution for MCFP Ex. 3 0.25 2 4 1 6 3 5 Slide 24
Optimal Solution for MCFP Ex. 3 (1,1) 2 4 (1,1) (0.75,1) 1 (0.75,1) 6 (0.75,1) (1,1) 3 5 (1,1) Slide 25
Residual Graph for MCFP Ex. 3 2 4 (0.25,) 1 (0.25) 6 (0.25) 3 5 Slide 26
Example Graph K2,3 dij = 1 uij = 1 2 4 (4)(1)z 2 z 0.5 z*= 3/7 1 3 5
Example Graph K2,3 dij = 1 uij = 1 2 4 (4)(1)z 2 Direct flow = 3/7 z 0.5 z*= 3/7 1 3 5
Example Graph K2,3 dij = 1 uij = 1 2 4 (4)(1)z 2 Direct flow = 3/7 z 0.5 2-i-4 flow = 1/7 z*= 3/7 odd-even-odd flow = 3/14 1 3 5
Maximum Concurrent Flow Problem (MCFP) Provides a way of finding a fair flow in a congested network Generalization of the standard s-t Maximum Flow Problem The maximum value of the concurrent flow is less than or equal to the density of the sparsest cut where the density of a cut is defined as the capacity of the cut divided by the demand across the cut Slide 30