22_Multicommodity_Flows_2

22_Multicommodity_Flows_2 - 15.082 and 6.855 Multicommodity...

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1 15.082 and 6.855 Multicommodity Flows 2
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2 On the Multicommodity Flow Problem O-D version K origin-destination pairs of nodes (s 1 , t 1 ), (s 2 , t 2 ), …, (s K , t K ) Network G = (N, A) d k = amount of flow that must be sent from s k to t k . u ij = capacity on (i,j) shared by all commodities k ij c = cost of sending 1 unit of commodity k in (i,j) k ij x = flow of commodity k in (i,j)
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3 A Linear Multicommodity Flow Problem 1 2 3 4 5 6 5 units good 1 5 units good 1 3 units good 2 3 units good 2 $5 $1 $6 $1 $1 $1 $1 u 25 = 5 u 32 = 2
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4 The Multicommodity Flow LP 5 ( , ) k k ij ij i j A k c x = - = - if if 0 k k k k ij ji k k j j d i s x x d i t otherwise i i for all ( , ) k ij ij k x u i j A 2200 0 ( , ) , k ij x i j A k K Min Supply/ demand constraints Bundle constraints
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5 Assumptions (for now) Homogeneous goods. Each unit flow of commodity k on (i,j) uses up one unit of capacity on (i,j). No congestion. Cost is linear in the flow on (i,j) until capacity is totally used up. Fractional flows . Flows are permitted to be fractional. OD pairs . Usually a commodity has a single origin and single destination.
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6 Optimality Conditions: Partial Dualization Theorem. The multicommodity flow x = (x k ) is an optimal multicommodity flow for (17) if there exists non-negative prices w = (w ij ) on the arcs so that the following is true 1. = If 0, then k ij ij ij k w x u 2. The flow x k is optimal for the k-th commodity if c k is replaced by c w,k , where = + , w k k ij ij ij c c w Recall: x k is optimal for the k-th commodity if there is no negative cost cycle in the kth residual network.
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7 Another approach: path-based approach Represent flows from s k to t k as the sum of flows in paths. The resulting LP may have an exponential number of columns Use “column generation” to solve the LP.
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8 A Linear Multicommodity Flow Problem 1 2 3 4 5 6 5 units good 1 5 units good 1 3 units good 2 3 units good 2 $5 $1 $6 $1 $1 $1 $1 u 25 = 5 u 32 = 2 P 1 = set of paths from node 1 to node 4. 1
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9 A path based formulation f(P) = flow in path P c(P) = cost of path P c(1-4) = 5 c(1-2-5-4) = 3 c(3-6) = 6 c(3-2-5-6) = 3 Minimize 5 f(1-4) + 3 f(1-2-5-4) + 6 f(3-6) + 3 f(3-2-5-6) subject to f(1-4) + f(1-2-5-4) = 5 f(3-6) + f(3-2-5-6) = 3 f(1-2-5-4) + f(3-2-5-6) u 25 = 5 f(3-2-5-6) u 32 = 2 f(P) 0 for all paths P
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10 Optimal solution for the path based version 1 2 3 4 5 6 5 units good 1 5 units good 1 3 units good 2 3 units good 2 $5 $1 $6 $1 $1 $1 $1 u 25 = 5 u 32 = 2 f(1-4) = 2 f(1-2-5-4) = 3 f(3-6) = 1 f(3-2-5-6) = 2 The path based LP can be solved using the simplex method.
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11 Master Problem General formulation for the path based version Let P k = set of directed paths from s
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22_Multicommodity_Flows_2 - 15.082 and 6.855 Multicommodity...

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