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1 Outline LP formulation of minimal cost flow problem useful results from shortest path problem optimality condition residual network potentials of nodes reduced costs of arcs (with respect to a given set of potentials) the optimality conditions for networks network simplex and existence of negative cycle transportation simplex method transportation by eliminating negative cycles determining an initial feasible solution for a minimal flow problem

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2 Minimum Cost Flow Models G: (connected) network, i.e., G = (N, A) N: the collection of nodes in G, i.e., N = {j} A: the collection of arcs in G, i.e., A = {(i, j)} c ij : the cost of arc (i, j) u ij : the capacity of arc (i, j) b(i): the amount of flow out of node i b(i) > (resp. <) 0 for out (resp. in) flow Assume direct arcs. How to deal with an undirected arc?

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3 Minimum Cost Flow Models N = {1, 2, 3, 4, 5, 6} A = {(1, 2), (1, 3), (2, 3), (2, 4), (2, 5), (3, 5), (4, 6), (5, 4), (5, 6)} b(1) = 9, b(6) = -9, and b(i) = 0 for i = 2 to 5 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (2, 3) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (c ij, u ij )

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4 A Minimum Cost Flow LP Model 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (2, 3) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9

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5 Useful Results from Shortest-Path Algorithms

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6 optimality condition for shortest distances residual network potentials of nodes reduced costs of arcs with respect to a given set of potentials

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7 Optimality Condition for Shortest Distances Theorem. Let G = (N, A) be a directed graph; d( ) be a function defined on N. Then d( ) is the shortest distance of node ( ) from node 1 iff d(1) = 0; d(i) is the length of a path from node 1 to node i; and d(j) d(i) + c ij for all (i, j) A.

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8 Optimality Condition for Shortest Distances LHS figure: {d(i)} = {0, 2, 3, 7} for all i, d(i) being shortest distance of node i from node 1 satisfying the three conditions RHS figure: {d(i)} = {0, 2, 3, 8} d(4) = the distance of a path from node 1 to node 4 d(3) + c 34 < d(4), i.e., this set cannot be the collection of shortest distances from nodes 1 2 3 4 2 7 1 6 4 d(1) = 0 d(2) = 2 d(3) = 3 d(4) = 7 1 2 3 4 2 7 1 6 4 d(1) = 0 d(2) = 2 d(3) = 3 d(4) = 8

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(-8,3) (-2, 3) (6, 4) (8, 2) (2, 4) 9 Residual Network (6,4) (8,5) s t 1 (2,7) Figure 1. Costs and Capacities of Arcs Figure 3. The Residual Network Corresponding to Figure 1 t (6, 4) (8, 5) s 1 (2, 7) 3 3 3 3 Figure 2. Flows on arcs 0 s t 1 Figure 4. The Residual Network Corresponding to Figure 2 s t 1

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0 0 3 3 3 (6, 1) (8, 5) (2, 7) (-6, 3) actual meaning of sending flow in cycle s-t-1-s (Figure 4) a different way to send flow from s to t flow along t-1-s reduction of flow in s-1-t 10 A Cycle in a Residual Network (-8,3) (-2, 3) (6, 4) (8, 2) (2, 4) Figure 4. The Residual Network Corresponding to Figure 2 s t 1 Figure 5. The New Residual Network After Adding a Flow of 3 units in cycle s-t-1-s s t 1 s t 1 Figure 6. Flows on arcs

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11 Potentials of Nodes and the Corresponding Reduced Costs of Arcs first the idea call any arbitrary set of numbers { i }, one for a each node, a set of potentials of nodes define the reduced cost with respect to this set of potentials { i } by interesting properties for this set of reduced costs

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12 Potentials of Nodes and the Corresponding Reduced Costs of Arcs C(P) be the total cost of path P C (P) be the total reduced cost of path P i = potential of node i P = a path from node s to node t then C (P) = C(P) s + t

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13 Potentials of Nodes and the Corresponding Reduced Costs of Arcs C(P) be the total cost of path P C (P) be the total reduced cost of path P i = potential of node i P = a path from node s to node t, then C (P) = C(P) s + t a shortest path from s to t for C( ) is also a shortest path from s to t for C ( ) 1 2 3 4 5 6 8 3 2 3 7 3 4 5 6 1 = 3 2 = -5 3 = 2 4 = 6 5 = -3 6 = 5 1 2 3 4 5 6 0 2 4 10 18 -2 13 4 14 C(P)C(P) C(P)C(P)

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Potentials of Nodes and the Corresponding Reduced Costs of Arcs C(P) be the total cost of path P C (P) be the total reduced cost of path P i = potential of node i P = a path from node s to node t, then if j = -d(j), then for all (i, j) A arcs with forming a tree 1 2 3 4 5 6 8 3 2 3 7 3 4 5 6 1 = 3 2 = -5 3 = 2 4 = 6 5 = -3 6 = 5 j = -d(j) 1 2 3 4 5 6 1 = 0 2 = -8 3 = -3 4 = -10 5 = -6 6 = -12 0 0 4 8 5 0 0 3 0

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15 Spanning Tree and BFS in a minimal cost flow network problem, a BFS a spanning tree of the network # of constraints = # of nodes # of basic variables = # of nodes - 1

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16 The optimality Conditions for Networks

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17 The Optimality Conditions for Networks Theorem 1. (Theorem 9.1, Negative Cycle Optimality Conditions) A feasible solution x * is optimal iff there is no negative cost (direct) cycle in the corresponding residual network G(x * ). Theorem 2. (Theorem 9.3, Reduced Cost Optimality Conditions) A feasible solution x * is optimal iff there exists dual variable such that for every arc (i, j) in the residual network G(x * ). Theorem 3. (Theorem 9.4, Complementary Slackness Optimality Conditions) A feasible solution x * is optimal iff there exists dual variables such that if then = 0; if 0 < < u ij, then if then = u ij.

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18 Ideas for Theorem 1 a cycle in a residual network: a different way to send flow across a network existing of a negative cycle existence of another flow pattern with lower cost no negative cycle no other flow pattern with lower cost

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19 Ideas for Theorem 3 Theorem 3. (Theorem 9.4, Complementary Slackness Optimality Conditions) A feasible solution x * is optimal iff there exists dual variables such that optimality conditions of simplex method for bounded variables

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20 Ideas for Theorem 2 contradictory statements of Theorem 2 & Theorem 3? Theorem 2: all reduced costs non-negative (for minimization) Theorem 3: some reduced costs positive, some negative Theorem 2: on reduced costs for (variables representing) arcs of a residual network Theorem 3: on reduced costs of all variables

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21 Ideas for Theorem 2 Theorem 2. (Theorem 9.3, Reduced Cost Optimality Conditions) A feasible solution x * is optimal iff there exists dual variable such that for every arc (i, j) in the residual network G(x * ). Theorem 2 minimum for any potential and cycle: C (P) = C(P) for every arc in G(x) no negative cycle in G minimal

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22 Ideas for Theorem 2 Theorem 2. (Theorem 9.3, Reduced Cost Optimality Conditions) A feasible solution x * is optimal iff there exists dual variable such that for every arc (i, j) in the residual network G(x * ). minimum Theorem 2 conditions of Theorem 3 satisfied by a minimum solution a forward arc (i, j) with by Theorem 3 no reversed arc in G(x * ) a forward arc (i, j) with = u ij, in the original network no such forward arc in G(x * ) for the reversed arc in G(x * ): a forward arc (i, j) with in the original network for the forward arc in G(x * ): for the reversed arc in G(x * ):

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23 Equivalence Between Network Simplex and Negative Cost Cycle

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24 Equivalence Between Network Simplex Method and Existence of a Negative Cost Cycle at a certain point, an extended basic feasible solution is shown below, where the dotted lines show arcs at upper bounds. 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (2, 3) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (c ij, u ij ) 1 2 3 4 5 6 7 2 2 0 5 2 5 4 9 -9 0 Flow

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25 An Iteration of the Network Simplex to find the entering variable x 46 entering reduction of flow from upper bound 1 2 3 4 5 6 7 2 2 0 5 2 5 4 9 -9 0 Flow 1 2 3 4 5 6 3 2 5 4 4 dual variables 1 = 0 2 = 1 – c 12 = -3 4 =-8 3 =-2 5 =-6 6 =-10 beneficial to reduce flow in (4, 6 ) How about (2, 3), (2, 5), (5, 4)?

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26 An Iteration of the Network Simplex to find the leaving variable reduction of flow from x 46 leaving variable, x 13 or x 35, for = 1 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (2, 3) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (c ij, u ij ) 1 2 3 4 5 6 0 0 Change of Flow 1 2 3 4 5 6 7 2 2 0 5 2 5 4 9 -9 0 Current Flow Pattern 0 2+ 3 0 7 8 0 5 7 0 5 5 0 2+ 3 0 4+ 6

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27 An Iteration of the Network Simplex = 1 arbitrarily making x 35 leaving 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (2, 3) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (c ij, u ij ) 1 2 3 4 5 6 0 7 8 0 0 Change of Flow 0 5 7 0 5 5 0 2+ 3 0 4+ 6 1 2 3 4 5 6 6 3 2 0 4 3 4 5 9 -9 0 New Flow Pattern

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(c ij, x ij ) 28 An Iteration of the Negative Cycle Algorithm negative cycle 1-3-5-6-4-2-1 cost = -1 maximum allowable flow = 1 unit 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (2, 3) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (c ij, u ij ) 1 2 3 4 5 6 7 2 2 0 5 2 5 4 9 -9 0 Current Flow Pattern Residual Network

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29 An Iteration of the Negative Cycle Algorithm negative cycle 1-3-5-6-4-2-1 cost = -1 maximum allowable flow = 1 unit 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (2, 3) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (c ij, u ij ) (c ij, x ij )

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30 Same Result from Network Simplex Method and Negative Cost Cycle 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (2, 3) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (c ij, u ij ) 1 2 3 4 5 6 6 3 2 0 4 3 4 5 9 -9 0 New Flow Pattern residual network (c ij, x ij )

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31 Transportation Simplex Method

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32 Transportation Simplex Method balanced transportation problem 2 sources, 3 destinations 5 constraints, with one degree of redundancy 4 basic variables in a BFS

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33 Transportation Simplex Method four basic variables in a BFS

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34 Transportation Simplex Method reduced cost (i, j) = c ij – u i – v j

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35 Transportation Simplex Method

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36 Transportation Problem by Eliminating Negative Cycles

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37 Transportation Problem by the Negative Cycle Approach

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38 Transportation Problem by the Negative Cycle Approach

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39 Transportation Problem by the Negative Cycle Approach

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40 Transportation Problem by the Negative Cycle Approach no more negative cycle optimal flow: x 13 = 100, x 14 = 500, x 15 = 200, x 23 = 300, x 24 = 0, x 25 = 0, minimum cost = 7600

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41 To Determine an Initial Feasible Solution

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42 To Find an Initial Feasible Solution to solve a maximal flow problem

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43 To Find an Initial Feasible Solution to solve a minimum cost flow problem with x 1t = 800, x 2t = 300, x ts = 1100, x s3 = 400, x s4 = 500, x s5 = 200 as the initial solution of the minimal cost flow problem

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