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Published byAntwan Rassel Modified over 3 years ago

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**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 1

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**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)} cij: the cost of arc (i, j) uij: 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 How to deal with an undirected arc? Assume direct arcs. 2

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**Minimum Cost Flow Models**

1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (cij, uij) 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 3

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**A Minimum Cost Flow LP Model**

1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 4

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**Useful Results from Shortest-Path Algorithms**

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**Useful Results from Shortest-Path Algorithms**

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

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**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) + cij for all (i, j) A. 7

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**Optimality Condition for Shortest Distances**

1 2 3 4 7 6 d(1) = 0 d(2) = 2 d(3) = 3 d(4) = 7 1 2 3 4 7 6 d(1) = 0 d(2) = 2 d(3) = 3 d(4) = 8 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) + c34 < d(4), i.e., this set cannot be the collection of shortest distances from nodes 8

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**Residual Network (6,4) (8,5) s t 1 (2,7)**

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

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**A Cycle in a Residual Network**

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 (-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 (6, 1) (8, 5) (2, 7) (-6, 3) s t 1 3 Figure 6. Flows on arcs 10

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**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 11

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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 C(P) = C(P) s + t 12

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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 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 7 1 = 3 2 = -5 3 = 2 4 = 6 5 = -3 6 = 5 C(P) C(P) 1 2 3 4 5 6 10 18 -2 13 14 13

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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 j = -d(j) 1 = 0 2 = -8 3 = -3 4 = -10 5 = -6 6 = -12 4 8 5 3 1 2 3 4 5 6 1 2 3 4 5 6 8 7 1 = 3 2 = -5 3 = 2 4 = 6 5 = -3 6 = 5 14

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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 15

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**The optimality Conditions for Networks**

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**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 < < uij, then if then = uij. 17

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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 18

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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 19

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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 20

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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 21

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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 = uij, 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*): 22

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**Equivalence Between Network Simplex and Negative Cost Cycle**

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**Equivalence Between Network Simplex Method and Existence of a Negative Cost Cycle**

1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (cij, uij) 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 7 9 -9 Flow 24

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**An Iteration of the Network Simplex**

1 2 3 4 5 6 7 9 -9 Flow to find the entering variable x46 entering reduction of flow from upper bound dual variables 1 = 0 2 = 1 – c12 = -3 4 =-8 3 =-2 5 =-6 6 =-10 1 2 3 4 5 6 How about (2, 3), (2, 5), (5, 4)? beneficial to reduce flow in (4, 6) 25

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**An Iteration of the Network Simplex**

1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (cij, uij) to find the leaving variable reduction of flow from x46 leaving variable, x13 or x35, for = 1 1 2 3 4 5 6 Change of Flow 0 2+ 3 0 7 8 0 5 7 0 5 5 0 4+ 6 1 2 3 4 5 6 7 9 -9 Current Flow Pattern 26

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**An Iteration of the Network Simplex**

1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (cij, uij) = 1 arbitrarily making x35 leaving 1 2 3 4 5 6 0 7 8 Change of Flow 0 5 7 0 5 5 0 2+ 3 0 4+ 6 1 2 3 4 5 6 9 -9 New Flow Pattern 27

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**An Iteration of the Negative Cycle Algorithm**

cost = -1 maximum allowable flow = 1 unit 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (cij, uij) 1 2 3 4 5 6 7 9 -9 Current Flow Pattern Residual Network (cij, xij) 28

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**An Iteration of the Negative Cycle Algorithm**

cost = -1 maximum allowable flow = 1 unit 1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (cij, uij) (cij, xij) (cij, xij) 29

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**Same Result from Network Simplex Method and Negative Cost Cycle**

1 2 3 4 5 6 (3, 8) (2, 3) (2, 2) (5, 7) (4, 3) (5, 4) (3, 5) (4, 6) 9 -9 (cij, uij) residual network (cij, xij) 1 2 3 4 5 6 9 -9 New Flow Pattern 30

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**Transportation Simplex Method**

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**Transportation Simplex Method**

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

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**Transportation Simplex Method**

four basic variables in a BFS 33

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**Transportation Simplex Method**

reduced cost (i, j) = cij – ui – vj 34

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**Transportation Simplex Method**

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**Transportation Problem by Eliminating Negative Cycles**

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**Transportation Problem by the Negative Cycle Approach**

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**Transportation Problem by the Negative Cycle Approach**

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**Transportation Problem by the Negative Cycle Approach**

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**Transportation Problem by the Negative Cycle Approach**

no more negative cycle optimal flow: x13 = 100, x14 = 500, x15 = 200, x23 = 300, x24 = 0, x25= 0, minimum cost = 7600 40

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**To Determine an Initial Feasible Solution**

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**To Find an Initial Feasible Solution**

to solve a maximal flow problem 42

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**To Find an Initial Feasible Solution**

to solve a minimum cost flow problem with x1t = 800, x2t = 300, xts = 1100, xs3 = 400, xs4 = 500, xs5 = 200 as the initial solution of the minimal cost flow problem 43

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