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Part 3 Linear Programming

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Presentation on theme: "Part 3 Linear Programming"— Presentation transcript:

1 Part 3 Linear Programming
3.5 Flows in Networks

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3 The Transportation Model (A Special Case of Transshipment Model)

4 Incidence Matrix of Directed Graph

5 Material Balance Around a Node

6 Minimum Cost Flow Problem

7 Transshipment Problem
The transportation problem is a special case of the minimum cost flow problem, corresponding to a network with arcs going only from supply nodes (origins) to demand nodes (destinations). The more general problem allows for arbitrary network configuration, so that flow from a supply node may progress through several intermediate nodes before reaching its destination. The more general problem is often termed the transshipment problem.

8 Problem Structure The coefficient matrix A of the flow constraints is the node-arc incidence matrix of the network. The column corresponding to arc (i, j) has a +1 entry in row i and a -1 entry in row j. Since the sum of all rows of A is the zero vector, the matrix A has a rank of at most n-1.

9 Incidence Matrix of Directed Graph

10 Spanning Tree A graph is connected if there is a chain between any two nodes. A graph is a tree if it is connected and has no cycles. Consider a tree within a graph G which is made of a subset of arcs from G. Such a tree is a spanning tree if it touches all nodes of G. A graph is connected iff it contains a spanning tree.

11 Structure of a Basis The is a direct one-to-one correspondence between the arcs in a basis and a spanning tree. Any basis (i.e., spanning tree) is triangular. Given a basis, the corresponding basic solution can be found by back substitution using the triangular structure. In terms of network concepts, one first look for an end of the spanning tree corresponding to the basis, i.e., an node touched by only one arc of the tree. The flow in this arc is then determined by the supply or demand at the node. Back substitution corresponds to solving for flows along the arcs of the spanning tree, starting from an end and successively eliminating arcs.

12 The Simplex Method

13 Step 3

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