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Chapter 5: Transportation, Assignment and Network Models © 2007 Pearson Education

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Network Flow Models Consist of a network that can be represented with nodes and arcs 1.Transportation Model 2.Transshipment Model 3.Assignment Model 4.Maximal Flow Model 5.Shortest Path Model 6.Minimal Spanning Tree Model

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Characteristics of Network Models A node is a specific location An arc connects 2 nodes Arcs can be 1-way or 2-way

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Types of Nodes Origin nodes Destination nodes Transshipment nodes Decision Variables X AB = amount of flow (or shipment) from node A to node B

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Flow Balance at Each Node (total inflow) – (total outflow) = Net flow Node TypeNet Flow Origin< 0 Destination> 0 Transshipment= 0

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The Transportation Model Decision: How much to ship from each origin to each destination? Objective: Minimize shipping cost

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Data Decision Variables X ij = number of desks shipped from factory i to warehouse j

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Objective Function: (in $ of transportation cost) Min 5X DA + 4X DB + 3X DC + 8X EA + 4X EB + 3X EC + 9X FA + 7X FB + 5X FC Subject to the constraints: Flow Balance For Each Supply Node (inflow) - (outflow) = Net flow - (X DA + X DB + X DC ) = -100 (Des Moines) OR X DA + X DB + X DC = 100 (Des Moines)

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Other Supply Nodes X EA + X EB + X EC = 300 (Evansville) X FA + X FB + X FC = 300 (Fort Lauderdale) Flow Balance For Each Demand Node X DA + X EA + X FA = 300 (Albuquerque) X DB + X EB + X FB = 200 (Boston) X DC + X EC + X FC = 200 (Cleveland) Go to File 5-1.xls

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Unbalanced Transportation Model If (Total Supply) > (Total Demand), then for each supply node: (outflow) < (supply) If (Total Supply) < (Total Demand), then for each demand node: (inflow) < (demand)

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Transportation Models With Max-Min and Min-Max Objectives Max-Min means maximize the smallest decision variable Min-Max mean to minimize the largest decision variable Both reduce the variability among the X ij values Go to File 5-3.xls

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The Transshipment Model Similar to a transportation model Have “Transshipment” nodes with both inflow and outflow Node TypeFlow Balance Net Flow (RHS) Supplyinflow < outflowNegative Demandinflow > outflowPositive Transshipmentinflow = outflowZero

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Revised Transportation Cost Data Note: Evansville is both an origin and a destination

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Objective Function: (in $ of transportation cost) Min 5X DA + 4X DB + 3X DC + 2X DE + 3X EA + 2X EB + 1X EC + 9X FA + 7X FB + 5X FC + 2X FE Subject to the constraints: Supply Nodes (with outflow only) - (X DA + X DB + X DC + X DE ) = -100 (Des Moines) - (X FA + X FB + X FC + X FE ) = -300 (Ft Lauderdale)

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Evansville (a supply node with inflow) (X DE + X FE ) – (X EA + X EB + X EC ) = -300 Demand Nodes X DA + X EA + X FA = 300 (Albuquerque) X DB + X EB + X FB = 200 (Boston) X DC + X EC + X FC = 200 (Cleveland) Go to File 5-4.xls

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Assignment Model For making one-to-one assignments Such as: –People to tasks –Classes to classrooms –Etc.

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Fit-it Shop Assignment Example Have 3 workers and 3 repair projects Decision: Which worker to assign to which project? Objective: Minimize cost in wages to get all 3 projects done

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Estimated Wages Cost of Possible Assignments

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Can be Represented as a Network Model The “flow” on each arc is either 0 or 1

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Decision Variables X ij = 1 if worker i is assigned to project j 0 otherwise Objective Function(in $ of wage cost) Min 11X A1 + 14X A2 + 6X A3 + 8X B1 + 10X B2 + 11X B3 + 9X C1 + 12X C2 + 7X C3 Subject to the constraints: (see next slide)

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One Project Per Worker (supply nodes) - (X A1 + X A2 + X A3 ) = -1(Adams) - (X B1 + X B2 + X B3 ) = -1(Brown) - (X C1 + X C2 + X C3 ) = -1(Cooper) One Worker Per Project (demand nodes) X A1 + X B1 + X C1 = 1(project 1) X A2 + X B2 + X C2 = 1(project 2) X A3 + X B3 + X C3 = 1(project 3) Go to File 5-5.xls

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The Maximal-Flow Model Where networks have arcs with limited capacity, such as roads or pipelines Decision: How much flow on each arc? Objective: Maximize flow through the network from an origin to a destination

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Road Network Example Need 2 arcs for 2-way streets

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Modified Road Network

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Decision Variables X ij = number of cars per hour flowing from node i to node j Dummy Arc The X 61 arc was created as a “dummy” arc to measure the total flow from node 1 to node 6

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Objective Function Max X 61 Subject to the constraints: Flow Balance At Each Node Node (X 61 + X 21 ) – (X 12 + X 13 + X 14 ) = 0 1 (X 12 + X 42 + X 62 ) – (X 21 + X 24 + X 26 ) = 0 2 (X 13 + X 43 + X 53 ) – (X 34 + X 35 )= 0 3 (X 14 + X 24 + X 34 + X 64 )–(X 42 + X 43 + X 46 ) = 0 4 (X 35 ) – (X 53 + X 56 ) = 0 5 (X 26 + X 46 + X 56 ) – (X 61 + X 62 + X 64 ) = 0 6

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Flow Capacity Limit On Each Arc X ij < capacity of arc ij Go to File 5-6.xls

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The Shortest Path Model For determining the shortest distance to travel through a network to go from an origin to a destination Decision: Which arcs to travel on? Objective: Minimize the distance (or time) from the origin to the destination

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Ray Design Inc. Example Want to find the shortest path from the factory to the warehouse Supply of 1 at factory Demand of 1 at warehouse

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Decision Variables X ij = flow from node i to node j Note: “flow” on arc ij will be 1 if arc ij is used, and 0 if not used Roads are bi-directional, so the 9 roads require 18 decision variables

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Objective Function (in distance) Min 100X X X X X X X X X X X X X X X X X X 65 Subject to the constraints: (see next slide)

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Flow Balance For Each Node Node (X 21 + X 31 ) – (X 12 + X 13 ) = -1 1 (X 12 +X 32 +X 42 +X 52 )–(X 21 +X 23 +X 24 +X 25 )=0 2 (X 13 + X 23 + X 53 ) – (X 31 + X 32 + X 35 ) = 0 3 (X 24 + X 54 + X 64 ) – (X 42 + X 45 + X 46 ) = 0 4 (X 25 +X 35 +X 45 +X 65 )–(X 52 +X 53 +X 54 +X 56 )=0 5 (X 46 + X 56 ) – (X 64 + X 65 ) = 1 6 Go to file 5-7.xls

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Minimal Spanning Tree For connecting all nodes with a minimum total distance Decision: Which arcs to choose to connect all nodes? Objective: Minimize the total distance of the arcs chosen

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Lauderdale Construction Example Building a network of water pipes to supply water to 8 houses (distance in hundreds of feet)

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Characteristics of Minimal Spanning Tree Problems Nodes are not pre-specified as origins or destinations So we do not formulate as LP model Instead there is a solution procedure

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Steps for Solving Minimal Spanning Tree 1.Select any node 2.Connect this node to its nearest node 3.Find the nearest unconnected node and connect it to the tree (if there is a tie, select one arbitrarily) 4.Repeat step 3 until all nodes are connected

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Steps 1 and 2 Starting arbitrarily with node (house) 1, the closest node is node 3

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Second and Third Iterations

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Fourth and Fifth Iterations

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Sixth and Seventh Iterations After all nodes (homes) are connected the total distance is 16 or 1,600 feet of water pipe

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