Chapter 5: Transportation, Assignment and Network Models © 2007 Pearson Education.
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Transcript of Chapter 5: Transportation, Assignment and Network Models © 2007 Pearson Education.
![Page 1: Chapter 5: Transportation, Assignment and Network Models © 2007 Pearson Education.](https://reader033.fdocuments.us/reader033/viewer/2022061505/56649cd95503460f949a3183/html5/thumbnails/1.jpg)
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
XAB = 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 Type Net Flow
Origin < 0
Destination > 0
Transshipment = 0
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The Transportation ModelDecision: How much to ship from each
origin to each destination?
Objective: Minimize shipping cost
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Data
Decision VariablesXij = number of desks shipped from factory i
to warehouse j
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Objective Function: (in $ of transportation cost)
Min 5XDA + 4XDB + 3XDC + 8XEA + 4XEB + 3XEC + 9XFA + 7XFB + 5XFC
Subject to the constraints:
Flow Balance For Each Supply Node
(inflow) - (outflow) = Net flow
- (XDA + XDB + XDC) = -100 (Des Moines) OR
XDA + XDB + XDC = 100 (Des Moines)
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Other Supply Nodes
XEA + XEB + XEC = 300 (Evansville)
XFA + XFB + XFC = 300 (Fort Lauderdale)
Flow Balance For Each Demand Node
XDA + XEA + XFA = 300 (Albuquerque)
XDB + XEB + XFB = 200 (Boston)
XDC + XEC + XFC = 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 WithMax-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 Xij 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 Type Flow BalanceNet Flow
(RHS)
Supply inflow < outflow Negative
Demand inflow > outflow Positive
Transshipment inflow = outflow Zero
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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 5XDA + 4XDB + 3XDC + 2XDE + 3XEA + 2XEB + 1XEC + 9XFA + 7XFB + 5XFC + 2XFE
Subject to the constraints:
Supply Nodes (with outflow only) - (XDA + XDB + XDC + XDE) = -100 (Des Moines)
- (XFA + XFB + XFC + XFE) = -300 (Ft Lauderdale)
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Evansville (a supply node with inflow)
(XDE + XFE) – (XEA + XEB + XEC) = -300
Demand Nodes
XDA + XEA + XFA = 300 (Albuquerque)
XDB + XEB + XFB = 200 (Boston)
XDC + XEC + XFC = 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 ExampleHave 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
Xij = 1 if worker i is assigned to project j
0 otherwise
Objective Function (in $ of wage cost)
Min 11XA1 + 14XA2 + 6XA3 + 8XB1 + 10XB2 + 11XB3 + 9XC1 + 12XC2 + 7XC3
Subject to the constraints:(see next slide)
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One Project Per Worker (supply nodes)
- (XA1 + XA2 + XA3) = -1 (Adams)
- (XB1 + XB2 + XB3) = -1 (Brown)
- (XC1 + XC2 + XC3) = -1 (Cooper)
One Worker Per Project (demand nodes)
XA1 + XB1 + XC1 = 1 (project 1)
XA2 + XB2 + XC2 = 1 (project 2)
XA3 + XB3 + XC3 = 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
Xij = number of cars per hour flowing from node i to node j
Dummy Arc
The X61 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 X61
Subject to the constraints:
Flow Balance At Each Node Node
(X61 + X21) – (X12 + X13 + X14) = 0 1
(X12 + X42 + X62) – (X21 + X24 + X26) = 0 2
(X13 + X43 + X53) – (X34 + X35) = 0 3
(X14+ X24 + X34 + X64)–(X42+ X43 + X46) = 0 4
(X35) – (X53 + X56) = 0 5
(X26 + X46 + X56) – (X61 + X62 + X64) = 0 6
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Flow Capacity Limit On Each Arc
Xij < capacity of arc ij
Go to File 5-6.xls
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The Shortest Path ModelFor 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
Xij = 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 100X12 + 200X13 + 100X21 + 50X23 + 200X24 + 100X25 + 200X31 + 50X32 + 40X35 + 200X42 + 150X45 + 100X46 + 40X53 + 100X52 + 150X54 + 100X56 + 100X64 + 100X65
Subject to the constraints:
(see next slide)
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Flow Balance For Each Node Node
(X21 + X31) – (X12 + X13) = -1 1
(X12+X32+X42+X52)–(X21+X23+X24+X25)=0 2
(X13 + X23 + X53) – (X31 + X32 + X35) = 0 3
(X24 + X54 + X64) – (X42 + X45 + X46) = 0 4
(X25+X35+X45+X65)–(X52+X53+X54+X56)=0 5
(X46 + X56) – (X64 + X65) = 1 6
Go to file 5-7.xls
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Minimal Spanning TreeFor 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