Transportation, Assignment, and Network Models
9
To accompany Quantitative Analysis for Management, Twelfth Edition,
by Render, Stair, Hanna and Hale
Power Point slides created by Jeff Heyl
Copyright ©2015 Pearson Education, Inc.
After completing this chapter, students will be able to:
LEARNING OBJECTIVES
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Structure LP problems for the transportation, transshipment, and assignment models.
Solve facility location and other application problems with transportation models.
Use LP to model shortest-route and maximal-flow problems.
Solve minimal-spanning tree problems.
9.1 Introduction
9.2 The Transportation Problem
9.3 The Assignment Problem
9.4 The Transshipment Problem
9.5 Maximal-Flow Problem
9.6 Shortest-Route Problem
9.7 Minimal-Spanning Tree Problem
CHAPTER OUTLINE
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Introduction
LP problems modeled as networks
Helps visualize and understand problems
Transportation problem
Transshipment problem
Assignment problem
Maximal-flow problem
Shortest-route problem
Minimal-spanning tree problem
Specialized algorithms available
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Introduction
Common terminology for network models
Points on the network are referred to as nodes
Typically circles
Lines on the network that connect nodes are called arcs
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The Transportation Problem
Deals with the distribution of goods from several points of supply (sources) to a number of points of demand (destinations)
Usually given the capacity of goods at each source and the requirements at each destination
Typically objective is to minimize total transportation and production costs
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Linear Program for Transportation
Executive Furniture Corporation transportation problem
Minimize transportation cost
Meet demand
Not exceed supply
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Linear Program for Transportation
Let Xij = number of units shipped from source i to destination j
Where
i = 1, 2, 3, with 1 = Des Moines, 2 = Evansville, and 3 = Fort Lauderdale
j = 1, 2, 3, with 1 = Albuquerque, 2 = Boston, and 3 = Cleveland
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Linear Program for Transportation
Minimize total cost = 5X11 + 4X12 + 3X13 + 8X21 + 4X22 + 3X23 + 9X31 +7X32 + 5X33
Subject to:
X11 + X12 + X13 ≤ 100 (Des Moines supply)
X21 + X22 + X23 ≤ 300 (Evansville supply)
X31 + X32 + X33 ≤ 300 (Fort Lauderdale supply)
X11 + X21 + X31 = 300 (Albuquerque demand)
X12 + X22 + X32 = 200 (Boston demand)
X13 + X23 + X33 = 200 (Cleveland demand)
Xij ≥ 0 for all i and j
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Linear Program for Transportation
Optimal solution
100 units from Des Moines to Albuquerque
200 units from Evansville to Boston
100 units from Evansville to Cleveland
200 units from Ft. Lauderdale to Albuquerque
100 units from Ft. Lauderdale to Cleveland
Total cost = $3,900
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Linear Program for Transportation
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FIGURE 9.1 – Network Representation of a Transportation Problem
100
300
300
Supply
$5
$4
$3
$8
$4
$3
$9
$7
$5
200
200
300
Demand
Source
Des Moines
(Source 1)
Evansville
(Source 2)
Fort Lauderdale
(Source 3)
Albuquerque
(Destination 1)
Boston
(Destination 2)
Cleveland
(Destination 3)
Destination
Using Excel QM
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PROGRAM 9.1 – Executive Furniture Corporation Solution in Excel 2013 Using Excel QM
A General LP Model for Transportation Problems
Let
Xij = number of units shipped from source i to destination j
cij = cost of one unit from source i to destination j
si = supply at source i
dj = demand at destination j
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A General LP Model for Transportation Problems
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Subject to:
Facility Location Analysis
Transportation method especially useful
New location is major financial importance
Several alternative locations evaluated
Subjective factors are considered
Final decision also involves minimizing total shipping and production costs
Alternative facility locations analyzed within the framework of one overall distribution system
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Facility Location Analysis
Hardgrave Machine Company produces computer components in Cincinnati, Salt Lake City, and Pittsburgh
Four warehouses in Detroit, Dallas, New York, and Los Angeles
Two new plant sites being considered – Seattle and Birmingham
Which of the new locations will yield the lowest cost for the firm in combination with the existing plants and warehouses?
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Facility Location Analysis
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WAREHOUSE MONTHLY DEMAND (UNITS) PRODUCTION PLANT MONTHLY SUPPLY COST TO PRODUCE ONE UNIT ($)
Detroit 10,000 Cincinnati 15,000 48
Dallas 12,000 Salt Lake City 6,000 50
New York 15,000 Pittsburgh 14,000 52
Los Angeles 9,000 35,000
46,000
Supply needed from a new plant = 46,000 – 35,000 = 11,000 units per month
TABLE 9.1 – Hardgrave’s Demand and Supply Data
ESTIMATED PRODUCTION COST PER UNIT AT PROPOSED PLANTS
Seattle $53
Birmingham $49
Facility Location Analysis
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TO FROM DETROIT DALLAS NEW YORK LOS ANGELES
CINCINNATI $25 $55 $40 $60
SALT LAKE CITY 35 30 50 40
PITTSBURGH 36 45 26 66
SEATTLE 60 38 65 27
BIRMINGHAM 35 30 41 50
TABLE 9.2 – Hardgrave’s Shipping Costs
Solve two transportation problems – one for each combination
Facility Location Analysis
Xij = number of units shipped from source i to destination j
Where
i = 1, 2, 3, 4 with 1 = Cincinnati, 2 = Salt Lake City, 3 = Pittsburgh, and 4 = Seattle
j = 1, 2, 3, 4 with 1 = Detroit, 2 = Dallas, 3 = New York, and 4 = Los Angeles
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Facility Location Analysis
Minimize total cost = 73X11 + 103X12 + 88X13 + 108X14 + 85X21 + 80X22 + 100X23 + 90X24 + 88X31 + 97X32 + 78X33 + 118X34 + 113X41 + 91X42 + 118X43 + 80X44
Subject to:
X11 + X21 + X31 + X41 = 10,000 Detroit demand
X12 + X22 + X32 + X42 = 12,000 Dallas demand
X13 + X23 + X33 + X43 = 15,000 New York demand
X14 + X24 + X34 + X44 = 9,000 Los Angeles demand
X11 + X12 + X13 + X14 ≤ 15,000 Cincinnati supply
X21 + X22 + X23 + X24 ≤ 6,000 Salt Lake City supply
X31 + X32 + X33 + X34 ≤ 14,000 Pittsburgh supply
X41 + X42 + X43 + X44 ≤ 11,000 Seattle supply
All variables Xij ≥ 0
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Minimize total cost = 73X11 + 103X12 + 88X13 + 108X14 + 85X21 + 80X22 + 100X23 + 90X24 + 88X31 + 97X32 + 78X33 + 118X34 + 113X41 + 91X42 + 118X43 + 80X44
Subject to:
X11 + X21 + X31 + X41 = 10,000 Detroit demand
X12 + X22 + X32 + X42 = 12,000 Dallas demand
X13 + X23 + X33 + X43 = 15,000 New York demand
X14 + X24 + X34 + X44 = 9,000 Los Angeles demand
X11 + X12 + X13 + X14 ≤ 15,000 Cincinnati supply
X21 + X22 + X23 + X24 ≤ 6,000 Salt Lake City supply
X31 + X32 + X33 + X34 ≤ 14,000 Pittsburgh supply
X41 + X42 + X43 + X44 ≤ 11,000 Seattle supply
All variables Xij ≥ 0
Facility Location Analysis
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The total cost for the Seattle alternative = $3,704,000
Minimize total cost = 73X11 + 103X12 + 88X13 + 108X14 + 85X21 + 80X22 + 100X23 + 90X24 + 88X31 + 97X32 + 78X33 + 118X34 + 113X41 + 91X42 + 118X43 + 80X44
Subject to:
X11 + X21 + X31 + X41 = 10,000 Detroit demand
X12 + X22 + X32 + X42 = 12,000 Dallas demand
X13 + X23 + X33 + X43 = 15,000 New York demand
X14 + X24 + X34 + X44 = 9,000 Los Angeles demand
X11 + X12 + X13 + X14 ≤ 15,000 Cincinnati supply
X21 + X22 + X23 + X24 ≤ 6,000 Salt Lake City supply
X31 + X32 + X33 + X34 ≤ 14,000 Pittsburgh supply
X41 + X42 + X43 + X44 ≤ 11,000 Seattle supply
All variables Xij ≥ 0
Facility Location Analysis
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The total cost for the Seattle alternative = $3,704,000
Reformulating the problem for the Birmingham alternative and solving, the total cost = $3,741,000
Using Excel QM
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PROGRAM 9.2 – Facility Location (Seattle) Solution in Excel 2013 Using Excel QM
Using Excel QM
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PROGRAM 9.3 – Facility Location (Birmingham) Solution in Excel 2013 Using Excel QM
The Assignment Problem
This class of problem determines the most efficient assignment of people or equipment to particular tasks
Objective is typically to minimize total cost or total task time
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Linear Program for Assignment Example
The Fix-it Shop has just received three new repair projects that must be repaired quickly
A radio
A toaster oven
A coffee table
Three workers with different talents are able to do the jobs
Owner estimates wage cost for workers on projects
Objective – minimize total cost
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Linear Program for Assignment Example
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FIGURE 9.2 – Assignment Problem in a Transportation Network Format
1
1
1
Supply
$11
$14
$6
$8
$10
$11
$9
$12
$7
1
1
1
Demand
Person
Adams
(Source 1)
Brown
(Source 2)
Cooper
(Source 3)
Project 1
(Destination 1)
Project 2
(Destination 2)
Project 3
(Destination 3)
Project
Linear Program for Assignment Example
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Let
where
i = 1, 2, 3, with 1 = Adams, 2 = Brown, and 3 = Cooper
j = 1, 2, 3, with 1 = Project 1, 2 = Project 2, and 3 = Project 3
Linear Program for Assignment Example
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Minimize total cost = 11X11 + 14X12 + 6X13 + 8X21 + 10X22 + 11X23 + 9X31 + 12X32 + 7X33
subject to
X11 + X12 + X13 = 1
X21 + X22 + X23 = 1
X31 + X32 + X33 = 1
X11 + X21 + X31 = 1
X12 + X22 + X32 = 1
X13 + X23 + X33 = 1
Xij = 0 or 1 for all i and j
Linear Program for Assignment Example
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Minimize total cost = 11X11 + 14X12 + 6X13 + 8X21 + 10X22 + 11X23 + 9X31 + 12X32 + 7X33
subject to
X11 + X12 + X13 ≤ 1
X21 + X22 + X23 ≤ 1
X31 + X32 + X33 ≤ 1
X11 + X21 + X31 = 1
X12 + X22 + X32 = 1
X13 + X23 + X33 = 1
Xij = 0 or 1 for all i and j
Solution
X13 = 1, Adams assigned to Project 3
X22 = 1, Brown assigned to Project 2
X31 = 1, Cooper is assigned to Project 1
Total cost = $25
Using Excel QM
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PROGRAM 9.4 – Mr. Fix-It Shop Assignment Solution in Excel 2013 Using Excel QM
The Transshipment Problem
Items are being moved from a source to a destination through an intermediate point (a transshipment point)
Transshipment problem
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The Transshipment Problem
Frosty Machines manufactures snow blowers in Toronto and Detroit
Shipped to regional distribution centers in Chicago and Buffalo
Then shipped to supply houses in New York, Philadelphia, and St. Louis
Shipping costs vary by location and destination
Snow blowers cannot be shipped directly from the factories to the supply houses
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The Transshipment Problem
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FIGURE 9.3 – Network Representation of Transshipment Example
800
700
Supply
300
350
450
Demand
Source
Toronto
(Node 1)
Detroit
(Node 2)
New York City
(Node 5)
Philadelphia
(Node 6)
St. Louis
(Node 7)
Destination
Chicago
(Node 3)
Transshipment Point
Buffalo
(Node 4)
The Transshipment Problem
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TABLE 9.3 – Frosty Machine Transshipment Data
TO
FROM CHICAGO BUFFALO NEW YORK CITY PHILADELPHIA ST. LOUIS SUPPLY
Toronto $4 $7 — — — 800
Detroit $5 $7 — — — 700
Chicago — — $6 $4 $5 —
Buffalo — — $2 $3 $4 —
Demand — — 450 350 300
Minimize transportation costs associated with shipping snow blowers subject to demands and supplies
The Transshipment Problem
Minimize cost subject to
The number of units shipped from Toronto is not more than 800
The number of units shipped from Detroit is not more than 700
The number of units shipped to New York is 450
The number of units shipped to Philadelphia is 350
The number of units shipped to St. Louis is 300
The number of units shipped out of Chicago is equal to the number of units shipped into Chicago
The number of units shipped out of Buffalo is equal to the number of units shipped into Buffalo
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The Transshipment Problem
Decision variables
Xij = number of units shipped from location (node) i to location (node) j
where
i = 1, 2, 3, 4
j = 3, 4, 5, 6, 7
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The Transshipment Problem
Minimize cost = 4X13 + 7X14 + 5X23 + 7X24 + 6X35 + 4X36 + 5X37 + 2X45 + 3X46 + 4X47
subject to
X13 + X14 ≤ 800 (Supply at Toronto [node 1])
X23 + X24 ≤ 700 (Supply at Detroit [node 2])
X35 + X45 = 450 (Demand at New York [node 5])
X36 + X46 = 350 (Demand at Philadelphia [node 6])
X37 + X47 = 300 (Demand at St. Louis [node 7])
X13 + X23 = X35 + X36 + X37 (Shipping through Chicago [node 3])
X14 + X24 = X45 + X46 + X47 (Shipping through Buffalo [node 4])
Xij ≥ 0 for all i and j (nonnegativity)
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Minimize cost = 4X13 + 7X14 + 5X23 + 7X24 + 6X35 + 4X36 + 5X37 + 2X45 + 3X46 + 4X47
subject to
X13 + X14 ≤ 800 (Supply at Toronto [node 1])
X23 + X24 ≤ 700 (Supply at Detroit [node 2])
X35 + X45 = 450 (Demand at New York [node 5])
X36 + X46 = 350 (Demand at Philadelphia [node 6])
X37 + X47 = 300 (Demand at St. Louis [node 7])
X13 + X23 = X35 + X36 + X37 (Shipping through Chicago [node 3])
X14 + X24 = X45 + X46 + X47 (Shipping through Buffalo [node 4])
Xij ≥ 0 for all i and j (nonnegativity)
The Transshipment Problem
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Ship 650 units from Toronto to Chicago
Ship 150 units from Toronto to Buffalo
Ship 300 units from Detroit to Buffalo
Ship 350 units from Chicago to Philadelphia
Ship 300 units form Chicago to St. Louis
Ship 450 units from Buffalo to New York
Total cost = $9,550
Using Excel QM
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PROGRAM 9.5 – Excel QM Solution to Frosty Machine Transshipment Problem
Maximal-Flow Problem
Determining the maximum amount of material that can flow from one point (the source) to another (the sink) in a network
Two common methods
Linear programming
Maximal-flow technique
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Maximal-Flow Problem
Determine maximum number of cars from east to west for Waukesha WI road system
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1
2
3
4
5
6
West Point
East Point
Capacity in Hundreds of Cars per Hour
3
2
10
1
2
1
0
1
1
0
3
2
2
1
0
1
6
1
FIGURE 9.4 – Road Network for Waukesha Maximal-Flow Example
Maximal-Flow Problem
Variables
Xij = flow from node i to node j
where
i = 1, 2, 3, 4, 5, 6
j = 1, 2, 3, 4, 5, 6
Constraints necessary for
Capacity of each arc
Equal flows into and out of each arc
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Maximal-Flow Problem
Maximize flow = X61
subject to
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(X21 + X61) – (X12 + X13 + X14) = 0 Flows into = flows out of node 1
(X12 + X42 + X62) – (X21 + X24 + X26) = 0 Flows into = flows out of node 2
(X13 + X43 + X53) – (X34 + X35) = 0 Flows into = flows out of node 3
(X14 + X24 + X34 + X64) – (X42 + X43 + X46) = 0 Flows into = flows out of node 4
(X35) – (X56 + X53) = 0 Flows into = flows out of node 5
(X26 + X46 + X56) – (X61 + X62 + X64) = 0 Flows into = flows out of node 6
Xij ≥ 0
X12 ≤ 3 X13 ≤ 10 X14 ≤ 2 Capacities for arcs from node 1
X21 ≤ 1 X24 ≤ 1 X26 ≤ 2 Capacities for arcs from node 2
X34 ≤ 3 X35 ≤ 2 Capacities for arcs from node 3
X42 ≤ 1 X43 ≤ 1 X46 ≤ 1 Capacities for arcs from node 4
X53 ≤ 1 X56 ≤ 1 Capacities for arcs from node 5
X62 ≤ 2 X64 ≤ 1 Capacities for arcs from node 6
Using Excel QM
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PROGRAM 9.6 –
Waukesha Maximal-Flow
Solution
Shortest-Route Problem
Find the shortest distance from one location to another
Can be modeled as
A linear programming problem with 0-1 variables
A special type of transshipment problem
Using specialized algorithm
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Shortest-Route Problem
Ray Design transports beds, chairs, and other furniture items from the factory to the warehouse
Travel through several cities
No direct interstate highways
Find the route with the shortest distance
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Plant
Warehouse
1
2
3
4
5
6
100
50
40
150
200
100
100
100
200
FIGURE 9.5 –
Roads from
Ray’s Plant
to Warehouse
Shortest-Route Problem
Variables
Xij = 1 if arc from node i to node j is selected and Xij = 0 otherwise
where
i = 1, 2, 3, 4, 5
j = 2, 3, 4, 5, 6
Constraints specify the number of units going into a node must equal the number of units going out of the node
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Variables
Xij = 1 if arc from node i to node j is selected and Xij = 0 otherwise
where
i = 1, 2, 3, 4, 5
j = 2, 3, 4, 5, 6
Constraints specify the number of units going into a node must equal the number of units going out of the node
Shortest-Route Problem
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Origin point must ship one unit
X12 + X13 = 1
Final destination must have one unit shipped into it
X46 + X56 = 1
Intermediate nodes must have same amounts entering and leaving
X12 + X32 = X23 + X24 + X25
or
X12 + X32 – X23 – X24 – X25 = 0
Shortest-Route Problem
Minimize distance = 100X12 + 200X13 + 50X23 + 50X32 + 200X24 + 200X42 + 100X25 + 100X52 + 40X35 + 40X53 + 150X45 + 150X54 + 100X46 + 100X56
subject to
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X12 + X13 = 1 Node 1
X12 + X32 – X23 – X24 – X25 = 0 Node 2
X13 + X23 – X32 – X35 = 0 Node 3
X24 + X54 – X42 – X45 – X46 = 0 Node 4
X25 + X35 + X45 – X52 – X53 – X54 – X56 = 0 Node 5
X46 + X56 = 1 Node 6
All variables = 0 or 1
Using Excel QM
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PROGRAM 9.7 –
Ray Designs, Inc.
Solution
Using Excel QM
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PROGRAM 9.7 –
Ray Designs, Inc.
Solution
Solution
X12 = X23 = X35 = X56 = 1
Route is City 1 to City 2 to City 3 to City 5 to City 6
Total distance traveled = 290 miles
Minimal-Spanning Tree Problem
Connecting all points of a network together while minimizing the total distance of the connections
Linear programming can be used but is complex
Minimal-spanning tree technique is quite easy
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Minimal-Spanning Tree Problem
Steps for the Minimal-Spanning Tree Technique
Select any node in the network.
Connect this node to the nearest node that minimizes the total distance.
Considering all of the nodes that are now connected, find and connect the nearest node that is not connected. If there is a tie for the nearest node, select one arbitrarily. A tie suggests there may be more than one optimal solution.
Repeat the third step until all nodes are connected.
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Minimal-Spanning Tree Problem
Lauderdale Construction
Housing project in Panama City Beach
Determine the least expensive way to provide water and power to each house
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1
2
4
6
Gulf
3
5
7
8
3
2
7
5
6
1
4
3
3
3
2
2
5
FIGURE 9.6 – Network for Lauderdale Construction
Minimal-Spanning Tree Problem
Step 1 – Arbitrarily select node 1
Step 2 – Connect node 1 to node 3 (nearest)
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1
2
4
6
Gulf
3
5
7
8
3
2
7
5
6
1
4
3
3
3
2
2
5
FIGURE 9.7 – First Iteration
Minimal-Spanning Tree Problem
Step 3 – Connect next nearest unconnected node, node 4
Continue for other unconnected nodes
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1
2
4
6
3
5
7
8
3
2
7
5
6
1
4
3
3
3
2
2
5
FIGURE 9.8 – Second and Third Iterations
1
2
4
6
3
5
7
8
3
2
7
5
6
1
4
3
3
3
2
2
5
(a) Second Iteration
(b) Third Iteration
Minimal-Spanning Tree Problem
Step 4 – Repeat the process
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1
2
4
6
3
5
7
8
3
2
7
5
6
1
4
3
3
3
2
2
5
FIGURE 9.9 – Last Four Iterations
1
2
4
6
3
5
7
8
3
2
7
5
6
1
4
3
3
3
2
2
5
(a) Fourth Iteration
(b) Fifth Iteration
Step 4 – Repeat the process
Minimal-Spanning Tree Problem
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1
2
4
6
3
5
7
8
3
2
7
5
6
1
4
3
3
3
2
2
5
FIGURE 9.9 – Last Four Iterations
1