render_qam13e_ppt_09
Quantitative Analysis for Management
Thirteenth Edition
Chapter 9
Transportation, Assignment, and Network Models
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved.
Copyright © 2018, 2015, 2012 Pearson Education, Inc. All Rights Reserved.
Learning Objectives
After completing this chapter, students will be able to:
9.1 Construct LP problems for the transportation, assignment, and transshipment models.
9.2 Solve facility location and other application problems with transportation models.
9.3 Use LP to model and solve maximal-flow problems.
9.4 Use LP to model and solve shortest route problems.
9.5 Solve minimal-spanning tree problems.
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Chapter Outline
9.1 The Transportation Problem
9.2 The Assignment Problem
9.3 The Transshipment Problem
9.4 Maximal-Flow Problem
9.5 Shortest-Route Problem
9.6 Minimal-Spanning Tree Problem
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Introduction (1 of 2)
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 (2 of 2)
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 (1 of 5)
Executive Furniture Corporation transportation problem
Minimize transportation cost
Meet demand
Not exceed supply
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Linear Program for Transportation (2 of 5)
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 (3 of 5)
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 (4 of 5)
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 = $3,900
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Linear Program for Transportation (5 of 5)
FIGURE 9.1 Network Representation of a Transportation Problem, with Costs, Demands, and Supplies
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Using Excel QM
PROGRAM 9.1 Executive Furniture Corporation Solution in Excel 2016 Using Excel QM
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A General LP Model for Transportation Problems (1 of 2)
Let
Xij = number of units shipped from source i to destination j
cij = cost of one unit from source i to destination j
si = supply at source I
dj = demand at destination j
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A General LP Model for Transportation Problems (2 of 2)
Subject to:
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Facility Location Analysis (1 of 7)
Transportation method especially useful
New location is major financial importance
Several alternative locations evaluated
Subjective factors are considered
Final decision also involves minimizing total shipping and production costs
Alternative facility locations analyzed within the framework of one overall distribution system
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Facility Location Analysis (2 of 7)
Hardgrave Machine Company produces computer components in Cincinnati, Salt Lake City, and Pittsburgh
Four warehouses in Detroit, Dallas, New York, and Los Angeles
Two new plant sites being considered – Seattle and Birmingham
Which of the new locations will yield the lowest cost for the firm in combination with the existing plants and warehouses?
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Facility Location Analysis (3 of 7)
TABLE 9.1 Hardgrave’s Demand and Supply Data
WAREHOUSE | MONTHLY DEMAND (UNITS) | PRODUCTION PLANT | MONTHLY SUPPLY | COST TO PRODUCE ONE UNIT ($) |
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 | Blank | 35,000 | Blank |
Blank | 46,000 | Blank | Blank | Blank |
Supply needed from a new plant = 46,000 − 35,000 = 11,000 units per month
ESTIMATED PRODUCTION COST PER UNIT AT PROPOSED PLANTS |
Seattle | $53 |
Birmingham | $49 |
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Facility Location Analysis (4 of 7)
TABLE 9.2 Hardgrave’s Shipping Costs
FROM | TO | 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 |
Solve two transportation problems – one for each combination
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Facility Location Analysis (5 of 7)
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 (6 of 7)
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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Facility Location Analysis (7 of 7)
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
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Using Excel QM (1 of 2)
PROGRAM 9.2 Facility Location (Seattle) Solution in Excel 2016 Using Excel QM
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Using Excel QM (2 of 2)
PROGRAM 9.3 Facility Location (Birmingham) Solution in Excel 2016 Using Excel QM
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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 (1 of 5)
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 (2 of 5)
FIGURE 9.2 Example of an Assignment Problem in a Transportation Network Format
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Linear Program for Assignment Example (3 of 5)
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
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Linear Program for Assignment Example (4 of 5)
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
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Linear Program for Assignment Example (5 of 5)
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
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Using Excel QM
PROGRAM 9.4 Mr. Fix-It Shop Assignment Solution in Excel 2016 Using Excel QM
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The Transshipment Problem (1 of 8)
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 (2 of 8)
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 (3 of 8)
FIGURE 9.3 Network Representation of a Transshipment Example
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The Transshipment Problem (4 of 8)
TABLE 9.3 Frosty Machines Transshipment Data
Blank | Blank | Blank | TO | Blank | Blank | Blank |
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 | Blank |
Minimize transportation costs associated with shipping snow blowers subject to demands and supplies
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The Transshipment Problem (5 of 8)
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 (6 of 8)
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 (7 of 8)
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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The Transshipment Problem (8 of 8)
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
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Using Excel QM
PROGRAM 9.5 Excel QM Solution to Frosty Machines Transshipment Problem in Excel 2016
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Maximal-Flow Problem (1 of 4)
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 (2 of 4)
Determine maximum number of cars from east to west for Waukesha WI road system
FIGURE 9.4 Road Network for Waukesha Maximal-Flow Example
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Maximal-Flow Problem (3 of 4)
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 (4 of 4)
Maximize flow = X61
subject to
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 | Blank | Blank | Blank | Capacities for arcs from node 3 |
X42 | ≤ | 1 | X43 | ≤ | 1 | X46 | ≤ | 1 | Capacities for arcs from node 4 |
X53 | ≤ | 1 | X56 | ≤ | 1 | Blank | Blank | Blank | Capacities for arcs from node 5 |
X62 | ≤ | 2 | X64 | ≤ | 1 | Blank | Blank | Blank | Capacities for arcs from node 6 |
(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 | Blank |
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Using Excel QM
PROGRAM 9.6 Waukesha Maximal-Flow Solution in Excel 2016 Using Excel QM
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Shortest-Route Problem (1 of 5)
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 (2 of 5)
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
FIGURE 9.5 Roads from Ray’s Plant to Warehouse
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Shortest-Route Problem (3 of 5)
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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Shortest-Route Problem (4 of 5)
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
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Shortest-Route Problem (5 of 5)
Minimize distance = 100X12 + 200X13 + 50X23 + 50X32
+ 200X24 + 200X42 + 100X25
+ 100X52 + 40X35 + 40X53 + 150X45
+ 150X54 + 100X46 + 100X56
subject to
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 |
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Using Excel QM (1 of 2)
PROGRAM 9.7 Ray Designs, Inc., Solution in Excel 2016 Using Excel QM
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Using Excel QM (2 of 2)
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
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Minimal-Spanning Tree Problem (1 of 8)
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 (2 of 8)
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 (3 of 8)
Lauderdale Construction
Housing project in Panama City Beach
Determine the
least expensive
way to provide
water and power
to each house
FIGURE 9.6 Network for Lauderdale Construction
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Minimal-Spanning Tree Problem (4 of 8)
Step 1 – Arbitrarily select node 1
Step 2 – Connect node 1 to node 3 (nearest)
FIGURE 9.7 First Iteration for Lauderdale Construction
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Minimal-Spanning Tree Problem (5 of 8)
Step 3 – Connect next nearest unconnected node, node 4
Continue for other unconnected nodes
FIGURE 9.8 Second and Third Iterations for Lauderdale Construction
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Minimal-Spanning Tree Problem (6 of 8)
Step 4 – Repeat the process
FIGURE 9.9 Last Four Iterations for Lauderdale Construction
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Minimal-Spanning Tree Problem (7 of 8)
Step 4 – Repeat the process
FIGURE 9.9 Last Four Iterations for Lauderdale Construction
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Minimal-Spanning Tree Problem (8 of 8)
TABLE 9.4 Summary of Steps in Lauderdale Construction Minimal-Spanning Tree Problem
STEP | CONNECTED NODES | UNCONNECTED NODES | CLOSE UNCONNECTED NODES | ARC SELECTED | ARC LENGTH | TOTAL DISTANCE |
1 | 1 | 2, 3, 4, 5, 6, 7, 8 | 3 | 1–3 | 2 | 2 |
2 | 1, 3 | 2, 4, 5, 6, 7, 8 | 4 | 3–4 | 2 | 4 |
3 | 1, 3, 4 | 2, 5, 6, 7, 8 | 2 or 6 | 3–2 | 3 | 7 |
4 | 1, 2, 3, 4 | 5, 6, 7, 8 | 5 or 6 | 2–5 | 3 | 10 |
5 | 1, 2, 3, 4, 5 | 6, 7, 8 | 6 | 3–6 | 3 | 13 |
6 | 1, 2, 3, 4, 5, 6 | 7, 8 | 8 | 6–8 | 1 | 14 |
7 | 1, 2, 3, 4, 5, 6, 8 | 7 | 7 | 8–7 | 2 | 16 |