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Transcript of Chapter 12 Network Models
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To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna12-1
2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Chapter 12
Network Models
Prepared by Lee Revere and John Large
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To accompany Quantitative Analysis
for Management, 9e
by Render/Stair/Hanna12-3
2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Chapter Outline
12.1 Introduction
12.2 Minimal-Spanning Tree
Technique
12.3 Maximal-Flow Technique12.4 Shortest-Route Technique
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Introduction
The presentation will cover threenetwork models that can be used to
solve a variety of problems:
1. the minimal-spanning tree technique,
2. the maximal-flow technique,
and
3. the shortest-route technique.
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Minimal-SpanningTree Technique
Definition:
The minimal-spanning tree technique
determines the path through the network
that connects all the points while
minimizing total distance.
For example:
If the points represent houses in asubdivision, the minimal spanning treetechnique can be used to determine thebest way to connect all of the houses toelectrical power, water systems, etc.
in a way thatminimizes the totaldistance or length of power lines orwater pipes.
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The Maximum FlowTechnique
Definition:
The maximal-flow technique finds the
maximum flow of any quantity or
substance through a network.
For example:
This technique can determine the
maximum number of vehicles (cars,trucks, etc.) that can go through a
network of roadsfrom one location
to another.
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To accompany Quantitative Analysis
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2006 by Prentice Hall, Inc.
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Shortest RouteTechnique
Definition:
Shortest route technique can find the
shortest path through a network.
For example:
This technique can find the shortest
route from one city to another through a
network of roads.
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2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Minimal-Spanning TreeSteps
1. Selecting any node in the network.
2. Connecting this node to the nearest
node minimizing the total distance.
3. Finding and connecting the nearestunconnected node.
If there is a tie for the nearest node, one
can be selected arbitrarily.
A tie suggests that there may be more than
one optimal solution.
4. Repeating the third step until all nodes
are connected.
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Minimal-SpanningTree Technique
Solving the network for Melvin
Lauderdale construction
Start by arbitrarily selecting node 1.
Since the nearest node is the third node at
a distance of 2 (200 feet), connect node 1
to node 3.
Shown in Figure 12.2 (2 slides hence)
Considering nodes 1 and 3, look for the
next-nearest node.
This is node 4, which is the closest to node 3
with a distance of 2 (200 feet).
Once again, connect these nodes (Figure
12.3a (3 slides hence).
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Figure 12.1: Network for
Lauderdale Construction
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Figure 12.2: First Iteration
Lauderdale Construction
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Fig 12.3a:Second Iteration
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Fig 12.3b:Third Iteration
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Summarize: Minimal-Spanning Tree Technique
Step 1: Select node 1
Step 2: Connect node 1 to node 3
Step 3: Connect the next nearest
nodeStep 4: Repeat the process
The total number of iterations to
solve this example is 7.This final solution is shown in the
following slide.
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Fig 12.5b:Third Iteration
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Final Solution to theMinimal-Spanning
Tree ExampleNodes 1, 2, 4, and 6 are all connected to node3. Node 2 is connected to node 5.
Node 6 is connected to node 8, and node 8 is
connected to node 7.
All of the nodes are now connected.
The total distance is found by adding the
distances for the arcs used in the spanning tree.
In this example, the distance is:
2 + 2 + 3 + 3 + 3 + 1 + 2 = 16 (or 1,600 feet).
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Maximal-FlowTechnique
Themaximal-flow techniqueallows themaximum amount of a material that can
flow through a network to be determined.
For example:
It has been used to find the maximumnumber of automobiles that can flow
through a state highway system.
An example:
Waukesha is in the process of developing
a road system for downtown.
City planners would like to determine the
maximum number of cars that can flow
through the town from west to east.
The road network is shown in Figure 12.6
(next slide).
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Road Network forWaukesha
Traffic can flow in both directions.
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Maximal-FlowTechnique (continued)
The Four Maximal-Flow Technique Steps:
1. Pick any path from the start (source) to thefinish (sink) with some flow.
If no path with flow exists, then the
optimal solution has been found.2. Find the arc on this path with the smallestflow capacity available.
Call this capacity C.
This represents the maximum additionalcapacity that can be allocated to thisroute.
3. For each node on this path, decrease the flowcapacity in the direction of flow by theamount C.
For each node on this path, increase the
flow capacity in the reverse direction bythe amount C.
4. Repeat these steps until an increase in flow isno longer possible.
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Solving the WaukeshaExample
Start by arbitrarily picking the path
126, at the top of the network.
What is the maximum flow from west
to east? It is 2 because only 2 units (200cars) can flow from node 2 to node 6.
Now we adjust the flow capacities
(Figure 12.7). As you can see, we
subtracted the maximum flow of 2along the path 126 in the direction of
the flow (west to east) and added 2 to
the path in the direction against the
flow (east to west). The result is the new path in Figure
12.7 (next slide).
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Capacity Adjustment
1
2
6
1
2
2
3
East
PoinWestPoint
Add 2
Subtract 2
Iteration 1
1
2
6
30
4
1EastPoin
West
Point
New path
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Solving the WaukeshaExample
The New Path reflects the new relative capacityat this stage.
The flow number by any node represents two
factors.
One factor is the flow that can comefromthat node.
The second factor is flow that can bereduced
coming intothe node.
The number 1 by node 1 indicates that 100 carscan flowfrom node 1 to node 2.
1
2
6
30
4
1EastPoint
West
Point
New path
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Solving the WaukeshaExample
The number 0 by node 2 on the path fromnode 2 to node 6 indicates that 0 cars can flow
from node 2 to node 6.
1
2
6
30
4
1
East
PointWest
Point
New path
The number 4 by node 6, on the path from node
6 to node 2, indicates that we can reduce the
flow into node 6 by 2 (or 200 cars) and that
there is a capacity of 2 (or 200 cars) that cancomefrom node 6.
These two factors total 4.
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Solving the
Waukesha Example
On the path from node 2 to node 1, thenumber 3 by node 2 shows that we can reducethe flow into node 2 by 2 (or 200 cars) andthat there is a capacity of 1 (or 100 cars)fromnode 2 to node 1
1
2
6
3 0
4
1
East
PointWest
Point
New path
At this stage, there is a flow of 200 cars
through the network from node 1 to node 2 to
node 6.
The new relative capacity reflects this.
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Repeat the Process
Now, repeat the process by picking another
path with existing capacity.
Can arbitrarily pick path 1246.
The maximum capacity along this path is 1.
In fact, the capacity at every node along this
path (1246) going from west to east is 1.
Remember, the capacity of branch 12 is
now 1 because 2 units (200 cars per hour)
are now flowing through the network.
So, need to increase the flow along path
1246 by 1 and adjust the capacity flow
(see next slide).
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Fig-12.8a: SecondIteration for Waukesha
1
2
4
6
1
3
1
11
1
Subtract 1
Old Path
Add 1
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Fig-12.8b: SecondIteration for Waukesha
1
2
4
3
5
6
40
0
2 0
2
4
0
6
1
2
30
10
2
0
0
1
EastPoint
West
Point
New Network
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Now, there is a flow of 3 units (300 cars):
200 cars per hour along path 126
plus
100 cars per hour along path 1246
Can the flow be further increased?
Yes, along path 1356.
This is the bottom path.
The maximum flow is 2 because this is the
maximum from node 3 to node 5.
The increased flow along this path is shown
in the next slide.
Continuing the Process
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Third Iteration
1
2
4
3
5
6
40
0
2 0
2
4
0
6
1
2
30
10
2
0
0
1
EastPoint
West
Point
Add 2
Subtract 2
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Again, repeat the process.
Try to find a path with any unusedcapacity through the network.
Carefully checking the third iteration inthe last slide reveals that there are nomore paths from node 1 to node 6 withunused capacity,
even though several other branches in the
network do have unused capacity.
The final network appears on the nextslide.
Continuing the Process
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Final Iteration
1
2
4
3
5
6
40
0
2 0
2
4
2
4
3
0
30
10
2
0
0
1
East
PointWest
Point
New Path
Path = 1, 3, 5, 6
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Final Network Flow
(Cars per Hour)PATH FLOW
1-2-6 200
1-2-4-6 1001-3-5-6 200
Total =500
The maximum flow of 500 cars per hour is
summarized in the following table:
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The Shortest-RouteTechnique
The shortest-route technique minimizes
the distance through a network.
The shortest-route technique finds how a
person or item can travel from onelocation to another while minimizing the
total distance traveled.
The shortest-route technique finds the
shortest route to a series of destinations.
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Example: From RaysPlant to Warehouse
For example,
Every day, Ray Design, Inc., must
transport beds, chairs, and other furniture
items from the factory to the warehouse.
This involves going through several
cities.
Ray would like to find the route with the
shortest distance. The road network is shown on the next
slide.
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Steps of the Shortest-Route Technique
1. Find the nearest node to the origin(plant). Put the distance in a box by the
node.
2. Find the next-nearest node to the origin
(plant), and put the distance in a box
by the node. In some cases, several
paths will have to be checked to find
the nearest node.
3. Repeat this process until you have
gone through the entire network.
The last distance at the ending node will
be the distance of the shortest route.
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Ray Design: 1st Iteration
Shortest-Route
Technique (continued)
The nearest node to the plant is node
2, with a distance of 100 miles. Thus, connect these two nodes.
1
2
3
4
5
615050
40
200
Warehouse
Plant
100
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Ray Design: 3rd Iteration
Shortest-Route
Technique (continued)
The nearest node to the plant is node
5, with a distance of 40 miles.
Thus, connect these two nodes.
1
2
3
4
5
615050
40
200
100
150 190
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4th and Final Iteration
Shortest Route
Technique (continued)
1
2
3
4
5
615050
40
200
100
150 190
290
Total Shortest Route =
100 + 50 + 40 + 100 = 290 miles.