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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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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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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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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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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


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


1

2

3

4

5

615050

40

200

100

150 190

290

Total Shortest Route =

100 + 50 + 40 + 100 = 290 miles.

Chapter 12 Network Models

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