24316 Networks – applications

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Networks – applications16

Achievement Standard

Mathematics and Statistics 2.5 – Use networks in solving problems

Mathematics and Statistics in the New Zealand Curriculum

Mathematics: Patterns and relationships

Level 7• M7-5Chooseappropriatenetworkstofindoptimalsolutions

Network inspection and minimising retracingNetworks can be classified by whether they are traversable.

In general, a network is traversable if it is possible to travel along every arc once, and once only. Traversable networks have either no or two odd vertices. In the case where there are two odd vertices, the start and end points are each of the two odd vertices.

Example

This network is traversable – for example, follow the path A → B → C → A → D → C → E → D. There are two odd vertices, A and D.

If a network is traversable and you can start and end at the same point, then the network is called an Eulerian circuit. An Eulerian circuit has no odd vertices.

Example

This network is an Eulerian circuit – for example, follow the path A → B → C → F → E → D → C → A. All vertices are even.

Most networks in real life are not traversable so, in order to travel along every arc (or line), some arcs have to be retraced.

A

B C

E

D

A

B

EC

F

D

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Example

(This is the network diagram for the Bridges of Königsberg problem on page 232.)

IslandIsland

Bank of river

Bank of river

a Explain why the network is not traversable.

b Show that the network can be made traversable by adding an arc (or bridge).

Answer

a The network is not traversable because there are four odd vertices. The orders of the vertices are 3, 3, 3 and 5.

b Adding an arc anywhere will make the network traversable. One solution (the arc drawn in orange) is shown below. Note that the start and end points each have to be a remaining odd vertex.

IslandIsland

Bank of riverStart

End

186

5

23

4

7

Bank of river

Added arc (or bridge)

Retraced travel is necessary whenever there are more than two odd vertices. Note that the number of odd vertices in a network must be even. This means it is impossible, for example, to draw a network with 1, 3, 5, etc. odd vertices.

We can modify a network and, in particular, change it from non-traversable to traversable, by adding arcs. If a duplicate path is added between two odd vertices, it changes each of the vertices from odd to even.

9 Travel along all arcsThe shortest closed path of a network is a route that gives the shortest distance required to travel along all arcs. We use the term ‘closed’ here to mean that the travel must be confined to the network – short-cuts are not possible.

If the network is traversable, then determining the length of the shortest closed path is trivial – we just add up the lengths of all the arcs.

However, if the network is not traversable (this is indicated by its having more than two odd vertices) then, in order to travel along all arcs, some travel will have to be repeated. Repeated travel will be along arcs that connect a pair of odd vertices. We first identify all possible pairs of odd vertices, and then select the combination that gives the shortest repeated travel.

Example

a Determine the length of the shortest closed path for this network.b Determine a pathway that is a minimum and allows travel back to the

starting point.

Answer

a First, note that there are four odd vertices (B, C, E and F) so the network is not traversable.

13

109

12

57

6 8

A

B

C

DE

F

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Arcs or pathways between odd vertices are:BC, length 13BE, length 6 + 5 = 11BF, length 8CE, length 7 + 10 = 17CF, length 10EF, length 7.

The shortest of these paths is EF, which has length 7 units. This is the path that has to be repeated, shown by adding the orange arc EF to the upper diagram at right.The shortest closed path is the sum of all the arc lengths plus the repeated one.Shortest closed path = (5 + 6 + 7 + 8 + 9 + 10 + 12 + 13) + 7 = 77 units.

The question doesn’t ask for the pathway that gives the shortest closed path, but let’s show how adding a repeated arc for EF (shown in orange) makes the network traversable:

Start at one of the odd vertices (B or C).A possible pathway is B → A → E → D → C → F → E → F → B → C.

The nine arrows show the eight original arcs plus the added one.b Because it is additionally required that the pathway had to return to

the starting point, we examine all possible pairs of arcs joining odd vertices.The possibilities are:

BC and EF (total length 20)BE and CF (total length 21)BF and CE (total length 25).

The minimum (best) choice is BC and EF.We add repeated BC and EF arcs (shown in orange) to the network. Now there are no odd vertices, so the network is an Eulerian circuit and you can traverse it and return to the starting point.

A possible pathway is B → A → E → D → C → F → E → F → B → C → B.

13

109

12

5 77

6 8

A

B

C

DE

F

13

13

109

12

5 77

6 8

A

B

C

DE

F

DID YOU KNOW?

The problem of traversing all arcs in a network and returning to the starting point while covering the shortest possible distance is famously known as the Chinese Postman problem.

A postman delivering mail wants to make the delivery round as short as possible. The postman has to return to the starting point. Every street (arc) has to have mail delivered.

The problem is described by this name because it was originally studied by the Chinese mathematician, M K Kwan, in 1962.

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EXERCISE 16.01

In this exercise, all the diagrams are available on blackline masters so that you can trace various paths as you investigate different routes through a network. These blackline masters are provided in the Theta Mathematics Teaching Resource.

1 The network shown here has lengths marked in kilometres.

a Explain why the network is an Eulerian circuit.

b What is the total distance travelled when traversing all arcs in the network, once only?

2 This network below has lengths marked in metres.

a Explain why the network is traversable.

b Where would the starting and ending points be for traversing the entire network without travelling along any arc more than once?

c What is the length of the path in part b?

d What part of the network would you have to travel twice if you wanted to start and finish at point D?

3 Determine shortest closed paths for each of these networks, and give the length of this path. All distances are in kilometres.

a b

c A

B

CD

E

G

F

74129

59

68

3762

43 115

77

d

D

C

BA

F

E

5063

72 9294 42

11588

182

4 The simplified network map shows the main roads between New Plymouth and towns in Taranaki. The distances are given in kilometres.

43

30

42

3864

22

21

17

18NewPlymouth

Waitara

InglewoodStratford

Mt Taranaki(Egmont)

Hawera

Opunake

18

21

20

35

26

27

20

A B

C

DE

42

3

8

65

7

5

6

LM

B

A

B

CD

E8

10

7

6

9

45

20

40

3025

35

45

55

E

DB

C

A

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A road inspection team has to check the condition of all the roads. Describe a shortest closed cycle, starting and finishing in New Plymouth. Which roads will be checked twice? What is the total distance the team will travel?

5 The diagram shows a newspaper delivery round. The times in minutes taken to bike along each street are shown. Determine the shortest closed cycle if the round starts and ends at point A, and state its duration in minutes.

5

46

1513

6 11

9

B

C

D F

E

A7

6 A district council operates a recycling collection with a single truck once every two weeks. The simplified network diagram shows the routes along which recycled material has to be collected. The distances are given in kilometres. The collection starts and ends at the depot (marked orange on the diagram). Determine the minimum distance covered by the truck.

201418

19Depot

16

21

15

12

13

11

710

8

9

PUZZLE Visiting the zoo(Some of you may remember this problem from Alpha Mathematics. Now you can use network principles to answers parts of it.)

This diagram shows how long it takes to walk around a large zoo and see some of the animals. Times are in minutes.

1 Is it possible to walk from the entrance back to the exit along all the paths and not retrace your footsteps? Explain.

2 The zoo has bought a mechanical sweeper to clean the paths. Where should the sweeper be located (name the animal enclosure) to keep the total distance the sweeper travels to a minimum? Which path will the sweeper have to clean twice?

3 Is it possible to see all these animals and visit the kiosk in under an hour? Explain.

Crocodiles

Lions (Times in minutes) Entrance/Exit

Kiwis

Giraffes

Snakes

Kiosk10

4

6

6

68

10

10

10

8

8 12

10

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INVESTIGATION Kangaroo AirKangaroo Air sells an air pass valid for travel, once only, along each route in their network in Australia. The map shows the distances between the cities the airline flies to. All distances are in kilometres.

A traveller wants to fly every route on the network. In some cases, the traveller may have to pay for individual flights in addition to using the air pass, and he wants this cost to be as low as possible. The cost of flights is proportional to their distance.

3277

752

705642

2701

2115PerthAdelaide

Melbourne

Sydney

Brisbane

N

E

S

W

1 Explain why it is not possible to fly, once only, on every route in this network.

2 Determine the best way for a traveller to fly at least once on every route while keeping the cost as low as possible. Give the route, and where the journey starts and ends.

3 Another traveller wants to fly the entire network and return to her starting point. What is the minimum total distance she will fly?

4 A third traveller just wants to visit all five cities on the network. The traveller has to return to his starting point. Find the shortest possible route combination.

5 A fourth traveller is visiting from outside Australia, so can arrive on another airline at any of the airports and depart from a different airport. What is the shortest distance the traveller will fly on Kangaroo Air if she visits all five cities?

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Many networks offer a choice of routes from one point to another. We often want to find the route with the shortest distance, or the one that takes the least time.

A ‘shortest path’ problem can be solved either by inspection or by following an algorithm (set sequence of checking steps).

Shortest path

STARTER

Look at this map of the North Island. It shows several possible routes for travel between Auckland and Wellington, with distances given in kilometres.

1 What is the shortest route shown between Auckland and Wellington?

2 Give two reasons why the shortest route in question 1 may not be the one most people use.

Hamilton

Auckland

TaupoTeKuiti Napier

Masterton

Woodville

Wellington

NewPlymouth

Wanganui

Palmerston North

126

82 158

162160 250

71 28

143

102

80

151248

142

N

Example

Determine the shortest path between A and F for the network below.

AD

BC

FE

74

5 610

913

12

15

TIP

When listing paths, it is easiest to keep track of them and make sure all are included by listing them in alphabetical order.

Answer

The paths and their lengths are:ABCDF = 26ABCF = 21ABDCF = 25ABDF = 18ADBCF = 35ADCF = 29ADF = 22AEF = 27.By inspection, the shortest path is ABDF with length 18.

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