Theta Dimensions - Uncorrected Sample Pages
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16 Networks – applications
243
Networks – applications
Mathematics and Statistics in the New Zealand Curriculum
Mathematics: Patterns and relationships
Level 7
• M7-5 Choose appropriate networks to find optimal solutions
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Network inspection and minimising retracing
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Achievement Standard
Mathematics and Statistics 2.5 – Use networks in solving problems
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Networks 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.
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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.
C
E
A
D
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B
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.
D
B
A
C
E
F
Most networks in real life are not traversable so, in order to travel along every arc (or line), some
arcs have to be retraced.
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244
2.5 Networks
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.
Example
(This is the network diagram for the Bridges of
Königsberg problem on page 232.)
Bank of river
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Island
Island
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.
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a Explain why the network is not
traversable.
b Show that the network can be made
traversable by adding an arc (or bridge).
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Travel along all arcs
Bank of river
Start
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1
8
Island
Island
6
End
5 4
3
Added arc
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(or bridge)
Bank of river
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Bank of river
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The 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.
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B
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A
F
7
5
E
13
8
C
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9
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D
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16 Networks – applications
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Arcs or pathways between odd vertices are:
BC, length 13
BE, length 6 + 5 = 11
BF, length 8
CE, length 7 + 10 = 17
CF, length 10
EF, 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
B
path, but let’s show how adding a repeated arc for EF (shown in orange)
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makes the network traversable:
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Start at one of the odd vertices (B or C).
A
F
C
A possible pathway is B → A → E → D → C → F → E → F → B → C.
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The nine arrows show the eight original arcs plus the added one.
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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
E
D
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vertices.
The possibilities are:
B
13
BC and EF (total length 20)
13
BE and CF (total length 21)
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8
BF and CE (total length 25).
A
F
C
The minimum (best) choice is BC and EF.
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We add repeated BC and EF arcs (shown in orange) to the network.
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7
5
9
Now there are no odd vertices, so the network is an Eulerian circuit and
you can traverse it and return to the starting point.
E
D
12
A possible pathway is B → A → E → D → C → F → E → F → B → C → B.
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DID YOU KNOW?
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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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2.5 Networks
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E
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F
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77
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D
4 The simplified network map shows the
main roads between New Plymouth and
towns in Taranaki. The distances are given
in kilometres.
Waitara
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2 This network below has
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lengths marked in metres.
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a Explain why the
network is traversable.
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b Where would
the starting and
ending points be for E
20
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?
c
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1 The network shown
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here has lengths
marked in kilometres.
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2
a Explain why the
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3
network is an
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Eulerian circuit.
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b What is the total
distance travelled
when traversing all
A
20
B
arcs in the network,
once only?
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In this exercise, all the diagrams are available on
blackline masters so that you can trace various
paths as you investigate different routes
B
through a network. These blackline
L
masters are provided in the Theta
M
Mathematics Teaching Resource.
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EXERCISE
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Inglewood
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New
Plymouth
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Stratford
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Mt Taranaki
(Egmont) 30
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3 Determine shortest closed paths for each of
these networks, and give the length of this
path. All distances are in kilometres.
a b
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Opunake
Hawera
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10
E
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A
6
5
C
A
E
45
7
4
D
25
B
B
9
40
35
20
D
C
30
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16 Networks – applications
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.
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.
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A
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Depot 7
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PUZZLE
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Visiting the zoo
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(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.
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1 Is it possible to walk from the
entrance back to the exit along
all the paths and not retrace
your footsteps? Explain.
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Giraffes
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 Crocodiles
total distance the sweeper
travels to a minimum? Which
path will the sweeper have to
clean twice?
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12
8
8
6
Kiwis
10
4
6
10
3 Is it possible to see all these
animals and visit the kiosk in
under an hour? Explain.
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Snakes
10
8
10
Lions
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Kiosk
(Times in minutes) Entrance/Exit
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2.5 Networks
INVESTIGATION
Kangaroo Air
Kangaroo 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.
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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.
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Brisbane
2115
Adelaide
752
Sydney
705
Melbourne
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1 Explain why it is not possible to fly, once only, on every route in this network.
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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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16 Networks – applications
249
Shortest path
ST
Look at this map of
the North Island. It
shows several possible
routes for travel
between Auckland
and Wellington, with
distances given in
kilometres.
ARTER
N
Auckland
126
Hamilton
1 What is the shortest
route shown
between Auckland
and Wellington?
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Taupo
Te
142
New
162 Kuiti
Napier
Plymouth
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160
Wanganui
71 28 151
Woodville
Palmerston North
80
143
Masterton
Wellington
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2 Give two reasons
why the shortest
route in question 1
may not be the one
most people use.
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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).
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Example
Determine the shortest path between A and F
for the network below.
D
10
9
15
F
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 = 26
ABCF = 21
ABDCF = 25
ABDF = 18
ADBCF = 35
ADCF = 29
ADF = 22
AEF = 27.
By inspection, the shortest path is ABDF with
length 18.
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