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CENTRE FOR EDUCATIONAL DEVELOPMENT

Developing practice around the realigned Level 2 Mathematics and Statistics standardsWorkshop Four Anne Lawrence, Alison Fagan , Cami SawyerAdvisers in Secondary Numeracy & Mathematics

http://ced-mxteachers-news-site.wikispaces.com/

Level 3 Consultation In small groups pick 2-3 standards and

discuss what has changed

• Think about the pathways from last time• Changes made to level 1 & 2

Resources• Draft level 3 standards• Draft Matrix• Summary of what has changed

2


Level 3 Consultation Manawatu:3.6 too much in it3.8 evaluation – concern at higher literacy

needed3.8 Using existing data sets – but this seems

to conflict with using each component of PPDAC ie posing problem, collecting data?

3.9 assume that making a prediction3


Level 3 Consultation Hawkes Bay feedback:

4


AO M 8-7 (trigonometric, polynomial, and other non-linear equations)

What is new/changed?* Manipulating logs will be new

AO M 8-11 (differentiation, integration, and anti-differentiation techniques)

What is new/changed? * There is no integration at level 7

* This does not include related rates of change, integration of relations, or volume of revolution.

What will happen to Simultaneous Equations and Linear Programming?

5


TKI Senior Secondary Teaching and Learning Guides

Achievement Objective M7- 5In a range of meaningful contexts, students will be engaged in thinking mathematically and statistically. They will solve problems and model situations that require them to:Choose appropriate networks to find optimal solutions.

Indicators• Solves problems that can be modelled by networks• Uses trial-and-improve methods to develop

algorithms for solving network problems

6


Network Definitions

7


Spot the FeaturesIdentify the features or terminology from the last activity that are shown in this tramping network of huts linked by tracks.

Do the US 5249 Tasks8

Ace

Becks Caps

DavidFreddy

Gum

HappyEddy


NetworksAS2.5 –Use networks in solving problems

Look at the standard:

– What are the understandings required? – What do you think should be the step up from

achieve to merit? merit to excellence?

9



10



11


Minimum Spanning Tree

Kruskal’s algorithm

1. Select the shortest edge in a network

2. Select the next shortest edge which does not create a cycle

3. Repeat step 2 until all vertices have been connected

Prim’s algorithm

1. Select any vertex

2. Select the shortest edge connected to that vertex

3. Select the shortest edge connected to any vertex already connected

4. Repeat step 3 until all vertices have been connected


A cable company want to connect five villagesto their network which currently extends to the

market town of Avonford. What is the minimum length of cable needed?

Avonford Fingley

Brinleigh Cornwell

Donster

Edan

2

7

45

8 64

5

3

8


First model the situation as a network, then the problem is to find the minimum connector for the network

A F

B C

D

E

2

7

45

8 64

5

3

8


AF

BC

D

E

2

7

45

8 64

5

3

8

ED 2AB 3AE 4CD 4BC 5EF 5CF 6AF 7BF 8CF 8

Kruskal’s Algorithm


List the edges in order of size:

Select the shortestedge in the network

ED 2


AF

BC

D

E

2

7

45

8 64

5

3

8


Select the next shortest edge which does not create a cycle

ED 2AB 3


AF

BC

D

E

2

7

45

8 64

5

3

8


Select the next shortest edge which does notcreate a cycle

ED 2AB 3CD 4 (or AE 4)


AF

BC

D

E

2

7

45

8 64

5

3

8


Select the next shortest edge which does not create a cycle

ED 2AB 3CD 4 AE 4


AF

BC

D

E

2

7

45

8 64

5

3

8

Select the next shortest edge which does notcreate a cycle

ED 2AB 3CD 4 AE 4BC 5 forms a cycle

EF 5


AF

BC

D

E

2

7

45

8 64

5

3

8


All vertices have been connected.The solution isED 2AB 3CD 4 AE 4EF 5

Total weight of tree: 18


AF

BC

D

E

2

7

45

8 64

5

3

8


AF

BC

D

E

2

7

45

8 64

5

3

8

Select any vertex

A

Select the shortest edge connected to that vertex

AB 3

Prim’s Algorithm


AF

BC

D

E

2

7

45

8 64

5

3

8

Select the shortestedge connected to any vertex already connected.

AE 4

Prim’s Algorithm



ED 2

Prim’s Algorithm

AF

BC

D

E

2

7

45

8 64

5

3

8



DC 4

Prim’s Algorithm

AF

BC

D

E

2

7

45

8 64

5

3

8



CB 5 forms a cycle

EF 5

Prim’s Algorithm

AF

BC

D

E

2

7

45

8 64

5

3

8


Prim’s Algorithm

AF

BC

D

E

2

7

45

8 64

5

3

8

All vertices have beenconnected.The solution is

ED 2AB 3CD 4 AE 4EF 5Total weight of tree: 18http://ced-mxteachers-news-site.wikispaces.com/

• Both algorithms will always give solutions with the same length.

• They will usually select edges in a different order – students need to show this in their working.

• Occasionally these algorithms will use different edges – this may happen when you have to choose between edges with the same length. In this case there is more than one minimum connector for the network.

Prim’s and Kruskal’s Algorithms


Dijkstra’s Algorithmfinds the shortest path from the start vertex to every other

vertex in the network. We will find the shortest path from A to G

4

3

7

1

4

2 4

7

25

3 2

A

C

D

B F

E

G


Dijkstra’s Algorithm

Order in which vertices are labelled.

Permanent label = Distance from A to vertex

Working label

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

1st 0

Label vertex A 1st as it is the first vertex labelled


A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

4

3

7

We update each vertex adjacent to A with a ‘working value’ for its distance from A.

1st 0



A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

4

3

7

2nd 3

Vertex C is closest to A so we give it a permanent label 3. C is the 2nd vertex to be permanently labelled.

1st 0


Order in which vertices are labelled.

Permanent label = Distance from A to vertex

Working label

Look at ALL the working labels (no ordinal yet). Which is smallest?


We update each vertex adjacent to C with a ‘working value’ for its total distance from A, by adding its distance from C to C’s permanent label of 3.

6

8

1st 0

4

7

2nd 33

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

6 < 7 so replace the t-label here



6

8

1st 0

4

7

2nd 33

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

The vertex with the smallest temporary label is B, so make this label permanent. B is the 3rd vertex to be permanently labelled.

3rd 4




6

8

4

7

3

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

3rd 4

We update each vertex adjacent to B with a ‘working value’ for its total distance from A, by adding its distance from B to B’s permanent label of 4.

5



1st 0

2nd 3


6

8

4

7

3

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

5

8

The vertex with the smallest temporary label is D, so make this label permanent. D is the 4th vertex to be permanently labelled.

4th 5


1st 0

2nd 3

3rd 4


6

8

4

7

3

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

5

8

4th 5

We update each vertex adjacent to D with a ‘working value’ for its total distance from A, by adding its distance from D to D’s permanent label of 5.


12

7


7


1st 0

2nd 3

3rd 4


6

8

4

7

3

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

5

8

12

7

7

The vertices with the smallest temporary labels are E and F, so choose one and make the label permanent. E is chosen - the 5th vertex to be permanently labelled.

5th 7


1st 0

2nd 3

3rd 4

4th 5



6

8

4

7

3

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

5

8

12

7

7

5th 7

We update each vertex adjacent to E with a ‘working value’ for its total distance from A, by adding its distance from E to E’s permanent label of 7.


9


1st 0

2nd 3

3rd 4

4th 5


6

8

4

7

3

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

5

8

12

7

7

The vertex with the smallest temporary label is F, so make this label permanent.F is the 6th vertex to be permanently labelled.

9

6th 7


1st 0

2nd 3

3rd 4

4th 5

5th 7



6

8

4

7

3

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

5

8

12

7

7

9

6th 7

We update each vertex adjacent to F with a ‘working value’ for its total distance from A, by adding its distance from F to F’s permanent label of 7.

11 > 9 so do not replace the t-label here


1st 0

2nd 3

3rd 4

4th 5

5th 7


6

8

4

7

3

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

5

8

12

7

7

9

6th 7

G is the final vertex to be permanently labelled.

7th 9


1st 0

2nd 3

3rd 4

4th 5

5th 7

Can you SEE the shortest path from A to G?


6

8

4

7

3

A

C

D

B F

E

G

4

3

7

1

4

2 4

7

25

3 2

5

8

12

7

7

9

6th 7

7th 9

To find the shortest path from A to G, start from G and work backwards, choosing arcs for which the difference between the permanent labels is equal to the arc length.

The shortest path is ABDEG, with length 9.


1st 0

2nd 3

3rd 4

4th 5

5th 7

Assessment JudgementsUsing the assessment activity 2.5A

• Complete the task• Examine the Assessment Schedule• Compare with your own solution

44


Discussion• Where will this learning fit in your curriculum

level 7 (NCEA level 2) courses?

• What prior knowledge will students need to access this AO and standard?

• What are some of the new ideas in this standard that you think are important?

• Where will it lead – careers and pathways?

45


http://myjazmanie.com/wp-content/uploads/2009/06/discussion.jpg

Where does this go?

46

http://ced-mxteachers-news-site.wikispaces.com/http://www.newton.ac.uk/wmy2kposters/june/

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