MATHematics CMatrices

Lillian Tey

TERM 2, 2016

1 ROUND-ROBIN BOWLING TOURNAMENT

a. Represent this information as a diagraph.

b. Represent this information as a dominance matrix called ‘D’. Use the order of names in the first sentence.

D=[defeats J A M D SJ 0 0 1 0 1A 1 0 0 0 0M 0 1 0 1 0D 1 1 0 0 1S 0 1 1 0 0

]*Note that the leading diagonal is zero as players cannot ‘win’ against themselves.

c. Calculate the Dominance Vector ‘V’. Explain the rank order of the competitors so far. Can you be sure of a definite rank order?

Dominance Vector:

V D=[0+0+1+0+11+0+0+0+00+1+0+1+01+1+0+0+10+1+1+0+0

] ¿ [21232]

From the dominance vector, it is evident that: David is in first place with 3 points, Juliet, Mat and Sam are tied for fourth with 2 points, and Anne is in last place with 1 point.

This can be represented as:

{D } { JMS }{ A }

Based on the evidence above, it can be said that the Dominance Vector does not give a definite rank order as 3 of the competitors are ranked equally. The competitors must be distinguished between in order to establish a rank order between all of them.

d. Find D + D2. Find a new Dominance vector based on D + D2 and explain the rank order of the competitors and why it also fails to yield a definite rank order.

D+D2=[0 0 1 0 11 0 0 0 00 1 0 1 01 1 0 0 10 1 1 0 0

]+[0 0 1 0 11 0 0 0 00 1 0 1 01 1 0 0 10 1 1 0 0

]2

D+D2=[0 0 1 0 11 0 0 0 00 1 0 1 01 1 0 0 10 1 1 0 0

]+[0 0 1 0 11 0 0 0 00 1 0 1 01 1 0 0 10 1 1 0 0

] [0 0 1 0 11 0 0 0 00 1 0 1 01 1 0 0 10 1 1 0 0

]D+D2=[0 0 1 0 1

1 0 0 0 00 1 0 1 01 1 0 0 10 1 1 0 0

]+[0 2 1 1 00 0 1 0 12 1 0 0 11 1 2 0 11 1 0 1 0

]D+D2=[0 2 2 1 1

1 0 1 0 12 2 0 1 12 2 2 0 21 2 1 1 0

]V D+D2=[0+2+2+1+11+0+2+0+1

2+2+0+1+12+2+2+0+21+2+1+1+0

]=[64685]

It is clear that while the dominance vector creates a more distinguished representation of the ranking of each competitor, it still does not determine a definite rank order. This is due to the fact that two of the competitors (Juliette and Mat) still have an equal ranking, and therefore, a distinction must be made between the two. Although this is so, a clear first (David) and fourth (Sam) place has now been established in this tournament. This is shown below:

{D } { JM }{S }{ A }

e. Find D + 0.5D2 + 0.3D3 and its associated Dominance Vector. Has the rank been established? Make some suggestions as to what you could do mathematically to establish a rank order.

D+0.5D2+0.3D3=[0 0 1 0 11 0 0 0 00 1 0 1 01 1 0 0 10 1 1 0 0

]+0.5[0 2 1 1 00 0 1 0 12 1 0 0 11 1 2 0 11 1 0 1 0

]+0.3[0 0 1 0 11 0 0 0 00 1 0 1 01 1 0 0 10 1 1 0 0

][0 2 1 1 00 0 1 0 12 1 0 0 11 1 2 0 11 1 0 1 0

]D+0.5D2+0.3D3=[0 0 1 0 1

1 0 0 0 00 1 0 1 01 1 0 0 10 1 1 0 0

]+[ 0 1 0.5 0.5 00 0 0.5 0 0.51 0.5 0 0 0.50.5 0.5 1 0 0.50.5 0.5 0 0.5 0

]+0.3 [3 2 0 1 10 2 1 1 01 1 3 0 21 3 2 2 12 1 1 0 2

]

D+0.5D2+0.3D3=[ 0 1 1.5 0.5 11 0 0.5 0 0.51 1.5 0 1 0.51.5 1.5 1 0 1.50.5 1.5 1 0.5 0

]+[0.9 0.6 0 0.3 0.30 0.6 0.3 0.3 00.3 0.3 0.9 0 0.60.3 0.9 0.6 0.6 0.30.6 0.3 0.3 0 0.6

]D+0.5D2+0.3D3=[0.9 1.6 1.5 0.8 1.3

1 0.6 0.8 0.3 0.51.3 1.8 0.9 1 1.11.8 2.4 1.6 0.6 1.81.1 1.8 1.3 0.5 0.6

]V D+0.5D2+0.3 D3=[0.9 1.6 1.5 0.8 1.3

1 0.6 0.8 0.3 0.51.3 1.8 0.9 1 1.11.8 2.4 1.6 0.6 1.81.1 1.8 1.3 0.5 0.6

]=[6.13.26.18.25.3

]After finding the associated Dominance Vector of D + 0.5D2 + 0.3D3, a definite rank order has again not been yielded as two competitors both have the same score, as determined previously. The current ranking is as follows:

{D } { JM }{S }{A }

In order to mathematically establish a rank order, the fourth-order influence may be investigated. Generally, the influences are investigated depending on the amount of players, i.e. If there are m players, Mm−1 is investigated. Arbitrary constants are applied accordingly to weight the influences; i.e. M +xM 2+ yM 3. In this case, since constants of 0.5 and 0.3 have been substituted respectively, a constant lower than 0.3 will be applied.

Another suggestion involves subtracting total losses from total wins, determining the initial order rank (r1). This can be determined by using the formula M 1−M T 1 where ‘1’ represents a column matrix of 1s (QSA, 2010). This therefore produces a matrix that determines whether a competitor has won more than they have lost, or vice versa (Total Wins – Total Losses). This method, though, does not take into account the weighting of the seed. In theory, Anne should be able to defeat anyone who has been defeated by Juliet, though this is not the case as Anne lost to both Mat and Sam. Although this is so, this allows for more information to be gained and therefore increases the accuracy of a ranking.

2 AUSTRALIAN WOMEN’S SOCCER COMPETITION

1. Use dominance matrices to try to obtain a rank order for the teams using the order: M, Q, F, S, P, W in your matrix. Allocate 2 for a win, 1 for a draw and 0 for a loss or has not yet played, in your dominance matrices.

D=[defeats M Q F S P WM 0 0 2 2 0 0Q 0 0 0 2 0 0F 0 0 0 1 0 0S 0 0 1 0 0 0P 2 2 0 0 0 2W 0 2 2 0 0 0

] V D=[0+0+2+2+0+00+0+0+2+0+00+0+0+1+0+00+0+1+0+0+02+2+0+0+0+20+2+2+0+0+0

]=[421164]

D+D2=[0 0 2 2 0 00 0 0 2 0 00 0 0 1 0 00 0 1 0 0 02 2 0 0 0 20 2 2 0 0 0

] + [0 0 2 2 0 00 0 0 2 0 00 0 0 1 0 00 0 1 0 0 02 2 0 0 0 20 2 2 0 0 0

] [0 0 2 2 0 00 0 0 2 0 00 0 0 1 0 00 0 1 0 0 02 2 0 0 0 20 2 2 0 0 0

] D+D2=[

0 0 2 2 0 00 0 0 2 0 00 0 0 1 0 00 0 1 0 0 02 2 0 0 0 20 2 2 0 0 0

] + [0 0 2 2 0 00 0 2 0 0 00 0 1 0 0 00 0 0 1 0 00 4 8 8 0 00 0 0 6 0 0

] D+D2=[

0 0 4 4 0 00 0 2 2 0 00 0 1 1 0 00 0 1 1 0 02 6 8 8 0 20 2 2 6 0 0

]V D+D2=[

0+0+4+4+0+00+0+2+2+0+00+0+1+1+0+00+0+1+1+0+02+6+8+8+0+20+2+2+6+0+0

]=[84222610

]

D+0.5D2+0.25D3=[0 0 2 2 0 00 0 0 2 0 00 0 0 1 0 00 0 1 0 0 02 2 0 0 0 20 2 2 0 0 0

] + 0.5 [0 0 2 2 0 00 0 2 0 0 00 0 1 0 0 00 0 0 1 0 00 4 8 8 0 00 0 0 6 0 0

] + 0.25 [0 0 2 2 0 00 0 2 0 0 00 0 1 0 0 00 0 0 1 0 00 4 8 8 0 00 0 0 6 0 0

] [0 0 2 2 0 00 0 0 2 0 00 0 0 1 0 00 0 1 0 0 02 2 0 0 0 20 2 2 0 0 0

]D+0.5D2+0.25D3=[

0 0 2 2 0 00 0 0 2 0 00 0 0 1 0 00 0 1 0 0 02 2 0 0 0 20 2 2 0 0 0

] + [0 0 1 1 0 00 0 1 0 0 00 0 0.5 0 0 00 0 0 0.5 0 00 2 4 4 0 00 0 0 3 0 0

] + 0.25 [0 0 2 2 0 00 0 0 2 0 00 0 0 1 0 00 0 1 0 0 00 0 8 16 0 00 0 6 0 0 0

]D+0.5D2+0.25D3=[

0 0 2 2 0 00 0 0 2 0 00 0 0 1 0 00 0 1 0 0 02 2 0 0 0 20 2 2 0 0 0

] + [0 0 1 1 0 00 0 1 0 0 00 0 0.5 0 0 00 0 0 0.5 0 00 2 4 4 0 00 0 0 3 0 0

] + [0 0 0.5 0.5 0 00 0 0 0.5 0 00 0 0 0.25 0 00 0 0.25 0 0 00 0 2 5 0 00 0 1.5 0 0 0

]D+0.5D2+0.25D3=[

0 0 4.5 4.5 0 00 0 2 2.5 0 00 0 1 1.25 0 00 0 1.25 1 0 02 6 10 12 0 20 2 3.5 6 0 0

]V D+0.5D2+0.25D 3=[

0+0+4.5+4.5+0+00+0+2+2.5+0+00+0+1+1.25+0+00+0+1.25+1+0+02+6+10+12+0+20+2+3.5+6+0+0

]=[94.52.252.253211.5

]2. Use the ranking you have calculated to discuss the outcomes of the next three rounds for each game as outlined.

In the first game of round 4, the Wests defeated the Fremantle Stripes. This is predictable due to the fact that the Stripes are currently tied for last place. In the second game of the 4th round, the Kangaroo Paws defeated the Quokkas. In relation to the ranking, this is also a predictable result as the Paws are the highest ranking team in comparison to the Quokkas, who happen to be ranked 4 th. In the Saint Mary’s vs. Swans game, the same occurrence was repeated as with the previous games; the higher ranking team (in this case, Saint Mary’s) defeated the lower ranking team (Swans). In round 5, the Wests drew with Saint Mary’s. From the ranking determined, both teams were comparatively even as they had both defeated two other teams in the first three rounds. In the Quokkas vs. Fremantle Stripes game, the Stripes were again defeated, as predicted from the rankings. The third game of round 5, the Paws defeated the Swans – it was a match between the highest ranking and lowest ranking teams. In round 6, the first game again represented a higher rank defeating the lower rank, as the Swans lost to the Wests. The Mary’s defeated the Quokkas in round 6, showing that again the result was predictable due to the ranking produced from the first three rounds. This was also evident in the final game of round 6, as the highest ranking team (the Paws) defeated the tied lowest team (Stripes).

Overall, the results of each game were predictable as they were consistent with the results from the first three rounds of the competition.

The rankings for the next three rounds are outlined below.

D2=[defeats M Q F S P WM 0 2 0 2 0 1Q 0 0 2 0 0 0F 0 0 0 0 0 0S 0 0 0 0 0 0P 0 2 2 2 0 0W 1 0 2 2 0 0

]V D2=[

0+2+0+2+0+10+0+2+0+0+00+0+0+0+0+00+0+0+0+0+00+2+2+2+0+01+0+2+2+0+0

]=[520065]

D2+D22=[0 2 0 2 0 10 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 2 2 0 01 0 2 2 0 0

] + [0 2 0 2 0 10 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 2 2 0 01 0 2 2 0 0

][0 2 0 2 0 10 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 2 2 0 01 0 2 2 0 0

]D2+D2

2=[0 2 0 2 0 10 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 2 2 0 01 0 2 2 0 0

] + [1 0 6 2 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 4 0 0 00 2 0 2 0 1

]D2+D2

2=[1 2 6 4 0 10 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 6 2 0 01 2 2 4 0 1

]V D2+D2

2=[1+2+6+4+0+10+0+2+0+0+00+0+0+0+0+00+0+0+0+0+00+2+6+2+0+01+2+2+4+0+1

]=[142001010

]

D2+0.5D22+0.25D 2

3=[0 2 0 2 0 10 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 2 2 0 01 0 2 2 0 0

] + 0.5 [1 0 6 2 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 4 0 0 00 2 0 2 0 1

] + 0.25 [1 0 6 2 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 4 0 0 00 2 0 2 0 1

] [0 2 0 2 0 10 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 2 2 0 01 0 2 2 0 0

]D2+0.5D2

2+0.25D 23=[0 2 0 2 0 10 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 2 2 0 01 0 2 2 0 0

] + 0.5 [1 0 6 2 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 4 0 0 00 2 0 2 0 1

] + 0.25 [0 2 0 2 0 10 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 01 0 6 2 0 0

]D2+0.5D2

2+0.25D 23=[0 2 0 2 0 10 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 2 2 0 01 0 2 2 0 0

]+¿ [0.5 0 3 1 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 2 0 0 00 1 0 1 0 0.5

] + [0 0.5 0 0.5 0 0.250 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 00.25 0 1.5 0.5 0 0

]D2+0.5D2

2+0.25D23=[

0.5 2.5 3 3.5 0 1.250 0 2 0 0 00 0 0 0 0 00 0 0 0 0 00 2 4 2 0 01.25 1 3.5 3.5 0 0

]V D2+0.5 D2

2+0.25D23=[0.5+2.5+3+3.5+0+1.25

0+0+2+0+0+00+0+0+0+0+00+0+0+0+0+00+2+4+2+0+0

1.25+1+3.5+3.5+0+0]=[10.7520089.25

]When comparing the two results, it is evident that while the Kangaroo Paws, Fremantle Stripes and the Swans remain in 4th and tied 6th places respectively, the three other competing teams’ positions have changed. Although this is so, in general, the games mirror the rankings in the way that a higher ranked team will beat a lower ranked team. Also, from both results, it is clear that the Stripes and the Swans remain with the same ranking due to the fact that they only gain points from tying with each other, and are do not gain points from any other teams. The ranking is therefore as follows:

{M } {W } {P }{Q }{FS }

3 FUTSAL COMPETITION

W=[defeats CCC RSHS SC SSHS BPSH WPSH MSHS PRSH MPSHCCC 0 0 0 0 1 1 1 0 1RSHS 0 0 1 0 1 0 1 0 1SC 1 0 0 0 0 0 0 0 0SSHS 0 0 0 0 1 1 1 1 1BPSH 0 0 0 0 0 0 1 0 0WPSH 0 0 1 0 0 0 0 1 0MSHS 0 0 0 0 0 0 0 0 0PRSH 0 1 1 0 0 0 0 0 0MPSH 0 0 0 0 0 0 1 0 0

]L=[

lost CCC RSHS SC SSHS BPSH WPSH MSHS PRSH MPSHCCC 0 0 1 0 0 0 0 0 0RSHS 0 0 0 0 0 0 0 1 0SC 0 1 0 0 0 1 0 1 0SSHS 0 0 0 0 0 0 0 0 0BPSH 1 1 0 1 0 0 0 0 0WPSH 1 0 0 1 0 0 0 0 0MSHS 1 1 0 1 1 0 0 0 1PRSH 0 0 0 1 0 1 0 0 0MPSH 1 1 0 1 0 0 0 0 0

]W 2=[

0 0 0 0 1 1 1 0 10 0 1 0 1 0 1 0 11 0 0 0 0 0 0 0 00 0 0 0 1 1 1 1 10 0 0 0 0 0 1 0 00 0 1 0 0 0 0 1 00 0 0 0 0 0 0 0 00 1 1 0 0 0 0 0 00 0 0 0 0 0 1 0 0

]×[0 0 0 0 1 1 1 0 10 0 1 0 1 0 1 0 11 0 0 0 0 0 0 0 00 0 0 0 1 1 1 1 10 0 0 0 0 0 1 0 00 0 1 0 0 0 0 1 00 0 0 0 0 0 0 0 00 1 1 0 0 0 0 0 00 0 0 0 0 0 1 0 0

]W 2=[

0 0 1 0 0 0 2 1 01 0 0 0 0 0 2 0 00 0 0 0 1 1 1 0 10 1 2 0 0 0 2 1 00 0 0 0 0 0 0 0 01 1 1 0 0 0 0 0 00 0 0 0 0 0 0 0 01 0 1 0 1 0 1 0 10 0 0 0 0 0 0 0 0

]

L2=[0 0 1 0 0 0 0 0 00 0 0 0 0 0 0 1 00 1 0 0 0 1 0 1 00 0 0 0 0 0 0 0 01 1 0 1 0 0 0 0 01 0 0 1 0 0 0 0 01 1 0 1 1 0 0 0 10 0 0 1 0 1 0 0 01 1 0 1 0 0 0 0 0

]×[0 0 1 0 0 0 0 0 00 0 0 0 0 0 0 1 00 1 0 0 0 1 0 1 00 0 0 0 0 0 0 0 01 1 0 1 0 0 0 0 01 0 0 1 0 0 0 0 01 1 0 1 1 0 0 0 10 0 0 1 0 1 0 0 01 1 0 1 0 0 0 0 0

]L2=[

0 1 0 0 0 1 0 1 00 0 0 1 0 1 0 0 01 0 0 2 0 1 0 1 00 0 0 0 0 0 0 0 00 0 1 0 0 0 0 1 00 0 1 0 0 0 0 0 02 2 1 2 0 0 0 1 01 0 0 1 0 0 0 0 00 0 1 0 0 0 0 1 0

]W+W 2−L−L2

¿ [0 0 0 0 1 1 1 0 10 0 1 0 1 0 1 0 11 0 0 0 0 0 0 0 00 0 0 0 1 1 1 1 10 0 0 0 0 0 1 0 00 0 1 0 0 0 0 1 00 0 0 0 0 0 0 0 00 1 1 0 0 0 0 0 00 0 0 0 0 0 1 0 0

]+[0 0 1 0 0 0 2 1 01 0 0 0 0 0 2 0 00 0 0 0 1 1 1 0 10 1 2 0 0 0 2 1 00 0 0 0 0 0 0 0 01 1 1 0 0 0 0 0 00 0 0 0 0 0 0 0 01 0 1 0 1 0 1 0 10 0 0 0 0 0 0 0 0

]−[0 0 1 0 0 0 0 0 00 0 0 0 0 0 0 1 00 1 0 0 0 1 0 1 00 0 0 0 0 0 0 0 01 1 0 1 0 0 0 0 01 0 0 1 0 0 0 0 01 1 0 1 1 0 0 0 10 0 0 1 0 1 0 0 01 1 0 1 0 0 0 0 0

]−[0 1 0 0 0 1 0 1 00 0 0 1 0 1 0 0 01 0 0 2 0 1 0 1 00 0 0 0 0 0 0 0 00 0 1 0 0 0 0 1 00 0 1 0 0 0 0 0 02 2 1 2 0 0 0 1 01 0 0 1 0 0 0 0 00 0 1 0 0 0 0 1 0

]W+W 2−L−L2=[

0 −1 0 0 1 0 3 0 11 0 1 −1 1 −1 3 −1 10 −1 0 −2 1 −1 1 −2 −10 1 2 0 1 1 3 2 1

−1 −1 −1 −1 0 0 1 −1 00 1 1 −1 0 0 0 1 0

−3 −3 −1 −3 −1 0 0 −1 −10 1 2 −2 1 −1 1 0 1

−1 −1 −1 −1 0 0 1 −1 0

]

V

W +W 2−L−L2=¿

CCCRSHSSCSSHSBPSHWPSHMSHSPRSHMPSH

[44

−511−42

−133

−4]¿

The ranking is shown as follows:

{ SSHS } {CCCRSHS}{PRSH } {WPSH } {BPSHMPSH}{SC } {MSHS }

This shows that a definite ranking has not been established for some of the competing teams (CCC, RSHS, BPSH and MPSH). To solve this, the three-stage win and loss matrices (W 3∧L3¿ are established and added into the initial equation formulated (W+W 2−L−L2 ¿ to produce the

matrix W+W 2+W 3−L−L2−L3.

W 3=[0 0 0 0 1 1 1 0 10 0 1 0 1 0 1 0 11 0 0 0 0 0 0 0 00 0 0 0 1 1 1 1 10 0 0 0 0 0 1 0 00 0 1 0 0 0 0 1 00 0 0 0 0 0 0 0 00 1 1 0 0 0 0 0 00 0 0 0 0 0 1 0 0

]×[0 0 0 0 1 1 1 0 10 0 1 0 1 0 1 0 11 0 0 0 0 0 0 0 00 0 0 0 1 1 1 1 10 0 0 0 0 0 1 0 00 0 1 0 0 0 0 1 00 0 0 0 0 0 0 0 00 1 1 0 0 0 0 0 00 0 0 0 0 0 1 0 0

]×[0 0 0 0 1 1 1 0 10 0 1 0 1 0 1 0 11 0 0 0 0 0 0 0 00 0 0 0 1 1 1 1 10 0 0 0 0 0 1 0 00 0 1 0 0 0 0 1 00 0 0 0 0 0 0 0 00 1 1 0 0 0 0 0 00 0 0 0 0 0 1 0 0

]W 3=[

1 1 1 0 0 0 0 0 00 0 0 0 1 1 1 0 10 0 1 0 0 0 2 1 02 1 2 0 1 0 1 0 10 0 0 0 0 0 0 0 01 0 1 0 2 1 2 0 20 0 0 0 0 0 0 0 01 0 0 0 1 1 3 0 10 0 0 0 0 0 0 0 0

]W+W 2+W 3−L−L2−L3

For comparative measures, the second-order rank vector has been calculated.

Let: W+W 2−L−L2=D1and VW +W 2−L−L2=¿V 1 ¿

Through the Win-Loss method, the ranking of each team has been determined as follows:

Team

Place

CCC 5RSHS

4

SC 6SSHS

1

BPSH

7

WPSH

3

MSHS

9

PRSH

2

MPSH

8

This therefore shows that the Win-Loss method has been successful and effective in establishing a prediction of the definite rank order of each of the teams for the Futsal Competition.

From the predicted rankings, it is almost certain that SSHS will win the competition as their amount of points is considerably higher than any of the other teams. While this is so, it is predicted that there will be close competition between PRSH and WPSH as there is a 1-point margin between the two. Similarly, the ranking between BPSH and MPSH could be interchangeable as the competition continues due to their 1-point difference between each of them. Also, it is clear that MSHS is the most-likely team to be in final place as the point variance between them and the team in second-last place (BPSH) is considerably large with a difference of 13 points.

To calculate the second-order rank vector, D1+D12=D2 is first calculated and D2 is then substituted into D1+D2 where the vector is

calculated (the second-order rank vector (V 2 ¿).

D1+D12=[

0 −1 0 0 1 0 3 0 11 0 1 −1 1 −1 3 −1 10 −1 0 −2 1 −1 1 −2 −10 1 2 0 1 1 3 2 1

−1 −1 −1 −1 0 0 1 −1 00 1 1 −1 0 0 0 1 0

−3 −3 −1 −3 −1 0 0 −1 −10 1 2 −2 1 −1 1 0 1

−1 −1 −1 −1 0 0 1 −1 0

]+[0 −1 0 0 1 0 3 0 11 0 1 −1 1 −1 3 −1 10 −1 0 −2 1 −1 1 −2 −10 1 2 0 1 1 3 2 1

−1 −1 −1 −1 0 0 1 −1 00 1 1 −1 0 0 0 1 0

−3 −3 −1 −3 −1 0 0 −1 −10 1 2 −2 1 −1 1 0 1

−1 −1 −1 −1 0 0 1 −1 0

]2

D1+D12=[

0 −1 0 0 1 0 3 0 11 0 1 −1 1 −1 3 −1 10 −1 0 −2 1 −1 1 −2 −10 1 2 0 1 1 3 2 1

−1 −1 −1 −1 0 0 1 −1 00 1 1 −1 0 0 0 1 0

−3 −3 −1 −3 −1 0 0 −1 −10 1 2 −2 1 −1 1 0 1

−1 −1 −1 −1 0 0 1 −1 0

]+[−12 −11 −6 −10 −4 1 −1 −4 −4−11 −16 −10 −10 −3 −1 2 −10 −5−4 −8 −11 3 −6 1 −11 −5 −6−10 −10 1 −21 2 −5 9 −9 −2−4 −3 −6 2 −6 2 −11 0 −41 −1 1 −5 2 −4 2 −5 0

−1 2 −9 9 −11 2 −31 1 −9−4 −10 −7 −9 0 −5 1 −13 −4−4 −3 −6 2 −6 2 −11 0 −4

]D1+D1

2=[−12 −12 −6 −10 −3 1 2 −4 −3−10 −16 −9 −11 −2 −2 5 −11 −4−4 −9 −11 1 −5 0 −10 −7 −7−10 −9 3 −21 3 −4 12 −7 −1−5 −4 −7 1 −6 2 −10 −1 −41 0 2 −6 2 −4 2 −4 0

−4 −1 −10 6 −12 2 −31 0 −10−4 −9 −5 −11 1 −6 2 −13 −3−5 −4 −7 1 −6 2 −10 −1 −4

]=D2 D1+D2=[

0 −1 0 0 1 0 3 0 11 0 1 −1 1 −1 3 −1 10 −1 0 −2 1 −1 1 −2 −10 1 2 0 1 1 3 2 1

−1 −1 −1 −1 0 0 1 −1 00 1 1 −1 0 0 0 1 0

−3 −3 −1 −3 −1 0 0 −1 −10 1 2 −2 1 −1 1 0 1

−1 −1 −1 −1 0 0 1 −1 0

] + [−12 −12 −6 −10 −3 1 2 −4 −3−10 −16 −9 −11 −2 −2 5 −11 −4−4 −9 −11 1 −5 0 −10 −7 −7−10 −9 3 −21 3 −4 12 −7 −1−5 −4 −7 1 −6 2 −10 −1 −41 0 2 −6 2 −4 2 −4 0

−4 −1 −10 6 −12 2 −31 0 −10−4 −9 −5 −11 1 −6 2 −13 −3−5 −4 −7 1 −6 2 −10 −1 −4

]

D1+D2=[−12 −13 −6 −10 −2 1 5 −4 −2−9 −16 −8 −12 −1 −3 8 −12 −3−4 −10 −11 −1 −4 −1 −9 −9 −8−10 −8 5 −21 4 −3 15 −5 0−6 −5 −8 0 −6 2 −9 −2 −41 1 3 −7 2 −4 2 −3 0

−7 −4 −11 3 −13 2 −31 −1 −11−4 −8 −3 −13 2 −7 3 −13 −2−6 −5 −8 0 −6 2 −9 −2 −4

]V 2=[

−43−56−57−23−38−5−73−45−38

]Although the ranking has been broken down even further, two teams still remain with the same score. Due to this, the second-order ranking vector is then weighted by an arbitrary constant of 0.5 and if this does not determine a definite ranking, the third-order ranking vector is then found.

D1+0.5D2=[0 −1 0 0 1 0 3 0 11 0 1 −1 1 −1 3 −1 10 −1 0 −2 1 −1 1 −2 −10 1 2 0 1 1 3 2 1

−1 −1 −1 −1 0 0 1 −1 00 1 1 −1 0 0 0 1 0

−3 −3 −1 −3 −1 0 0 −1 −10 1 2 −2 1 −1 1 0 1

−1 −1 −1 −1 0 0 1 −1 0

] +

0.5 [−12 −12 −6 −10 −3 1 2 −4 −3−10 −16 −9 −11 −2 −2 5 −11 −4−4 −9 −11 1 −5 0 −10 −7 −7−10 −9 3 −21 3 −4 12 −7 −1−5 −4 −7 1 −6 2 −10 −1 −41 0 2 −6 2 −4 2 −4 0

−4 −1 −10 6 −12 2 −31 0 −10−4 −9 −5 −11 1 −6 2 −13 −3−5 −4 −7 1 −6 2 −10 −1 −4

]

D1+0.5D2=[−6 −6.5 −3 −5 −1 0.5 2.5 −2 −1

−4.5 −8 −4 −6 −0.5 −1.5 4 −6 −1.5−2 −5 −5.5 −0.5 −2 −0.5 −4.5 −4.5 −4−5 −4 2.5 −10.5 2 −1.5 7.5 −2.5 0−3 −2.5 −4 0 −3 1 −4.5 −1 −20.5 0.5 1.5 −3.5 1 −2 1 −1.5 0

−3.5 −2 −5.5 1.5 −6.5 1 −15.5 −0.5 −5.5−2 −4 −1.5 −6.5 1 −3.5 1.5 −6.5 −1−3 −2.5 −4 0 −3 1 −4.5 −1 −2

] = V 3

V 3=[−21.5−28

−28.5−11.5−19−2.5−36.5−22.5−19

]As shown above, the weighting of the second-order ranking vector was insufficient in providing a definite rank order for the Futsal Competition and therefore, to continue the

D1+D12+D1

3=D3 substituted into:

D1+D2+D3

D3=D1+D12+D1

3

D3=[0 −1 0 0 1 0 3 0 11 0 1 −1 1 −1 3 −1 10 −1 0 −2 1 −1 1 −2 −10 1 2 0 1 1 3 2 1

−1 −1 −1 −1 0 0 1 −1 00 1 1 −1 0 0 0 1 0

−3 −3 −1 −3 −1 0 0 −1 −10 1 2 −2 1 −1 1 0 1

−1 −1 −1 −1 0 0 1 −1 0

]+[0 −1 0 0 1 0 3 0 11 0 1 −1 1 −1 3 −1 10 −1 0 −2 1 −1 1 −2 −10 1 2 0 1 1 3 2 1

−1 −1 −1 −1 0 0 1 −1 00 1 1 −1 0 0 0 1 0

−3 −3 −1 −3 −1 0 0 −1 −10 1 2 −2 1 −1 1 0 1

−1 −1 −1 −1 0 0 1 −1 0

]2

+[0 −1 0 0 1 0 3 0 11 0 1 −1 1 −1 3 −1 10 −1 0 −2 1 −1 1 −2 −10 1 2 0 1 1 3 2 1

−1 −1 −1 −1 0 0 1 −1 00 1 1 −1 0 0 0 1 0

−3 −3 −1 −3 −1 0 0 −1 −10 1 2 −2 1 −1 1 0 1

−1 −1 −1 −1 0 0 1 −1 0

]3

V 4=[−227−310509

−527139

−129505

−259139

] It is evident that through this process, the Win-Loss method is comparatively more effective as is able to determine a definite ranking, as opposed to the order method shown here, which fails to yield a definite rank order as two teams maintain the exact same score in each calculation of first-order, second-order, and third-order ranking vectors.

The fact that even after determining the third-order vector, the definite ranking has not been determined, reflects that each of the teams, especially in the middle-placings, will be facing great competition for each game they play.

Overall, the Win-Loss method is quite successful in determining a prediction for the ranking of each of the teams in the competition, and it allows for an accurate representation of the current state of

References/Resources:

References:1. Dominance Matrices. (2008). Retrieved May 11, 2016, from https://www.qcaa.qld.edu.au/downloads/senior/snr_maths_c_08_sai_modelling.pdf

Resources used:1. Math Quest Math C Year 11 For Queensland - Simpson