MPS 1

Part A: Determining fair rank between teams(no ties)

Let win=3, draw=1 and lose=0. The reason for this weighting method is because it is natural for the loser not to get a point. Also, there must be a visible gap between winners and draw players. This can be proved by contradiction proof. Blue| B|

Crimson| C|

Green| G|

Orange| O|

Red| R|

Yellow| Y|

Assume that win=2 draw=1 and lose=0. If the supremacy matrix is calculated, tie always occurs, which proves that if there is a small gap between win and draws, that supremacy matrix isn’t valid.

This is the teams and their initial letters, arranged by alphabetical order.

Then, the diagraph below is converted into the following matrix. M=

In order to find the supremacy matrix with no ties, three cases with different coefficients are examined. Let coefficient of M is Mc, and coefficient of M^2 is M^2c; Mc>M^2c (S=M+0.5M^2), Mc=M^2c (S=M+M^2), and Mc<M^2c (S=M+2M^2). The result is followed by calculation.

S=M+0.5M^2

=

The rank order is Oranges, Blues, Reds, Yellows, Crimsons and Greens. As it can be seen, there is no tie amongst the teams. This means, the supremacy S=M+0.5M^2 is valid. To identify that this matrix is valid, other cases are examined. S=M+M^2

=

The rank order is Oranges, Blues, Reds, Yellows, Crimsons=Greens. As it can be seen in the vector, there is tie between Crimson and Green. This shows that the Supremacy Matrix S=M+M^2 is invalid.

S=M+2M^2

=

The rank order is Oranges, Blues, Reds, Yellows, Crimsons and Greens. As it can be seen, there is no tie amongst the teams. However, this matrix is less reliable than S=M+0.5M^2 because in most of matrices including this matrix, the first order matrix is more reliable than second order matrix because M^2 has lesser reliability because M is more related to the original data. M is the matrix derived from the actual data, whilst M^2 is the data generated from M through calculation.

Therefore, the most appropriate Supremacy Matrix is s=M+0.5M^2.

Part B – Escalating the rank of Team Red

In part A, the ranking of team was found as following: Oranges, Blues, Reds, Yellows, Crimsons and Greens. However, in this task, it is required to escalate the rank of Red. In the current situation, Red stays at third rank. In order to escalate it, third order Matrix, M^3, is used. M^3=M^2*M

=

=

Let’s call M+M^2 as ‘N’, the coefficient of M+0.5M^2 ‘cN’, and coefficient of M^3 cM^3 The third order matrix, M3, is examined in 3 cases to determine its validity; M+0.5M^2+o.5M^3, M+0.5M^2+M^3 and M+0.5M^2+2M^3 Case 1: cN>cM

S= M+0.5M2+0.5M3.

=

=

The result rank was as following: Oranges, Reds, Blues, Yellows, Crimsons and Greens. This shows the rank change compared to the result of S=M+0.5M2 with the result of Oranges, Blues, Reds, Yellows, Crimsons and Greens. In this result, the Red Team’s rank was escalated from 3rd to 2nd, which corresponds the Red team coach’s requirement. If M3 is smaller than M+0.5M2, the supremacy matrix becomes valuable, because it doesn’t make any ties and also escalates the Red team’s rank. To verify that this answer is correct, other two cases are examined.

Case 2: cN=cM^3

S=M+0.5M2+M3

=

=

If the coefficient of N is equal to M^3, the rank is same as if cN>cM3. The result was found as Oranges, Reds, Blues, Yellows, Crimsons and Greens, which is same as the case above. It provides the validity of cN=cM^3 graph, because it suits all the conditions; The Red team’s rank is escalated, and There is no tie. Case 3: cN<cM3

S=M+0.5M2+2M3

=

=

If the coefficient of N is bigger than M^3, the rank is same as if cN>cM^3 or cN=cM^3. The result was found as Oranges, Reds, Blues, Yellows, Crimsons and Greens, which is same as the case above. It provides the validity of S= M+0.5M2+M3 graph, because it suits all the conditions; The Red team’s rank is escalated, and There is...