Let's go through each part step by step: --- ## (h) **System of Equations Using the Colley Method** ### **Step 1: Understanding the Graph** - Each node (A, B, C, D) represents a team. - Each directed edge (arrow) from one team to another means a win for the team at the tail of the arrow against the team at the head. From the graph: - **A → B**: A beats B - **A → D**: A beats D - **B → C**: B beats C - **D → C**: D beats C - **D → B**: D beats B ### **Step 2: Set Up the Colley Matrix** #### **Colley Method Basics** For each team \( i \): - \( n_i \): Total games played by team \( i \) - \( w_i \): Total wins by team \( i \) - \( l_i \): Total losses by team \( i \) - The Colley matrix \( C \) is defined as: - Diagonal entry: \( C_{ii} = 2 + n_i \) - Off-diagonal entry: \( C_{ij} = - \) (number of games played between teams \( i \) and \( j \)) - The right side \( b \) is: \( b_i = 1 + (w_i - l_i)/2 \) #### **Counting Games** - **A:** vs B (win), vs D (win) → 2 games, 2 wins - **B:** vs A (loss), vs C (win), vs D (loss) → 3 games, 1 win - **C:** vs B (loss), vs D (loss) → 2 games, 0 wins - **D:** vs A (loss), vs B (win), vs C (win) → 3 games, 2 wins #### **Games Played Between Pairs** - A–B: 1 game - A–D: 1 game - B–C: 1 game - B–D: 1 game - C–D: 1 game #### **Build the Colley Matrix \( C \):** Arrange teams as (A, B, C, D). \[ C = \begin{bmatrix} 2 + 2 & -1 & 0 & -1 \\ -1 & 2 + 3 & -1 & -1 \\ 0 & -1 & 2 + 2 & -1 \\ -1 & -1 & -1 & 2 + 3 \\ \end{bmatrix} = \begin{bmatrix} 4 & -1 & 0 & -1 \\ -1 & 5 & -1 & -1 \\ 0 & -1 & 4 & -1 \\ -1 & -1 & -1 & 5 \\ \end{bmatrix} \] #### **Build the \( b \) vector:** \[ b_A = 1 + \frac{2 - 0}{2} = 2 \] \[ b_B = 1 + \frac{1 - 2}{2} = 0.5 \] \[ b_C = 1 + \frac{0 - 2}{2} = 0 \] \[ b_D = 1 + \frac{2 - 1}{2} = 1.5 \] So, \[ b = \begin{bmatrix} 2 \\ 0.5 \\ 0 \\ 1.5 \\ \end{bmatrix} \] --- ### **Final System of Equations** \[ \begin{bmatrix} 4 & -1 & 0 & -1 \\ -1 & 5 & -1 & -1 \\ 0 & -1 & 4 & -1 \\ -1 & -1 & -1 & 5 \\ \end{bmatrix} \begin{bmatrix} r_A \\ r_B \\ r_C \\ r_D \\ \end{bmatrix} = \begin{bmatrix} 2 \\ 0.5 \\ 0 \\ 1.5 \\ \end{bmatrix} \] Where \( r_A, r_B, r_C, r_D \) are the ratings for teams A, B, C, and D. --- ## (i) **Solve for Ratings and Rank Teams** Let's solve the system step by step. ### **Write Out the Equations:** From the matrix: 1. \( 4r_A - r_B - r_D = 2 \) 2. \( -r_A + 5r_B - r_C - r_D = 0.5 \) 3. \( -r_B + 4r_C - r_D = 0 \) 4. \( -r_A - r_B - r_C + 5r_D = 1.5 \) #### **Solving the System:** Let's do this step by step. **Equation 3:** \( -r_B + 4r_C - r_D = 0 \implies r_B = 4r_C - r_D \) **Let's substitute \( r_B \) in the other equations:** Now use equation 1: \( 4r_A - r_B - r_D = 2 \) Substitute \( r_B \): \( 4r_A - (4r_C - r_D) - r_D = 2 \) \( 4r_A - 4r_C + r_D - r_D = 2 \) \( 4r_A - 4r_C = 2 \) \( r_A - r_C = 0.5 \) So, \( r_A = r_C + 0.5 \) (Equation A) Now equation 2: \( -r_A + 5r_B - r_C - r_D = 0.5 \) Substitute \( r_B \): \( -r_A + 5(4r_C - r_D) - r_C - r_D = 0.5 \) \( -r_A + 20r_C - 5r_D - r_C - r_D = 0.5 \) \( -r_A + 19r_C - 6r_D = 0.5 \) (Equation B) Now equation 4: \( -r_A - r_B - r_C + 5r_D = 1.5 \) Substitute \( r_B \): \( -r_A - (4r_C - r_D) - r_C + 5r_D = 1.5 \) \( -r_A - 4r_C + r_D - r_C + 5r_D = 1.5 \) \( -r_A - 5r_C + 6r_D = 1.5 \) (Equation C) Now substitute \( r_A = r_C + 0.5 \) into Equations B and C. Equation B: \( -(r_C + 0.5) + 19r_C - 6r_D = 0.5 \) \( -r_C - 0.5 + 19r_C - 6r_D = 0.5 \) \( 18r_C - 6r_D = 1 \) \( 18r_C - 6r_D = 1 \) \( 9r_C - 3r_D = 0.5 \) Equation C: \( -(r_C + 0.5) - 5r_C + 6r_D = 1.5 \) \( -r_C - 0.5 - 5r_C + 6r_D = 1.5 \) \( -6r_C + 6r_D = 2 \) \( -3r_C + 3r_D = 1 \) \( -3r_C + 3r_D = 1 \) Now we have: 1. \( r_A = r_C + 0.5 \) 2. \( 9r_C - 3r_D = 0.5 \) 3. \( -3r_C + 3r_D = 1 \) Equation 3: \( -3r_C + 3r_D = 1 \implies r_D = r_C + \frac{1}{3} \) Substitute \( r_D \) into equation 2: \( 9r_C - 3(r_C + \frac{1}{3}) = 0.5 \) \( 9r_C - 3r_C - 1 = 0.5 \) \( 6r_C - 1 = 0.5 \) \( 6r_C = 1.5 \) \( r_C = 0.25 \) Now substitute \( r_C \) into \( r_D \): \( r_D = 0.25 + \frac{1}{3} = \frac{3}{12} + \frac{4}{12} = \frac{7}{12} \approx 0.583 \) \( r_A = r_C + 0.5 = 0.25 + 0.5 = 0.75 \) Now, \( r_B = 4r_C - r_D = 4(0.25) - 0.583 = 1 - 0.583 = 0.417 \) ## **Final Ratings** - \( r_A = 0.75 \) - \( r_B = 0.417 \) - \( r_C = 0.25 \) - \( r_D = 0.583 \) ### **Ranking the Teams** 1. **A:** 0.75 (1st) 2. **D:** 0.583 (2nd) 3. **B:** 0.417 (3rd) 4. **C:** 0.25 (4th) --- ## (j) **Strengths and Weaknesses of the Colley Method** ### **Strengths** - **Objective:** Based only on win-loss records, not margin of victory or subjective criteria. - **Adjusts for Schedule Strength:** Teams are rated more fairly if they played tougher opponents. - **Solvable System:** Always yields a unique solution. ### **Weaknesses** - **Ignores Score Margin:** Does not consider how decisively a team won or lost. - **No Head-to-Head Emphasis:** Treats all games equally, even if one team beat another directly. - **Limited with Small Data:** Can be less effective with very few games or teams. --- ## **Summary Table** | Team | Colley Rating | Rank | |------|--------------|------| | A | 0.75 | 1 | | D | 0.583 | 2 | | B | 0.417 | 3 | | C | 0.25 | 4 | --- **Let me know if you need further explanations or calculations!**