## 4. Mathematical Optimization Formulation

#### Decision Variables
Let \( x_{p,c} \) be a binary decision variable where:
- \( x_{p,c} = 1 \) if player \( p \) is assigned to coach \( c \),
- \( x_{p,c} = 0 \) otherwise.

#### Objective Function
Maximize the total points earned:
\[
\text{Maximize } \sum_{p \in P} \sum_{c \in C} (\text{player.Votes}_p \times \text{coach.Rank}_c) \times x_{p,c}
\]
Where:
- \( P \) is the set of players,
- \( C \) is the set of coaches,
- \( \text{player.Votes}_p \) is the number of votes for player \( p \),
- \( \text{coach.Rank}_c \) is the rank of coach \( c \).

#### Constraints
1. **Single Assignment Constraint**: Each player must be assigned to exactly one coach.
\[
\sum_{c \in C} x_{p,c} = 1 \quad \forall p \in P
\]

2. **Coach Capacity Constraint**: Each coach can handle a maximum of four players.
\[
\sum_{p \in P} x_{p,c} \leq 4 \quad \forall c \in C
\]

#### Data Source Verification
- **Coefficients in Objective Function**:
  - \( \text{player.Votes}_p \) comes from `player.Votes`.
  - \( \text{coach.Rank}_c \) comes from `coach.Rank`.

- **Constraints**:
  - The constant `1` in the Single Assignment Constraint ensures each player is assigned to exactly one coach.
  - The constant `4` in the Coach Capacity Constraint ensures no coach exceeds their maximum capacity.

### Numerical Example Based on Provided Data
Given the data:
- Players: \( P = \{1, 2, 3\} \)
  - Votes: \( \text{player.Votes}_1 = 12 \), \( \text{player.Votes}_2 = 18 \), \( \text{player.Votes}_3 = 9 \)
- Coaches: \( C = \{1, 2, 3\} \)
  - Ranks: \( \text{coach.Rank}_1 = 3 \), \( \text{coach.Rank}_2 = 2 \), \( \text{coach.Rank}_3 = 1 \)

The objective function becomes:
\[
\text{Maximize } 12 \times 3 \times x_{1,1} + 12 \times 2 \times x_{1,2} + 12 \times 1 \times x_{1,3} + 18 \times 3 \times x_{2,1} + 18 \times 2 \times x_{2,2} + 18 \times 1 \times x_{2,3} + 9 \times 3 \times x_{3,1} + 9 \times 2 \times x_{3,2} + 9 \times 1 \times x_{3,3}
\]

The constraints are:
1. Single Assignment Constraint:
\[
x_{1,1} + x_{1,2} + x_{1,3} = 1 \\
x_{2,1} + x_{2,2} + x_{2,3} = 1 \\
x_{3,1} + x_{3,2} + x_{3,3} = 1
\]

2. Coach Capacity Constraint:
\[
x_{1,1} + x_{2,1} + x_{3,1} \leq 4 \\
x_{1,2} + x_{2,2} + x_{3,2} \leq 4 \\
x_{1,3} + x_{2,3} + x_{3,3} \leq 4
\]

This is a complete, immediately solvable Mixed-Integer Linear Programming (MIP) model.

## 5. Gurobipy Implementation

```python
#!/usr/bin/env python3
"""
Gurobipy 12.0.2 Implementation for Riding Club Optimization Problem
"""

import gurobipy as gp
from gurobipy import GRB

def riding_club_optimization():
    """Optimize player-coach assignments to maximize total points."""
    
    # 1. MODEL & DATA SETUP
    model = gp.Model("riding_club")
    
    # Data from the problem
    players = [1, 2, 3]
    coaches = [1, 2, 3]
    
    player_votes = {1: 12, 2: 18, 3: 9}
    coach_ranks = {1: 3, 2: 2, 3: 1}
    
    # CRITICAL: Validate array lengths before loops
    assert len(players) == len(player_votes), "Player data length mismatch"
    assert len(coaches) == len(coach_ranks), "Coach data length mismatch"
    
    # 2. VARIABLES
    # Binary decision variables: x[p, c] = 1 if player p is assigned to coach c
    x = model.addVars(players, coaches, vtype=GRB.BINARY, name="x")
    
    # 3. OBJECTIVE FUNCTION
    # Maximize total points: sum(player_votes[p] * coach_ranks[c] * x[p, c])
    model.setObjective(
        gp.quicksum(player_votes[p] * coach_ranks[c] * x[p, c] for p in players for c in coaches),
        GRB.MAXIMIZE
    )
    
    # 4. CONSTRAINTS
    
    # Single Assignment Constraint: Each player must be assigned to exactly one coach
    for p in players:
        model.addConstr(
            gp.quicksum(x[p, c] for c in coaches) == 1,
            name=f"single_assignment_{p}"
        )
    
    # Coach Capacity Constraint: Each coach can handle a maximum of four players
    for c in coaches:
        model.addConstr(
            gp.quicksum(x[p, c] for p in players) <= 4,
            name=f"coach_capacity_{c}"
        )
    
    # 5. SOLVING & RESULTS
    model.optimize()
    
    if model.status == GRB.OPTIMAL:
        print(f"Optimal value: {model.objVal}")
        for p in players:
            for c in coaches:
                if x[p, c].x > 0.5:  # Check if assignment is active
                    print(f"Player {p} is assigned to Coach {c}")
    elif model.status == GRB.INFEASIBLE:
        print("Problem is infeasible")
    elif model.status == GRB.UNBOUNDED:
        print("Problem is unbounded")
    
    return model

# Execute the optimization
if __name__ == "__main__":
    riding_club_optimization()
```

### Execution Results
**Status**: OPTIMAL
**Optimal Value**: 117.0
**Execution Time**: 0.19 seconds
**Reliability**: high
**Analysis**: Gurobipy found the optimal solution quickly with the lowest execution time.

## 6. DOCplex Implementation

```python
#!/usr/bin/env python3
"""
DOCplex implementation for the riding club optimization problem
"""

from docplex.mp.model import Model

def riding_club_optimization():
    """Optimize player assignments to coaches to maximize total points"""
    
    # 1. MODEL & DATA SETUP
    mdl = Model(name="riding_club_optimization")
    
    # Player data
    players = [1, 2, 3]
    player_votes = {1: 12, 2: 18, 3: 9}
    
    # Coach data
    coaches = [1, 2, 3]
    coach_ranks = {1: 3, 2: 2, 3: 1}
    
    # CRITICAL: Validate array lengths to prevent IndexError
    assert len(players) == len(player_votes), "Player data length mismatch"
    assert len(coaches) == len(coach_ranks), "Coach data length mismatch"
    
    # 2. VARIABLES
    # Binary decision variables: x[p, c] = 1 if player p is assigned to coach c
    x = {(p, c): mdl.binary_var(name=f"x_{p}_{c}") for p in players for c in coaches}
    
    # 3. OBJECTIVE FUNCTION
    # Maximize total points: sum(player_votes[p] * coach_ranks[c] * x[p, c])
    objective = mdl.sum(player_votes[p] * coach_ranks[c] * x[p, c] for p in players for c in coaches)
    mdl.maximize(objective)
    
    # 4. CONSTRAINTS
    
    # Single Assignment Constraint: Each player must be assigned to exactly one coach
    for p in players:
        mdl.add_constraint(mdl.sum(x[p, c] for c in coaches) == 1, ctname=f"single_assignment_{p}")
    
    # Coach Capacity Constraint: Each coach can handle a maximum of four players
    for c in coaches:
        mdl.add_constraint(mdl.sum(x[p, c] for p in players) <= 4, ctname=f"coach_capacity_{c}")
    
    # 5. SOLVING & RESULTS
    solution = mdl.solve()
    
    if solution:
        print(f"Optimal value: {solution.objective_value}")
        for p in players:
            for c in coaches:
                if solution.get_value(x[p, c]) > 0.5:
                    print(f"Player {p} assigned to Coach {c}")
    else:
        print("No solution found")
        print(f"Status: {mdl.solve_details.status}")
    
    return mdl

# Execute the optimization
if __name__ == "__main__":
    riding_club_optimization()
```

### Execution Results
**Status**: OPTIMAL
**Optimal Value**: 117.0
**Execution Time**: 1.05 seconds
**Reliability**: high
**Analysis**: DOCplex also found the optimal solution but took longer compared to Gurobipy.

## 7. Pyomo Implementation

```python
#!/usr/bin/env python3
"""
Pyomo 6.9.2 Implementation for Riding Club Optimization Problem
"""

import pyomo.environ as pyo
from pyomo.opt import SolverFactory

def riding_club_optimization():
    """Optimization model for maximizing total points in a riding club."""
    
    # 1. MODEL CREATION
    model = pyo.ConcreteModel()
    
    # 2. DATA SETUP
    # Players data
    players = {1: 12, 2: 18, 3: 9}  # Player_ID: Votes
    # Coaches data
    coaches = {1: 3, 2: 2, 3: 1}    # Coach_ID: Rank
    
    # CRITICAL: Validate array lengths before indexing
    assert len(players) > 0 and len(coaches) > 0, "Players or coaches data is empty"
    
    # 3. SETS (Pyomo way to define indices)
    model.P = pyo.Set(initialize=players.keys())  # Set of players
    model.C = pyo.Set(initialize=coaches.keys())  # Set of coaches
    
    # 4. PARAMETERS (data containers)
    model.player_votes = pyo.Param(model.P, initialize=players)
    model.coach_rank = pyo.Param(model.C, initialize=coaches)
    
    # 5. VARIABLES
    # Binary decision variable: x[p, c] = 1 if player p is assigned to coach c
    model.x = pyo.Var(model.P, model.C, within=pyo.Binary)
    
    # 6. OBJECTIVE FUNCTION
    def obj_rule(model):
        return sum(model.player_votes[p] * model.coach_rank[c] * model.x[p, c] for p in model.P for c in model.C)
    model.objective = pyo.Objective(rule=obj_rule, sense=pyo.maximize)
    
    # 7. CONSTRAINTS
    
    # Single Assignment Constraint: Each player must be assigned to exactly one coach
    def single_assignment_rule(model, p):
        return sum(model.x[p, c] for c in model.C) == 1
    model.single_assignment = pyo.Constraint(model.P, rule=single_assignment_rule)
    
    # Coach Capacity Constraint: Each coach can handle a maximum of four players
    def coach_capacity_rule(model, c):
        return sum(model.x[p, c] for p in model.P) <= 4
    model.coach_capacity = pyo.Constraint(model.C, rule=coach_capacity_rule)
    
    # 8. SOLVING WITH GUROBI (your available solver)
    solver = SolverFactory('gurobi')
    
    # Optional: Set solver options
    solver.options['TimeLimit'] = 300  # 5 minutes
    solver.options['MIPGap'] = 0.01    # 1% gap
    
    # Solve the model
    results = solver.solve(model, tee=True)  # tee=True shows solver output
    
    # 9. RESULT PROCESSING
    # Check solver status
    if results.solver.termination_condition == pyo.TerminationCondition.optimal:
        print(f"Optimal value: {pyo.value(model.objective)}")
        
        # Extract variable values
        print("\nAssignment details:")
        for p in model.P:
            for c in model.C:
                if pyo.value(model.x[p, c]) > 0.5:  # Only print assignments
                    print(f"Player {p} is assigned to Coach {c}")
        
    elif results.solver.termination_condition == pyo.TerminationCondition.infeasible:
        print("Problem is infeasible")
    elif results.solver.termination_condition == pyo.TerminationCondition.unbounded:
        print("Problem is unbounded")
    else:
        print(f"Solver terminated with condition: {results.solver.termination_condition}")
    
    return model

# Execute the optimization
if __name__ == "__main__":
    riding_club_optimization()
```

### Execution Results
**Status**: OPTIMAL
**Optimal Value**: 117.0
**Execution Time**: 1.02 seconds
**Reliability**: high
**Analysis**: Pyomo found the optimal solution with a similar execution time to DOCplex.

## 8. Cross-Solver Analysis and Final Recommendation

### Solver Results Comparison

| Solver | Status | Optimal Value | Execution Time | Decision Variables | Retry Attempt |
|--------|--------|---------------|----------------|-------------------|---------------|
| Gurobipy | OPTIMAL | 117.00 | 0.19s | N/A | N/A |
| Docplex | OPTIMAL | 117.00 | 1.05s | N/A | N/A |
| Pyomo | OPTIMAL | 117.00 | 1.02s | N/A | N/A |

### Solver Consistency Analysis
**Result**: All solvers produced consistent results ✓
**Consistent Solvers**: gurobipy, docplex, pyomo
**Majority Vote Optimal Value**: 117.0

### Final Recommendation
**Recommended Optimal Value**: 117.0
**Confidence Level**: HIGH
**Preferred Solver(s)**: gurobipy
**Reasoning**: Gurobipy is preferred due to its significantly faster execution time while still providing the optimal solution.

### Business Interpretation
**Overall Strategy**: The optimal assignment of players to coaches maximizes the total points earned based on player votes and coach ranks.
**Objective Value Meaning**: The optimal objective value of 117 represents the maximum total points achievable given the constraints.
**Resource Allocation Summary**: Players should be assigned to coaches in a way that maximizes the total points, ensuring no coach exceeds their capacity of four players.
**Implementation Recommendations**: Implement the assignment as determined by the solver, ensuring each player is assigned to exactly one coach and no coach is overloaded.