## 4. Mathematical Optimization Formulation

#### Decision Variables
- Let \( x_{m,r} \) be a binary decision variable where:
  - \( x_{m,r} = 1 \) if manager \( m \) is assigned to railway \( r \),
  - \( x_{m,r} = 0 \) otherwise.

#### Objective Function
Minimize the total cost of assigning managers to railways:
\[
\text{Minimize} \quad \sum_{m} \sum_{r} c_m \cdot x_{m,r}
\]
where \( c_m \) is the cost per level of manager \( m \).

#### Constraints
1. **Railway Management Constraint**: Each railway must be managed by at least one manager:
\[
\sum_{m} x_{m,r} \geq 1 \quad \forall r
\]
2. **Manager Capacity Constraint**: No manager can be assigned to more railways than their specified capacity:
\[
\sum_{r} x_{m,r} \leq k_m \quad \forall m
\]
where \( k_m \) is the capacity of manager \( m \).

#### Data Source Verification
- **Objective Function Coefficient \( c_m \)**: Comes from `manager.cost_per_level`.
- **Manager Capacity \( k_m \)**: Comes from `manager_capacity.capacity`.
- **Decision Variable \( x_{m,r} \)**: Represents the assignment status in `manager_railway_assignment.assignment_status`.

This formulation provides a complete, immediately solvable LINEAR mathematical model with all numerical coefficients derived from the provided data.

## 5. Gurobipy Implementation

```python
#!/usr/bin/env python3
"""
Gurobipy Implementation for Railway Management Optimization
"""

import gurobipy as gp
from gurobipy import GRB

def railway_optimization():
    """Optimize manager assignments to railways to minimize total cost."""
    
    # 1. MODEL & DATA SETUP
    model = gp.Model("railway_management")
    
    # Data from the database
    managers = [1, 2, 3]
    railways = [1, 2, 3]
    
    # Manager costs and capacities
    cost_per_level = {1: 120, 2: 150, 3: 200}
    manager_capacity = {1: 4, 2: 6, 3: 8}
    
    # CRITICAL: Validate array lengths before loops
    assert len(managers) == len(cost_per_level) == len(manager_capacity), "Array length mismatch"
    
    # 2. VARIABLES
    # Binary decision variables: x[m, r] = 1 if manager m is assigned to railway r
    x = model.addVars(managers, railways, vtype=GRB.BINARY, name="x")
    
    # 3. OBJECTIVE FUNCTION
    # Minimize the total cost of assigning managers to railways
    model.setObjective(gp.quicksum(cost_per_level[m] * x[m, r] for m in managers for r in railways), GRB.MINIMIZE)
    
    # 4. CONSTRAINTS
    
    # Railway Management Constraint: Each railway must be managed by at least one manager
    for r in railways:
        model.addConstr(gp.quicksum(x[m, r] for m in managers) >= 1, name=f"railway_{r}_managed")
    
    # Manager Capacity Constraint: No manager can be assigned to more railways than their capacity
    for m in managers:
        model.addConstr(gp.quicksum(x[m, r] for r in railways) <= manager_capacity[m], name=f"manager_{m}_capacity")
    
    # 5. SOLVING & RESULTS
    model.optimize()
    
    if model.status == GRB.OPTIMAL:
        print(f"Optimal value: {model.objVal}")
        for m in managers:
            for r in railways:
                if x[m, r].x > 0.5:
                    print(f"Manager {m} is assigned to Railway {r}")
    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__":
    railway_optimization()
```

### Execution Results
**Status**: OPTIMAL
**Optimal Value**: 360.0
**Execution Time**: 0.22 seconds
**Reliability**: high
**Analysis**: Gurobipy achieved the optimal solution with the lowest execution time, indicating high efficiency and reliability.

## 6. DOCplex Implementation

```python
# Complete DOCPLEX implementation

from docplex.mp.model import Model

def railway_optimization():
    """Optimize manager to railway assignments to minimize total cost."""
    
    # 1. MODEL & DATA SETUP
    mdl = Model(name="railway_management")
    
    # Data from the database
    managers = [1, 2, 3]
    railways = [1, 2, 3]
    
    # Manager costs and capacities
    manager_costs = {1: 120, 2: 150, 3: 200}
    manager_capacities = {1: 4, 2: 6, 3: 8}
    
    # CRITICAL: Validate array lengths to prevent IndexError
    assert len(managers) == len(manager_costs) == len(manager_capacities), "Array length mismatch"
    
    # 2. VARIABLES
    # Binary decision variables: x[m][r] = 1 if manager m is assigned to railway r
    x = {(m, r): mdl.binary_var(name=f"x_{m}_{r}") for m in managers for r in railways}
    
    # 3. OBJECTIVE FUNCTION
    # Minimize the total cost of assigning managers to railways
    objective = mdl.sum(manager_costs[m] * x[(m, r)] for m in managers for r in railways)
    mdl.minimize(objective)
    
    # 4. CONSTRAINTS
    
    # Railway Management Constraint: Each railway must be managed by at least one manager
    for r in railways:
        mdl.add_constraint(mdl.sum(x[(m, r)] for m in managers) >= 1, ctname=f"railway_{r}_managed")
    
    # Manager Capacity Constraint: No manager can be assigned to more railways than their capacity
    for m in managers:
        mdl.add_constraint(mdl.sum(x[(m, r)] for r in railways) <= manager_capacities[m], ctname=f"manager_{m}_capacity")
    
    # 5. SOLVING & RESULTS
    solution = mdl.solve()
    
    if solution:
        print(f"Optimal value: {solution.objective_value}")
        for m in managers:
            for r in railways:
                if solution.get_value(x[(m, r)]) > 0:
                    print(f"Manager {m} is assigned to Railway {r}")
    else:
        print("No solution found")
        print(f"Status: {mdl.solve_details.status}")
    
    return mdl

# Run the optimization
railway_optimization()
```

### Execution Results
**Status**: OPTIMAL
**Optimal Value**: 360.0
**Execution Time**: 1.23 seconds
**Reliability**: high
**Analysis**: DOCplex also achieved the optimal solution but with a significantly higher execution time compared to Gurobipy.

## 7. Pyomo Implementation

```python
# Complete PYOMO implementation

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

def railway_optimization():
    """Optimization model for assigning managers to railways."""
    
    # 1. MODEL CREATION
    model = pyo.ConcreteModel()
    
    # 2. DATA SETUP
    # Managers data
    managers = [1, 2, 3]
    cost_per_level = {1: 120, 2: 150, 3: 200}
    capacities = {1: 4, 2: 6, 3: 8}
    
    # Railways data
    railways = [1, 2, 3]
    
    # CRITICAL: Validate array lengths before indexing
    assert len(managers) == len(cost_per_level) == len(capacities), "Array length mismatch"
    
    # 3. SETS (Pyomo way to define indices)
    model.M = pyo.Set(initialize=managers)  # Set of managers
    model.R = pyo.Set(initialize=railways)  # Set of railways
    
    # 4. PARAMETERS (data containers)
    model.cost = pyo.Param(model.M, initialize=cost_per_level)
    model.capacity = pyo.Param(model.M, initialize=capacities)
    
    # 5. VARIABLES
    # Binary decision variables: x[m, r] = 1 if manager m is assigned to railway r
    model.x = pyo.Var(model.M, model.R, within=pyo.Binary)
    
    # 6. OBJECTIVE FUNCTION
    # Minimize the total cost of assigning managers to railways
    def obj_rule(model):
        return sum(model.cost[m] * model.x[m, r] for m in model.M for r in model.R)
    model.objective = pyo.Objective(rule=obj_rule, sense=pyo.minimize)
    
    # 7. CONSTRAINTS
    
    # Railway Management Constraint: Each railway must be managed by at least one manager
    def railway_management_rule(model, r):
        return sum(model.x[m, r] for m in model.M) >= 1
    model.railway_management_constraint = pyo.Constraint(model.R, rule=railway_management_rule)
    
    # Manager Capacity Constraint: No manager can be assigned to more railways than their capacity
    def manager_capacity_rule(model, m):
        return sum(model.x[m, r] for r in model.R) <= model.capacity[m]
    model.manager_capacity_constraint = pyo.Constraint(model.M, rule=manager_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("Optimal solution found!")
        print(f"Optimal value: {pyo.value(model.objective)}")
        
        # Extract variable values
        print("\nVariable values:")
        for m in model.M:
            for r in model.R:
                if pyo.value(model.x[m, r]) > 0.5:  # Only print assignments
                    print(f"Manager {m} is assigned to Railway {r}")
        
    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

# Run the optimization
if __name__ == "__main__":
    railway_optimization()
```

### Execution Results
**Status**: OPTIMAL
**Optimal Value**: 360.0
**Execution Time**: 0.87 seconds
**Reliability**: high
**Analysis**: Pyomo achieved the optimal solution with an execution time between Gurobipy and DOCplex, indicating moderate efficiency.

## 8. Cross-Solver Analysis and Final Recommendation

### Solver Results Comparison

| Solver | Status | Optimal Value | Execution Time | Decision Variables | Retry Attempt |
|--------|--------|---------------|----------------|-------------------|---------------|
| Gurobipy | OPTIMAL | 360.00 | 0.22s | N/A | N/A |
| Docplex | OPTIMAL | 360.00 | 1.23s | N/A | N/A |
| Pyomo | OPTIMAL | 360.00 | 0.87s | N/A | N/A |

### Solver Consistency Analysis
**Result**: All solvers produced consistent results ✓
**Consistent Solvers**: gurobipy, docplex, pyomo
**Majority Vote Optimal Value**: 360.0

### Final Recommendation
**Recommended Optimal Value**: 360.0
**Confidence Level**: HIGH
**Preferred Solver(s)**: gurobipy
**Reasoning**: Gurobipy is recommended due to its optimal solution and the fastest execution time, indicating superior efficiency and reliability.

### Business Interpretation
**Overall Strategy**: The optimal total cost of assigning managers to railways is $360, ensuring all railways are managed without exceeding any manager's capacity.
**Objective Value Meaning**: The optimal objective value of $360 represents the minimum total cost of assigning managers to railways while satisfying all constraints.
**Resource Allocation Summary**: Managers should be assigned to railways in a way that minimizes total cost, ensuring each railway is managed and no manager exceeds their capacity.
**Implementation Recommendations**: Implement the assignment plan derived from the optimal solution, ensuring all railways are managed and manager capacities are respected. Monitor the execution to ensure adherence to the plan.