### 问题重述 针对窗边框生产中的最优切割问题，需要解决以下两个子问题： - **问题2**：假设每个原材料存在不规则缺陷，缺陷的大小和位置是已知的，缺陷会影响切割的部分材料，即加工的成品不能包含缺陷区域。根据这些缺陷，重新制定最优切割方案，并计算调整后切割方案的整体切割损失率及利用率。 - **问题3**：基于附件中的缺陷数据和新的订单需求，建立数学模型，制定最优切割方案。 ### 给出答案（答案为带注释的代码块） ```python import pandas as pd from ortools.linear_solver import pywraplp # 读取订单需求数据 orders_data = [ {"order_id": 1, "quantity": 20, "width_target": 1.6, "height_target": 2.2, "price_per_set": 480}, {"order_id": 2, "quantity": 40, "width_target": 1.8, "height_target": 2.4, "price_per_set": 680}, {"order_id": 3, "quantity": 40, "width_target": 1.7, "height_target": 2.3, "price_per_set": 550}, {"order_id": 4, "quantity": 30, "width_target": 1.5, "height_target": 2.0, "price_per_set": 420} ] # 读取原材料数据及缺陷信息 materials_data = [ {"material_id": 1, "length": 5.5, "defects": [(1.0, 0.03), (2.5, 0.04)], "cost_per_meter": 18}, {"material_id": 2, "length": 6.2, "defects": [(0.5, 0.02), (1.8, 0.05)], "cost_per_meter": 22}, {"material_id": 3, "length": 7.8, "defects": [(3.0, 0.03)], "cost_per_meter": 28} ] # 锯口宽度 kerf_width = 0.005 # 创建线性规划求解器 solver = pywraplp.Solver.CreateSolver('SCIP') # 定义变量 num_materials = len(materials_data) num_orders = len(orders_data) # 定义决策变量 cut_vars = {} for m in range(num_materials): for o in range(num_orders): cut_vars[(m, o)] = solver.IntVar(0, solver.infinity(), f'cut_{m}_{o}') # 目标函数：最大化利润 objective = solver.Objective() for m in range(num_materials): material_length = materials_data[m]['length'] cost_per_meter = materials_data[m]['cost_per_meter'] for o in range(num_orders): order_quantity = orders_data[o]['quantity'] width_target = orders_data[o]['width_target'] height_target = orders_data[o]['height_target'] price_per_set = orders_data[o]['price_per_set'] # 计算每个切割方案的成本和收益 total_length_needed = (width_target + kerf_width) * 2 + (height_target + kerf_width) * 2 num_sets_cut = cut_vars[(m, o)] objective.SetCoefficient(num_sets_cut, price_per_set - (total_length_needed * cost_per_meter)) objective.SetMaximization() # 添加约束条件 for o in range(num_orders): solver.Add(solver.Sum([cut_vars[(m, o)] for m in range(num_materials)]) >= orders_data[o]['quantity']) for m in range(num_materials): material_length = materials_data[m]['length'] defects = materials_data[m]['defects'] valid_lengths = [] last_end = 0 for start, length in defects: if start > last_end: valid_lengths.append(start - last_end) last_end = start + length if last_end < material_length: valid_lengths.append(material_length - last_end) solver.Add(solver.Sum([cut_vars[(m, o)] * ((orders_data[o]['width_target'] + kerf_width) * 2 + (orders_data[o]['height_target'] + kerf_width) * 2) for o in range(num_orders)]) <= sum(valid_lengths)) # 求解 solver.Solve() # 输出结果 for m in range(num_materials): print(f'Material {m + 1}:') for o in range(num_orders): if cut_vars[(m, o)].solution_value() > 0: print(f' Cut {cut_vars[(m, o)].solution_value()} sets for Order {o + 1}') print() # 计算切割损失率和利用率 total_material_used = sum(material['length'] for material in materials_data) total_waste = total_material_used - sum(cut_vars[(m, o)].solution_value() * ((orders_data[o]['width_target'] + kerf_width) * 2 + (orders_data[o]['height_target'] + kerf_width) * 2) for m in range(num_materials) for o in range(num_orders)) utilization_rate = (total_material_used - total_waste) / total_material_used * 100 waste_rate = total_waste / total_material_used * 100 print(f'Total utilization rate: {utilization_rate:.2f}%') print(f'Total waste rate: {waste_rate:.2f}%') ``` ### 代码解析 1. **导入必要的库**：使用`pandas`处理数据，使用`ortools`进行线性规划求解。 2. **定义订单需求和原材料数据**：包括订单的数量、目标尺寸、价格，以及原材料的长度、缺陷信息和成本。 3. **创建线性规划求解器**：使用`ortools`创建一个求解器实例。 4. **定义决策变量**：表示从每种原材料中切割多少套窗框。 5. **设置目标函数**：最大化利润，即总收益减去总成本。 6. **添加约束条件**：确保每个订单的需求得到满足，并考虑缺陷对原材料的有效长度的影响。 7. **求解线性规划问题**：调用求解器进行求解。 8. **输出结果**：打印每种原材料的切割方案，并计算切割损失率和利用率。 ### 知识点 1. **线性规划**：用于优化资源分配，使得在给定约束条件下最大化或最小化某个目标函数。 2. **整数规划**：确保切割数量为整数，避免非整数解。 3. **约束条件建模**：准确表达原材料长度、缺陷位置和订单需求之间的关系。