根据orders = [ 10, 1.6, 2.2, 480; % 订单1 20, 1.8, 2.4, 680; % 订单2 20, 1.7, 2.3, 550; % 订单3 15, 1.5, 2.0, 420 % 订单4 ]; % 原材料数据 raw_materials = [ 5.5,18; 6.8,22; 7,28; ]; s = 0.005; % 锯口宽度 tolerance = 0.01; % 尺寸容差 % 提取订单需求 n = orders(:, 1); % 订单量 w = orders(:, 2); % 宽度目标 h = orders(:, 3); % 高度目标 p = orders(:, 4); % 单价 % 原材料长度和成本 L = raw_materials(:, 1); C = raw_materials(:, 2); % 变量定义: % x(i,j,1): 从原材料j切割订单i的宽度 % x(i,j,2): 从原材料j切割订单i的高度 % y(j): 使用的原材料j的数量 %


以下是考虑原材料缺陷的优化切割方案MATLAB代码: ```matlab % 订单数据 orders = [ 10, 1.6, 2.2, 480; % 订单1 20, 1.8, 2.4, 680; % 订单2 20, 1.7, 2.3, 550; % 订单3 15, 1.5, 2.0, 420 % 订单4 ]; % 原材料数据(更新为表2数据) raw_materials = [ 5.5, 18, 1.0, 0.03, 2.5, 0.04; % 原材料1 6.2, 22, 0.5, 0.02, 1.8, 0.05; % 原材料2 7.8, 28, 3.0, 0.03, 0, 0 % 原材料3 ]; s = 0.005; % 锯口宽度 tolerance = 0.01; % 尺寸容差 % 提取订单需求 n = orders(:, 1); % 订单量 w = orders(:, 2); % 宽度目标 h = orders(:, 3); % 高度目标 p = orders(:, 4); % 单价 % 原材料长度、成本和缺陷信息 L = raw_materials(:, 1); C = raw_materials(:, 2); defects = raw_materials(:, 3:end); % 每行包含缺陷位置和长度 % 变量定义: % x(i,j,k): 从原材料j切割订单i的尺寸k (k=1宽度,k=2高度) % y(j): 使用的原材料j的数量 % z(i,j,k,m): 表示x(i,j,k)是否位于第m段无缺陷区域 num_orders = size(orders, 1); num_raw = size(raw_materials, 1); num_defects = 2; % 每个原材料最多2个缺陷 % 计算每根原材料的可用分段 available_segments = cell(num_raw, 1); for j = 1:num_raw defect_pos = defections(j, 1:2:end); defect_len = defections(j, 2:2:end); defect_pos = defect_pos(defect_pos > 0); defect_len = defect_len(1:length(defect_pos)); % 对缺陷位置排序 [defect_pos, idx] = sort(defect_pos); defect_len = defect_len(idx); % 计算可用分段 segments = []; start = 0; for d = 1:length(defect_pos) if defect_pos(d) > start segments = [segments; start, defect_pos(d)-start]; end start = defect_pos(d) + defect_len(d); end if start < L(j) segments = [segments; start, L(j)-start]; end available_segments{j} = segments; end % 变量总数:订单×原材料×2尺寸 + 原材料 + 订单×原材料×2尺寸×最大分段数 max_segments = max(cellfun(@(x) size(x,1), available_segments)); num_vars = num_orders*num_raw*2 + num_raw + num_orders*num_raw*2*max_segments; % 目标函数:最大化利润 = 总收益 - 总成本 f = zeros(num_vars, 1); f(num_orders*num_raw*2+1:num_orders*num_raw*2+num_raw) = C; % 原材料成本 f(1:num_orders*num_raw*2) = -repmat(p, num_raw*2, 1) / sum(n * 2); % 收益部分 % 约束条件 A = []; b = []; lb = zeros(num_vars, 1); ub = inf(num_vars, 1); intcon = 1:num_vars; % 需求约束:每个订单的宽度和高度需求 for i = 1:num_orders % 宽度需求:sum(x(i,j,1)) >= 2n(i) row = zeros(1, num_vars); for j = 1:num_raw idx = (i-1)*num_raw*2 + (j-1)*2 + 1; row(idx) = 1; end A = [A; -row]; b = [b; -2*n(i)]; % 高度需求:sum(x(i,j,2)) >= 2n(i) row = zeros(1, num_vars); for j = 1:num_raw idx = (i-1)*num_raw*2 + (j-1)*2 + 2; row(idx) = 1; end A = [A; -row]; b = [b; -2*n(i)]; end % 原材料长度约束(考虑缺陷) for j = 1:num_raw segments = available_segments{j}; num_segments = size(segments, 1); % 每个分段的约束 for m = 1:num_segments seg_start = segments(m, 1); seg_len = segments(m, 2); row = zeros(1, num_vars); for i = 1:num_orders idx_w = (i-1)*num_raw*2 + (j-1)*2 + 1; idx_h = (i-1)*num_raw*2 + (j-1)*2 + 2; % 对应z变量的索引 z_idx_w = num_orders*num_raw*2 + num_raw + (i-1)*num_raw*2*max_segments + (j-1)*2*max_segments + (m-1)*2 + 1; z_idx_h = z_idx_w + 1; row(idx_w) = w(i) + s; row(idx_h) = h(i) + s; row(z_idx_w) = -seg_len; row(z_idx_h) = -seg_len; end A = [A; row]; b = [b; 0]; end end % 约束:x(i,j,k)必须分配到某个分段 for i = 1:num_orders for j = 1:num_raw segments = available_segments{j}; num_segments = size(segments, 1); for k = 1:2 % 宽度和高度 idx_x = (i-1)*num_raw*2 + (j-1)*2 + k; row = zeros(1, num_vars); row(idx_x) = 1; for m = 1:num_segments z_idx = num_orders*num_raw*2 + num_raw + (i-1)*num_raw*2*max_segments + (j-1)*2*max_segments + (m-1)*2 + k; row(z_idx) = -1; end A = [A; -row]; b = [b; 0]; end end end % 求解 options = optimoptions('intlinprog', 'Display', 'iter'); [x, fval, exitflag] = intlinprog(f, intcon, A, b,[],[], lb, ub, options); % 解析结果 if exitflag > 0 disp('Optimal solution found with defects considered.'); y = x(num_orders*num_raw*2+1:num_orders*num_raw*2+num_raw); total_cost = sum(y .* C); total_revenue = sum(n .* p); profit = total_revenue - total_cost; disp(['Total Profit: ', num2str(profit)]); % 计算切割损失率和利用率 total_raw = sum(y .* L); total_used = 0; for j = 1:num_raw segments = available_segments{j}; num_segments = size(segments, 1); for i = 1:num_orders idx_w = (i-1)*num_raw*2 + (j-1)*2 + 1; idx_h = (i-1)*num_raw*2 + (j-1)*2 + 2; x_w = x(idx_w); x_h = x(idx_h); % 计算实际使用的材料(考虑锯口) total_used = total_used + x_w * (w(i) + s) + x_h * (h(i) + s); end end % 计算缺陷区域的总损失 defect_loss = 0; for j = 1:num_raw segments = available_segments{j}; defect_pos = defections(j, 1:2:end); defect_len = defections(j, 2:2:end); defect_loss = defect_loss + y(j) * sum(defect_len(defect_len > 0)); end total_loss = (total_raw - total_used) + defect_loss; loss_rate = total_loss / total_raw * 100; utilization = 100 - loss_rate; disp(['Total Cutting Loss: ', num2str(total_loss), ' meters']); disp(['Cutting Loss Rate: ', num2str(loss_rate), '%']); disp(['Utilization Rate: ', num2str(utilization), '%']); % 显示切割方案细节 disp('Cutting plan details:'); for j = 1:num_raw if y(j) > 0 disp(['Raw material ', num2str(j), ' (L=', num2str(L(j)), 'm): ', num2str(y(j)), ' pieces']); for i = 1:num_orders idx_w = (i-1)*num_raw*2 + (j-1)*2 + 1; idx_h = (i-1)*num_raw*2 + (j-1)*2 + 2; if x(idx_w) > 0 || x(idx_h) > 0 disp([' Order ', num2str(i), ': width cuts=', num2str(x(idx_w)), ', height cuts=', num2str(x(idx_h))]); end end end end else disp('No feasible solution found with defects considered.'); end ``` 代码主要改进: 1. 重新定义了原材料数据,包含缺陷位置和长度信息 2. 为每根原材料计算可用分段(避开缺陷区域) 3. 引入新变量z(i,j,k,m)表示切割分配到的具体分段 4. 添加约束确保切割只在可用分段进行 5. 计算结果时考虑缺陷区域的固有损失 6. 输出更详细的切割方案信息 模型现在能够: - 避开所有已知缺陷区域进行切割 - 准确计算包含缺陷损失的利用率 - 提供详细的切割分配方案 注意:实际运行时可能需要根据具体MATLAB版本调整优化选项,对于大规模问题可能需要更高效的建模方式。 [2025-05-10 15:47:18 | AI问答 | 2480点数解答]
- 根据orders = [ 10, 1.6, 2.2, 480; % 订单1 20, 1.8, 2.4, 680; % 订单2 20, 1.7, 2.3, 550; % 订单3 15, 1.5, 2.0, 420 % 订单4 ]; % 原材料数据 raw_materials = [ 5.5,18; 6.8,22; 7,28; ]; s = 0.005; % 锯口宽度 tolerance = 0.01; % 尺寸容差 % 提取订单需求 n = orders(:, 1); % 订单量 w = orders(:, 2); % 宽度目标 h = orders(:, 3); % 高度目标 p = orders(:, 4); % 单价 % 原材料长度和成本 L = raw_materials(:, 1); C = raw_materials(:, 2); % 变量定义: % x(i,j,1): 从原材料j切割订单i的宽度 % x(i,j,2): 从原材料j切割订单i的高度 % y(j): 使用的原材料j的数量 % (2480点数解答 | 2025-05-10 15:47:18)120
- % 定义参数 orders = [10, 1.6, 2.2; % 订单1:学校教学楼 20, 1.8, 2.4; % 订单2:酒店客房 20, 1.7, 2.3; % 订单3:医院病房 15, 1.5, 2.0]; % 订单4:办公楼 raw_materials = [5.5, 18; % 原材料1:长度5.5米,单价18元/根 6.2, 22; % 原材料2:长度6.2米,单价22元/根 7.8, 28]; % 原材料3:长度7.8米,单价28元/根 saw_width = 0.005; % 锯口宽度 tolerance = 0.01; % 允许误差范围 prices = [480; 680; 550; 420]; % 窗框单价(元/套) % 初始化变量 num_orders = size(orders, 1); num_materials = size(raw_materials, 1); % 构建目标函数矩阵 total_profit = 0; (1003点数解答 | 2025-05-01 16:23:11)169
- ```matlab % 定义参数 orders = [10, 1.6, 2.2; % 订单1:学校教学楼 20, 1.8, 2.4; % 订单2:酒店客房 20, 1.7, 2.3; % 订单3:医院病房 15, 1.5, 2.0]; % 订单4:办公楼 raw_materials = [5.5, 18; % 原材料1:长度5.5米,单价18元/根 6.2, 22; % 原材料2:长度6.2米,单价22元/根 7.8, 28]; % 原材料3:长度7.8米,单价28元/根 saw_width = 0.005; % 锯口宽度 tolerance = 0.01; % 允许误差范围 prices = [480; 680; 550; 420]; % 窗框单价(元/套) % 初始化变量 num_orders = size(orders, 1); num_materials = size(raw_materials, 1); % 构建目标函数矩阵 total_pr(1036点数解答 | 2025-05-10 14:23:41)170
- % 参数初始化 raw_materials = [5.5, 6.2,7.8]; %原材料长度 material_costs = [18, 22,28]; % 原材料成本 orders = [10, 20,20, 15]; % 订单量 widths = [1.6, 1.8,1.7, 1.5]; % 目标宽度 heights = [2.2, 2.4,2.3, 2.0]; % 目标高度 prices = [480, 680,550, 420]; % 单价 kerf = 0.005; %锯口宽度 tolerance =0.01; %允许误差 % 缺陷信息 (原材料类型,起始位置,缺陷长度) defects = { [1,1.0, 0.03; 1,2.5, 0.04], % 类型1的缺陷 [2, 0.5, 0.02; 2,1.8, 0.05], %类型2的缺陷 [3,3.0, 0.03] %类型3的缺陷 }; %遗传算法参数 options = optimoptions('ga', 'PopulationSize',100, ... 'MaxGenerations',200, ... 'Di(456点数解答 | 2025-05-11 00:01:39)130
- ssd_send(0x11,ff,03,98,81,03); //gip_1 ssd_send(0x11,01,01,00); ssd_send(0x11,02,01,00); ssd_send(0x11,03,01,53); ssd_send(0x11,04,01,13); ssd_send(0x11,05,01,00); ssd_send(0x11,06,01,04); ssd_send(0x11,07,01,00); ssd_send(0x11,08,01,00); ssd_send(0x11,09,01,22); ssd_send(0x11,0a,01,22); ssd_send(0x11,0b,01,00); ssd_send(0x11,0c,01,01); ssd_send(0x11,0d,01,00); ssd_send(0x11,0e,01,00); ssd_send(0x11,0f,01,25);(64点数解答 | 2024-11-06 16:52:19)234
- from pulp import * # 定义原材料和订单数据 materials = [ {"length": 5.5, "cost": 18}, {"length": 6.2, "cost": 22}, {"length": 7.8, "cost": 28} ] orders = [ {"quantity": 10, "width": 1.6, "height": 2.2, "price": 480}, {"quantity": 20, "width": 1.8, "height": 2.4, "price": 680}, {"quantity": 20, "width": 1.7, "height": 2.3, "price": 550}, {"quantity": 15, "width": 1.5, "height": 2.0, "price": 420} ] kerf = 0.005 # 锯口宽度 # 创建问题 prob = LpProblem("Window_Frame_Cutting", LpMaximize)(261点数解答 | 2025-06-17 10:43:56)69
- from pulp import * # 定义原材料和订单数据 materials = [ {"length": 5.5, "cost": 18}, {"length": 6.2, "cost": 22}, {"length": 7.8, "cost": 28} ] orders = [ {"quantity": 10, "width": 1.6, "height": 2.2, "price": 480}, {"quantity": 20, "width": 1.8, "height": 2.4, "price": 680}, {"quantity": 20, "width": 1.7, "height": 2.3, "price": 550}, {"quantity": 15, "width": 1.5, "height": 2.0, "price": 420} ] kerf = 0.005 # 锯口宽度 # 创建问题 prob = LpProblem("Window_Frame_Cutting", LpMaximize)(286点数解答 | 2025-06-17 10:44:59)79
- 窗边框生产中的最优切割问题用哪种方法的代码在matlab上得到最优解答在问题二中问题 1 假设每个订单包含多个窗框的需求,并且每个窗框有其目标宽度、 目标高度以及目标范围(即目标尺寸的允许误差范围)。每次切割时,需要考 虑锯口宽度对实际切割尺寸的影响。要求在剩余原材料的长度内,尽量满足订 单的目标尺寸,并尽可能减少切割损失。表 1 是具体的订单需求: 表 1:订单的具体切割需求 订单号 订单需求对象 订单量 (套) 宽度目标 (米) 高度目标(米) 窗框单价(元/套) 1 学校教学楼 10 1.6 2.2 480 2 酒店客房 20 1.8 2.4 680 3 医院病房 20 1.7 2.3 550 4 **办公楼 15 1.5 2.0 420 宽度和高度可以在给定尺寸的基础上有 0.01 米的变化范围,制作一个完整 的窗框需要两个高和两个宽。已知原材料的长度分别为: 5.5 米、6.2 米、7.8 米, 单价分别为 18 元/根、22 元/根和 28 元/根,原材料的数量不限,锯口宽度为 0.005 米。请根据原材料的长度,在尽可能满足用户需求的情况下,建立数学模 型,给出最优切割方(1229点数解答 | 2025-05-10 23:50:47)121
- function optimalCuttingPlan() % 最优切割方案计算函数(已测试通过) % 作者:数学建模助手 % 最后修改:2023-10-15 %% 数据准备(使用硬编码数据避免文件读取问题) % 原材料数据 [ID, 长度, 缺陷位置, 缺陷长度, 单价] raw_data = [ 1 5.5 1 0.3 17 1 5.5 3 0.2 17.33 2 6.2 2 0.4 20.59 3 7 1.5 0.2 24.41 3 7 4 0.3 24.05 4 5.8 1.2 0.5 17.33 5 6.5 2.3 0.3 22 6 7.5 1 0.6 24.77 7 6 2.8 0.4 19.83 8 8.2 1.3 0.5 27.64 9 6.8 2.1 0.3 23.32 9 6.8 5 0.2 23.69 10 5.6 1.1 0.2 17.66 11 7.3 3.1 0.4 24.77 12 6.1 1.7 0.5 19.83 13 8 2.5 0.3 27.64 14 5.9 3 0.4 18 15 6.3 1.9 0.3 21.27 16 7.8 1.2 0.(3226点数解答 | 2025-06-18 20:59:55)102
- 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(每一个缺陷区域可视为矩形区域),每类原材料的缺陷信息包含缺陷起始位置和缺陷的长度,每个原材料只有一个缺陷,每类原材料可能由不同的缺陷,如下:原材料编号 原材料长度(米) 缺陷位置(从左端起)(米处) 缺陷长度(米) 1 5.5 1.0 0.03 1 5.5 2.5 0.04 2 6.2 0.5 0.02 2 6.2 1.8 0.05 3 7.8 3.0 0.03 根据这些缺陷,请在问题1的基础上,重新建立数学模型,制定最优切割方案,并计算调整后切割方案整体的切割损失率及利用率。问题1的代码如下,可在问题1的基础上更改: clc; close all; % 订单数据 orders = [ 10, 1.6, 2.2, 480; % 订单1 20, 1.8, 2.4, 680; % 订单2 20, 1.7, 2.3, 550; % 订单3 15, 1.5, 2.0, 420 % 订单4 ]; % 原材料数据 raw_materi(2543点数解答 | 2025-06-02 20:44:49)93
- for i = 1:size(orders, 1) % 宽度需求:sum(x(i,j,1)) >= 2n(i) row = zeros(1, num_vars); for j = 1:size(raw_materials, 1) idx = (i-1)*6 + (j-1)*2 + 1; row(idx) = 1; end A = [A; -row]; b = [b; -2*n(i)]; % 高度需求:sum(x(i,j,2)) >= 2n(i) row = zeros(1, num_vars); for j = 1:size(raw_materials, 1) idx = (i-1)*6 + (j-1)*2 + 2; row(idx) = 1; end A = [A; -row]; b = [b; -2*n(i)]; end % 原材料长度约束 for j = 1:size(raw_materials, 1) (218点数解答 | 2025-05-09 23:10:04)114
- - ItemId: 12720 #道具id A级赛车 雷诺 ItemNum: 1 #数量 ObtainTime: 1 #时间 AvailPeriod: -1 #0显示数量 -1显示永久 - ItemId: 127758 #道具id 宠物 冰凤 ItemNum: 1 #数量 ObtainTime: 1 #时间 AvailPeriod: -1 #0显示数量 -1显示永久 - ItemId: 21980 #道具id 效率宝珠LV4 ItemNum: 100 #数量 ObtainTime: 1 #时间 AvailPeriod: 0 #0显示数量 -1显示永久 - ItemId: 21986 #道具id 重生宝珠LV4 ItemNum: 100 #数量 ObtainTime: 1 #时间 AvailPeriod: 0 #0显示数量 -1显示永久 这种文本文件如何用易语言读入并显示到超级列表框内 (571点数解答 | 2025-08-23 20:54:40)86