优化石料运输路线：基于数学模型的铲位-卸点运量分配

% 从i号铲位到j号卸点的石料运量\n\n\nx = optimvar('x', 5, 10, 'Type', 'integer', 'LowerBound', 0);\n\n\n% 从i号铲位到j号卸点的距离\nc = [5.26 5.19 4.21 4.00 2.95 2.74 2.46 1.90 0.64 1.27; 1.90 0.99 1.90 1.13 1.27 2.25 1.48 2.04 3.09 3.51; 5.89 5.61 5.61 4.56 3.51 3.65 2.46 2.46 1.06 0.57; 0.64 1.76 1.27 1.83 2.74 2.60 4.21 3.72 5.05 6.10; 4.42 3.86 3.72 3.16 2.25 2.81 0.78 1.62 1.27 0.50];\nprob = optimproblem;\nprob.Objective = sum(sum(c .* x));\n\n\n% 从i号铲位到j号卸点路线上运行一个周期平均所需时间(分钟)\nT = (c .* 2) / 28 * 60 + 3 + 5;\n\n\n% 从i号铲位到j号卸点最多能同时运行的卡车数\nA = T ./ 5;\n\n\n% 从i号铲位到j号卸点路线上最多可以运行的次数\n% B = (8 * 60 - (A - 1) * 5) ./ T;\nB = (8 * 60) ./ T;\n\n\n% i号铲位矿石的铁含量\np = [30 28 29 32 31 33 32 31 33 31];\n\n\n% j号卸点任务需求\nq = [1.2 1.3 1.3 1.9 1.3];\n\n\n% i号铲位的铁矿石储量\nck = [0.95 1.05 1.00 1.05 1.10 1.25 1.05 1.30 1.35 1.25];\n\n\n% i号铲位的岩石储量\ncy = [1.25 1.10 1.35 1.05 1.15 1.35 1.05 1.15 1.35 1.25];\n\n\n% 描述第i号铲位是否使用的0-1开关变量\ny = optimvar('y', 1, 10, 'Type', 'integer', 'LowerBound', 0, 'UpperBound', 1)\n\n\nprob.Constraints.cons1 = x <= 154 .* A .* B;\nprob.Constraints.cons2 = sum(x) <= y .* 8 .* 60 ./ 5 .* 154;\nprob.Constraints.cons3 = sum(x, 2) <= 8 .* 20 .* 154;\nprob.Constraints.cons4 = x(1, :) + x(2, :) + x(5, :) <= ck .* 10000;\nprob.Constraints.cons5 = x(3, :) + x(4, :) <= cy .* 10000;\nprob.Constraints.cons6 = sum(x, 2) >= q';\ns1 = sum(x, 2) * (p - 30.5);\ns2 = sum(x, 2) * (p - 28.5);\nprob.Constraints.cons7 = s1(1:5) <= zeros(1, 5);\nprob.Constraints.cons8 = s2(1:5) >= zeros(1, 5);\n% prob.Constraints.cons9 = fix(double(x / 154)) == x / 154;\nprob.Constraints.cons10 = sum(y) <= 7;\nprob.Constraints.cons11 = sum(sum(x ./ (154 .* B))) <= 20;\n[sol, fval] = solve(prob)\nsol.x