我们丢代码吧：
## This file is part of the Cardinal Optimizer, all rights reserved.#import coptpy as cpfrom coptpy import COPTimport mathimport itertools# Optimization data for multicommodity problemORIG = ['zhenzhou', 'shanghai', 'tianjin']DEST = ['xi‘an', 'kunming', 'shenzhen',]PROD = ['bands',]supply_list = [19,20,12, ]supply = dict(zip(itertools.product(ORIG, PROD), supply_list))demand_list = [19,25,7,]demand = dict(zip(itertools.product(DEST, PROD), demand_list))limit_list = [625.0] * len(ORIG) * len(DEST)limit = dict(zip(itertools.product(ORIG, DEST), limit_list))cost_list = [3500, 13500,10500,24500, 24000,10500,17500, 16500,25500,]cost = dict(zip(itertools.product(ORIG, DEST, PROD), cost_list))# Create COPT environmentenv = cp.Envr()# Create COPT problemmmulti = env.createModel()# Add variables to problemvtrans = mmulti.addVars(ORIG, DEST, PROD, nameprefix='trans')# Add constraints to problemmmulti.addConstrs((vtrans.sum(i, '*', k) == supply[i, k] for i in ORIG for k in PROD), nameprefix='supply')mmulti.addConstrs((vtrans.sum('*', j, k) == demand[j, k] for j in DEST for k in PROD), nameprefix='demand')mmulti.addConstrs((vtrans.sum(i, j, '*') <= limit[i, j] for i in ORIG for j in DEST), nameprefix='multi')# Set objective functionmmulti.setObjective(vtrans.prod(cost), COPT.MINIMIZE)# Set optimization parametersmmulti.setParam(COPT.Param.TimeLimit, 100)# Solve the problemmmulti.solve()# Display solutionif mmulti.status == COPT.OPTIMAL:print('\nObjective value: {}'.format(mmulti.objval))print('Variable solution: ')multvars = mmulti.getVars()for mvar in multvars:if math.fabs(mvar.x) >= 1e-6:print(' {0:s} = {1:.6f}'.format(mvar.name, mvar.x))
说明一下，这个案例是可以推广的，比如更复杂一点的。
第一：可以试试看混合调度，就是两个公司调度。
第二：可以试试看加一个天数的约束。
3行3列表示的是对应关系，消耗的情况，其实我个人认为如果没有天数约束，可以直接把天数×在成本里面。可以尝试一下的。
上结果吧：
Cardinal Optimizer v7.1.3. Build date Apr 29 2024Copyright Cardinal Operations 2024. All Rights ReservedSetting parameter 'TimeLimit' to 100Model fingerprint: ab1d09dbUsing Cardinal Optimizer v7.1.3 on WindowsHardware has 4 cores and 8 threads. Using instruction set X86_AVX512_E1 (14)Minimizing an LP problemThe original problem has:15 rows, 9 columns and 27 non-zero elementsThe presolved problem has:6 rows, 9 columns and 18 non-zero elementsStarting the simplex solver using up to 8 threadsMethod Iteration Objective Primal.NInf Dual.NInf TimeDual 0 0.0000000000e+00 6 0 0.01sDual 4 6.5000612544e+05 0 0 0.01sPostsolvingDual 4 6.5000000000e+05 0 0 0.01sSolving finishedStatus: Optimal Objective: 6.5000000000e+05 Iterations: 4 Time: 0.01sObjective value: 650000.0Variable solution:= 19.000000= 13.000000= 7.000000= 12.000000
结果很明显了吧，很典型的混合整数规划，我们终于初步尝试了代码，感谢杉数官方案例。
