纸浆在路线上的运输问题，而不是数量问题

Question

纸浆厂的运输问题涉及到运输的每一个项目。然而，在我的情况下，您使用的每一条路线/车道都有成本，而不是每一个装运的项目，即目标函数是最小化使用的路线（卡车）数量

*即，如果任何路线变量（数量）选择为>；通过优化器0，我希望将相同的成本附加到它，而不考虑数量，否则忽略它（0成本）。 概率+=lpSum（[np.最小值（路线变量[w][b]，1）路线中（w，b）的成本[w][b]），“总车道”

我尝试使用np.minimum，但解决方案似乎没有考虑到它。替代方案是什么

supply=pd.DataFrame.from_dict({38893: {'location_code': '2025', 'excess_cases': 18.0},
 43872: {'location_code': '1580', 'excess_cases': 16.0},
 43929: {'location_code': '1036', 'excess_cases': 16.0},
 62403: {'location_code': '1607', 'excess_cases': 10.0},
 67220: {'location_code': '1983', 'excess_cases': 9.0}}).T

demand=pd.DataFrame.from_dict({12223: {'location_code': '3321', 'deficit_cases': 12.0},
 15682: {'location_code': '3077', 'deficit_cases': 9.0},
 16147: {'location_code': '1264', 'deficit_cases': 9.0},
 18964: {'location_code': '3208', 'deficit_cases': 7.0},
 19389: {'location_code': '1031', 'deficit_cases': 7.0}}).T


VendorStores = supply['location_code']
excess = supply.set_index(['location_code'])['excess_cases'].to_dict()
deficitStores = demand['location_code']
deficit = demand.set_index(['location_code'])['deficit_cases'].to_dict()
costs = makeDict((VendorStores, deficitStores),[[1]*len(deficitStores)]*len(VendorStores))

prob = LpProblem("LP Problem",LpMinimize)
Routes = [(w,b) for w in VendorStores for b in deficitStores]
route_vars = LpVariable.dicts("Route",(VendorStores,deficitStores),0,None,LpInteger)
prob += lpSum([np.minimum(route_vars[w][b],1)*costs[w][b] for (w,b) in Routes]), "Total Lanes"
for w in VendorStores:
    prob += lpSum([route_vars[w][b] for b in deficitStores]) <= excess[w], "Sum of Cases out of VendorStore {0}".format(str(w))
for b in deficitStores:
    prob += lpSum([route_vars[w][b] for w in VendorStores]) >= deficit[b]

Answer 1

你的代码不是很清楚，也不完整；我们不知道成本是多少。例如，请通过解释或使用更清晰的变量名来改进它

要添加一个参数，表示您在何处使用路由，在LP术语中，应满足以下条件：

Let c_r = cost of using route r
    r   = whether route r is being used
    d_r = total quantity shipped over route r
    M   = a very big number (at least the sum of all quantities or the capacity of a truck)

min  sum(c_r * r)
s.t. Mr >= d_r
     d_r >= 0
     r in {0, 1}

在这里，如果没有通过路由r运送任何东西，那么r将为零以最小化目标函数，如果d_r > 0，那么r将为1，Mr = M，如果d_r <= M，这将起作用。因此，这完全取决于您为M选择的值

用python术语来说：

prob = LpProblem("LP Problem", LpMinimize)

route_vars = LpVariable.dicts("Route",(VendorStores, deficitStores), 0, None, LpInteger)  # d_r in the example
route_used = LpVariable.dicts("Route",(VendorStores, deficitStores), 0, 1, LpInteger)  # d_r in the example
a_very_large_number = 10000 # replace or calculate

# Objective function
prob += lpSum([(route_used[w][b],1)*costs[w][b] for (w,b) in Routes])

for w in VendorStores:
    prob += lpSum([route_vars[w][b] for b in deficitStores]) <= excess[w], "Sum of Cases out of VendorStore {0}".format(str(w))
for b in deficitStores:
    prob += lpSum([route_vars[w][b] for w in VendorStores]) >= deficit[b]

for (w, b) in routes:
    prob += a_very_large_number * route_used[w][b] >= route_vars[w][b], "Route used"