如何使用 Pulp 解决网络流量问题？

问题描述

考虑以下网络交通模型并假设有 1000 名司机，我们如何通过分配使用每条可能路线的司机数量来计算模型的社会最优值。

我知道这个问题可以通过使用 Pulp 的线性编程来解决。但是，我的代码坏了。

from pulp import *

prob = LpProblem("Network",LpMinimize)
a = LpVariable('A',lowBound=0,upBound=None,cat=LpInteger)
b = LpVariable('B',cat=LpInteger)
c = LpVariable('C',cat=LpInteger)

# objective function
prob +=  a * (((a+b)/100) + 20 + 5) + b * (((a+b)/100) + 10 + ((b+c)/100)) + c * (10 + 20 + ((b+c)/100))

#constraints
prob += a + b + c == 1000

prob.solve()

print("Status: ",LpStatus[prob.status])
for v in prob.variables():
    print(v.name,'=',v.varValue)

错误：

TypeError: Non-constant expressions cannot be multiplied

解决方案：
ABCF = 201
ABEF = 798
ADEF = 1

解决方法

删除非线性项的一种选择是将非线性成本弧上的整数决策变量替换为沿该弧的每个可能数量的驱动程序的二元决策变量（将严重扩展 - 但没有问题小问题）。

下面给出的示例代码。请注意，此返回的解决方案与您在问题中给出的解决方案不同（我得到 ABCF=250，ABEF=750），我将此解决方案的目标值为 29375 VS 您引用的那个成本略高于 29399 - 除非我误解了？...

from pulp import *

prob = LpProblem("Network",LpMinimize)

nodes = ['A','B','C','D','E','F']

arc_costs = [[('A','D'),10],[('B','C'),20],'E'),[('D',[('C','F'),5]]

ab_cost = LpVariable('ab_cost',lowBound=0,upBound=None,cat=LpContinuous)
ef_cost = LpVariable('ef_cost',cat=LpContinuous)

# Binary variable == 1 iff number of drivers along arc = index value
ab_flow = LpVariable.dicts('ab_flow',range(1001),cat=LpBinary)
ef_flow = LpVariable.dicts('ef_flow',cat=LpBinary)

# Variables to contain the selected Number of drivers
ab_flow_val = LpVariable('ab_flow_val',cat=LpContinuous)
ef_flow_val = LpVariable('ef_flow_val',cat=LpContinuous)


arc_flow_val = LpVariable.dicts('arc_flow_val',[i[0] for i in arc_costs],cat=LpInteger)

# objective function
prob +=  lpSum([arc_flow_val[i]*j for i,j in arc_costs] + ab_cost + ef_cost)

#constraints

# costs for the non-linear costed arcs:
prob += ab_cost == lpSum([ab_flow[x]*((x**2)/100) for x in range(1001)])
prob += ef_cost == lpSum([ef_flow[x]*((x**2)/100) for x in range(1001)])

# only one of binary variables can be true for each of the non-linear arcs
prob += lpSum([ab_flow[i] for i in range(1001)]) == 1
prob += lpSum([ef_flow[i] for i in range(1001)]) == 1

# set flow values from the binary variables:
prob += ab_flow_val == lpSum([ab_flow[i]*i for i in range(1001)])
prob += ef_flow_val == lpSum([ef_flow[i]*i for i in range(1001)])

# 1000 must leave and 1000 must arrive
prob += ab_flow_val + arc_flow_val[('A','D')] == 1000
prob += arc_flow_val[('C','F')] + ef_flow_val == 1000

# Non terminal nodes must have flow balance
for n in nodes[1:-1]:
    if n == 'B':
        prob += ab_flow_val == arc_flow_val[('B','C')] + arc_flow_val[('B','E')]
    elif n == 'E':
        prob +=  arc_flow_val[('B','E')] + arc_flow_val[('D','E')] == ef_flow_val
    else:
        # flow into node == flow out of node
        prob += lpSum([arc_flow_val[i] for i in arc_flow_val.keys() if i[0] == n]) == lpSum([arc_flow_val[j] for j in arc_flow_val.keys() if j[1] == n])

prob.solve()

print("Status: ",LpStatus[prob.status])
for v in prob.variables():
    if ('_val' in v.name) or ('cost' in v.name):
        print(v.name,'=',v.varValue)