basic_python_coding.html

供水问题

1问题

供水公司有三个水库分别为A,B,C向四个小区甲乙丙丁供水，A和B向所有小区供水，C仅向甲乙丙供水，水库最大供水量（千吨）

水库	A	B	C
最大供水量（千吨）	50	60	50

小区用水情况为

小区	甲	乙	丙	丁
基本用水量（千吨）	30	70	10	10
额外用水量（千吨）	50	70	20	40

水库供水收入900元/千吨，支出费用为：其他费用450/千吨，引水管理费：

由水库到小区引水费用：元/千吨	甲	乙	丙	丁
A	160	130	220	170
B	140	130	190	150
C	190	200	230

问如何分配水库供水量，公司才能获利最多。

2 解题思路

使引水管理费最小，则利润最高。

设：水库i向小区j的供水量为,总引水管理费为Z

目标函数为：z = 160.0*x11 + 130.0*x12 + 220.0*x13 + 170.0*x14 + 140.0*x21 + 130.0*x22 + 190.0*x23 + 150.0*x24 + 190.0*x31 + 200.0*x32 + 230.0*x33

求的值，使Z最小。

约束条件为：

python 之pulp 线性规划介绍及举例-LMLPHP

 # coding=utf-8

 from pulp import *

 def get_re():

     pass

 def getresult(c, con):

 # 设置对象

     prob = LpProblem('myPro', LpMinimize)

 # 设置三个变量，并设置变量最小取值

     x11 = LpVariable("x11", lowBound=0)

     x12 = LpVariable("x12", lowBound=0)

     x13 = LpVariable("x13", lowBound=0)

     x14 = LpVariable("x14", lowBound=0)

     x21 = LpVariable("x21", lowBound=0)

     x22 = LpVariable("x22", lowBound=0)

     x23 = LpVariable("x23", lowBound=0)

     x24 = LpVariable("x24", lowBound=0)

     x31 = LpVariable("x31", lowBound=0)

     x32 = LpVariable("x32", lowBound=0)

     x33 = LpVariable("x33", lowBound=0)

     X = [x11, x12, x13, x14, x21, x22, x23, x24, x31, x32, x33]

     #c = [160, 130, 220, 170, 140, 130, 190, 150, 190, 200, 230]

 # 目标函数

     z = 0

     for i in range(len(X)):

         z += X[i]*c[i]

     #print(z)

     prob += z

 # 载入约束变量

     prob += x11+x12+x13+x14 == con[0]# 约束条件1

     prob += x21+x22+x23+x24 == con[1]

     prob += x31+x32+x33 == con[2]

     prob += x11+x21+x31 <= con[3]

     prob += x11+x21+x31 >= con[4]

     prob += x12 + x22 + x32 <= con[5]

     prob += x12 + x22 + x32 >= con[6]

     prob += x13 + x23 + x33 <= con[7]

     prob += x13 + x23 + x33 >= con[8]

     prob += x14 + x24 <= con[9]

     prob += x14 + x24 >= con[10]

 # 求解

     status = prob.solve()

     print(status)

     print(LpStatus[status])

     print(value(prob.objective))  # 计算结果

 # 显示结果

 #     for i in prob.variables():

 #         print(i.name + "=" + str(i.varValue))

     for i in prob.variables():

         print(i.varValue)

 if __name__ == '__main__':

     c = [160, 130, 220, 170, 140, 130, 190, 150, 190, 200, 230]

     con = [50, 60, 50, 80, 30, 140, 70, 30,10, 50, 10]

     getresult(mubiao, yueshu)

输出结果：

1
Optimal
24400.0
0.0
50.0
0.0
0.0
0.0
50.0
0.0
10.0
40.0
0.0
10.0