数学建模 1 线性规划

可以解决线性规划中的

一. 线性规划

参考《数学建模算法与应用-司守奎》第一章线性规划

求解最普通的线性规划问题：

PULP调用包

import pulp
#目标函数的系数
z = [2, 3, 1]
#约束
a = [[1, 4, 2], [3, 2, 0]]
b = [8, 6]
#确定最大化最小化问题，最大化只要把Min改成Max即可
m = pulp.LpProblem(sense=pulp.LpMinimize)
#定义三个变量放到列表中
x = [pulp.LpVariable(f'x{i}', lowBound=0) for i in [1,2,3]]
#定义目标函数，lpDot可以将两个列表的对应位相乘再加和
#相当于z[0]*x[0]+z[1]*x[0]+z[2]*x[2]
m += pulp.lpDot(z, x)

#设置约束条件
for i in range(len(a)):
    m += (pulp.lpDot(a[i], x) >= b[i])
#求解
m.solve()
#输出结果
print(f'优化结果：{pulp.value(m.objective)}')
print(f'参数取值：{[pulp.value(var) for var in x]}')

#output:
#优化结果：7.0
#参数取值：[2.0, 0.0, 3.0]

如果存在类似 x_1+2x_2+4x_3=101 这种情况，可以

A_eq = [1,2,4]
b_eq = 101
m += (pulp.lpDot(A_eq, x) == b_eq)

二、运输问题

pulp调包代码

import pulp
import numpy as np
from pprint import pprint

def transportation_problem(costs, x_max, y_max):

    row = len(costs)
    col = len(costs[0])

    prob = pulp.LpProblem('Transportation Problem', sense=pulp.LpMaximize)

    var = [[pulp.LpVariable(f'x{i}{j}', lowBound=0, cat=pulp.LpInteger) for j in range(col)] for i in range(row)]

    flatten = lambda x: [y for l in x for y in flatten(l)] if type(x) is list else [x]

    prob += pulp.lpDot(flatten(var), costs.flatten())

    for i in range(row):
        prob += (pulp.lpSum(var[i]) <= x_max[i])

    for j in range(col):
        prob += (pulp.lpSum([var[i][j] for i in range(row)]) <= y_max[j])

    prob.solve()

    return {'objective':pulp.value(prob.objective), 'var': [[pulp.value(var[i][j]) for j in range(col)] for i in range(row)]}


if __name__ == '__main__':
    costs = np.array([[500, 550, 630, 1000, 800, 700],
                       [800, 700, 600, 950, 900, 930],
                       [1000, 960, 840, 650, 600, 700],
                       [1200, 1040, 980, 860, 880, 780]])

    max_plant = [76, 88, 96, 40]
    max_cultivation = [42, 56, 44, 39, 60, 59]
    res = transportation_problem(costs, max_plant, max_cultivation)

    print(f'最大值为{res["objective"]}')
    print('各变量的取值为：')
    pprint(res['var'])

#output:
#最大值为284230.0
#各变量的取值为：
#[[0.0, 0.0, 6.0, 39.0, 31.0, 0.0],
# [0.0, 0.0, 0.0, 0.0, 29.0, 59.0],
# [2.0, 56.0, 38.0, 0.0, 0.0, 0.0],
# [40.0, 0.0, 0.0, 0.0, 0.0, 0.0]]