多目标线性规划——pulp_pulp python

一、背景

因为工作原因需要了解一些运筹优化相关的算法，所以在同事的介绍下找了《运筹学导论（第10版）》（Introduction to Operations Research）看。最近又在微信群中看到有些同学在讨论，于是乎便把用到的相关工具，以及简单用法做一些记录。（后续可能还会逐步更新）

二、多目标优化问题

优化目标1： $ma{x}_{x_1,x_2}2x_1+3x_2$
优化目标2： $ma{x}_{x_1,x_2}4x_1-2x_2$
约束：s.t.
$$\begin{gathered} x_1+x_2\le 10 \\ 2x_1+x_2\le 15 \\ {x}_1,x_2\ge 0 \end{gathered}$$

三、解

3.1 Pulp包解

3.1.1 两步解

最大化一个目标，然后将其添加为约束并解决另一个目标

import pulp
from pulp import LpVariable, LpProblem, LpMaximize
# 最大化一个目标，然后将其添加为约束并解决另一个目标
linear_prob = LpProblem('为第一目标最大化', LpMaximize)
# variable
x1 = LpVariable('x1', lowBound=0)
x2 = LpVariable('x2', lowBound=0)
# max_target
linear_prob += 2*x1 + 3*x2
# s.t.
linear_prob += x1 + x2 <= 10
linear_prob += 2 * x1 + x2 <= 15
# solve
solution = linear_prob.solve()
print('Obj: {Obj}; \nmax = {max_} \nSolution:  x1={x1} x2={x2}'.format(
        Obj = linear_prob.objective,
        max_=pulp.value(linear_prob.objective),
        x1 = pulp.value(x1),
        x2 = pulp.value(x2)
    )
)
"""
Obj: 2*x1 + 3*x2;
max = 30.0
Solution:  x1=0.0 x2=10.0
"""
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基于第一个解，将其作为限制解决第二个目标

linear_prob_sec = LpProblem('为第二目标最大化', LpMaximize)
# max_target
linear_prob_sec += 4*x1 - 2*x2
# s.t.
linear_prob_sec += x1 + x2 <= 10
linear_prob_sec += 2 * x1 + x2 <= 15
linear_prob_sec += 2*x1 + 3*x2 >= 30
# solve
solution = linear_prob_sec.solve()
print('Obj: {Obj}; \nmax = {max_}\nSolution:  x1={x1} x2={x2}'.format(
        Obj = linear_prob_sec.objective,
        max_=pulp.value(linear_prob_sec.objective),
        x1 = pulp.value(x1),
        x2 = pulp.value(x2)
    )
)
"""
Obj: 4*x1 - 2*x2;
max = -19.999999999995993
Solution:  x1=1.0018653e-12 x2=10.0
"""
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3.1.2 使用采样的权重和具有定义步长的迭代组合目标

现在的问题是如何选择α。
在这种情况下，典型的方法是确定有效边界。在经济学中，例如被称为“最佳最优”。

问题重构

优化目标： $ma{x}_{x_1,x_2}\alpha (2x_1+3x_2)+(1-\alpha )(4x_1-2x_2)$
约束：s.t.
$$\begin{gathered} x_1+x_2\le 10 \\ 2x_1+x_2\le 15 \\ {x}_1,x_2\ge 0 \end{gathered}$$

import matplotlib.pyplot as plt
import pulp
from pulp import LpVariable, LpProblem, LpMaximize
import numpy as np
import pandas as pd
plt.style.use('ggplot')


x1 = LpVariable('x1', lowBound=0)
x2 = LpVariable('x2', lowBound=0)
step_size = 0.01
sove_df = pd.DataFrame(columns=['alpha', 'x1', 'x2', 'obj_v', 'org_v'])
for i in range(0, 101, int(step_size * 100)):
    lp = LpProblem('多目标优化', LpMaximize)
    # obj
    lp += (i/100)*(2*x1 + 3*x2) + (1 - i/100)*(4*x1 - 2*x2)
    # s.t.
    lp += x1 + x2 <= 10
    lp += 2 * x1 + x2 <= 15
    res = lp.solve()
    x1_, x2_ = pulp.value(x1), pulp.value(x2)
    org_v = (2*x1_ + 3*x2_) + (4*x1_ - 2*x2_)
    sove_df.loc[int(i/(step_size*100))] = [i/100, x1_, x2_, pulp.value(lp.objective), org_v]


sv_max = sove_df['obj_v'].max()
sv_min = sove_df['obj_v'].min()


plt.title('$ \\alpha \  With \  max_{x_1,x_2} \\alpha(2x_1 + 3x_2) + (1-\\alpha)(4x_1 - 2x_2)$')
plt.plot(sove_df["alpha"], sove_df["obj_v"],color="darkred", alpha=0.7, label="obj_v")

for idx, row in sove_df[sove_df['obj_v']==sv_max].iterrows():
    plt.text(row.alpha, row.obj_v, f'alpha={row.alpha:.3f} obj={row.obj_v:.2f}\n(x1={row.x1}, x2={row.x2})')

for idx, row in sove_df[sove_df['obj_v']==sv_min].iterrows():
    plt.text(row.alpha, row.obj_v, f'alpha={row.alpha:.3f} obj={row.obj_v:.2f}\n(x1={row.x1}, x2={row.x2})')


plt.plot(sove_df["alpha"], sove_df["org_v"],color="steelblue", alpha=0.7, linestyle='--', label="org_v")
# plt.ylim(15, 35)
plt.xlim(0, 1.5)
plt.legend()
plt.xlabel("alpha", size=12)
plt.ylabel("obj_value", size=12)
plt.show()
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