线性规划

scipy库求解

from scipy import optimize
import numpy as np
#求解函数
res = optimize.linprog(c, A, b, Aeq, beq, LB, UB, X0, OPTIONS)
#目标函数最小值
print(res.fun)
#最优解
print(res.x)

样例

样例1

#导入包
from scipy import optimize
import numpy as np
#确定c, A, b, Aeq, beq
c = np.array([2, 3, -5])
A = np.array([[-2, 5, -1], [1,3,1]]) #方向是反的，取负
B = np.array([-10, 12])  
Aeq = np.array([[1,1,1]])  # A和Aeq都是二维数组
Beq = np.array([7])
#求解
res = optimize.linprog(-c, A, B, Aeq, Beq) #求最大值，加个负号
print(res)

注意A和Aeq里面有两个括号（矩阵运算的规则，两个括号表示是在一行的）

结果

pulp库求解

样例2

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]
aeq = [[1,2,4]]
beq = [101]
#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.LpMaximize)
#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[1] + 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] )
for i in range(len(aeq)):
  m += (pulp.lpDot(a[i], x) == beq[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]}')

结果

运输问题

样例

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'])

整数规划

分支定界代码

import math
from scipy.optimize import linprog
import sys
def integerPro(c, A, b, Aeq, beq, t=1.0E-12):
  res = linprog(c, A_ub = A, b_ub = b, A_eq = Aeq, b_eq = beq)
  if (type(res.x) is float):
    bestX = [sys.maxsize]*len(c)
  else:
    bestX = res.x
  bestVal = sum([x*y for x,y in zip(c, bestX)])
  if all(((x-math.floor(x)) < t or (math.ceil(x)-x) < t) for x in bestX):
    return (bestVal, bestX) 
  else:
    ind = [i for i, x in enumerate(bestX) if (x-math.floor(x)) > t and (math.ceil(x) - x) > t][0]
    newCon1 = [0]*len(A[0])
    newCon2 = [0]*len(A[0])
    newCon1[ind] = -1
    newCon2[ind] = 1
    newA1 = A.copy()
    newA2 = A.copy()
    newA1.append(newCon1)
    newA2.append(newCon2)
    newB1 = b.copy()
    newB2 = b.copy()
    newB1.append(-math.ceil(bestX[ind]))    
    newB2.append(math.floor(bestX[ind]))
    r1 = integerPro(c, newA1, newB1, Aeq, beq)
    r2 = integerPro(c, newA2, newB2, Aeq, beq)
    if r1[0] < r2[0]:
      return r1
    else:
      return r2    

非线性规划

样例

1.计算的最小值

from scipy.optimize import minimize
import numpy as np
#计算1/x+x的最小值
def fun(args):
  a = args
  v = lambda x : a / x[0] + x[0]
  return v

if __name__ == "__main__":
  args = (1)    #a
  x0 = np.asarray((2))  #初始猜测值
  res = minimize(fun(args), x0, method = 'SLSQP')
  print(res.fun)
  print(res.success)
  print(res.x)   

2.计算的最小值，其中范围在到之间

from scipy.optimize import minimize
import numpy as np
def fun(args):
  a,b,c,d = args
  v = lambda x : (a + x[0]) / (b + x[1]) - c*x[0] + d*x[2]
  return v

def con(args):
  #约束条件分为eq和ineq
  #eq表示函数结果等于0；ineq表示表达式大于等于0
  x1min, x1max, x2min, x2max, x3min, x3max = args
  cons = ({'type': 'ineq', 'fun': lambda x : x[0] - x1min}, 
          {'type': 'ineq', 'fun': lambda x : -x[0] + x1max}, 
          {'type': 'ineq', 'fun': lambda x : x[1] - x2min}, 
          {'type': 'ineq', 'fun': lambda x : -x[1] + x2max},  
          {'type': 'ineq', 'fun': lambda x : x[2] - x3min}, 
          {'type': 'ineq', 'fun': lambda x : -x[2] + x3max})
  return cons

if __name__ == "__main__":
  #定义常量值
  args = (2, 1, 3, 4)  #abcd
  #设置参数范围/约束条件
  argsl = (0.1, 0.9, 0.1, 0.9, 0.1, 0.9) #x1min, x1max, x2min, x2max……
  cons = con(argsl)
  #设置初始猜测值
  x0 = np.asarray((0.5, 0.5, 0.5))
  res = minimize(fun(args), x0, method = 'SLSQP', constraints = cons)
  print(res.fun)
  print(res.success)
  print(res.x)