Python 线性规划 运输问题

运输问题-农田产出最大线性规划问题

from scipy import optimize as op 
import numpy as np
c=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])
A=np.zeros((6,24),dtype=np.float64)
B=np.zeros((6,1),dtype=np.float64)
Aeq=np.array([[1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1],[1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0],[0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0],[0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0],[0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0],[0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0],[0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1]])
Beq=np.array([76,88,96,40,42,56,44,39,60,59])
print(op.linprog(-c,A,B,Aeq,Beq))

 sicpy.optimize运行结果为：

           con: array([0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])
 crossover_nit: 0
         eqlin:  marginals: array([   -0.,  -100.,  -210.,  -410.,  -790.,  -750.,  -630., -1000.,
        -800.,  -830.])
  residual: array([0., 0., 0., 0., 0., 0., 0., 0., 0., 0.])
           fun: -284230.0
       ineqlin:  marginals: array([-0., -0., -0., -0., -0., -0.])
  residual: array([0., 0., 0., 0., 0., 0.])
         lower:  marginals: array([290., 200.,   0.,   0.,   0., 130.,  90., 150., 130., 150.,   0.,
         0.,   0.,   0.,   0., 560., 410., 340.,   0., 120.,  60., 550.,
       330., 460.])
  residual: array([ 0.,  0.,  6., 39., 31.,  0.,  0.,  0.,  0.,  0., 29., 59.,  2.,
       56., 38.,  0.,  0.,  0., 40.,  0.,  0.,  0.,  0.,  0.])
       message: 'Optimization terminated successfully. (HiGHS Status 7: Optimal)'
           nit: 15
         slack: array([0., 0., 0., 0., 0., 0.])
        status: 0
       success: True
         upper:  marginals: array([0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,
       0., 0., 0., 0., 0., 0., 0.])
  residual: array([inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf,
       inf, inf, inf, inf, inf, inf, inf, inf, inf, inf, inf])
             x: array([ 0.,  0.,  6., 39., 31.,  0.,  0.,  0.,  0.,  0., 29., 59.,  2.,
       56., 38.,  0.,  0.,  0., 40.,  0.,  0.,  0.,  0.,  0.])

于是，可得：

目标函数最大值为：284230元。

目标函数最大时，决策变量取值为：[第1块地小麦，第2块地小麦，第3块地小麦，第4块地小麦，第5块地小麦，第6块地小麦，第1块地玉米，第2块地玉米，第3块地玉米，第4块地玉米，第5块地玉米，第6块地玉米，第1块地蔬菜，第2块地蔬菜，第3块地蔬菜，第4块地蔬菜，第5块地蔬菜，第6块地蔬菜，第1块地水果，第2块地水果，第3块地水果，第4块地水果，第5块地水果，第6块地水果]=[ 0.,  0.,  6., 39., 31.,  0.,  0.,  0.,  0.,  0., 29., 59.,  2.,56., 38.,  0.,  0.,  0., 40.,  0.,  0.,  0.,  0.,  0.]。