线性规划的例题

盘荣博/线性规划/optimiaze模块/5.4投资的收益和风险模型1.py

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+from matplotlib import pyplot  as plt
+from numpy import ones , diag , c_ , zeros
+from scipy.optimize import  linprog
+import time
+start = time.time()
+c=list( [-0.05,-0.27,-0.19,-0.185,-0.185])
+A=c_[zeros(4),diag([0.025,0.015,0.055,0.026])]
+Aeq = [[1,1.01,1.02,1.045,1.065]];beq=[[1]]
+a=0;aa=[];ss = []
+while a<=0.05:
+    b=list(ones(4)*a)
+    res= linprog(c,A,b,Aeq,beq)
+    aa.append(a);ss.append(-res.fun)
+    a=a+0.001
+end = time.time()
+print("花费时间：",end-start)
+plt.plot(aa,ss,"r*")
+plt.xlabel("$a$");plt.ylabel("$Q$",rotation=90)
+plt.savefig("figure5_1_1.png",dpi=500) ;plt.show()

盘荣博/线性规划/optimiaze模块/5.4投资的收益和风险模型2.py

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+import numpy as np
+from numpy import ones , zeros , c_,diag
+from scipy.optimize import linprog
+import matplotlib.pyplot as plt
+c = np.append(zeros(5).tolist(),[1]).tolist()
+print(c)
+A=np.append(zeros(4).reshape(4,1),diag([0.025,0.015,0.055,0.026]),axis=1)
+A=np.append(A,ones(4).reshape(4,1)*-1,axis=1).tolist()
+Aeq =[[1,1.01,1.02,1.045,1.065,0]] ;beq=[1]
+A.append([-0.05,-0.27,-0.19,0.185,-0.185,0])
+print(A)
+k=0.05;step = 0.005
+b=([0]*4);b.append(-k)
+print(b)
+kk=[];ss=[]
+while k<0.28:
+    res= linprog(c,A,b,Aeq,beq)
+    kk.append(k)
+    ss.append(res.fun)
+    print(res.fun)
+    k+=step
+    b[4]=-k
+plt.plot(kk,ss,'r*')
+plt.xlabel("$k$");plt.ylabel('$R$')
+plt.savefig("figures5_1_2.png",dpi=500);plt.show()
+

盘荣博/线性规划/optimiaze模块/scipy函数.py

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+from scipy.optimize import linprog
+import time
+import numpy as np
+'''
+目标函数：
+min z= -x1 + 4 x2 
+
+约束条件：
+-3x1 + x2 < = 6
+x1 + 2x2 <= 4 
+x2 >= -3 
+
+'''
+begin = time.time()
+c=[-1,4]
+A=[[-3,1],[1,2]]
+b=[[6],[4]]
+bounds=((None , None),(-3,None))
+res= linprog(c,A,b,None,None,bounds)
+end= time.time()
+print("结果为：",res.fun)
+print("x的值位",res.x," ",type(res.x))
+print(end-begin,"\n",'----------------------------------------')
+'''
+线性规划问题
+ max z = x1  - 2x2 - 3x3 
+ 约束条件
+ -2x1 + x2 + x3 <= 9
+ -3x1 + x2 + 2x3 >= 4 
+ 4x1 -2x2 -x3 = -6 
+ x1 >= -10 ,x2 >=0
+'''
+'''
+scipy标准形式：
+min w= -x1 + 2x2  + 3x3 
+
+约束： 
+-2x1 + x2 + x3 <= 9
+3x1 - x2 - 2x3 <= -4
+'''
+c= (-np.array([1,-2,-3])).tolist()
+A=[[-2,1,1],[-3,1,2]]
+A[1]= (-np.array(A)[1]).tolist()
+b= [[9],[4]]
+b[1]=(-np.array(b[1])).tolist()
+Aeq=[[4,-2,-1]]
+beq =[[-6]]
+LB =[-10,0,None]
+UB = [None]*len(c)
+bounds= tuple(zip(LB,UB))
+res= linprog(c,A,b,Aeq,beq,bounds)
+end= time.time()
+print("结果为：",res.fun)
+print("x的值位",res.x," ",type(res.x))
+print(end-begin,"\n",'----------------------------------------')
+print(res)

盘荣博/线性规划/optimiaze模块/例 5.3.py

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+import time
+
+import numpy as np
+from scipy.optimize import linprog
+start = time.time()
+c = [-110,-120,-130,-110,-115,150]
+c = (-np.array(c)).tolist()
+A=[[1,1,0,0,0,0],
+   [0,0,1,1,1,0],
+   [8.8,6.1,2.0,4.2,5.0,-6],
+   [8.8,6.1,2.0,4.2,5.0,-3]
+   ]
+A[3]=(-np.array(A[3])).tolist()
+b=[[200],[250],[0],[0]]
+Aeq =[[1,1,1,1,1,-1]]
+beq =  [[0]]
+LB= [0]*len(c)
+UB= [None]*len(c)
+bounds= tuple(zip(LB,UB))
+res = linprog(c,A,b,Aeq,beq,bounds)
+end = time.time()
+print("最优解：\n",res.x)
+print("目标函数最小值：\n",-res.fun)