MATLAB和Python解线性规划

Posted 2020-11-27 逸枚

MATLAB和Python解线性规划

1. 作可行域的方法

//画可行域的方法
[X,Y]=meshgrid(0:0.1:100,0:0.1:100);
idx=(X+Y>=10)&(-2*X+2*Y<=10)&(-4*X+2*Y<=20)&(X+4*Y>=20);
x=X(idx);
y=Y(idx);
k=convhull(x,y);
fill(x(k),y(k),'c');

//m文件,自定义函数的保存路径
G:MATLAB	oolboxsharedmaputils
画完图后再调用xyplot函数

1. 使y轴放置在中间, 也就是上面的xyplot函数

   %作用：将Y坐标轴放在中间
   function xyplot(x,y)
   % PLOT
   if nargin>0
   if nargin == 2
   plot(x,y);
   else
   display(' Not 2D Data set !')
   end
   end
   hold on;
   % GET TICKS
   X=get(gca,'Xtick');
   Y=get(gca,'Ytick');
   % GET LABELS
   XL=get(gca,'XtickLabel');
   YL=get(gca,'YtickLabel');
   % GET OFFSETS
   Xoff=diff(get(gca,'XLim'))./40;
   Yoff=diff(get(gca,'YLim'))./40;
   % DRAW AXIS LINEs
   plot(get(gca,'XLim'),[0 0],'k');
   plot([0 0],get(gca,'YLim'),'k');
   % Plot new ticks
   for i=1:length(X)
   plot([X(i) X(i)],[0 Yoff],'-k');
   end;
   for i=1:length(Y)
   plot([Xoff, 0],[Y(i) Y(i)],'-k');
   end;
   % ADD LABELS
   text(X,zeros(size(X))-2.*Yoff,XL);
   text(zeros(size(Y))-3.*Xoff,Y,YL);
   box off;
   % axis square;
   axis off;
   set(gcf,'color','w');
   set(gca,'FontSize',20);

2. MATLAB直接求解线性规划

   >> f=[2,-1];
   >> A=[-1,-1;-2,2;-4,2;-1,-4];
   >> b=[-10;10;20;-20];
   >> lb=zeros(2,1);
   >> [x,fval]=linprog(f,A,b,[],[],lb,[])

3. Python代码实现单纯形法

   # coding=utf-8
   # 单纯形法的实现，只支持最简单的实现方法
   # 且我们假设约束矩阵A的最后m列是可逆的
   # 这样就必须满足A是行满秩的（m*n的矩阵）
   
   import numpy as np
   
   
   class Simplex(object):
       def __init__(self, c, A, b):
           # 形式 minf(x)=c.Tx
           # s.t. Ax=b
           self.c = c
           self.A = A
           self.b = b
   
       def run(self):
           c_shape = self.c.shape
           A_shape = self.A.shape
           b_shape = self.b.shape
           assert c_shape[0] == A_shape[1], "Not Aligned A with C shape"
           assert b_shape[0] == A_shape[0], "Not Aligned A with b shape"
   
           # 找到初始的B，N等值
           end_index = A_shape[1] - A_shape[0]
           N = self.A[:, 0:end_index]
           N_columns = np.arange(0, end_index)
           c_N = self.c[N_columns, :]
           # 第一个B必须是可逆的矩阵，其实这里应该用算法寻找，但此处省略
           B = self.A[:, end_index:]
           B_columns = np.arange(end_index, A_shape[1])
           c_B = self.c[B_columns, :]
   
           steps = 0
           while True:
               steps += 1
               print("Steps is {}".format(steps))
               is_optim, B_columns, N_columns = self.main_simplex(B, N, c_B, c_N, self.b, B_columns, N_columns)
               if is_optim:
                   break
               else:
                   B = self.A[:, B_columns]
                   N = self.A[:, N_columns]
                   c_B = self.c[B_columns, :]
                   c_N = self.c[N_columns, :]
   
       def main_simplex(self, B, N, c_B, c_N, b, B_columns, N_columns):
           B_inverse = np.linalg.inv(B)
           P = (c_N.T - np.matmul(np.matmul(c_B.T, B_inverse), N)).flatten()
           if P.min() >= 0:
               is_optim = True
               print("Reach Optimization.")
               print("B_columns is {}".format(B_columns))
               print("N_columns is {}".format(sorted(N_columns)))
               best_solution_point = np.matmul(B_inverse, b)
               print("Best Solution Point is {}".format(best_solution_point.flatten()))
               print("Best Value is {}".format(np.matmul(c_B.T, best_solution_point).flatten()[0]))
               print("
   ")
               return is_optim, B_columns, N_columns
           else:
               # 入基
               N_i_in = np.argmin(P)
               N_i = N[:, N_i_in].reshape(-1, 1)
               # By=Ni， 求出基
               y = np.matmul(B_inverse, N_i)
               x_B = np.matmul(B_inverse, b)
               N_i_out = self.find_out_base(y, x_B)
               tmp = N_columns[N_i_in]
               N_columns[N_i_in] = B_columns[N_i_out]
               B_columns[N_i_out] = tmp
               is_optim = False
   
               print("Not Reach Optimization")
               print("In Base is {}".format(tmp))
               print("Out Base is {}".format(N_columns[N_i_in]))  # 此时已经被换过去了
               print("B_columns is {}".format(sorted(B_columns)))
               print("N_columns is {}".format(sorted(N_columns)))
               print("
   ")
               return is_optim, B_columns, N_columns
   
       def find_out_base(self, y, x_B):
           # 找到x_B/y最小且y>0的位置
           index = []
           min_value = []
           for i, value in enumerate(y):
               if value <= 0:
                   continue
               else:
                   index.append(i)
                   min_value.append(x_B[i] / float(value))
   
           actual_index = index[np.argmin(min_value)]
           return actual_index
   
   
   if __name__ == "__main__":
       '''
       c = np.array([-20, -30, 0, 0]).reshape(-1, 1)
       A = np.array([[1, 1, 1, 0], [0.1, 0.2, 0, 1]])
       b = np.array([100, 14]).reshape(-1, 1)
       c = np.array([-4, -1, 0, 0, 0]).reshape(-1, 1)
       A = np.array([[-1, 2, 1, 0, 0], [2, 3, 0, 1, 0], [1, -1, 0, 0, 1]])
       b = np.array([4, 12, 3]).reshape(-1, 1)'''
       c = np.array([-3, -5, -4, 0, 0, 0]).reshape(-1, 1)
       A = np.array([[2, 3, 0, 1, 0, 0], [0, 2, 5, 0, 1, 0], [3, 2, 4, 0, 0, 1]])
       b = np.array([8, 10, 16]).reshape(-1, 1)
       simplex = Simplex(c, A, b)
       simplex.run()
   #转载自https://blog.csdn.net/cpluss/article/details/102596890