二次規劃

在MATLAB中，quadprog是具有線性約束的二次目標函數的求解器。
quadprog 求由下式指定的問題的最小值

H、A 和 Aeq 是矩陣，f、b、beq、lb、ub 和 x 是向量。
其具體用法為：

x = quadprog(H,f)
x = quadprog(H,f,A,b)
x = quadprog(H,f,A,b,Aeq,beq)
x = quadprog(H,f,A,b,Aeq,beq,lb,ub)
x =quadprog(H,f,A,b,Aeq,beq,lb,ub,x0)
x =quadprog(H,f,A,b,Aeq,beq,lb,ub,x0,options)
x = quadprog(problem)
[x,fval] = quadprog()
[x,fval,exitflag,output] = quadprog()
[x,fval,exitflag,output,lambda] = quadprog(___)

當解決一個具有線性約束的二次規劃問題時：


要求解此問題，首先輸入系數矩陣。


使用MATLAB語句為：

H = [1 -1; -1 2]; f = [-2; -6]; A = [1 1; -1 2; 2 1]; b = [2; 2; 3]; [x,fval,exitflag,output,lambda] = ...quadprog(H,f,A,b); x,fval %打印結果

當使用python進行二次型求解時，采用qpsolvers庫中的solve_qp函數進行求解：

進行庫文件的安裝

pip install cvxopt==1.2.6 cvxpy==1.1.11 Cython==0.29.22 ecos==2.0.7.post1 numpy osqp==0.6.2.post0 qdldl==0.1.5.post0 qpsolvers==1.7.0 quadprog==0.1.8 scipy scs==2.1.4 -i https://pypi.tuna.tsinghua.edu.cn/simple from numpy import array from qpsolvers import solve_qpH=array([[1.,-1.],[-1.,2.]]) f=array([[-2.],[-6.]]).reshape((2,)) L=array([[1.,1.],[-1.,2.],[2.,1.]]) k=array([[2.],[2.],[3.]]).reshape((3,))x = solve_qp(H, f, L,k) print("QP solution: x = {}".format(x))

輸出

QP solution: x = [0.66666667 1.33333333]

也可參考這位大佬的博客

(https://blog.csdn.net/u013421629/article/details/108358409)

import numpy as np import cvxoptdef quadprog(H, f, L=None, k=None, Aeq=None, beq=None, lb=None, ub=None):"""Input: Numpy arrays, the format follows MATLAB quadprog function: https://www.mathworks.com/help/optim/ug/quadprog.htmlOutput: Numpy array of the solution"""n_var = H.shape[1]P = cvxopt.matrix(H, tc='d')q = cvxopt.matrix(f, tc='d')if L is not None or k is not None:assert(k is not None and L is not None)if lb is not None:L = np.vstack([L, -np.eye(n_var)])k = np.vstack([k, -lb])if ub is not None:L = np.vstack([L, np.eye(n_var)])k = np.vstack([k, ub])L = cvxopt.matrix(L, tc='d')k = cvxopt.matrix(k, tc='d')if Aeq is not None or beq is not None:assert(Aeq is not None and beq is not None)Aeq = cvxopt.matrix(Aeq, tc='d')beq = cvxopt.matrix(beq, tc='d')sol = cvxopt.solvers.qp(P, q, L, k, Aeq, beq)return np.array(sol['x'])if __name__ == '__main__':H=np.array([[1,-1],[-1,2]])print(H)f=np.array([[-2],[-6]])print(f)L=np.array([[1,1],[-1,2],[2,1]])print(L)k=np.array([[2],[2],[3]])print(k)res=quadprog(H, f, L,k)print(res)

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