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使用 fmincon() 求解一个特殊的非线性程序

[英]Solving a special nonlinear program using fmincon()

 原文  2019-12-02 08:41:36  4   1    matlab/ nonlinear-optimization

问题描述

我的问题是以下优化问题：

min J=X'*E*X+U'*E*U
s.t. X'- X0'-(X')*D1*Q*P+(X')*D1*Q*Z=0,
     I*I'*(D2')*U-Q*I=0

其中X和U是2*r+1 by 1列矩阵， X0是2*r+1 by 1已知列矩阵， E, D1, D2, P and Z是2*r+1 by 2*r+1已知列矩阵矩阵和I是一个2*r+1 by 1已知列矩阵。 此外， Q 是满足I*I'*(D2')*U=Q*I的矩阵。

已知矩阵： X0, E, D1, D2, Z, P, I

提前致谢。

我的尝试：

目标函数文件名： objective_function.m

function objective = objective_function(Unknown,r, E) 
Unknown = ones(2*r+3, 2*r+1);
X = Unknown(1, :);
U = Unknown(2, :);
Q = Unknown(3:end, :);
objective = (X')*E*X +(U')*E*U;
end

约束函数文件名： constraint.m

function [inequality, equality] = constraint(Unknown, r,  X0, Z, I, P, D1, D2)
Unknown = ones(2*r+3, 2*r+1);
X = Unknown(1, :);
U = Unknown(2, :);
Q = Unknown(3:end, :);
% No inequality constraint 
inequality = [];
equality = [X'- X0'-(X')*D1*Q*P+(X')*D1*Q*Z ; I*I'*(D2')*U-Q*I];
end

优化文件名： main.m

   clear all
    clc
    r=2;

    % X0
      X0(1)=1;
    for i=2:2*r+1
        X0(i)=0;
    end
    X0=X0';
    % P
    P1=[1/2];
    P2=zeros(1,r);
    for i=1:r
        P3(i)=(-1)/(i*pi);
    end
    P4=zeros(r,1);
    P5=zeros(r,r);
    for i=1:r
        V1(i)=1/(2*i*pi);
    end
    P6=diag(V1);
    for i=1:r
        W(i)=1/(2*i*pi);
    end
    P7=W';
    for i=1:r
        V2(i)=(-1)/(2*i*pi);
    end
    P8=diag(V2);
    P9=zeros(r,r);
    P=3*[P1 P2 P3 ; P4 P5 P6 ; P7 P8 P9];
    % D1
    M1=[1];
    M2=zeros(1,r);
    M3=zeros(1,r);
    M4=zeros(r,1);
    for i=1:r
        V4(i)=cos((2*i*pi)/3);
    end
    M5=diag(V4);
    for i=1:r
        V5(i)=sin((2*i*pi)/3);
    end
    M6=diag(V5);
    M7=zeros(r,1);
    for i=1:r
        V6(i)=-sin((2*i*pi)/3);
    end
    M8=diag(V6);
    for i=1:r
        V7(i)=cos((2*i*pi)/3);
    end
    M9=diag(V7);
    D1=[M1 M2 M3 ; M4 M5 M6 ; M7 M8 M9];
  
    % D2
    N1=[1];
    N2=zeros(1,r);
    N3=zeros(1,r);
    N4=zeros(r,1);
    for i=1:r
        VV4(i)=cos((2*i*pi*2)/3);
    end
    N5=diag(VV4);
    for i=1:r
        VV5(i)=sin((2*i*pi*2)/3);
    end
    N6=diag(VV5);
    N7=zeros(r,1);
    for i=1:r
        VV6(i)=-sin((2*i*pi*2)/3);
    end
    N8=diag(VV6);
    for i=1:r
        VV7(i)=cos((2*i*pi*2)/3);
    end
    N9=diag(VV7);
    D2=[N1 N2 N3 ; N4 N5 N6 ; N7 N8 N9];
   
    % Z
    Z1=[1];
    Z2=zeros(1,2*r);
    for i=1:r
        Z3(i)=(3/(2*i*pi))*sin((2*i*pi)/3);
    end
    Z3=Z3';
    Z4=zeros(r,2*r);
    for i=1:r
        Z5(i)=(3/(2*i*pi))*(1-cos((2*i*pi)/3));
    end
    Z5=Z5';
    Z6=zeros(r,2*r);
    Z=[Z1 Z2 ; Z3 Z4 ;Z5 Z6];

     % E
    V3(1)=2;
    for i=2:2*r+1
        V3(i)=1;
    end
    E=diag(V3);

    % PHi
     R1=@(x) arrayfun(@(i)cos(i*pi*x),1:r);
     R2=@(x) arrayfun(@(i)sin(i*pi*x),1:r);
     R = @(t) [1, R1(t), R2(t)];
     I=R(1);
     I=I';
     

     A=[];
    b=[];
    Aeq=[];
    beq=[];
    lb=[];
    ub=[];

      initial=ones(2*r+3, 2*r+1);
   
% Objective function
J =@(decision_variable)objective_function(decision_variable, r, E);

% Constraint
equality = @(decision_variable)constraint(decision_variable, r, X0, Z, I, P, D1, D2);



solution = fmincon(J,initial,A,b,Aeq,beq,lb,ub,equality);

% X and U extraction 
 X_sol = solution(1:2*r+1);
 U_sol = solution(2*r + 2:end);

1 个解决方案

解决方案1
 1 已采纳  2019-12-05 04:54:14

U和X是列向量描述如下r = 1

X = [x11; x12; x13] --> X is 3 by 1

U = [u11; u12; u13] --> Y is 3 by 1

Q = [ q11 q12 q13; 
      q21 q22 q23; 
      q31 q32 q33] ---> Q is 3 by 3

Unknown = [ x11 x12 x13; 
            u11 u12 u13; 
            q11 q12 q13; 
            q21 q22 q23; 
            q31 q32 q33]--> Unknown is 5 by 3

Unknown(1, :) is [ x11 x12 x13] --> 1 by 3 --> X = transpose(Unknown(1, :))

% Variable identification
---> X = Unknown(1, :)'
---> U = Unknown(2, :)'
---> Q = Unknown(3:end, :)

在constraint.m ，连接时存在尺寸问题

equality = [equality1; equality2]

equality1 = X'-X0'-(X')*D1*Q*P+(X')*D1*Q*Z

equality1_dimension = 1*3-1*3-(1*3)*(3*3)*(3*3)*(3*3)+(1*3)*(3*3)*(3*3)*(3*3) = 1*3

equality2 = I*I'*(D2')*U-Q*I

equality2_dimension = (3*1)*(1*3)*(3*3)*(3*1) - (3*3)*(3*1) = 3*1
  • 两个等式应具有相同的列数。
  • 只需调换其中之一即可。
equality = [equality1; equality2']

简而言之，更新constraint.m和objective_function.m但请先阅读以了解原因。

目标函数文件名：objective_function.m

function objective = objective_function(Unknown,r, E) 
X = Unknown(1, :)';
U = Unknown(2, :)';
Q = Unknown(3:end, :);
objective = (X')*E*X +(U')*E*U;
end

约束函数文件名：constraint.m

function [inequality, equality] = constraint(Unknown, r,  X0, Z, I, P, D1, D2)
X = Unknown(1, :)';
U = Unknown(2, :)';
Q = Unknown(3:end, :);
% No inequality constraint 
inequality = [];
equality = [X'- X0'-(X')*D1*Q*P+(X')*D1*Q*Z ; (I*I'*(D2')*U-Q*I)'];
end

主文件

clear all
    clc
    r=2;

    % X0
      X0(1)=1;
    for i=2:2*r+1
        X0(i)=0;
    end
    X0=X0';
    % P
    P1=[1/2];
    P2=zeros(1,r);
    for i=1:r
        P3(i)=(-1)/(i*pi);
    end
    P4=zeros(r,1);
    P5=zeros(r,r);
    for i=1:r
        V1(i)=1/(2*i*pi);
    end
    P6=diag(V1);
    for i=1:r
        W(i)=1/(2*i*pi);
    end
    P7=W';
    for i=1:r
        V2(i)=(-1)/(2*i*pi);
    end
    P8=diag(V2);
    P9=zeros(r,r);
    P=3*[P1 P2 P3 ; P4 P5 P6 ; P7 P8 P9];
    % D1
    M1=[1];
    M2=zeros(1,r);
    M3=zeros(1,r);
    M4=zeros(r,1);
    for i=1:r
        V4(i)=cos((2*i*pi)/3);
    end
    M5=diag(V4);
    for i=1:r
        V5(i)=sin((2*i*pi)/3);
    end
    M6=diag(V5);
    M7=zeros(r,1);
    for i=1:r
        V6(i)=-sin((2*i*pi)/3);
    end
    M8=diag(V6);
    for i=1:r
        V7(i)=cos((2*i*pi)/3);
    end
    M9=diag(V7);
    D1=[M1 M2 M3 ; M4 M5 M6 ; M7 M8 M9];

    % D2
    N1=[1];
    N2=zeros(1,r);
    N3=zeros(1,r);
    N4=zeros(r,1);
    for i=1:r
        VV4(i)=cos((2*i*pi*2)/3);
    end
    N5=diag(VV4);
    for i=1:r
        VV5(i)=sin((2*i*pi*2)/3);
    end
    N6=diag(VV5);
    N7=zeros(r,1);
    for i=1:r
        VV6(i)=-sin((2*i*pi*2)/3);
    end
    N8=diag(VV6);
    for i=1:r
        VV7(i)=cos((2*i*pi*2)/3);
    end
    N9=diag(VV7);
    D2=[N1 N2 N3 ; N4 N5 N6 ; N7 N8 N9];

    % Z
    Z1=[1];
    Z2=zeros(1,2*r);
    for i=1:r
        Z3(i)=(3/(2*i*pi))*sin((2*i*pi)/3);
    end
    Z3=Z3';
    Z4=zeros(r,2*r);
    for i=1:r
        Z5(i)=(3/(2*i*pi))*(1-cos((2*i*pi)/3));
    end
    Z5=Z5';
    Z6=zeros(r,2*r);
    Z=[Z1 Z2 ; Z3 Z4 ;Z5 Z6];

     % E
    V3(1)=2;
    for i=2:2*r+1
        V3(i)=1;
    end
    E=diag(V3);

    % PHi
     R1=@(x) arrayfun(@(i)cos(i*pi*x),1:r);
     R2=@(x) arrayfun(@(i)sin(i*pi*x),1:r);
     R = @(t) [1, R1(t), R2(t)];
     I=R(1);
     I=I';


     A=[];
    b=[];
    Aeq=[];
    beq=[];
    lb=[];
    ub=[];
% intial contains X, U and Q
initial = ones(2*r+3, 2*r+1);

% Objective function
J =@(decision_variable)objective_function(decision_variable, r, E);

% Constraint
equality = @(decision_variable)constraint(decision_variable, r, X0, Z, I, P, D1, D2);



solution = fmincon(J,initial,A,b,Aeq,beq,lb,ub,equality);

% X U and Q extraction 
 X_sol = solution(1, :)
 U_sol = solution(2, :)
 Q_sol = solution(3:end, :)

解决方案

X_sol =

    0.2905   -0.2881   -0.2905    0.3984   -0.2893


U_sol =

   -0.0097   -0.0097    0.0097    0.0167    0.0167


Q_sol =

    2.6741    6.1937    3.4713    1.9752    4.3157
    0.1062    0.7722    0.7143    0.4598    0.6802
    1.7584    3.6335    1.8268    0.7934    1.3207
   -1.2972   -3.2643   -1.9671    0.7630   -0.1526
    2.2088    5.4597    3.2508    1.3825    0.4301

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