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我有一个线性不等式/等式系统要在 Matlab 中求解，我使用linprog. 由于某些不等式是严格的，因此我使用非常小的成本eps来获得严格的包含，如此处所述

下面的函数solve在为 提供值后求解系统eps。

function pj=solve(eps)

%Inequalities
%x(1)-x(5)-x(9)-x(10)-x(11)-x(15)-x(16)-x(17)-x(19)<=0;
%x(2)-x(6)-x(9)-x(12)-x(13)-x(15)-x(17)-x(18)-x(19)<=0;
%x(3)-x(7)-x(10)-x(12)-x(14)-x(16)-x(17)-x(18)-x(19)<=0;
%x(4)-x(8)-x(11)-x(13)-x(14)-x(15)-x(16)-x(18)-x(19)<=0;

%x(1)+x(2)-x(5)-x(9)-x(10)-x(11)-x(15)-x(16)-x(17)-x(19)-...
%              x(6)-x(12)-x(13)-x(18)<=0;
%x(1)+x(3)-x(5)-x(9)-x(10)-x(11)-x(15)-x(16)-x(17)-x(19)-...
%              x(7)-x(12)-x(14)-x(18)<=0;
%x(1)+x(4)-x(5)-x(9)-x(10)-x(11)-x(15)-x(16)-x(17)-x(19)-...
%             x(8)-x(13)-x(14)-x(18)<=0;
%x(2)+x(3)-x(6)-x(9)-x(12)-x(13)-x(15)-x(17)-x(18)-x(19)-...
%              x(7)-x(10)-x(14)-x(16)<=0;
%x(2)+x(4)-x(6)-x(9)-x(12)-x(13)-x(15)-x(17)-x(18)-x(19)-...
%              x(8)-x(11)-x(14)-x(16)<=0;
%x(3)+x(4)-x(7)-x(10)-x(12)-x(14)-x(16)-x(17)-x(18)-x(19)-...
%              x(8)-x(11)-x(13)-x(15)<=0;


%x(1)+x(2)+x(3)-x(5)-x(9)-x(10)-x(11)-x(15)-x(16)-x(17)-x(19)-...
%                  x(6)-x(12)-x(13)-x(18)-...
%                  x(7)-x(14)<=0;  
%x(1)+x(2)+x(4)-x(5)-x(9)-x(10)-x(11)-x(15)-x(16)-x(17)-x(19)-...
%                  x(6)-x(12)-x(13)-x(18)-...
%                  x(8)-x(14)<=0;   
%x(1)+x(3)+x(4)-x(5)-x(9)-x(10)-x(11)-x(15)-x(16)-x(17)-x(19)-...
%                  x(7)-x(12)-x(14)-x(18)-...
%                  x(8)-x(13)<=0; 
%x(2)+x(3)+x(4)-x(6)-x(9)-x(12)-x(13)-x(15)-x(17)-x(18)-x(19)-...
%                  x(7)-x(10)-x(14)-x(16)-...
%                  x(8)-x(11)<=0; 


%Equalities
%x(1)+x(2)+x(3)+x(4)=1;
%x(5)+x(6)+x(7)+x(8)+x(9)+x(10)+x(11)+x(12)+x(13)+x(14)+x(15)+x(16)+x(17)+x(18)+x(19)=1;


%I also want each component of x to be different from 1 and 0 (strictly included between 1 and 0 given the equalities constraints)
%x(1)>0 ---> x(1)>=eps ---> -x(1)<=-eps
%...
%x(19)>0
%x(1)<1 ---> x(1)<=1-eps
%...
%x(19)<1

%52 inequalities (14+19+19)
%2 equalities
%19 unknowns

A=[1  0  0  0  -1  0  0  0  -1  -1  -1  0  0  0  -1  -1  -1  0  -1;...
   0  1  0  0   0 -1  0  0  -1   0   0 -1 -1  0  -1   0  -1  -1 -1;...
   0  0  1  0   0  0 -1  0   0  -1   0 -1  0 -1   0  -1  -1  -1 -1;...
   0  0  0  1   0  0  0 -1   0   0  -1  0 -1 -1  -1  -1   0  -1 -1;...

   1  1  0  0  -1 -1  0  0  -1  -1  -1 -1 -1  0  -1  -1  -1  -1 -1;...
   1  0  1  0  -1  0 -1  0  -1  -1  -1 -1  0 -1  -1  -1  -1  -1 -1;...
   1  0  0  1  -1  0  0 -1  -1  -1  -1  0 -1 -1  -1  -1  -1  -1 -1;...
   0  1  1  0   0 -1 -1  0  -1  -1   0 -1 -1 -1  -1  -1  -1  -1 -1;...
   0  1  0  1   0 -1  0 -1  -1   0  -1 -1 -1 -1  -1  -1  -1  -1 -1;...
   0  0  1  1   0  0 -1 -1   0  -1  -1 -1 -1 -1  -1  -1  -1  -1 -1;...

   1  1  1  0  -1 -1 -1  0  -1  -1  -1 -1 -1 -1  -1  -1  -1  -1 -1;...
   1  1  0  1  -1 -1  0 -1  -1  -1  -1 -1 -1  0  -1  -1  -1  -1 -1;...
   1  0  1  1  -1  0 -1 -1  -1  -1  -1 -1 -1 -1  -1  -1  -1  -1 -1;...
   0  1  1  1   0 -1 -1 -1  -1  -1  -1 -1 -1 -1  -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 -1  0  0   0  0  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0 -1  0   0  0  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0 -1   0  0  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0  -1  0  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0 -1  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0 -1  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0 -1   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0  -1   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0  -1   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0  -1  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0 -1  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0 -1  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0 -1   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0  -1   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0   0  -1   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0   0   0  -1   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0   0   0   0  -1  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0   0   0   0   0 -1;...

   1  0  0  0   0  0  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  1  0  0   0  0  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  1  0   0  0  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  1   0  0  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   1  0  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  1  0  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  1  0   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  1   0   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   1   0   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   1   0  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   1  0  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  1  0  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  1  0   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  1   0   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0   1   0   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0   0   1   0   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0   0   0   1   0  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0   0   0   0   1  0;...
   0  0  0  0   0  0  0  0   0   0   0  0  0  0   0   0   0   0  1]; %52x19

b=[zeros(1,14) -eps*ones(1,19) (1-eps)*ones(1,19)]; %1x52

Aeq=[1 1 1 1 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 1 1 1 1 1 1 1 1]; %2x19
beq=[1 1]; %1x2

f=zeros(1,19); %1x19

x=linprog(f,A,b,Aeq,beq); 

if ~isempty(x)
    pj=x;
else
    pj=NaN;


end

我相信（但我不知道如何分析地表明它linprogr）是我在函数内部的算法中放入的不等式/等式solve使得 Matlab 产生的解决方案将满足另一个不等式，如下所示：

clear
rng default

%solve system 
p1=solve(unifrnd(0,0.05));

%solve system
p2=solve(unifrnd(0,0.05));


%solve system
p3=solve(unifrnd(0,0.05));



if ~isnan(p1) & ~isnan(p2) & ~isnan(p3) %#ok<AND2>

%LHS
lhs=(p1(2)+p1(3)+p1(4))*1*1+...
     p1(1)*(p2(1)+p2(4))*1+...
     p1(1)*p2(2)*(p3(2)+p3(3)+p3(4))+...
     p1(1)*p2(3)*(p3(1)+p3(2)+p3(3)); 


%RHS 
 rhs=(1-(p1(5)+p1(9)+p1(10)+p1(11)+p1(15)+p1(16)+p1(17)+p1(19)))*...
      1*...
      1+... 
      ... +
     (p1(5)+p1(9)+p1(10)+p1(11)+p1(15)+p1(16)+p1(17)+p1(19))*...
     (p2(5)+p2(8)+p2(11))*...
      1+...
      ... +
     (p1(5)+p1(9)+p1(10)+p1(11)+p1(15)+p1(16)+p1(17)+p1(19))*...
     (p2(7)+p2(10)+p2(14)+p2(16))*...
     ((p3(5)+p3(9)+p3(10)+p3(17))+(p3(6)+p3(7)+p3(12)))+...
     ... +
     (p1(5)+p1(9)+p1(10)+p1(11)+p1(15)+p1(16)+p1(17)+p1(19))*...
     (p2(6)+p2(9)+p2(13)+p2(15))*...
     ((p3(8)+p3(13)+p3(14)+p3(18))+(p3(6)+p3(7)+p3(12)))+...
     ... +
     (p1(5)+p1(9)+p1(10)+p1(11)+p1(15)+p1(16)+p1(17)+p1(19))*...
     (p2(12)+p2(17)+p2(19))*...
     (p3(6)+p3(7)+p3(12)); 

check=(lhs>=rhs); %I expect check to be 1
else
end

我相信解决方案p1,p2,p3将提供check=1。

问题：如上所述，我不知道如何分析地表明这一点；有没有办法产生令人满意的数值论证？或者，你能扼杀我的信念并提供解决方案p1,p2,p3吗check=0？

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1 回答 1

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你可以做的是将问题从 19 个参数扩展到 3x19 = 57 个参数，编写一个计算 lhs 和 rhs 之间差异的函数，并尝试在给定约束条件下最小化这个函数。如果低于零，则意味着 rhs > lhs。扩展约束：

ACell = repmat({A}, 1, 3); % diagonal matrix with A repeated 3 times on diagonal
A = blkdiag(ACell{:}); % 156x57
b = [b b b]; % 1x156
AeqCell = repmat({Aeq}, 1, 3);
Aeq = blkdiag(AeqCell{:}); % 6x57
beq = [beq beq beq]; % 1x6

计算误差的函数：

function error = errorFun( p )

p1 = p(1:19);
p2 = p(20:38);
p3 = p(39:57);

if ~isnan(p1) & ~isnan(p2) & ~isnan(p3) %#ok<AND2>

    lhs= (...); 
    rhs= (...);
    error= lhs - rhs;

else
    error= 1;
end
end

现在您可以使用一些优化工具来找到最小的 errorFun，例如fmincon：

x = fmincon(@errorFun, zeros(57, 1), A, b, Aeq, beq, [], [], [], options);

我发现的解决方案是肯定的，这意味着您的假设应该是正确的。

于 2017-11-01T15:41:40.473 回答