我有一个线性不等式/等式系统要在 Matlab 中求解，我使用 linprog .由于一些不平等是严格的，我使用了一个非常小的成本eps来获得严格的包含，如此处解释
下面solve函数在为eps提供值后求解系统。
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显示它）是我在函数solve内放入linprogr算法中的不等式/等式使得 Matlab 生成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p1,p2,p3的解决方案将提供check=1。
问：如上所述，我不知道如何分析地证明这一点;有没有办法产生一个令人满意的数字论证？或者，你能扼杀我的信念并提供解决方案p1,p2,p3提供check=0吗？
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 之间差值的函数，并尝试在给定约束下最小化这个函数。如果你低于零，这意味着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
现在你可以使用一些优化工具来找到最小的错误乐趣，例如fmincon：
x = fmincon(@errorFun, zeros(57, 1), A, b, Aeq, beq, [], [], [], options);
我发现的解决方案是积极的，这意味着你的假设应该是正确的。