我有一个在Matlab中解决的线性不等式/等式系统，我使用linprog。
由于某些不等式是严格的，因此我使用了非常小的代价eps来获得严格的包含，如here所述
在为solve提供值之后，下面的eps函数解决了系统问题。
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(但我不知道如何解析地表示)的是，我在函数
linprogr中放入solve算法的不等式/等式使得Matlab产生的解决方案可以满足并满足另一个不等式作为如下所示: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，我不知道如何分析性地显示它。有没有办法产生令人满意的数值论证？或者，您可以杀死我的信念并提供
p1,p2,p3交付check=0的解决方案吗？ 最佳答案
您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之差的函数，并尝试在给定约束的情况下最小化此函数。如果低于零，则表示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
计算误差的功能:
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查找errorFun的最小值，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。
关于matlab - Matlab中的线性不等式/等式系统暗含的不等式: numerical arguments or counterexample?，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/46939502/