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Inequalities resulting from a system of linear inequalities / equalities in Matlab: numeric arguments or counterexample?

I have a system of linear inequalities / equalities solved in Matlab, and I use linprog . Since some of the inequalities are strict, I use a very small costant eps to get a strict inclusion, as described here

The solve function below solves the system after providing a value for 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

What I believe (but I don’t know how to show it analytically) is that the inequalities / equalities that I introduced into the linprogr algorithm inside the solve function are such that the solution that Matlab produces satisfies another inequality like indicated below:

 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

I believe that solutions p1,p2,p3 will deliver check=1 .

Question As mentioned above, I do not know how to show this analytically; is there any way to give a satisfactory quantitative argument? Or can you kill my faith and provide delivery solutions p1,p2,p3 delivery check=0 ?

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1 answer

What you can do is expand the problem of 19 parameters to 3x19 = 57 parameters, write a function that calculates the difference between lhs and rhs, and try to minimize this function with the given restrictions. If you go below zero, that means rhs> lhs. Extension of restrictions:

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

Error calculation function:

 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

Now you can use some optimization tool to find the minimum errorFun, for example fmincon :

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

The solution I found was positive, which means your assumption must be correct.

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