Using "fminimax" in Matlab to solve the Max-Min programming problem

Consider the following function defined in the following picture.

Define a finite set of points as follows:

I want to solve the following programing problem:

My idea is the following:

This programing is the first step of a general programing I am studying at. So it should not give trivial solution like 0 or something like this. However, when I attempted it in Matlab, the solution is not desirable.

The following is my code:

fun = @(x)[-log(max(1,0.1037))+x(1)*log(abs(poly1(0.1037)))+x(2)*log(abs(poly2(0.1037)))...

+x(3)*log(abs(poly3(0.1037)))+...

x(4)*log(abs(poly4(0.1037)));

-log(max(1,0.0259))+x(1)*log(abs(poly1(0.0259)))+x(2)*log(abs(poly2(0.0259)))...

+x(3)*log(abs(poly3(0.0259)))+...

x(4)*log(abs(poly4(0.0259)));

-log(max(1,0.2288))+x(1)*log(abs(poly1(0.2288)))+x(2)*log(abs(poly2(0.2288)))...

+x(3)*log(abs(poly3(0.2288)))+...

x(4)*log(abs(poly4(0.2288)));

-log(max(1,0.0938))+x(1)*log(abs(poly1(0.0938)))+x(2)*log(abs(poly2(0.0938)))...

+x(3)*log(abs(poly3(0.0938)))+...

x(4)*log(abs(poly4(0.0938)));

-log(max(1,0.0917))+x(1)*log(abs(poly1(0.0917)))+x(2)*log(abs(poly2(0.0917)))...

+x(3)*log(abs(poly3(0.0917)))+...

x(4)*log(abs(poly4(0.0917)));

-log(max(1,0.2386))+x(1)*log(abs(poly1(0.2386)))+x(2)*log(abs(poly2(0.2386)))...

+x(3)*log(abs(poly3(0.2386)))+...

x(4)*log(abs(poly4(0.2386)));

-log(max(1,0.2003))+x(1)*log(abs(poly1(0.2003)))+x(2)*log(abs(poly2(0.2003)))...

+x(3)*log(abs(poly3(0.2003)))+...

x(4)*log(abs(poly4(0.2003)));];

A= [];

b= [];

Aeq = [];

beq = [];

x0 = [0,0,0,0];

lb = [0,0,0,0];

up = [1000,1000,1000,1000];

[x,fval] = fminimax(fun,x0,A,b,Aeq,beq,lb,ub)

where in the code:

The solution that Matlab provides is the following:

I don't quite understand why this happens. Is there anything wrong in my idea or code? Thank you so much for your help!

Answer 1

[Torsten]

Yes, [0 0 0 0] is the correct solution to your problem.

For all other c vectors >= 0, min g(x,c) would be negative.

Since your problem is linear in the c's, here is an easier way to solve your problem:

poly1=@(x)x.^2-x-1;
poly2=@(x)x.^4-x.^3-3*x.^2+x+1;
poly3=@(x)x.^8-x.^7-7*x.^6+4*x.^5+13*x.^4-4*x.^3-7*x.^2+x+1;
poly4=@(x)x.^3+x.^2-2*x-1;
X = [0.1037;0.0259;0.2288;0.0938;0.0917;0.2386;0.2003];
A = [log(abs(poly1(X))),log(abs(poly2(X))),log(abs(poly3(X))),log(abs(poly4(X))),ones(numel(X),1)]  b = zeros(numel(X),1);
f = [0 0 0 0 -1].';
lb = [0 0 0 0 -Inf].';
ub = [1000 1000 1000 1000 Inf].';
c = linprog(f,A,b,[],[],lb,ub)

[JacobsonRadical]

I am sorry for the late reply. After reading the document of linear programing, I believe that your code is doing the following.

I ran the code and it gave 0,0,0,0,0. So bascially the solution of the min max is just 0 with every parameter being 0. Am I right?

[Torsten]

log(max(1,x)) can always be neglected because it does not depend on c1,...,c4.

c(5) is min_(x in X) g(x,c) which is to be maximized - therefore min: -c(5) in the code.

[JacobsonRadical]

Great. Thank you so much.