Example nonlinear integer programming

example 1

The title is as follows

Using Monte Carlo method （ Random sampling method ）
Although calculating the optimal solution is not realistic , But the application of probability theory can prove , In the case of a certain amount of calculation , A satisfactory solution can be obtained
The implementation of this algorithm is very simple , Easy to understand , Maintain the maximum value of random number of violent cycle

1） First write $M$ file $mengte.m$ Define the objective function $f$ and Constraint vector function $g$, The procedure is as follows
be aware $g$ The essence of $5\times 1$ The column vector

function [f,g]=mengte(x);
% Input parameters x,  Return value [f,g],  The name of the function mengte
f=x(1)^2+x(2)^2+3*x(3)^2+4*x(4)^2+2*x(5)^2-8*x(1)-2*x(2)-3*x(3)...
-x(4)-2*x(5);
g=[sum(x)-400
    x(1)+2*x(2)+2*x(3)+x(4)+6*x(5)-800
    2*x(1)+x(2)+6*x(3)-200
    x(3)+x(4)+5*x(5)-200];
end

2） To write $M$ file $work.m$ Answer the question

rand('state', sum(clock));
Max=-10000000;% The maximum value of the function is initialized to negative infinity 
tic% Show run time 
for i=1:10^6
    x=99*rand(5,1);% Generate 5*1 Random real matrix of 
    x1=floor(x);x2=ceil(x);% integer 
    [f,g]=mengte(x1);
    if sum(g<=0)==4% Input parameter is x1 when , If all inequalities are satisfied 
        if Max<=f% Maintain the maximum value of the function 
            Max=f;
            ans=x1;
        end
    end
    [f,g]=mengte(x2);% repeat 
    if sum(g<=0)==4
        if Max<=f
            Max=f;
            ans=x2;
            
        end
    end
end
ans,Max
toc

lingo solution

model:
sets:
row/1..4/:b;
col/1..5/:c1,c2,x;
link(row,col):a;
endsets
data:
c1=1,1,3,4,2;
c2=-8,-2,-3,-1,-2;
a=1 1 1 1 1
  1 2 2 1 6
  2 1 6 0 0
  0 0 1 1 5;
b=400,800,200,200;
enddata
max=@sum(col:c1*x^2+c2*x);
@for(row(i):@sum(col(j):a(i,j)*x(j))<b(i));
@for(col:@gin(x));
@for(col:@bnd(0,x,99));
end

Example 2

Explain ：
1） To write $M$ file $fun1.m$ Define the objective function

function f=fun1(x);
f=sum(x.^2)+8;

2） To write $M$ file $fun2.m$ Define nonlinear constraints

function [g,h]=fun2(x);
g=[-x(1)^2+x(2)-x(3)^2% Nonlinear inequality constraints 
    x(1)+x(2)^2+x(3)^3-20];
h=[-x(1)-x(2)^2+2% Nonlinear equality constraints 
    x(2)+2*x(3)^2-3];

3） Write main program

options=optimset('largescale', 'off');
[x,y]=fmincon('fun1',rand(3,1),[],[],[],[],...
    zeros(3,1),[],'fun2',options)

Can be found $x_1=0.5522,x_2=1.2033,x_3=0.9478$ when , minimum value $y=10.6511$.