Supplement - nonlinear programming

Example

1. Parameters with given values appear in the model

• Example

With the help of for loop , In the parameter a,b Solve the minimum value problem in the range

clc,clear
for a = 0:4
    for b = [2,4,6,7]
        f = @ (x) 4 * x(1) - a * x(1) -2 * x(2);
        aa = [1,1;2,1;1,-b]
        bb = [4;5;-2]
        [x,y] = fmincon(f,rand(2,1),aa,bb,[],[],zeros(2,1))
        fprintf('a = % d,b = % d when ,y = % f \n',a,b,y) % y The value is floating-point, so %f
        x
    end
end
%  Output is ：
a =  4,b =  7 when ,y = -7.999995 

2. The constraints appear both linear and nonlinear

• Example
  
  Linear constraints are written into the file of the main program

 Write the objective function ff1
function y = ff1(x)
c1 = [2,3,1];
c2 = [3,1,0];
y = c1 * x + c2 * x.^2;
y = -y;
 Writing nonlinear constraint functions ff2
function [f,g] = ff2(x)
f = [ x(1) + 2 * x(1)^2 + x(2) + 2 * x(2)^2 + x(3) - 10
      x(1) + x(1)^2 + x(2) + x(2)^2 - x(3) - 50
      2 * x(1) + x(1)^2 + 2 * x(2) + x(3) - 40];
g = x(1)^2 + x(3) - 2;
 Run the main program 
a = [-1,-2,0;-1,0,0];
b = [-1,0];
[x,y] = fmincon(@ff1,rand(3,1),a,b,[],[],[],[],@ff2)
x,y=-y

%  Output is 
x1 =2.3333
x2=0.1667
x3=-3.4444
y =18.0833