Optimization: Optimize multiple input variables to minimize the output

Question

Anand Ra 2021-10-3

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1465629-optimization-optimize-multiple-input-variables-to-minimize-the-output

评论： Anand Ra 2021-10-8

采纳的回答： Walter Roberson

Hello

I am looking to optimize multiple input variables to minimize the output using fminsearch.

Clearly, I am doing it wrong :( ( see below ) Below is my initial attempt.

Ultimately wanted to bound the predictions for all the variables ( x,y,z,p,q,r) from 0.1 to 100 in the step 0.1

Any help will be greatly appreciated.Thanks a ton!

%Objective: Attempting to Minimize function output with respect to multiple input variables

% Wanted to minimize function, Pow(X) = ((x*p) + (y*q) + (z*r) ) *l*w), by

% optimizing the variables, x, y,z,p ,q and r.

%l and w are constants

%Creating the objective function with its extra parameters( l,w) as extra arguments.

f =@(X, l,w)(X(1)*X(4) + X(2)*X(5) + X(3)*X(6))*l*w; %

%Declaring extra parameter values

l =2;

w=1;

%Create an anonymous function of x alone that includes the workspace value of the parameter.

fun =@(X)f(X,l,w)

%x0 = [-1,1.9];

X_guess = [1 1.5 1 2 1.25 1];

Xmin = fminsearch(fun,X_guess)

x1 = Xmin(1);

y1 = Xmin(2);

z1 = Xmin(3);

p1 = Xmin(4);

p2 = Xmin(5);

p3 = Xmin(6);

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Anand Ra 2021-10-3

编辑：Anand Ra 2021-10-3

在 MATLAB Online 中打开

@Matt J

Apologies.

Below is the code with output (P) and the optimized variables(x1,y1,z1,p1,q1,r1). Clearly they are off the charts.

If I simply execute the function with my initial guesses, the output P=.75. ( added those calculations as well below.)

I am guessing, that I might be incorrect with the sybtaxm but I am unable to determine whats wrong.

Also, guessing, I should bound it (if so, now sure how?) so it doesnt go off charts, espcially I need the optimized variables to remain positive numbers.

%Objective:  Attempting to Minimize function output with respect to multiple input variables
% Wanted to minimize function, Pow(X) = ((x*p) + (y*q) + (z*r) ) l*w), by
% optimizing the variables, x, y,z,p ,q and r.
%l and w are constants
%Creating the objective function with its extra parameters( l,w) as extra arguments.
f =@(X, l,w)(X(1)*X(4) + X(2)*X(5) + X(3)*X(6))*l*w; %
%Declaring extra parameter values
l =2;
w=1;
%Create an anonymous function of x alone that includes the workspace value of the parameter.
fun =@(X)f(X,l,w)
fun = function_handle with value:
    @(X)f(X,l,w)
X_guess = [1 1.5 1 2 1.25 1];
Xmin = fminsearch(fun,X_guess)
 
Exiting: Maximum number of function evaluations has been exceeded
         - increase MaxFunEvals option.
         Current function value: -302178148141371722286612819344701945253341106128754416021789277177203582000056690981679928524741379537028251648.000000 
Xmin = 1×6
	1.0e+55 *

   -0.1661    0.8198    0.0940   -1.8024   -2.2644    0.4914
x1 = Xmin(1);
y1 = Xmin(2);
z1 = Xmin(3);
p1 = Xmin(4);
q1 = Xmin(5);
r1 = Xmin(6);
P= ((x1*p1)+(y1*q1)+(z1*r1))*l*w
P = -3.0218e+110

Anand Ra 2021-10-3

在 MATLAB Online 中打开

If I simply execute the function with my initial guesses:

l =2;
w=1;
X = [1 1.5 1 2 1.25 1];
P = (X(1)*X(4) + X(2)*X(5) + X(3)*X(6))*l*w
P = 9.7500

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Answer 1

Walter Roberson 2021-10-3

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https://ww2.mathworks.cn/matlabcentral/answers/1465629-optimization-optimize-multiple-input-variables-to-minimize-the-output#answer_800639

Ultimately wanted to bound the predictions for all the variables ( x,y,z,p,q,r) from 0.1 to 100 in the step 0.1

fminsearch() cannot bound variables. fmincon() can bound variables though.

However, you have discrete variables. fminsearch() and fmincon() cannot handle discrete variables.

You have a few options:

use ga() with each of those variables being marked as having an integer constraint from 1 to 1000 (not 100), and divide each variable by 10 inside the objective function; or
Use ndgrid() to construct all of the possible combinations of inputs, and evaluate the function at all of them and take the minimum of all of the evaluations
recognize that multiplying positive values by positive values and summing them is always going to have its minima when the values are as small as possible, so just take the lower bounds of everything and do not bother optimizing.

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Walter Roberson 2021-10-4

在 MATLAB Online 中打开

format long g
%Declaring extra parameter values
l=2 ;
w=1;
fun = @(x)(x(1)*x(4) + x(2)*x(5) + x(3)*x(6))*l*w; ;
%Look in the region where  has positive values, , and .
lb = zeros(1,6);
ub = 100*ones(1,6);
%The problem has no linear constraints, so set those arguments to [].
A = [];
b = [];
Aeq = [];
beq = [];
%Intial guss
x0 = [1 2 1 1.5 2 2];
%Solving the problem using fmincon
[x, P] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub)
Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.
x = 1×6
    0.0000    0.0005    0.0000    0.2510    0.0005    0.7280
P = 
      1.27675251615493e-06
%% A different initial point can lead to a different solution.
x01 = x0/5;
[x0, P1] = fmincon(fun,x0,A,b,Aeq,beq,lb,ub)
Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.
x0 = 1×6
    0.0000    0.0005    0.0000    0.2510    0.0005    0.7280
P1 = 
      1.27675251615493e-06

But like I said, the real minima is at the lower bound

fun(lb)
ans = 
     0

Walter Roberson 2021-10-8

在 MATLAB Online 中打开

%defining optimization variables and an optimization problem object.
a = optimvar('a','LowerBound',0.1,"UpperBound",20);
b = optimvar('b','LowerBound',0.1,"UpperBound",20);
c = optimvar('c','LowerBound',0.1,"UpperBound",20);
d = optimvar('d','LowerBound',0.1,"UpperBound",20);
e = optimvar('e','LowerBound',0.1,"UpperBound",20);
prob = optimproblem;
k= 2;
w=1;
v=1.5;
%constraints
% cons1 = e >= (a+d);
% cons2 = d >=a ;
% cons3 = b <=c ;
% cons4 = (e-d) >= (d-a) ;
% cons5 = (c-b) <= b;
cons1 = e - a- d >= 0.1;
cons2 = d - a >= 0.1 ;
cons3 = c-b >= 0.1;
cons4 = b - a >= 0.1 ;
cons5 = (e-c) >=0.1;
prob.Constraints.cons1 = cons1;
prob.Constraints.cons2 = cons2;
prob.Constraints.cons3 = cons3;
prob.Constraints.cons4 = cons4;
prob.Constraints.cons5 = cons5;
x0.a = 4;
x0.b = 6;
x0.c = 8;
x0.d = 7;
x0.e = 12;
%new variables
AB = sqrt(a.^2 + b.^2);
BC = sqrt( c.^2 + ((e-d)/2).^2 );
CS = sqrt( c.^2 + ((e-d)/2).^2 );
VAB = sqrt(((((a.*v).^2/(((b.^2).*4))) + (v^2)/2 )));
% VBS = sqrt(((a*v)^2/((4*b*b)) + (v^2)/2 ));
VCS = ((2*c)./(e-d)).*sqrt(AB.^2);
VBC= CS.^2 + BC.^2;
%objective function as an expression in the optimization variables.
P =  (AB.*VAB + BC.*VBC + CS.*VCS).*k*w;
%the objective function in prob.
prob.Objective = P;
sol = solve(prob, x0)
Solving problem using fmincon.

Feasible point with lower objective function value found.


Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the value of the optimality tolerance,
and constraints are satisfied to within the value of the constraint tolerance.
sol = struct with fields:
    a: 0.1000
    b: 0.2000
    c: 0.3000
    d: 9.2995
    e: 9.6959

Anand Ra 2021-10-8

@Walter Roberson _/\_ Thank you!

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Optimization: Optimize multiple input variables to minimize the output

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