How to Set a Minimum Step Size for Variables in GA Solver (Global Optimization Toolbox)?

Hello,

I am using the GA solver from the Global Optimization Toolbox and I was wondering if it is possible to specify a minimum variation for the variables. I have already set the lower bound (lb) and upper bound (ub), but I would like to define a minimum step size. For example, let's say the variables range from -1 to 1 (ub=1 and lb=-1), but I would like them to take only values that are multiples of 0.1, such as -1, -0.9, -0.8, ..., 0.9, 1, and not values like 0.0X. Thank you in advance.

Answer 1

[John D'Errico]

Really easy.

Don't think of a step size for the GA solver. It does not take steps anyway, not like fmincon or other solvers of that class.

But GA can handle variables with an integer constraint. BUT.... You don't want integer variables. Even so, it is easy, once you start thinking the right way about your problem. Just constrain your variables to be integer, in this case, from -10 to 10. Then divide by 10 when you use them in your objective.

For example. Suppose we want to minimize the simple function -cos(x - pi/8), for x between -1 and 1?, BUT also subject to the constraint that x takes on only values from the set -1:0.1:1? SIMPLE!

intcon = 1; % one unknown, and it will be integer.
nvars = 1;
obj = @(x) -cos(x/10 - pi/8);

lb = -10;
ub = 10;

So x will vary from -10 to 10, in integer steps. But we use x/10 in the objective.

[xsol,fval,exitflag] = ga(obj,nvars,[],[],[],[],lb,ub,[],intcon)

ga stopped because the average change in the penalty function value is less than options.FunctionTolerance and 
the constraint violation is less than options.ConstraintTolerance.

xsol = 4

fval = -1.0000

exitflag = 1

At the end, we divide by 10, to get x in the steps we wanted to see.

xsol = xsol/10

xsol = 0.4000

If we had no constraint on x, what would the min have been?

[x2,fval2] = fminbnd(obj,-10,10);
x2 = x2/10

x2 = 0.3927

So ga found the minimum solution, subject to the constraint that x comes from the desired set. We are the only ones who care that ga thinks it is dealing with integers, but we know better.

[Torsten]

Yes, that's far easier than the way I suggested above.

[Mr. Pavl M.]

Correction to John D'Errico found and applied:
So ga found the minimum solution, subject to the constraint that x comes from the desired set. We are the only ones who care that ga thinks it is dealing with integers, but we know better.
Both Your and my solutions are fitted well, match orderly the specs/requirements.
In order to think optimally, optimal(which means incremental within symmetric time fractals) provisions of what is necessary to supply to us please.
What, which, wherefore, when, for what, who knows better?
What is the default discretization step?
Can we continue to work on the valued IT, ICT domain, Electric Power Steering,Steer-by-wire,Hydraulic-Power-Steering(HPS), motor-driven power steering and IT(for example to join our heads and hands together, to merge effort, to help to each other substantially, not to ignore financial aid/help requests and hiring/employing looking for a job request (what really does it take to fill commercial fin. form, online fields order to pay to us to support better thinking financial worries free, best thinking requires it) within the well defined limited circle to regulate Vehicular Systems (they think servomotor driven power steering will dominate, while HPS is more efficient, right?, >>>>> towards better future) Vickrey–Clarke–Groves auction, ML, Steer-by-wire, Data Science, Optimizations Engineering Design, Revamping and application, Machine Learning, codeworks primary duty projects projects+?

For what to devide by 10 ( /10) twice the regressant and the results if
Please accept my proposal solution above, since it doesn't require devide by 10 twice + more advantages(if you contact me in private). (less burden).

[Paolo]

For reasons of code readability and comprehensibility (since it will need to be modified and adapted by others later), this solution ( the one proposed by John D'Errico) seems to be the best. Anyway, thank you very much to everyone, this is my first question here, and I didn’t expect to receive answers so quickly.

[Mr. Pavl M.]

No, why?, sorry, my solutions is better, also since (because) it comprehends also more versatile solve(...) function and doesn't require 1/10 division twice. It is obviously better, I correct even the decision/selection steps here.

Please review your decision. My solution is indeed more readable, comprehensive, integral and adaptable.

[Paolo]

Given the structure of the problem I’m facing and how my objective function is defined (it’s not a "simple" mathematical function but a series of calculations to optimize a parallel kinematic system based on desired behaviors), I find this option more useful. The number of my variables is high (>50), and I would need to define x_val multiple times (thus adding variables to the workspace), with different upper and lower bounds for each variable. Since I have specific functions to extract everything through vectors, I only need to divide the entire variable vector once during the extraction phase and multiply the matrix containing my limits just once. So for my pourpose seem better and simpler to implement also if maybe this method is more "hard-coded". Anyway, I’m grateful for the time you took to respond to me even though my choice falls on the option proposed by John D'Errico.

Answer 2

[Mr. Pavl M.]

I solved it complete.

Solution AB1:

  
  close all force;
  clc;
  clear all
  clear global;
  
  tol = 2.7756e-1;
  format('long','g') %format of the numeric values: each number represented by about 15 digits

ub = 1;
lb = -1;
pop_size = 40;
step = 0.1;
A = 10;

xval = lb:step:ub;
yval = lb:step:ub;

x = optimvar('x',1,1,1,'Type','continuous','LowerBound',lb,'UpperBound',ub)

x = 
  OptimizationVariable with properties:

          Name: 'x'
          Type: 'continuous'
    IndexNames: {{}  {}  {}}
    LowerBound: -1
    UpperBound: 1

  See variables with show.
  See bounds with showbounds.

y = optimvar('y',1,1,1,'Type','continuous','LowerBound',lb,'UpperBound',ub)

y = 
  OptimizationVariable with properties:

          Name: 'y'
          Type: 'continuous'
    IndexNames: {{}  {}  {}}
    LowerBound: -1
    UpperBound: 1

  See variables with show.
  See bounds with showbounds.

f1 = A*sqrt(y.^2 + x.^2 + 5);

prob = optimproblem(Objective=f1);
vals = optimvalues(prob,x=xval,y=yval);

opts = optimoptions("ga",PopulationSize=400);
[sol,fv] = solve(prob,vals,Solver="ga",Options=opts)

Solving problem using ga.
ga stopped because the average change in the fitness value is less than options.FunctionTolerance.

sol = struct with fields:
    x: 0
    y: 0

fv = 
          22.3606797749979

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Answer 3

[Torsten]

Define 21 integer variables x_i that can take values 0 or 1 (by using the "intcon" input to "ga" and setting the lower and upper bounds for the x_i to 0 and 1, respectively)

Set the linear equality constraint that the sum over these 21 integer variables x_i equals 1 (by using Aeq and beq).

Then the variable

y = -1 + 0.1*sum_{i=0}^{i=20} i*x_i

is the variable you are after: it can only take values -1,-0.9,-0.8,...,1.9,2.