How to apply ga on single variable?

i have a equation, y = 0.009(1 + x)^0.278

i want to solve it through genetic algorithm

the output says to put atleast 2 arguments

how to operate ga on single variable, do tell

[Githin George]

Hi Shyam,

Does the parameter description in the documentation help. You can take a look at the documentation here

[Shyam pratap singh rathore]

nothing as such mentioned in the documentation

Answer 1

[Walter Roberson]

You do not define any range for x, so it could be less than -1, leading to a negative value raised to a fraction. It is not clear what the "optimal" value for a complex function is.

y = @(x) norm(0.009*(1 + x).^0.278)

y = function_handle with value:
    @(x)norm(0.009*(1+x).^0.278)

nvars = 1;
[bestx, fval] = ga(y, nvars)

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

bestx = -0.9942

fval = 0.0021

%cross-check

syms X real
Y = y(X)

Y = 

dY = diff(Y,X)

dY = 

fplot([Y, dY], [-2 2])

bestX = solve(dY,X)

 
bestX =
 
Empty sym: 0-by-1
 

vpa(bestX)

 
ans =
 
Empty sym: 0-by-1
 

No best value in calculus because it has a discontinuity in the derivative.

[Shyam pratap singh rathore]

please tell me how to put range of x for function to be solved by ga

i have x in the row vector form (1*84)

[Walter Roberson]

You earlier indicated that x is a single variable, but now you are talking about a 1 x 84 variable. I do not know how the 1 x 84 is intended to fit into this problem?

y = @(x) 0.009*(1 + x).^0.278

y = function_handle with value:
    @(x)0.009*(1+x).^0.278

lb = -1; ub = 10;
A = []; b = []; Aeq = []; beq = [];

nvars = 1;
[bestx, fval] = ga(y, nvars, A, b, Aeq, beq, lb, ub)

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

bestx = -1

fval = 0

[Shyam pratap singh rathore]

Thank you for your time,

my intentions are to get minimum value of function through ga

i need to check the function at some particular values of x (84 values i have) , out of these 84 values , if ga could tell me where function will be minimum and x value where it's minimum,

so isn't there any way where i can define the x values where ga should check and give me minimum value accordingly,

upper and lower bounds wouldnt help

[Walter Roberson]

If you have a vector of specific values, and there are no model parameters to change, then you would just evaluate the function at the 84 different values and use min()

You would only use ga() if there is something that has to be changed. For example if you had

y = @(x,p) 0.009*(1 + x).^(p-x)

y = function_handle with value:
    @(x,p)0.009*(1+x).^(p-x)

then it would make sense to define something like

nvars = 1;
lb = -1; ub = 10;
A = []; b = []; Aeq = []; beq = [];

x = sort(rand(1,15) * (ub-lb) + lb)

x = 1×15

-0.9596 -0.8781 -0.2031 0.0707 1.2581 3.4550 4.2471 4.3598 4.4943 5.6380 7.7537 7.7766 8.2198 9.6891 9.6978

pvals = arrayfun(@(X) ga(@(p) y(X,p), nvars, A, b, Aeq, beq, lb, ub), x).'

Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.
Optimization terminated: average change in the fitness value less than options.FunctionTolerance.

pvals = 15×1

10.0000 10.0000 10.0000 -1.0000 -1.0000 -1.0000 -1.0000 -1.0000 -1.0000 -1.0000

In the case of this particular function, the outcome can be predicted as being the ub if x is negative, and the lb if x is positive, but you could imagine other objectives in which the value of x played a more important role.