Discussing the article: "Population optimization algorithms: Micro Artificial immune system (Micro-AIS)" - page 2

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Vladimir Suslov #:

and why should we search for parameters for this function by optimisation?

To see if the optimisation algorithm can cope or not.

 
fxsaber #:

To see if the optimisation algorithm can cope or not.

Handle what?


This function has an infinite number of equal maxima.

ps: and minima
 
Vladimir Suslov #:

Handle what?

The problem of finding (given the discreteness) a set as close to one as possible. One of the simplest FFs by which algorithms can be compared.

 
fxsaber #:

With the problem of finding (given the discreteness) a set as close to one as possible. One of the simplest FFs by which algorithms can be compared.

in the series of articles there are FFs by which one can indeed compare and there is a comparison of algorithms.

To the author thank you.

I would like to apply these and other algorithms in optimising Expert Advisors.

 

To be fair, it should be noted that periodic functions are rarely found in practical problems (I mean those that make sense to solve with AO). Usually, if there is a recurrence, it is a cyclicity with a changing period.

One of the articles said that periodicity is not allowed in functions used as benchmarks because it can give false positives due to some peculiarities of search strategies, such as swarming behaviour, use of periodic oscillations, use of geometric regularities such as the golden ratio and many others, which can show excellent results on strictly periodic test functions but mediocre on other, more practical problems.

It is like throwing a wolf into the sea with killer whales and seeing which of them is stronger, or throwing a killer whale into the forest with wolves. That's why I gave up the Rastrigin function to equalise the possibilities of algorithms for comparison.

There is one more nuance that I did not foresee earlier, some types of algorithms may show overestimated results on benchmarks where multiple duplication is used (in order to simulate multidimensionality). I'm not ready to draw conclusions yet, more research is required, but it's possible that the testing methodology will change slightly in the near future.

This is a living series of articles in the sense that experience and knowledge are accumulated and the reader can go all the way, which is far from obvious at first sight, together with the author.


PS. If it is known that the NP-complete practical problem to be solved has periodicity, one should choose among algorithms those that perform better on periodic benches. Otherwise, periodic benchmarks should be avoided.

PPS. I do not claim to be the opinion of last resort, the theory of optimisation is so vast that it is hardly possible to really have a single correct view of the problem.

 
Vladimir Suslov #:

1- To the author, thank you.

2. I would like to apply these and other algorithms to optimise Expert Advisors.

1. Thank you.

2. Thanks @fxsaber, now your wish is closer to fulfilment.

 
Vladimir Suslov #:

There are an infinite number of equal maxima in this function.

ps: and minima

Any function has a finite number of extrema on a finite domain of definition with a non-infinitesimal increment. There seems to be a mathematical justification for this, I don't remember which one, though))) It is proved by limits, if I'm not mistaken.
 
Andrey Dik #:

Any function has a finite number of extrema on a finite domain of definition with a non-infinitesimal increment. There is a mathematical justification for this, I don't remember which one, though)))) It is proved by limits, if I am not mistaken. 
Res *= MathSin(Arg[i]);

it is obvious that the sine cannot be greater than +1
respectively the product of sines cannot be greater than 1.

without limits)


we can call the maxima of this function

max = pi/2 + n*2*pi

where n is any integer


there can be even number of values in the parameters with sin(x)=-1

where x = pi/2+pi + n*2*pi

which, when multiplied, give +1



 
Vladimir Suslov #:

it is obvious that the sine cannot be greater than +1
respectively the product of sines cannot be greater than 1.

without limits)


we can call the maxima of this function

max = pi/2 + n*2*pi

where n is any integer


there can be even number of values in the parameters with sin(x)=-1

where x = pi/2+pi + n*2*pi

which, when multiplied, give +1




И?))
There are a finite number of extrema in a finite domain of definition, for crying out loud. Not an unbounded number.
Let's keep in mind that the step has a finite value.
One more thing to keep in mind: we know the extrema of the test FF, but the algo does not. That's the point, that's why we can test the algo and not the algo us. Ugh, ugh.)))
A joke, but in every joke.....
 
Andrey Dik #:
AND?)))
There are a finite number of extrema on a finite domain of definition, for crying out loud. Not an unbounded number.
Keep in mind that the step has a finite value.

max = pi/2 + n*2*pi

where n is any integer.


where is the limit?

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