Info

Cette question est clôturée. Rouvrir pour modifier ou répondre.

How can i used GA toolbox?

1 vue (au cours des 30 derniers jours)
Ji-hwan Hwang
Ji-hwan Hwang le 13 Mai 2015
Clôturé : MATLAB Answer Bot le 20 Août 2021
I want operating GA toolbox. I use a GA tool box, I want to solve the next problem. y = exp((1/L)*x)^B I want L and B optimal value. constant 0.7<=y<=1 I will be waiting for your answer.

Réponses (1)

Walter Roberson
Walter Roberson le 13 Mai 2015
  1 commentaire
Walter Roberson
Walter Roberson le 13 Mai 2015
y = exp((1/L)*x)^B is the same as y = exp(B*(1/L)*x) which is the same as y = exp((B/L)*x) . Therefore the problem can be reduced to finding the optimal C for y = exp(C*x), after which any ratio of B/L that matches that C will be optimal. This is therefor not a problem of optimization in two variables, it is a problem of optimization in 1 variable, after which L = B*C for whatever value of B you choose.
Note that you have not defined what it means for "L and B optimal value". We cannot tell from your problem description what you have as data and what you intend to be variables. Is this really a fitting problem, that you have a list of y and a list of x and you are trying to find the coefficients that best fit the formula y = exp((1/L)*x)^B ? If so then just do a log reduction:
y = exp(C*x)
log(y) = C * x
and then C is just the best fit
log(y) / x
(or is it log(y) \ x ? I can never remember the order.)

Cette question est clôturée.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by