Optimisation with function handle
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Hi guys!
I have a non-convex optimisation problem which possibly can be solved using function handle and fmincon.
Lets start with the simple example
c = 5;
min c - lambda*p
ST: cb >= 0
I already have an "inner" function which is called revenue that computes the product of lambda and p (lambda*p) and takes the input (cb)
Thus,I am just creating a function handle,
revenue = @revenue;
And re-formulate my problem as
min c + revenue(cb)
ST: cb >= 0
This works, which is great!
However, as my problem is slightly different...
min c*p - lambda*p
ST: cb >= 0
p is now multiplied to the c as well.
My first thought was just to modify the revenue function in such way that it would return the p itself as well.
function [revenue,p] = revenue(cb)
and then make a function handle as
[revenue,p] = @revenue;
However, this is not working :/
Any help, ideas or suggestions are highly appreciated.
Cheers Fred
6 个评论
Frederik Skøtt
2018-2-2
Adam
2018-2-2
In general you do it just like any other function handle - e.g.
>> f = @min
f =
function_handle with value:
@min
>> [m, i] = f( [3 4] )
m =
3
i =
1
Just saying something is 'not working' if not much use to people aiming to help though - it doesn't tell us anything!
revenue = @revenue;
This is a bad idea, even if it works. You see, that the confusion is perfect, if you replace the handle of a function by a variable which is called like the function. I do not have an idea, what the purpose of this line is:
[revenue,p] = @revenue;
Do you mean:
fcn = @(x), revenue(x, p)
?
Please explain "this is not working" with any details. It is easier to solve a problem than to guess, what the problem is.
Frederik Skøtt
2018-2-2
编辑：Frederik Skøtt
2018-2-2
Jan
2018-2-2
Your notation is hard to read. Please use the standard formatting for text and code, see http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup
Frederik Skøtt
2018-2-2
采纳的回答
Nothing you've posted indicates why the approach wouldn't work. In my opinion, though, the code would be much clearer if revenue(cb) would return lambda and p. Then your objective function could be written, for the simple problem,
function fval=myObjective(cb)
[lambda,p]=revenue(cb);
fval=c-lambda*p;
end
and for the more complicated problem,
function fval=myObjective(cb)
[lambda,p]=revenue(cb);
fval=(c-lambda)*p;
end
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