Optimisation application: how do I create the fitness function?

I have two functions: EndCarb and EndCost.

These functions use inputs of: ID [1 2] and Q [0.000001:0.000001:0.01], and give an output of carbon quantity and cost respectively. (EndCarb [ID,Q]; EndCost[ID,Q])

I want to use the gamultiobj tool to optimise the inputs of ID and Q to give the pareto optimal cost and carbon. So far I have been attempting to write a script along the lines of:

fun(1) = @(ID,Q) EndCost(ID,Q);
%parameterised function
Q=0.000001:0.000001:0.01;
%parameter
fun1a=@(ID) fun(1)(ID,Q);
%function of ID alone
ID=[1,2];

fun(2) = @(ID,Q) EndCarbon(ID,Q);
%parameterised function
Q=0.000001:0.000001:0.01;
%parameter
fun2a=@(ID) fun(2)(ID,Q);
%function of ID alone
ID=[1,2];

% Combine two objectives 'fun1' and 'fun2' 
       fun1and2 = @(ID,Q) [fun(1)(ID,Q) fun(2)(ID,Q)];

I'm not sure if this is the right track, as I am unable to get it to work. Can anybody advise anything?

Answer 1

[Adam Barber]

Hey Howard,

It looks like you are on the right track to defining your fitness function for "gamultiobj". I think the issues you are running into is that:

1. The input to the fitness needs to be a single variable
2. MATLAB does not allow for arrays of function handles

You can resolve the first issue by following the example at: http://www.mathworks.com/help/gads/computing-objective-functions.html

The example function there has 3 values, but only accepts one input argument

function z = my_fun(x)
z = zeros(1,3); % allocate output
z(1) = x(1)^2 - 2*x(1)*x(2) + 6*x(1) + 4*x(2)^2 - 3*x(2);
z(2) = x(1)*x(2) + cos(3*x(2)/(2+x(1)));
z(3) = tanh(x(1) + x(2));

You will need to do something similar where you combine "ID" and "Q" into a single variable that you pass to your objective function.

For the second issue, you can either use a cell array to define your function handles:

fun{1} = @(ID,Q) EndCost(ID,Q);
fun{2} = @(ID,Q) EndCarbon(ID,Q);

or just store them as separate variables.

Now, you can define your fitness function as something like the following:

fun1and2 = @(X) [fun{1}(X(1:2),X(3:end)) fun{2}(X(1:2),X(3:end))]; % ID and Q are "stacked" into X

Hope this helps get you on your way, and if this does not help then post the error(s) that you are seeing.

-Adam

[Howard Clapp]

Hi Adam, thanks for the reply.

I'd like to ask a few questions to clarify what you are saying.

Firstly, for the first step am I looking for something along the lines of:

z(1)=ID=f(x)

z(2)=Q=f(x)

If this is the case, I am struggling to comprehend how to represent a variable that is either 1 or 2, and a variable that can be anything between 0.000001 to 0.01 using the term of X. Or am I approaching that incorrectly.

Secondly, once I succesfully complete the above, should I be substituting that into the

fun{1} = @(ID,Q) EndCost(ID,Q);
fun{2} = @(ID,Q) EndCarbon(ID,Q);

essentially re-writing it as:

fun{1} = @(x) EndCost(x);
fun{2} = @(x) EndCarbon(x);

Also, in the final section I am slightly unsure as to what the "X(1:2)" and "X(3:end)" are doing in the line "[fun{1}(X(1:2),X(3:end))"

and finally if "fun1and2" is my fitness function, does that mean my function should be something along the line of:

function fun1and2 = my_fun(x)

z(1)=ID=f(x)
z(2)=Q=f(x)

fun{1} = @(x) EndCost(x);
fun{2} = @(x) EndCarbon(x);

fun1and2 = @(X) [fun{1}(X(1:2),X(3:end)) fun{2}(X(1:2),X(3:end))];

Is this correct? And if so, how would I incorporate the z(1) and z(2)

Thanks again, and apologise for all the questions or if I have missed the point on some of what you have said.

[Howard Clapp]

Thanks for your help, managed to use what you said to approach the solution. Had a little issue but was just a simple error and a colleague spotted it, so it seems to be running smoothly now! Cheers again!