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Using maximum of optimisation variable as an upper bound for another variable

Asked by Michael-Allan Millar on 12 Aug 2019
Latest activity Commented on by Matt J
on 7 Oct 2019 at 11:28
Hello
I am trying to minimise a cost function where the investment cost is the the maximum value of an optimisation variable multiplied by the cost e.g.
Total cost=Cost per kW*MaxkW
However, matlab doesn't seem to like using an optimisation variable as a boundary condition.
Any solutions?
Thanks
Michael

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5 Answers

Answer by Matt J
on 7 Oct 2019 at 10:48
Edited by Matt J
on 7 Oct 2019 at 11:20
 Accepted Answer

Solve N different sub-problems
for j=1:N
prob.Constraint.TotalCost = sum(x)=CostPerkW*x(j);
prob.Constraint.Max = x(j)>=x; %x(j) is the maximum
[X{j} Cost{j}]=solve(prob)
end
and take the X{j} with the lowest Cost{j}.

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Answer by Ted Shultz
on 21 Aug 2019

you can do a bounded obtimization using this file from filecentral: fminsearchbnd

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Answer by Bruno Luong
on 21 Aug 2019

Not sure what you mean by "boundary condition" (there is no such thing in all optimization frameworks as far as I know), but take a look at fminmax

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Answer by Naveen Venkata Krishnan on 7 Oct 2019 at 10:07

Hello Micheal,
Based on the question, I understood the problem in this way , please correct me if am wrong :
"You want to minimize a cost function in such a way that optimization variable lies close to its upper boundary region ".
If this the requirment one way to do it to confine the search space of the optimizer close to upper boundary ( like giving the values of upper - ub and lower boundary - lb of the optimzation variable around the maximum value of the optimization variable ) .
Another way is by using an enforcement operator, i think the following IEEE paper may be of some help to you reg this.

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Answer by Michael-Allan Millar on 7 Oct 2019 at 10:57

Thanks very much for the help, everyone.
Basically, it's sizing a tank and heat pump. So the bigger the tank, the lower the operating cost but greater initial investment. I ended up doing it the way Matt J suggested, but was hoping there would be a way to give a single answer, rather than having to produce a list then manually picking the smallest cost.
Thanks again!
Michael

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Well, you don't have to "manually" pick. You would just do
[~,jmin]=min([Cost{:}]);
x=X{jmin};

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