(Custom) domain restrictions for NMaximize

Question

I'm new to mathematica, so please be forbearing with me :D

I would like to use NMaximize together with domain restriction (where the domain is a custom domain).

I've got an example:

dom = {1, 2, 3, 4};
NMaximize[{UP[fl, a, p], UA[fl, a, p] >= resutil, 
  D[UAS[fl, a, p], X[a]] == D[DA[a], a], MemberQ[dom, a] == True}, {a,
   p}]

I would like to maximize UP(.), subject to the given constraints (including that the result for a is a member of dom).

The maximization works fine if I set the constraint to a \[Element] Integers:

NMaximize[{UP[fl, a, p], UA[fl, a, p] >= resutil, 
  D[UAS[fl, a, p], X[a]] == D[DA[a], a], a \[Element] Integers}, {a, 
  p}]

Is there a way to include such a constraint into NMaximize?

Thank you very much for your help :)

Can't you combine Integers with adding UP>0&&UP<5 in the curly brackets? — Feyre, Jul 17, 2016 at 11:10
Thank you Feyre, maybe the example above isn't the best. Let's say the solution where a is restricted to Integers is a=4 and p=0.25; in a second step I would like to limit the possible values for a to dom={1,2,5,7,8} and find the optimal values for a and p where a \element dom — Stephan, Jul 17, 2016 at 11:14
If dom is a small list in fact, what's the trouble with brute-force enumeration? — J. M.'s eventual burnout, Jul 17, 2016 at 11:40
Thank you J.M :) .. dom is not necessarily a small list. Anyhow, even if dom were a small list, p can take any value (i.e., there are a lot possible combinations of a and p) — Stephan, Jul 17, 2016 at 11:47

Jack LaVigne · Accepted Answer · 2016-07-17 14:41:22Z

5

As I understand your question your are trying to use NMaximize where you want to constrain a parameter to belong to a custom domain.

One way to do it is to use Or on the individual elements of the domain list.

dom = {1, 2, 3, 4};

Or @@ Map[a == # &, dom]

(* a == 1 || a == 2 || a == 3 || a == 4 *)

Use this in NMaximize as a constraint.

NMaximize[{a^2, Or @@ Map[a == # &, dom]}, a]

(* {16., {a -> 4.}} *)

answered Jul 17, 2016 at 14:41

Jack LaVigne

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$\begingroup$ Thank you very much for your help :) That's exactly what I was trying to do :) $\endgroup$
– Stephan
Jul 17, 2016 at 14:55

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