How can I set an optimization variable to be an element of a set / categorical?

I am building an optimization problem and I am trying to work with problem based programming. I want to define decision variables using optimvar function. How can I set a variable to be an element of a set. For example I want my variable to be 5, 8 or 12.5.

Answer 1

[Alan Weiss]

Set your variable to be an integer, and use the integer value to select from a list of predefined values. For examples, see optimal dispatch or a genetic algorithm example (not in problem-based framework).

Alan Weiss

MATLAB mathematical toolbox documentation

[Robert van der kraan]

Hallo Alan,

Does your reply mean it is not possible to set an optimization variable to be an element of a specified vector or even to be one out of two specified values in the problem-based approach?

[Alan Weiss]

That is correct. Optimization variables can be real or integer, but not categorical. If you want to have them indicate specified values, have integer values that you then use to index. For example, if your predefined values are in the vector v = [-16,3,pi], and if x is an integer optimization variable with 1 <= x <= 3, then v(x) takes the values that you want.

For exactly two values in v, you can do the following. Set x to be integer-valued with lower bound 0 and upper bound 1.

y = x*v(1) + (1-x)*v(2)

takes the value v(1) for x = 1 and the value v(2) for x = 0.

Alan Weiss

MATLAB mathematical toolbox documentation

[Kendall Knight]

When I attmepted an implementation of you comment Alan I got the error.

"Unable to use a value of type 'optim.problemdef.OptimizationVariable' as an index."

when the following code was run.

v = [1,2,3,4];
po_kl_picker = optimvar('po_kl_picker','LowerBound',1,'UpperBound',4, 'Type', 'integer');
rate_func = func_choice(v(po_kl_picker),shackle_angle,travel);

[Ayush Modi]

Hi Kendall,

Optimization variables cannot be used as an index to get the value from an array. In the example given by @Alan Weiss, index for variable "v" is 1 and 2 and not x (which is the Optimization variable).